Answer:
According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Since Sharon's puck stops after the collision, all of its momentum is transferred to Kaylyn's puck. Therefore, the resulting momentum of Kaylyn's puck is equal to the initial momentum of Sharon's puck:
m1v1 = m2v2
where m1 and v1 are the mass and velocity of Sharon's puck before the collision, and m2 and v2 are the mass and velocity of Kaylyn's puck after the collision.
Substituting the given values:
0.3 kg × 6 m/s = 0.3 kg × v2
Simplifying:
1.8 kg m/s = 0.3 kg × v2
Dividing both sides by 0.3 kg:
v2 = 6 m/s
Therefore, the resulting velocity of Kaylyn's puck should be equal to the velocity of Sharon's puck before they collided, which is 6 m/s.
Answer: A. Equal to the velocity of Sharon's puck before they collided.
your challenge is to determine what factors affect the frequency and the period of a vibrating mass on a spring
The frequency and period of a vibrating mass on a spring are affected by the mass of the object, the spring constant, and the amplitude of the vibration. The frequency is the number of complete back-and-forth vibrations per second, while the period is the amount of time it takes for one complete vibration.
The mass of the object affects the frequency and period because it affects the amount of force exerted on the spring. A larger mass will require more force to be exerted on the spring to produce the same amount of displacement as a smaller mass. This means the frequency and period will increase as the mass increases.
The spring constant affects the frequency and period because it determines how stiff the spring is. A stiffer spring requires more force to produce the same amount of displacement as a looser spring. Therefore, the frequency and period will increase as the spring constant increases.
The amplitude of the vibration affects the frequency and period because it determines how much the mass will be displaced from its equilibrium position. A larger amplitude will require more force to produce the same amount of displacement as a smaller amplitude, meaning the frequency and period will increase as the amplitude increases.
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2020 State of the Bay Summary PLS HELP PLS ILYSM IF U DO!!!!
What are your first feelings or impressions when looking at the tables above?
Which indicators do you believe are most important for the health of the Bay? Explain.
Chesapeake Bay Report Grade vs. Year Graph
From your viewpoint, 1) what does the graph reveal about the current state of the Chesapeake Bay? 2) Do you believe your community contributes to the Chesapeake Bay’s condition? Explain your answer.
There is likely to be eutrophication in the water
There are less turbid substances deposited in the water within the time.
The decrease of water grasses means the primary productivity reduced.
What does increase in nitrogen and phosphorus content mean for Chesapeake Bay ?An increase in nitrogen and phosphorus content in Chesapeake Bay can have significant environmental impacts. Chesapeake Bay is an estuary that is highly sensitive to nutrient pollution, which can cause the growth of harmful algal blooms and oxygen depletion in the water.
When excess nitrogen and phosphorus from fertilizers, animal waste, and sewage enter the Bay, they can cause an overgrowth of algae. As these algae die and decompose, they consume oxygen in the water, leading to low oxygen conditions known as "hypoxia" or "dead zones". This can harm fish, shellfish, and other aquatic life, and can also have economic impacts on the fishing and tourism industries.
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The wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n = 2 to n = 1 state is :
The wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n = 2 to n = 1 state is 121.6 nm.
It is the most straightforward type of atom, with only one electron in its atomic shell. When an electron in a hydrogen atom moves from one energy level to another, it emits or absorbs a photon of light with a particular energy, E.
This energy difference can be found using the Rydberg formula for hydrogen atom wavelengths.
[tex]λ= 1/((Ry) × (1/ n1^2 - 1/ n2^2))[/tex]
where Ry = 1.097 x 107 m-1, and n1 and n2 are the initial and final quantum numbers of the electron, respectively.
In this instance, the electron goes from the n = 2 state to the n = 1 state.
The energy difference can be calculated as follows:
E = Rh (1/n2² - 1/n1²)
E = 2.18 × 10⁻¹⁸ J(1/12 - 1/22)
E = 1.63 × 10⁻¹⁸ J
The frequency of the photon emitted can be calculated
asv = E/hv = 1.63 × 10⁻¹⁸ J/6.63 × 10⁻³⁴
J.sv = 2.46 × 10¹⁴ Hz
Finally, we can use the formula c = λvc = λv
to find the wavelength of the photon emitted.
c/ v = λ121.6
nm = λ
Therefore, the wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n = 2 to n = 1 state is 121.6 nm.
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A disk is rotating at 2 rev/sec. The disk has a moment of inertia of 25 kg m2. If an identical, non-rotating disk, which has a moment of inertia exactly ½ as large, is dropped onto the rotating disk, what will be the new rotational speed of the combined rotating object?
Answer: 8.38 revolutions per second
Explanation:
Before the second disk is dropped, the initial angular momentum of the system is given by:
L = I1 * w1
where I1 is the moment of inertia of the first disk, and w1 is its angular velocity.
Substituting the given values, we have:
L = (25 kg m^2) * (2 rev/sec * 2π rad/rev) = 100π kg m^2/s
When the second disk is dropped onto the rotating disk, the total moment of inertia of the system will be the sum of the moment of inertia of the first disk and the moment of inertia of the second disk:
Itotal = I1 + I2/2
where I2/2 is the moment of inertia of the second disk, which is half as large as that of the first disk.
Substituting the given values, we have:
Itotal = (25 kg m^2) + (12.5 kg m^2) = 37.5 kg m^2
Conservation of angular momentum requires that the initial angular momentum of the system be equal to its final angular momentum, so:
L = Itotal * wf
where wf is the final angular velocity of the combined disk.
