The final temperature of the mixture of cold and hot water is 60°C.
To calculate the final temperature of the mixture of hot and cold water, we can use the principle of conservation of energy. The heat gained by the cold water is equal to the heat lost by the hot water.
Given:
Rate of hot water flow (m₁) = 20 kg/min
The initial temperature of hot water (T₁) = 80°C
Rate of cold water flow (m₂) = 10 kg/min
The initial temperature of cold water (T₂) = 20°C
Duration (t) = 3 minutes
Let's calculate the heat gained by the cold water and the heat lost by the hot water using the formula:
Heat gained = mass * specific heat capacity * temperature change
For the cold water:
Heat gained by cold water = m₂ * specific heat capacity * (final temperature - T₂)
For the hot water:
Heat lost by hot water = m₁ * specific heat capacity * (T₁ - final temperature)
Since the heat gained by the cold water is equal to the heat lost by the hot water, we can set up the equation:
m₂ * specific heat capacity * (final temperature - T₂) = m₁ * specific heat capacity * (T₁ - final temperature)
Let's substitute the given values and solve for the final temperature:
10 kg/min * specific heat capacity * (final temperature - 20°C) = 20 kg/min * specific heat capacity * (80°C - final temperature)
Simplifying the equation:
10 * (final temperature - 20) = 20 * (80 - final temperature)
10 * final temperature - 200 = 1600 - 20 * final temperature
30 * final temperature = 1800
final temperature = 1800 / 30
final temperature = 60°C
Therefore, the final temperature of the mixture of hot and cold water after 3 minutes will be 60°C.
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A bowling ball of mass m = 1.1 kg is resting on a spring compressed by a distance d = 0.35 m when the spring is released. At the moment the spring reaches its equilibrium point, the ball is launched from the spring into the air in projectile motion at an angle of θ = 39° measured from the horizontal. It is observed that the ball reaches a maximum height of h = 4.4 m, measured from the initial position of the ball. Let the gravitational potential energy be zero at the initial height of the bowling ball.
a) what is the spring constant, k, in newtons per meter?
(I got that the speed of the ball after the launch is 14.76)
The spring constant, k, in newtons per meter is 1,955.9 N/m.
Speed of the ball after the launchh = v²sin²θ/2g
v = √[(2gh)/sin²θ]
v = √[(2 x 9.8 x 4.4)/ (sin 39)²]
v = 14.76 m/s
Energy of the ball at topE = K.E + P.E
E = ¹/₂m(v cosθ)² + mgh
E = ¹/₂(1.1)(14.76 cos39)² + (1.1 x 9.8 x 4.4)
E = 119.8 J
Spring constantE = ¹/₂kx²
k = 2E/x²
k = (2 x 119.8)/(0.35²)
k = 1,955.9 N/m
Thus, the spring constant, k, in newtons per meter is 1,955.9 N/m.
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*Look at attachment for photo of object**
An object, whose mass is 0.520 kg, is attached to a spring with a force constant of 106 N/m. The object rests upon a frictionless, horizontal surface (shown in the figure below).
The object is pulled to the right a distance A = 0.150 m from its equilibrium position (the vertical dashed line) and held motionless. The object is then released from rest.
(a) At the instant of release, what is the magnitude of the spring force (in N) acting upon the object?
__________ N
(b) At that very instant, what is the magnitude of the object's acceleration (in m/s2)?
________ m/s2
(c)
In what direction does the acceleration vector point at the instant of release?
- The direction is not defined (i.e., the acceleration is zero).
- Toward the equilibrium position (i.e., to the left in the figure).
- Away from the equilibrium position (i.e., to the right in the figure).
- You cannot tell without more information.
A. The magnitude of the spring force (in N) acting upon the object is 15.9 N
B. The magnitude of the object's acceleration (in m/s²) is 30.58 m/s²
C. The direction of the acceleration vector points toward the equilibrium position (i.e., to the left in the figure).
A. How to determine the force Extension (e) = 0.150 mSpring constant (K) = 106 N/mForce (F) = ?F = Ke
F = 106 × 0.15
F = 15.9 N
B. How to determine the accelerationMass (m) = 0.52 KgForce (F) = 15. 9 NAcceleration (a) =?F = ma
Divide both sides by m
a = F / m
a = 15.9 / 0.52
a = 30.58 m/s²
C. How to determine the direction of the acceleration vectorConsidering the diagram, we can see that the spring was pulled away from the equilibrium point.