Solving for wf, we get:
wf = L / Itotal = (100π kg m^2/s) / (37.5 kg m^2) ≈ 8.38 rev/sec
Therefore, the new rotational speed of the combined rotating object is approximately 8.38 revolutions per second.
A uniform, 255-N rod that is 2.00m long carries a 225-N weight at its right end and an unknown weight W toward the left end. When W is placed 50.0cm from the left end of the rod, the system just balances horizontally when the fulcrum is located 75.0cm from the right end.
a.) Find W
b.) If W is now moved 25.0cm to the right, how far and in what direction must the fulcrum be moved to restore balance?
Answer:
a) To find W, we can use the principle of moments. At balance, the sum of the clockwise moments about the fulcrum is equal to the sum of the counterclockwise moments about the fulcrum. Let the fulcrum be at a distance x from the left end of the rod.
Clockwise moment = (225 N)(2.00 m - x)
Counterclockwise moment = W(0.50 m) + (255 N - W)(1.50 m + x)
Setting these two moments equal, we have:
(225 N)(2.00 m - x) = W(0.50 m) + (255 N - W)(1.50 m + x)
Solving for W, we get:
W = 81.7 N
Therefore, the unknown weight W is 81.7 N.
b) If W is moved 25.0 cm to the right, its new position is 75.0 cm from the left end of the rod. To restore balance, we need to find the new position of the fulcrum. Again, we can use the principle of moments:
Clockwise moment = (225 N)(2.00 m - 0.75 m) + W(0.25 m)
Counterclockwise moment = (255 N - W)(1.50 m + 0.75 m - x)
Setting these two moments equal, we have:
(225 N)(2.00 m - 0.75 m) + W(0.25 m) = (255 N - W)(1.50 m + 0.75 m - x)
Substituting the value of W we found in part (a), we get:
(225 N)(2.00 m - 0.75 m) + (81.7 N)(0.25 m) = (255 N - 81.7 N)(1.50 m + 0.75 m - x)
Simplifying and solving for x, we get:
x = 1.32 m
Therefore, to restore balance, the fulcrum must be moved 0.57 m (1.32 m - 0.75 m) to the left.
lonie is being pulled from a snake pit with a rope that breaks if tension in it exceeds 755N. If one has a mass of 70kg and the snake pit is 3 Am deep, what is the minimum time necessary to pull out lonie?
The minimum time required to pull out lonie from the snake pit is √(3/4) seconds.
Given: Mass, m = 70 kg Distance, d = 3 m Limiting tension, T = 755N
The minimum time required to pull out lonie from a snake pit, t
Given, mass, m = 70 kg Acceleration due to gravity, g = 9.8 m/s²Distance, d = 3 m
Let's assume the minimum time required to pull out lonie from a snake pit is t.
So, using the equation of motion,S = ut + 1/2 at²
Here, S = d = 3m (Distance),u = 0 m/s (Initial velocity),a = g = 9.8 m/s² (Acceleration) and t = time
Substituting the above values in the equation, we get3 = 0 + 1/2 × 9.8 × t² => t² = 3/4 => t = √(3/4) sec
Also, we know that the tension in the rope is given byT = mg + ma
Now, the rope will break when T exceeds 755 N.
So, substituting the values of m, g, and a in the above equation, we getT = mg + ma = 70 × 9.8 + 70 × a
Since the tension in the rope should be less than or equal to 755 N, we have70 × 9.8 + 70 × a ≤ 755 => a ≤ (755 - 70 × 9.8)/70=> a ≤ 3.29 m/s²
Therefore, the minimum time required to pull out lonie from the snake pit is √(3/4) seconds.
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jason weighs 150 n and sits on his big brother's shoulders. his big brother weighs 400 n. the support force supplied by the floor must be
Jason weighs 150 N and sits on his big brother's shoulders. Big brother weighs 400 N. the support force supplied by the floor must be 500N.
Force can be defined as any influence that causes an object to undergo a change in motion, direction, or shape. In physics, force is represented as a vector quantity, which means it has both magnitude and direction. The standard unit of force is the newton, which is defined as the force required to accelerate a mass of one kilogram at a rate of one meter per second squared.
Force can be categorized into two types: contact force and non-contact force. Contact force is a force that is exerted through direct contact between two objects, while non-contact force is a force that is exerted without any physical contact between two objects, such as gravity or electromagnetic force. Force plays a crucial role in the study of physics, as it governs the behavior of all objects in the universe.
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Complete Question:-
Jason weighs 150 N and sits on his big brother's shoulders. Big brother weighs 400 N. the support force supplied by the floor must be
A 440-V, three-phase, 2-pole, 50-Hz Y-connected wound rotor induction motor is rated at 75 KW. Its equivalent circuit components are:
R1=0.075 Ohms R2=0.065 Ohms Xm=7.2 Ohms
X1=0.17 Ohms X2=0.17 Ohms
Pmech=1.0 KW W Pstray=150 W Pcore=1.1 KW
Determine the slip at pullout torque, the pullout torque, and the rotor speed at the pullout torque of this motor.
If the same motor is to be driven from a 440-V, 60 Hz power supply, what will the pullout torque be? What will the slip be at pullout torque?
If the same motor is to be driven from a 440-V, 60 Hz power supply, The slip at pullout torque is 0.0455.
The slip at pullout torque, pullout torque, and rotor speed at the pullout torque of the motor are given by Slip
[tex]s = Pmech/ (Xm\times 1.5\times V^2/1000)[/tex]
Pullout torque is given by [tex](0.5 \times V^2 / X2) \times (R2 / (R1^2 + (X1 + X2)^2))[/tex]
Rotor speed at pullout torque is given by Ns = (120f/p)(1 - s)
The given parameters of the motor are R1 = 0.075 Ohms, R2 = 0.065 Ohms, Xm = 7.2 Ohms, X1 = 0.17 Ohms, X2 = 0.17 Ohms, Pmech = 1.0 KW W, Pstray = 150 W, Pcore = 1.1 KW.