Thus, when the spring is released, it will move toward the equilibrium point. This is also true about the acceleration.
Therefore, we can conclude that the direction of the acceleration vector is towards the equilibrium point.
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Help me <3 please
Thank you :)
Answer:
11,890
Explanation:
First we need to know what is considered a significant figure.
A significant figure is a value that is not a zero at the start OR end of a value.
Which means, the 0 in the value of 90 or 0.363 are not considered a significant figure.
The 0 in the value of 3056 is considered a significant figure.
So from the table, we can deduce:
0.275 has 3 significant figures
750 has 2 significant figures
[tex]10.4 \times {10}^{5} = 1040000[/tex]
has 3 significant figures.
11,890 has 4 significant figures.
320,050 has 5 significant figures.
So from the above, we can already see the answer.
A cart of mass 0.5 kg sits at rest on a table on which it can roll without friction. It is attached to an unstretched spring. You give the mass a push with a constant force over a distance of 5 cm in the direction that compresses the spring, after which the mass starts undergoing simple harmonic motion with a frequency of 0.5 complete oscillations per second and an amplitude of 15 cm.
A) What is the spring constant of the spring?
B) How fast was the cart moving at the instant when you finished pushing it?
C) What force did you exert on the cart?
(A) The spring constant of the spring is 4.94 N.
(B) The speed of the cart after pushing it is 0.47 m/s.
(C) The force applied to the cart is 0.75 N.
Spring constant
ω = √k/m
where;
ω is angular frequencyk is spring constantm is mass0.5 rev/s = 0.5(2π) rad/s = π rad/s = 3.142 rad/s
ω² = k/m
k = mω²
k = 0.5 x (3.142)²
k = 4.94 N/m
Energy stored in the springE = ¹/₂kA²
where;
A is amplitude
E = ¹/₂(4.94)(0.15)²
E = 0.056 J
Speed of the cartE = ¹/₂mv²
2E = mv²
v² = 2E/m
v² = (2 x 0.056)/(0.5)
v² = 0.224
v = √0.224
v = 0.47 m/s
Force exerted on the cartE = ¹/₂FA
2E = FA
F = 2E/A
F = (2 x 0.056)/(0.15)
F = 0.75 N
Thus, the spring constant of the spring is 4.94 N. The speed of the cart after pushing it is 0.47 m/s. The force applied to the cart is 0.75 N.
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According to the "Law of Increasing Opportunity Costs," what would be the opportunity cost of a student who is staying up all night to study for an exam that he has to take in the early morning?
A. Food or Drink.
B. Money or income.
C. Sleep or rest.
C. The opportunity cost of a student who is staying up all night to study for an exam that he has to take in the early morning is sleep or rest.
What is law of opportunity cost?
The law of increasing opportunity cost is an economic principle that describes how opportunity costs increase as resources are applied.
As the student gives up his sleep or night rest in the place of his exam preparation, we say that the opportunity cost is the sleep or rest.
Thus, the opportunity cost of a student who is staying up all night to study for an exam that he has to take in the early morning is sleep or rest.
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an 8 kilogram ball is fired horizontally form a 1 times 10^3 kilogram cannon initially at rest after having been fired the momentum of the ball is 2.4 time 10^3 kilogram meters per second east calculate the magnitude of the cannons velocity after the ball is fired
The magnitude of the cannons velocity after the ball is fired will be 2.4m/s.
To find the answer, we have to know more about the law of conservation of linear momentum.
How to find the magnitude of the cannons velocity after the ball is fired?The law of conservation of linear momentum states that, the initial momentum of a system will be equal to the final momentum.Here, then initial momentum of the system will be equal to zero, since the initial velocity of the cannon is equal to zero.Given that,[tex]M_B=8kg\\M_C=1*10^3 kg\\P_f=2.4*10^3 kgm/s.\\[/tex]
When the ball is fired from the canon, then the cannon will have a recoil velocity in the opposite direction of motion of the ball.Since it has a final momentum towards east, the recoil momentum will be in the west.Thus, the velocity of the cannon after when the ball is fired will be,[tex]P_f=M_CV_C\\V_C=\frac{P_f}{M_C}=\frac{2.4*10^3}{1*10^3} =2.4m/s \\west[/tex]
Thus, we can conclude that, the magnitude of the cannons velocity after the ball is fired is 2.4m/s.