The motor is rated at 75 KW with a power factor of 0.85.Assuming the motor is running at unity power factor.
The output power of the motor is Pout = 75 KW and the input power to the motor is Pin = 75 KW / 0.85 = 88.24 KW
The input current to the motor is [tex]Iin = 88.24 KW / (3 \times 440 V \times sqrt(3)) = 89.5 A[/tex]
Therefore, the torque developed by the motor is [tex]T = 1000 \times Pout / (2 \times pi \times N)[/tex]
The slip at the pullout torque is given by [tex]s = sqrt((Pcore + Pstray) / Pout) = sqrt((1.1 KW + 150 W) / 75 KW) = 0.0455.[/tex]
The pullout torque is given by [tex](0.5 \times 440^2 / 0.17) \times (0.065 / (0.075^2 + (0.17 + 0.17)^2)) = 411.7 Nm.[/tex]
The rotor speed at pullout torque is [tex]Ns = (120 \times 50 / 2) \times (1 - 0.0455) = 2846 rpm.[/tex]
The pullout torque of the motor if driven from a 440 V, 60 Hz power supply is given by
[tex](0.5 \times 440^2 / 0.17) \times (0.065 / (0.075^2 + (0.17 + 0.17)^2)) \times (60 / 50) = 548.4 Nm.[/tex]
The slip at pullout torque is given by [tex]s = sqrt((1.1 KW + 150 W) / 75 KW) = 0.0455.[/tex]
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A 65 kg-mass person stands at the end of a diving board, 1.5 m from the board's pivot point. Determine the torque the person is exerting on the board with respect to the pivot point. Show your work.
Answer:
Explanation:
To calculate the torque exerted by the person on the diving board, we need to know the force exerted and the lever arm.
The force exerted by the person is the weight of their body, which can be calculated as:
F = mg
F = 65 kg x 9.81 m/s^2
F = 637.65 N
note: The acceleration of gravity "g" is therefore the result of gravitation (gravitational attraction) between the Earth and other celestial bodies, and of the centrifugal acceleration, due to the movement of the earth's rotation and its average global value is 9.81 ms -2.
The lever arm is the distance from the person to the pivot point, which is given as 1.5 m.
The torque (τ) can then be calculated as:
τ = F x d
τ = 637.65 N x 1.5 m
τ = 956.47 Nm
Therefore, the torque exerted by the person on the diving board with respect to the pivot point is 956.47 Nm.
The torque exerted by a force F at a distance r from the pivot point is given by the formula:
τ = F x r x sin(θ)
where θ is the angle between the force vector and the vector from the pivot point to the point where the force is applied.
In this case, the person's weight is the force being exerted on the board, and its magnitude is:
F = m x g = 65 kg x 9.8 m/s^2 = 637 N
The distance from the pivot point to the person is r = 1.5 m. Since the person is standing vertically, the angle between the weight vector and the vector from the pivot point to the person is 90 degrees, so sin(θ) = 1. Substituting the values into the torque formula, we get:
τ = 637 N x 1.5 m x 1 = 955.5 Nm
Therefore, the person is exerting a torque of 955.5 Nm on the diving board with respect to the pivot point.
find the age for a rock for which you determine that 53 % % of the original uranium-238 remains, while the other 47 % % has decayed into lead.
The age of the rock is approximately 2.58 billion years.
The half-life of uranium-238 is approximately 4.47 billion years. This means that after 4.47 billion years, half of the original uranium-238 will have decayed into lead.
We can use this information to determine the age of a rock for which we have measured that 53% of the original uranium-238 remains.
If 53% of the original uranium-238 remains, then 47% has decayed into lead.
Since half of the original uranium-238 decays every 4.47 billion years, the ratio of remaining uranium-238 to original uranium-238 after t years is given by:
[tex](remaining uranium-238) / (original uranium-238) = 0.53 = e^(-t/4.47E9)[/tex]
Solving for t, we get:
[tex]t = -4.47E9 * ln(0.53) = 2.58 billion years[/tex]
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which statement most accurately captures what current evidence tell us about the habitability of mars?
"Mars may once have met the requirements for livability, such as having liquid ocean, but at the moment, it top orbit is too thin so it lacks the huge magnetic field needed to support life as we know it.
What being magnetic entails?having great aptitude or power to attract. a magnetic personality; of or pertaining to the a magnet or even to magnetism; of, pertaining to, and characterized by earth's magnetism; magnetized or able to be magnetized.
How can a magnet become more powerful?Yet, some substances have a high magnetic field, meaning that the majority of its electrons spin opposite direction. The strongest magnets are made of these materials because of their great magnetic permeability. They include the elements nickel, cobalt, and iron. The most potent type of magnet is one made of neodymium iron boron (NdFeb).
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Two trains ‘A’ and ‘B’ leave the same station on parallel lines. Train ‘A’ starts with a uniform acceleration of 1/6 m/s2 and attains the speed of 24 km/hr, then stem is reduced to keep the speed constant. Train ‘B’ leaves 40 seconds after, with uniform acceleration of 1/3 m/s2 to attain the maximum speed of 48 km/hr. When will it overtake the train ‘A’?
Answer:
Explanation:
Let's first convert the speeds of the trains from km/hr to m/s:
Train A: 24 km/hr = (24 x 1000) / (60 x 60) = 6.67 m/s
Train B: 48 km/hr = (48 x 1000) / (60 x 60) = 13.33 m/s
Now, let's find the distance covered by train A until it reaches its constant speed:
v = u + at
where
u = initial velocity = 0
a = acceleration = 1/6 m/s^2
t = time taken to reach constant speed
At constant speed, v = 6.67 m/s
So, 6.67 = (1/6)t + 0
t = 40 seconds
Using the formula for distance covered during uniform acceleration:
s = ut + (1/2)at^2
The distance covered by train A during the acceleration phase is:
s = (1/2)(1/6)(40^2) = 133.33 m
Now, let's find the equation of motion for train B:
s = ut + (1/2)at^2
where
u = initial velocity = 0
a = acceleration = 1/3 m/s^2
t = time taken to overtake train A
At the time of overtaking, both trains will cover the same distance. Let's call this distance "d". So we have:
d = 133.33 + 6.67t (distance covered by train A + distance covered by train B)
Setting the equations for both trains equal to each other, we get:
133.33 + 6.67t = (1/2)(1/3)t^2 + (1/3)t^2
Simplifying and solving for t, we get:
t = 180 seconds
Therefore, train B will overtake train A after 180 seconds or 3 minutes.
A 1500-kg car is traveling at 21 m/s around a curve with a radius of 120 m. Determine the angular momentum of the car. Show your work.
Answer:
Explanation:
Angular momentum of an object is given by L = Iω, where I is the moment of inertia and ω is the angular velocity.
To find the angular momentum of the car, we need to first find its moment of inertia about the axis of rotation (which in this case is the center of the curve).
The moment of inertia of a point mass m rotating around an axis at a distance r is given by I = mr^2.
The car can be thought of as a collection of point masses, so we need to integrate the moment of inertia of each point mass over the entire car.
Assuming the car is a uniform rectangular box with dimensions 4.5 m x 1.8 m x 1.5 m, we can divide it into smaller cubes of mass dm and volume dV. The moment of inertia of each cube is given by dI = dm(r^2), where r is the distance from the cube to the axis of rotation.
Integrating over the entire car, we get:
I = ∫(dm)(r^2) = ∫(ρdV)(r^2)
where ρ is the density of the car.
The volume of the car is V = (4.5 m)(1.8 m)(1.5 m) = 12.15 m^3, and assuming a density of 1000 kg/m^3 (typical for cars), the mass of the car is m = (1000 kg/m^3)(12.15 m^3) = 12150 kg.
We can express the distance r as the sum of the radius of the curve (120 m) and the distance between the center of mass of the car and the pivot point (which we can approximate as half the length of the car, or 0.75 m).
So, r = 120 m + 0.75 m = 120.75 m.
Substituting these values into the equation for I, we get:
I = ∫(ρdV)(r^2) = ρ∫dV(r^2) = ρV(r^2) = (1000 kg/m^3)(12.15 m^3)(120.75 m)^2 = 1.796 x 10^7 kg·m^2
To find the angular velocity of the car, we can use the formula v = ωr, where v is the linear velocity of the car and ω is the angular velocity.
The linear velocity of the car is given by v = 21 m/s, and the radius of the curve is 120 m. So,
ω = v/r = (21 m/s)/(120 m) = 0.175 rad/s
Finally, we can find the angular momentum of the car by multiplying the moment of inertia by the angular velocity:
L = Iω = (1.796 x 10^7 kg·m^2)(0.175 rad/s) = 3.145 x 10^6 kg·m^2/s
Therefore, the angular momentum of the car is 3.145 x 10^6 kg·m^2/s.
The angular momentum of the car in the given example is [tex]3,780,000 kg m^2/s.[/tex]
What is an Angular momentum?The amount of rotational motion an object has around a certain axis is measured by its angular momentum. It is the result of an object's angular velocity and moment of inertia. It is a vector quantity with a direction and magnitude.
We need to apply the following calculation to determine the car's angular momentum:
L = I * ω
where L is angular momentum, I is the moment of inertia, and ω is the angular velocity.
We must apply the following formula to determine the moment of inertia:
[tex]I = m * r^2[/tex]
where r is the curve's radius and m is the car's mass.
m = 1500 kg r = 120 m
[tex]I = m * r^2[/tex]
[tex]I = 1500 kg * (120 m)^2\\I = 21,600,000 kg m^2[/tex]
The following formula must be used to determine the angular velocity:
v = ω * r
where v denotes the car's speed.
v = 21 m/s r = 120 m
where,
v/r = 21 m/s/120 m = 0.175 rad/s
We can now determine the angular momentum:
L = I * ω
[tex]L = 21,600,000 kg m^2 * 0.175 rad/s\\L = 3,780,000 kg m^2/s[/tex]
Therefore, the angular momentum of the car is [tex]3,780,000 kg m^2/s.[/tex]
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a battleship simultaneously fires two shells at enemy ships. if the shells follow the parabolic trajectories shown, which ship gets hit first?
A battleship simultaneously fires two shells in parabolic projectile motion and no information about initial speeds at enemy ships. The ship B got hit first. So, the correct choice for answer is option (c).
Here is we have a battleship Which fires two shells simultaneously at the enemy ship along the two paths. The initial speed of projection may be same or different. See the above figure carefully, the angle of projection for ship A is more than ship B. Time of flight for ship A is
[tex]T_A = \frac{ 2u_{A} sinθ_{A}}{g }[/tex]
For ship B, [tex]T_B = \frac{2u_B sinθ_{B}}{g }[/tex]
We have no idea about the initial speed of projection, so we cannot consider it for comparison. As we know from above,
[tex]θ_{A} > θ_{B}[/tex]
=> [tex]sinθ_{A} > sinθ_{B}[/tex]
So, [tex]T_{A} > T_{B}[/tex]
That is time of flight for ship A is greater than for the ship B. Therefore, ship B gets hit first.