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After the ball is shot, the cannon will move at a speed of 2.4 m/s.
We must learn more about the rule of conservation of linear momentum in order to locate the solution.
How can I determine the cannon's post-ball velocity magnitude?
According to the law of conservation of linear momentum, a system's starting and ultimate velocities are equal.Since the cannon's initial velocity is 0 in this case, the system's initial momentum will be equal to zero as well.We have,[tex]m=8kg\\M=1000kg\\P_f=2.4*10^3kgm/s[/tex]
When a cannon fires a ball, the cannon will recoil quickly and in the opposite direction of the ball's motion.Given that it is now moving eastward, the recoil momentum is towards the west.As a result, when the ball is fired, the cannon's velocity will be,[tex]P_f=MV\\V=\frac{P_f}{M} =2.4m/s[/tex]
Thus, we can infer that the cannon's maximum speed once the ball is fired is 2.4 m/s.
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A traffic light hangs from a pole as shown in (Figure 1). The uniform aluminum pole AB is 7.60 m long and has a mass of 10.0 kg . The mass of the traffic light is 23.5 kg. 1.) Determine the tension in the horizontal massless cable CD. 2) Determine the vertical component of the force exerted by the pivot A on the aluminum pole. 3.) Determine the horizontal component of the force exerted by the pivot A on the aluminum pole.
There is a 446N force in the horizontal massless cable CD. The vertical component of the pivot A's force on the aluminum pole, which is 328.3N. The force that the pivot A applies to the aluminum pole has a horizontal component of 446N.
We need to be aware of the force in order to discover the solution.
How can I determine the tension in the CD's horizontal massless cable?The free body diagram of the masses must be drawn in order to determine the tension in the horizontal cable CD.To obtain the tension on CD in the free body diagram, let's balance all the vertical and horizontal forces. [tex]TH-mg\frac{l}{2}cos\alpha -Mglcos\alpha =0\\T=\frac{glcos\alpha (\frac{m}{2}+M )}{h} \\T=446N[/tex]where, H=3.8m, l=7.6m, m=10kg, M=23.5 kg and alpha= 37 degree,
How to calculate the force the pivot A exerted on aluminum's vertical and horizontal components?The overall force acting vertically is,[tex]F_V-mg-Mg=0\\F_v=328.2N\\[/tex]
The overall force acting horizontally is,[tex]F_H=T=446N[/tex]
In light of this, we may say that the tension in the horizontal massless cable CD and the horizontal component of the force applied by the pivot A to the aluminum pole are identical and each exert 466N of force, whereas the vertical component of that force is 328.3N.
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do you think an Electromagnet can be used for separating plastic bags from a garbage heap? explain
Answer:
No
Explanation:
Plastic bags are not magnetic materials, only magnetic materials (such as iron) can be attracted by the magnet.
Hope this helps.
in the absence of friction, the output power of a winding engine is 100kw but thus is reduced by friction to 90kw . how much oil initially at 120° is required per second to to keep the Temperature of the bearing down to 70°C ? specific heat capacity of oil is 2100 j/kg°C.
please I need it now Bosses ♀️
Answer:
Explanation:
The angular speed of the rotor is 200 rad/s.
The torque needed to be transmitted by the engine is 180 Nm.
The power of the rotor required to transmit energy to apply a torque τ to rotate a motor with angular speed ω,
P=τω
=180×200W
=36kW
A proton is moved from the negative to the positive plate of a parallel-plate arrangement. The plates are 1.50 cm apart, and the electric field is uniform with a magnitude of 1 500 N/C.
What is the proton’s potential energy change?
What is the potential difference between the plates?
What is the potential difference between the negative plate and a point midway between the plates?
If the proton is released from rest at the positive plate, what speed will it have just before it hits the negative plate?
(a) The proton’s potential energy change is 3.6 x 10⁻¹⁸ J.
(b) The potential difference between the negative plate and a point midway between the plates is 11.25 V.
(c) The speed of the proton just before it hits the negative plate is 6.57 x 10⁴ m/s.