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Complete question:
A battleship simultaneously fires two shells at enemy ships. if the shells follow the parabolic trajectories shown, which ship gets hit first?
a) A
b) both simultaneously
c) B
d) None
An automobile engine slows down from 4500 rpm to 1200 rpm in 2.5 s. Calculate:
a) Its angular acceleration, assumed constant.
b) the total number of revolutions the engine makes in this time
a. The angular acceleration of the engine is -1320 rad/s².
b. The engine makes approximately 118.7 revolutions during the deceleration.
Given:
Initial angular velocity (ω1) = 4500 rpm
Final angular velocity (ω2) = 1200 rpm
Time is taken (t) = 2.5 s
a) The formula for angular acceleration is:
α = (ω2 - ω1) / t
Substituting the given values, we get:
α = (1200 rpm - 4500 rpm) / 2.5 s = -1320 rad/s² (negative sign indicates a deceleration)
Therefore, the angular acceleration of the engine is -1320 rad/s².
b) To find the total number of revolutions, we need to convert the initial and final angular velocities from rpm to rad/s:
ω1 = 4500 rpm × 2π/60 = 471 rad/s
ω2 = 1200 rpm × 2π/60 = 126 rad/s
The average angular velocity (ω_avg) during the deceleration is given by:
ω_avg = (ω1 + ω2) / 2 = (471 rad/s + 126 rad/s) / 2 = 298.5 rad/s
The total angular displacement (θ) of the engine during the deceleration is given by:
θ = ω_avg × t = 298.5 rad/s × 2.5 s = 746.25 rad
Finally, the total number of revolutions (N) made by the engine is:
N = θ / 2π = 746.25 rad / 2π rad/rev = 118.7 rev (approximately)
Therefore, the engine makes approximately 118.7 revolutions during the deceleration.
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resistance of wire does not depend on ____current supplied area of cross-section of the wire length of the wire material of the wire
Resistance of wire does not depend on current supplied, area of cross-section of the wire, length of the wire, or material of the wire.
The resistance of a wire does not depend on the current supplied. The resistance of a wire is determined by the following factors:
Length of the wire: The longer the wire, the more resistance it will have. This is due to the fact that a longer wire has more atoms that obstruct the flow of electrons. The more time the electrons take to traverse the wire, the more collisions they will have, resulting in more resistance.
Area of cross-section of the wire: The cross-sectional area of a wire is inversely related to its resistance. This implies that if the cross-sectional area is reduced, the resistance will rise. This is due to the fact that a smaller cross-sectional area leads to more electron collisions with atoms.
Material of the wire: The resistivity of the wire's material determines its resistance. A material with a higher resistivity would have more resistance than one with a lower resistivity. It's important to note that the resistivity of a material varies with temperature. As the temperature rises, so does the resistivity of the substance.
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there is very little hydrogen or helium in the inner part of the solar system today. astronomers believe the reason for this is that
Astronomers believe that the reason for the low levels of hydrogen and helium in the inner part of the Solar System is that these elements were lost due to the Sun's strong solar wind
. Most of the gas in the solar nebula was hydrogen and helium. However, there is very little hydrogen or helium in the inner part of the solar system today. Astronomers believe this is because the intense solar winds from the young Sun blew the lighter gases out into space.
The solar winds also blew away any smaller particles that had not yet stuck together to form larger objects, which is why there are very few small objects close to the Sun. This is why there is more hydrogen and helium in the outer part of the solar system.
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The magnitudes of the current density and the diameters for wires 1 and 2 are given in the table. The current directions are indicated by the arrows.
Find the current I3 in wire 3.
Express your answer in amperes to two significant figures. Call current out of the junction positive and current into the junction negative.
The magnitudes of the current density and the diam
Wire Current density
(A/mm2) Diameter
(mm)
1 2.1 1.7
2 3.9 2.4
To solve for the current I3 in wire 3, we can apply the principle of conservation of charge, which states that the total current flowing into a junction must equal the total current flowing out of the junction. This can be expressed mathematically as:
I1 + I2 = I3
where I1 and I2 are the currents in wires 1 and 2, respectively, and I3 is the current in wire 3.
To find I1 and I2, we can use the formulas for current (I = J x A), where J is the current density and A is the cross-sectional area of the wire. The cross-sectional area can be calculated from the diameter using the formula A = π/4 x d^2, where d is the diameter.
For wire 1:
I1 = J1 x A1 = 2.1 A/mm^2 x π/4 x (1.7 mm)^2 = 4.52 A
For wire 2:
I2 = J2 x A2 = 3.9 A/mm^2 x π/4 x (2.4 mm)^2 = 13.21 A
Substituting these values into the conservation of charge equation, we get:
4.52 A + 13.21 A = I3
Simplifying, we get:
I3 = 17.73 A
Therefore, the current I3 in wire 3 is approximately 17.73 amperes (or amps) to two significant figures.
What is a current ?Current can be thought of as the rate at which electric charges flow through a circuit. It is caused by the movement of electrons, which are negatively charged particles, in response to an electric field. In a metallic conductor, such as a wire, electrons can move freely through the material, which allows current to flow easily.
Current can be either direct current (DC) or alternating current (AC). DC flows in one direction, whereas AC changes direction periodically. DC is commonly used in batteries and electronic devices, while AC is used for power transmission over long distances.
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which of the following describes the subtle gradation of light and shadow to create the effect of three dimensionality?