Potential energy of the proton
U = qΔV
where;
q is charge of the protonΔV is potential differenceU = q(Ed)
U = (1.6 x 10⁻¹⁹)(1500 x 1.5 x 10⁻²)
U = 3.6 x 10⁻¹⁸ J
Potential difference between the negative plate and a point midwayΔV = E(0.5d)
ΔV = 0.5Ed
ΔV = 0.5 (1500)(1.5 x 10⁻²)
ΔV = 11.25 V
Speed of the protonU = ¹/₂mv²
U = mv²
v² = 2U/m
where;
m is mass of proton = 1.67 x 10⁻²⁷ kgv² = (2 x 3.6 x 10⁻¹⁸) / ( 1.67 x 10⁻²⁷)
v² = 4.311 x 10⁹
v = √(4.311 x 10⁹)
v = 6.57 x 10⁴ m/s
Thus, the proton’s potential energy change is 3.6 x 10⁻¹⁸ J.
The potential difference between the negative plate and a point midway between the plates is 11.25 V.
The speed of the proton just before it hits the negative plate is 6.57 x 10⁴ m/s.
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How do I connect with my higher self?
Answer:
Create space
Watch your breath
Watch your thoughts
Be gentle with yourself
Affirm what you want
Explanation:
The figure below shows a motorcycle leaving the end of a ramp with a speed of 39.0 m/s and following the curved path shown. At the peak of the path, a maximum height h above the top of the ramp, the motorcycle's speed is 37.1 m/s. What is the maximum height h? Ignore friction and air resistance. (Enter your answer in m.)
_____m
The maximum height h of the curved path is 7.38 m.
Maximum height of the curved path
Apply the following kinematic equation;
v² = u² - 2gh
where;
v is the final velocity of the motorcycleu is initial velocity of the motorcycleh is the maximum height(u² - v²)/2g = h
(39² - 37.1²)/(2 x 9.8) = h
7.38 m = h
Thus, the maximum height h of the curved path is 7.38 m.
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A 0.80-kg mass attached to the end of a string swings in a vertical circle (radius = 2.0 m). When the mass is at the highest point of the circle, the speed of the mass is 9.0 m/s. What is the magnitude of the force of the string on the mass at this position?
Answer:
Approximately [tex]25\; {\rm N}[/tex] assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].
Explanation:
This mass is in a circular motion of radius [tex]r[/tex]. Hence, when the velocity of the mass is [tex]v[/tex], the acceleration of this mass should be [tex](v^{2} / r)[/tex]. The net force on this mass should be [tex](\text{net force}) = (m\, v^{2}) / r[/tex] towards the center of the circle.
When this [tex]m = 0.80\; {\rm kg}[/tex] mass is at the top of this circle, both gravitational pull and the force of the string (tension) point downwards. Hence, the net force on this mass would be:
[tex](\text{net force}) = (\text{weight}) + (\text{tension})[/tex].
Thus:
[tex]\begin{aligned} (\text{tension}) &= (\text{net force}) -(\text{weight})\\ &= \frac{m\, v^{2}}{r} - m\, g \\ &= m\, \left(\frac{v^{2}}{r} - g\right) \\ &= 0.80\; {\rm kg}\times \left(\frac{(9.0\; {\rm m\cdot s^{-1}})^{2}}{2.0\; {\rm m}} - 9.81\; {\rm m\cdot s^{-2}}\right) \\ &\approx 25\; {\rm kg \cdot m\cdot s^{-2}} \\ &= 25\; {\rm N}\end{aligned}[/tex].
When a scientific calculator shows the quantity below, what does it mean?
1.5E8
A 15.0-kg child descends a slide 2.40 m high and reaches the bottom with a speed of 1.10 m/s .
How much thermal energy due to friction was generated in this process?
Express your answer to three significant figures and include the appropriate units.
The thermal energy that is generated due to friction is 344J.
What is the thermal energy?Now we know that the total mechanical energy in the system is constant. The loss in energy is given by the loss in energy.
Thus, the kinetic energy is given as;
KE = 0.5 * mv^2 =0.5 * 15.0-kg * (1.10 m/s)^2 = 9.1 J
PE = mgh = 15.0-kg * 9.8 m/s^2 * 2.40 m = 352.8 J
The thermal energy is; 352.8 J - 9.1 J = 344J
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A 21.0 kg uniform beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.
If the angle between the beam and the cable is θ = 66.0° what is the tension in the cable?
Since upward forces must be equal to the downward forces, the tension in the cable is 225.3 N
What are the two conditions for equilibrium ?The two conditions are;
Sum of the upward forces must be equal to the sum of the downward forces.The sum of the clockwise moment must be equal to the sum of the anticlockwise moment.The given parameters are;
Mass m = 21 kgangle θ = 66°Tsinθ = mg
Tsin 66 = 21 x 9.8
T = 205.8 / 0.914
T = 225.3 N
Therefore, the tension in the cable is 225.3 N
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The small spherical planet called "Glob" has a mass of 7.88×10^18 kg and a radius of 6.32×10^4 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×10^3 m, above the surface of the planet, before it falls back down.
1. What was the initial speed of the rock as it left the astronaut's hand? (Glob has no atmosphere, so no energy is lost to air friction. G = 6.67×10^-11 Nm2/kg2.)
2. A 36.0 kg satellite is in a circular orbit with a radius of 1.45×10^5 m around the planet Glob. Calculate the speed of the satellite.
(a) The initial speed of the rock as it left the astronaut's hand is 19.35 m/s.
(b) The speed of the satellite is 50.24 m/s.
Acceleration due to gravity of the planet
g = GM/R²
where;
M is mass of the planetR is radius of the planetg = (6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(63200)²
g = 0.13 m/s²
Initial speed of the rockv² = u² - 2gh
where;
v is final velocityu is initial velocityat maximum height, v = 0
u² = 2gh
u = √2gh
u = √(2 x 0.13 x 1,440)
u = 19.35 m/s
Speed of the satellitev = √GM/r
M is mass of the planet Globr is the total distance from the center of the planet Globr = radius of planet Glob + radius of the satellite
r = 63200 m + 145,000 m = 208,200 m
v = √[(6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(208,200)]
v = 50.24 m/s
Thus, the initial speed of the rock as it left the astronaut's hand is 19.35 m/s.
The speed of the satellite is 50.24 m/s.
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using a weston cadmium cell of 1.0283 v and a standard resistance of .1ohm a potentiometer was adjusted so that 1.0183 m was equivalent to the emf of the cell: when a certain direct current was flowing through the standard resistance, the voltage across it corresponds to 150 cm. what was the value of current
Answer:
Explanation:
the value is -8 cm
A rectangle measuring 30.0 cm by 40.0 cm is located inside a region of a spatially uniform magnetic field of 1.65 T, with the field perpendicular to the plane of the coil (Figure 1). The coil is pulled out at a steady rate of 2.00 cm/s traveling perpendicular to the field lines. The region of the field ends abruptly as shown.
a) Find the emf induced in this coil when it is all inside the field.
b) Find the emf induced in this coil when it is partly inside the field.
c) Find the emf induced in this coil when it is all outside the field.
The solution for the questions below is mathematically given as
induced emf=0induced emf=0.0132Vinduced emf=0What is the emf induced in this coil when it is all inside the field.?(A)
Generally, the equation for the flux is mathematically given as
[tex]\Phi _{B}=B.A[/tex]
[tex]\Phi _{B}=B.A\\\\\Phi _{B}=B(40*30*10^{-4}) ,[/tex]
So, the magnetic flux through the coil is constant.
From faradays law,
[tex]\varepsilon =-\frac{\mathrm{d} \Phi _{B}}{\mathrm{d} t}[/tex]
---(1) for the induced emf. Since magnetic flux is constant, LHS. of (1) =0
induced emf=0
(B)
Let x be the length of the coil's magnetic field area.
[tex]then, \Phi _{B}=B.A=B(40*x*10^{-4}) \\\\\frac{\mathrm{d}\Phi _{B} }{\mathrm{d} t}=40\\B*10^{-4}*\frac{\mathrm{d} x}{\mathrm{d} t}[/tex]
induced emf=0.0132V
(C)
In conclusion, Therefore, there is no variation in the magnetic flux across the coil when magnetic flux=0.
induced emf=0
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converting 67 m•s¹ to km•h¹
Answer:
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I need to figure out how to extend a single paragraph's worth of answers that I can make into 3 paragraphs, requesting a LOT of help here.