The subtle gradation of light and shadow to create the effect of three-dimensionality is described as shading.
In the context of art, shading refers to the use of light and shadow to render a three-dimensional object or scene realistically.
When an object is illuminated, light falls on its surface in different ways, depending on the position of the light source, the object's orientation, and the object's shape.
In this way, shadows are formed on the object's surface, and some regions receive more or less light than others.
By accurately depicting these light and shadow areas, artists can create an illusion of depth and dimensionality in their works.
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given that the value of the bohr radius of hydrogen is 0.5 a, what is the radius of the first bohr orbit of positronium
The Bohr radius is denoted by a and is equal to 0.529 Å.
The radius of the first Bohr orbit of hydrogen is then equal to a0 which is equal to 0.529 Å. The first Bohr orbit of hydrogen has an energy of -13.6 eV.
The energy of a stationary electron in the nth Bohr orbit of a hydrogen-like atom is given by-13.6 / n² eV
where n is the principal quantum number for the atom. The value of the principal quantum number for the atom is given by the formula
Ry / E = (n² / Z)
where Ry is the Rydberg constant, E is the ionization energy of the atom, and Z is the atomic number of the atom.
The radius of the nth Bohr orbit of a hydrogen-like atom is given bya / n²
where a is the Bohr radius. The Bohr radius is equal to 0.529 Å for hydrogen.
The radius of the first Bohr orbit of positronium is then given by
0.529 / 1²=0.529 Å
therefore, The radius of the first Bohr orbit of positronium is 0.529 Å.
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sort each of the astronomical questions below into the appropriate bin based on the type of observation you would need to perform to answer it.
Imaging: How big is the Andromeda Galaxy? What are the prime surface features of Mars? Are the stars in the Orion Nebula surrounded by gas?
Spectroscopy: What is the chemical composition of Crab Nebula? Determine the temperature of Jupiter's atmosphere?
Timing: Does the star Mira vary in it's brightness?
Give a brief account on Spectroscopy.Spectroscopy is known to be the study of the absorption and emission of light and other radiation by matter. It is related that these processes depend on the wavelength of the radiation. Recently, the definition has been extended to include the study of interactions between particles like electrons, protons and ions, and interactions with other particles as a function of collision energy.
Spectroscopy has been essential to the development of some of the most fundamental theories of physics, including quantum mechanics, special and general relativity, and quantum electrodynamics. Spectroscopy applied to high-energy collisions has become an important tool in advancing the scientific understanding not only of electromagnetic forces, but also of strong and weak nuclear forces.
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I actually have 3 questions. >33
1. Write about a time when you felt very cold and did something to make yourself feel warm, or a time when you felt hot and did something to cool yourself down. What caused the heat to transfer from one place to another place? How did this transfer of heat cause a change in temperature?
2. Why is the temperature of the liquid in the flask on the previous page measured when the liquid in the thermometer has stopped rising?
3. How can the thermometer in the flask on the previous page be used to demonstrate the relationship between heat transfer and kinetic energy? Explain.
When you contact anything hot, the heat is transmitted from the object to your hand, making it feel hot. When you contact something cold, heat is transmitted from your hand to the object, making it feel chilly.
When heated the molecules of the liquid move faster causes them to get a little further apart?when heated, the molecules of the liquid in the thermometer move faster, causing them to get a little further apart. this results in movement up the thermometer. when cooled, the molecules of the liquid in the thermometer move slower, causing them to get a little closer together.
When the liquid in the thermometer is heated, the molecules move quicker, forcing them to move wider apart. This causes the thermometer to rise. When the liquid in the thermometer is chilled, the molecules travel slower, leading them to get closer together.
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Electric field lines are used to represent the vector electric field around point charges and charged objects. Which of the following statements are true about electric field lines. Select ALL that apply.
Select all that apply
A. Electric field lines cannot cross.
B. Lines of electric field only originate from positive charges.
C. Field lines point in the direction of the force the electric field creates on an electron.
D. The strength of the electric field is greater in regions where the field lines are closer together.
E. In an electric-field-line drawing with many point charges, the number of field lines originating or terminating on each charge is proportional to the charge. That is, bigger charges have proportionally more field lines. F. The true strength of an electric field at any point can be determined from an electric field representation.
Electric field lines are a powerful tool to understand and visualize electric fields. They help to represent the direction and magnitude of the electric field at various points around a charged object.
The following statements are true about electric field lines:
A. Electric field lines cannot cross: This is because at the point where two field lines cross, there would be two directions for the electric field, which is impossible. Hence, the lines do not cross, and this is one of the fundamental characteristics of electric field lines.
B. Lines of electric field only originate from positive charges: Electric field lines originate from positive charges and terminate at negative charges. This is because positive charges repel positive charges and attract negative charges. Therefore, the electric field lines originating from a positive charge terminate at a negative charge.
C. Field lines point in the direction of the force the electric field creates on an electron: Electric field lines point in the direction of the force that would be experienced by a positive charge placed at any point in the field. Electrons, being negatively charged, would experience a force in the opposite direction to the electric field.
D. The strength of the electric field is greater in regions where the field lines are closer together: The density of field lines indicates the strength of the electric field. The closer the lines are, the stronger the field at that point.
E. In an electric-field-line drawing with many point charges, the number of field lines originating or terminating on each charge is proportional to the charge. That is, bigger charges have proportionally more field lines: The number of field lines originating or terminating on each charge is directly proportional to the magnitude of the charge.
F. The true strength of an electric field at any point can be determined from an electric field representation: The strength of the electric field at a point can be determined by the density of electric field lines at that point. However, the actual strength of the field would require quantitative measurements using instruments such as a voltmeter or an electrometer.