Question:
Discuss one feature of a spacesuit that protects astronauts outside the spacecraft.
An astronaut's spacesuit is considerably more than just a pair of garments they put on during spacewalks. An entire spacesuit is actually a single-person spacecraft. The Extravehicular Mobility Unit, or EMU, is the official name for the spacesuit used on the International Space Station and Space Shuttle. "Extravehicular" refers to something that is outside of a vehicle or spaceship. To be "mobile" implies to be able to move around while wearing an astronaut suit. The astronaut is shielded by the spacesuit from the perils of being outdoors in space.
The human body cannot survive in space's hostile atmosphere without protection. A spacesuit must shield the astronaut from the vacuum of space, the drastic temperature changes in space, and, if at all possible, it must lessen the astronaut's exposure to radiation. Therefore, the suit's primary function is to act as a pressure vessel. It must keep an air environment close to the skin constant, deliver a constant supply of clean air to the lungs, and expel stale, carbon dioxide-rich air. Every 90 minutes or so, an astronaut in low Earth orbit (LEO) experiences day and night. They can fast chill to -250 F during the night and reach 250 F while the sun is shining on them (-156 C). The body's normal temperature of 98.6 F must be maintained by the suit (37 C).
To do all this, the current NASA suit has 14 layers. The liquid ventilation and cooling garment is the first three layers. It resembles a body-hugging spandex garment that has tubes inside that carry cool water across the body to dissipate extra heat. Air below the next layer cannot escape since it is a pressure vessel made of nylon coated with urethane. The following layer resembles a tent since it is constructed of Dacron. Its goal is to exert pressure on the pressure garment so that it keeps its form and doesn't expand. Neoprene coasted nylon ripstop is the following layer. It is quite resilient to tears. Seven layers of mylar film and foil blanket are used to cover it. These restrict warmth transfer into or out of the suit by acting like a thermos. Due to its several layers, it also provides defense against very small micrometeroids, which drain energy as they pierce and break each layer. The top layer is made of orthofabric, a goretex, nomex, and kevlar mixture. It is very impervious to tearing and thermally reflecting to help regulate temperature. The suits are made by ILC Dover. Incorporating materials that minimize radiation penetration through the suit is something they are always working on.
An electron moves at 0.130 c as shown in the figure (Figure 1). There are points: A, B, C, and D 2.10 μm from the electron.
a) Find the magnitude of the magnetic field this electron produces at the point A .
b) Find the magnitude of the magnetic field this electron produces at the point B .
c) Find the magnitude of the magnetic field this electron produces at the point C .
d) Find the magnitude of the magnetic field this electron produces at the point D
Hi there!
We can use Biot-Savart's Law for a moving particle:
[tex]B= \frac{\mu_0 }{4\pi}\frac{q\vec{v}\times \vec{r}}{r^2 }[/tex]
B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)
q = charge of particle (1.6 × 10⁻¹⁹ C)
μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)
r = distance from particle (2.10 μm)
There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:
[tex]B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }[/tex]
Where 'θ' is the angle between the velocity and radius vectors.
a)
To find the angle between the velocity and radius vector, we find the complementary angle:
θ = 90° - 60° = 30°
Plugging 'θ' into the equation along with our other values:
[tex]B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }\\\\B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(30)}{(2.1*10^{-5})^2 }[/tex]
[tex]B = \boxed{7.07 *10^{-10} T}[/tex]
b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.
c)
In this instance, the radius vector and the velocity vector are perpendicular so
'θ' = 90°.
[tex]B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(90)}{(2.1*10^{-5})^2 } = \boxed{1.415 * 10^{-9}T}[/tex]
d)
This point is ALONG the velocity vector, so there is no magnetic field produced at this point.
Aka, the radius and velocity vectors are parallel, and since sin(0) = 0, there is no magnetic field at this point.
[tex]\boxed{B = 0 T}[/tex]
A force of 100 newtons is applied to a box at an angle of 36° with the horizontal. If the mass of the box is 25 kilograms, what is the horizontal
acceleration of the box?
OA
1.52 meters/second²
OB. 3.24 meters/second²
OC. 5.48 meters/second²
O D.