In conclusion, electric field lines are an essential tool in understanding the behavior of electric fields. They provide a visual representation of the electric field, its direction, and its strength at various points in space.
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ball thrown upward from the top of a building 220 feet tall. The height of the ball is described by the function A is h()-162 + 20t + 220. where t is the time in seconds and t 0 corresponds to the moment the ball is thrown (a) Determine for which value of f the ball reaches the maximum height and determine this maximum height. Max Height: 905/4 (b) Determine when the ball reaches the ground. t(5+sqrt(905)/8 (c) With what velocity does the ball hit the ground?
The value of f is 905/4 feet, After 4 seconds the ball reaches the ground and the velocity of the ball hit the ground is -10 - 4sqrt(905) ft/s
step 1:
When the ball reaches the maximum height, it means that the velocity is zero, we use this fact to calculate the value of "f".
The height of the ball is described by the function A is
[tex]h(t) = -16t² + 20t + 220[/tex]
When the ball reaches the maximum height, its velocity is zero, therefore:
[tex]v = dh/dt = 0[/tex]
We take the derivative of the height function to get the velocity function:
[tex]v(t) = -32t + 20[/tex]
When the velocity is zero, the ball has reached its maximum height:
[tex]-32t + 20 = 0[/tex] => t = 5/8 seconds
Step 2:
Now we calculate the maximum height by plugging in t = 5/8 seconds into the height function:\
[tex]h(5/8) = -16(5/8)² + 20(5/8) + 220[/tex]
= 905/4 feet
Step 3:
To determine when the ball reaches the ground, we need to find the time when the ball reaches a height of 0:
[tex]0 = -16t² + 20t + 220= > 2t² - 5t - 55 = 0[/tex]
Using the quadratic formula:
[tex]t = [5 ± sqrt(5² - 4(2)(-55))] / [2(2)]= (5 ± sqrt(905)) / 4[/tex]
We take the positive root since time cannot be negative:
t = 4 seconds
Step 4:
To calculate the velocity at which the ball hits the ground,
we take the derivative of the height function and evaluate it at the time when the ball hits the ground:
[tex]v(t) = -32t + 20= > v((5 + sqrt(905)) / 4)[/tex]
= -32((5 + sqrt(905)) / 4) + 20
= -10 - 4sqrt(905) ft/s
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clock a remains in place and clock b is carried around the earth ( 40,000 km). by how many seconds will is clock b slower if carried on
Clock a remains in place and clock b is carried around the earth (40,000 km). According to Einstein's theory of relativity, The clock b is slower by approximately 44.6 seconds.
According to Einstein's theory of relativity, time dilation takes place when an object moves at a velocity close to the speed of light. The closer the velocity is to the speed of light, the more time slows down. This is why time on Earth is slower at high altitudes than it is on the ground.
According to the theory, the same effect happens when objects are moving at a high speed, which is why clocks that are taken on an airplane, for example, appear to be ticking more slowly.
1. The following equation is used to determine the time dilation:
t = t0 / √(1 – v²/c²),
where t is the time elapsed, t0 is the time at rest, v is the velocity, and c is the speed of light. When the earth rotates on its axis, every point on the planet's surface moves at a different velocity, with the highest velocity at the equator, and the velocity decreases as we move towards the poles. The earth's circumference at the equator is roughly 40,000 kilometers (24,901 miles).
As a result, a person standing on the equator would be traveling at a speed of around 1,674 kilometers per hour (1,040 miles per hour) because the earth spins once every 24 hours. We must first determine the velocity of a point on the earth's surface at the equator before we can use the equation to calculate time dilation.
2. We use the formula
v = 2πr / T,
where v is velocity, r is the radius of the earth, and T is the time it takes the earth to complete one rotation. The formula is as follows:
v = 2πr / Tv
= 2 x 3.14 x 6,378 km / 24 hv
= 1,674 km/h
3. Substituting these values into the equation, we get:
t = t0 / √(1 – v²/c²)t = t0 / √(1 – (1,674 m/s)² / (299,792,458 m/s)²)t = t0 / √(1 – 2.8 x 10^-8)t = t0 / 0.9999999714
This means that the clock on the equator will tick slightly slower than it would at rest. The difference in time can be calculated by subtracting the two values:
t – t0 = t0 / 0.9999999714 – t0t – t0 = t0 (1 – 0.9999999714)t – t0 = 0.0000000286 t0
4. We must first calculate the amount of time elapsed on the equator if a clock b is carried 40,000 km around the earth. It is easy to calculate the distance and speed, but we must also consider that the earth is rotating as well. As a result, we must determine the combined speed of the earth's rotation and the motion of clock b relative to the earth's surface.
5. To calculate this combined velocity, we can use the Pythagorean theorem, which states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides. If we imagine the velocity of the earth's rotation as the base of the triangle and the velocity of clock b as the height of the triangle, we can use this theorem to calculate the combined velocity as follows:
combined velocity = √(1,674² + vclock²)
where v clock is the velocity of clock b. Since clock b is being transported at the equator, it has the same velocity as the earth's rotation. As a result, we can substitute 1,674 km/h for v clock:
combined velocity = √(1,674² + 1,674²)
combined velocity = √(2 x 1,674²)
combined velocity = 2,367 km/h
6. Substituting the combined velocity into the equation for time dilation, we obtain:
t – t0 = t0 (1 – √(1 – v²/c²))t – t0 = t0 (1 – √(1 – (2,367 km/h)² / (299,792,458 m/s)²))t – t0
= t0 (1 – √(1 – 1.579 x 10^-11))t – t0
= t0 (1 – 0.999999999920215)t – t0
= 0.000000000079785 t0
Converting this value to seconds, we get:
0.000000000079785 t0 = 79.785 ns
Now we can combine the time dilation for the earth's rotation and the motion of clock b to obtain the total time dilation:
t – t0 = 0.0000000286 t0 + 0.000000000079785 t0t – t0 = 0.000000028679785 t0
Substituting the value of t0 (one second) into the equation, we get:
t – 1 = 0.000000028679785 seconds
Therefore, clock b will be approximately 44.6 seconds slower than clock a after being carried 40,000 km around the earth.
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Two balls collide and bounce off of each other. Ball A has a mass of 4.00kg and an initial velocity of 10.0m/s. Ball B has a mass of 6.00kg and an initial velocity of
-5.00m/s.
A. What is the initial momentum of the system?
B. Ball B bounces off with a velocity of 7.00m/s. What is the velocity of Ball A?
C. Prove if this is a perfectly elastic collision or not.
Based on observations (both images and spectra), which of the following statements are true about star-forming clouds?
A. Star-forming clouds are much hotter than most other interstellar material in the galaxy.
B. Star-forming clouds have the same overall chemical composition as the galaxy as a whole.
C. Most of the hydrogen in star-forming clouds is in the form of hydrogen molecules (H2).
D. The darkness of these clouds (in visible light) is due primarily to light absorption by tiny grains of interstellar dust.
E. The densest cloud regions appear dark to visible light telescopes but we can see into these regions with Infrared telescopes.
F. Young stars shine only in the infrared, which is why infrared observations are important.
G. Star-forming clouds glow with visible light in regions where the gas is heated by radiation from nearby stars.
Option B, C, D, E, and G: The chemical makeup of star-forming clouds is identical to that of the galaxy as a whole and the majority of the hydrogen in clouds that generate stars is present as hydrogen molecules (H₂).
Interstellar molecular clouds, which are opaque collections of extremely cold gas and dust, are where stars are formed. When some of those aggregates accumulate enough mass to collapse due to gravity alone, the process begins. The cause could even be as simple as random density changes within the cloud.
The correct statements are:
The chemical makeup of star-forming clouds is identical to that of the galaxy as a whole.
The majority of the hydrogen in clouds that create stars is present as hydrogen molecules (H₂).
The interstellar dust particles that make up these clouds' visual blackness absorb the majority of light.
Even though the densest cloud regions appear dark to telescopes using visible light, we can see inside them using infrared telescopes.
A region where the gas is heated by radiation from neighboring stars is where star-forming clouds are found.
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A large container holds 7.6 gallons of water. Use the following information to convert this amount of water to liters. Round your answer to the nearest tenth. 1 gallon = 4 quarts 1 L = 1.057 quarts
Answer:
You can use the same conversion factor to convert liters to gallons or you can use: 1 liter = 0.264 US gallons. To find how many gallons are in 4 liters, for example: gallons = 4 liters x 0.264 gallons/liter. The liters cancel out, leaving the gallon unit: 4 liters = 1.056 gallons. Keep this in mind: there are about 4 liters per US gallon.
Explanation:
Can someone check my answers? If I’m incorrect can you correct me? Thank you. Image below.
Explanation:
a) looks correct EXCEPT KE and GPE labels are reversed ( if the one on the L is initially and the R one is at max height)
b) looks good
c) incorrect total energy will be KE + GPE = 57.51 J at any point
d) correct
P2. Charges q and Q are placed on the x-y plane at (0,0) and at (0, 3) m, respectively.
Where q = 50 pC and Q = -40 pC.
a. Draw the situation to solve the next step.
b. Determine the net electric flux through a closed cylindrical surface that has a diameter of 5 ma
a height of 4 m, where the axis of the cylinder is the z axis and its mid-point is at the origin.
a. Here is a diagram of the situation:
Q (-40 pC)
|
|
| (0,3)
|
------ o-------- x-axis
|
|
| (0,0)
|
q (50 pC)
(b) The net electric flux through the closed cylindrical surface is -5.69×10⁵ Nm²/C.
To calculate this, we use Gauss's Law, which states that the net electric flux through any closed surface is proportional to the net charge enclosed by the surface. Mathematically, this is expressed as:
flux = E * A = (q_enclosed / ε0) * A
where E is the electric field, A is the area of the closed surface, q_enclosed is the net charge enclosed by the surface, and ε0 is the permittivity of free space.
In this case, we have a cylindrical surface with a height of 4 m and a diameter of 5 mA (which means a radius of 2.5 mA). The cylinder is centered at the origin and has the z-axis as its axis of symmetry. To apply Gauss's Law, we need to find the net charge enclosed by the cylinder.
Both charges q and Q are on the x-y plane, so they do not contribute to the net charge enclosed by the cylindrical surface. Therefore, the net charge enclosed by the surface is simply the sum of q and Q:
q_enclosed = q + Q = (50 pC) + (-40 pC) = 10 pC
Substituting this into Gauss's Law, we get:
flux = (q_enclosed / ε0) * A = (10 pC / 8.85×10⁻¹² F/m) * π (2.5×10⁻³ m)² (4 m) = -5.69×10⁵ Nm²/C
Therefore, the net electric flux through the closed cylindrical surface is -5.69×10⁵ Nm²/C.
What is an electric flux?
Electric flux is the measure of the number of electric field lines passing through a given surface. It is a scalar quantity and represents the amount of electric field passing through a surface per unit area. The SI unit of electric flux is volt-meter (V m) or newton-meter squared per coulomb (N m2/C).
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