6.87 meters/second²
OE
7.15 meters/second²
Answer:
See below
Explanation:
I will assume the force is in a DOWNWARD direction ( I believe it makes no answer difference)
Horizontal component is then 100 cos 36° =80.9 N
F = ma
80.9 = 25 kg *a
a = 3.24 m/s^2
Answer:
See image
Explanation:
Plato
b. Calculate the total resistance of the circuit below. (4 points)
c. In the circuit diagram above, the meters are labeled 1 and 2. Write 2 - 3 sentences identifying each type of meter and how it is connected with the 12 Ω resistor. (4 points)
d. In the circuit diagram above, predict which resistors (if any) will stop working when the switch is opened. Write 2 - 3 sentences explaining your reasoning. (4 points)
Please answer in complete sentences. Will mark brainliest.
The total resistance in the circuit, R is 4 Ω .
The meters connected to the 12 Ω resistance are:
Voltmeter - connected in parallelAmmeter - connected in seriesIf the switch is opened, only the 4 Ω and 2 Ω resistances will stop working.
What is the equivalent or total resistance in the circuit?The resistances in the circuit are connected both in series and in parallel
The resistances in series are the 4 Ω and the 2 Ω resistances.
Equivalent resistance = 4 + 2 = 6 Ω
The 4 Ω and the 2 Ω resistances are then connected in parallel with the 12 Ω resistance.
Total resistance, R is calculated as follows:
1/R = 1/12 + 1/6
1/R = 3/12
R = 12/3
R = 4Ω
The meters connected to the 12 Ω resistance are:
Voltmeter - connected in parallelAmmeter - connected in seriesIf the switch is opened, only the 4 Ω and 2 Ω resistances will stop working because the circuit connecting them to the cell is broken whereas the circuit to the 12 Ω resistance is continuous.
In conclusion, resistances can be connected in parallel or in series.
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The image shows a wheel that’s wound up and released. The wheel moves up and down as shown. Identify the position of the wheel when its potential energy is greatest.
The highest point of the wheel is the position of the wheel when its potential energy is greatest.
At what position of the wheel potential energy is greatest?The position of the wheel when its potential energy is greatest when it is at the highest point because potential energy depends on the height of an object. If the object is at more height then it has more potential energy and vice versa.
So we can conclude that the highest point of the wheel is the position of the wheel when its potential energy is greatest.
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A roller-coaster car shown in the figure below is pulled up to point 1 where it is released from rest. Take y = 39 m .
(Figure 1)
Assuming no friction, calculate the speed at point 2.
Express your answer to two significant figures and include the appropriate units.
Calculate the speed at point 3.
Express your answer to two significant figures and include the appropriate units.
Calculate the speed at point 4.
Express your answer to two significant figures and include the appropriate units.
The speed of the roller coaster at point 2 is 28 m/s, at point 3 is 16 m/s and at point 4 is 22 m/s.
Speed of the roller coasterThe speed of the roller coaster at any position is calculated by applying the principle of conservation of energy.
K.E = P.E
¹/₂mv² = mgh
v = √2gh
where;
h is vertical displacementSpeed at point 2v(2) = √[(2 x 9.8)(39 - 0)]
v(2) = 28 m/s
Speed at point 3v(3) = √[(2 x 9.8)(39 - 26)]
v(3) = 16 m/s
Speed at point 4v(4) = √[(2 x 9.8)(39 - 14)]
v(4) = 22 m/s
Thus, the speed of the roller coaster at point 2 is 28 m/s, at point 3 is 16 m/s and at point 4 is 22 m/s.
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A block with a mass of 33.0 kg is pushed with a horizontal force of 150 N. The block moves at a constant speed across a level, rough surface a distance of 4.95 m.
(a) What is the work done (in J) by the 150 N force?
_________J
(b) What is the coefficient of kinetic friction between the block and the surface?
________
The work done is given by 742.5 J while the coefficient of kinetic friction between the block and the surface is 0.46.
What is the work done?The work done is given by the use of the formula;
W = F * x
Where;
F = force applied
x = distance covered
W = 150 N * 4.95 m = 742.5 J
Now;
The coefficient of kinetic friction is given by;
μ = F/mg
μ = 150/ 33 * 9.8
μ = 0.46
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In physics, when a baseball player catches a ball, which one of newtons laws is it an example of?
A. 1st law
B. 2nd law
C. 3rd law
In Milikan’s experiment, a drop of radius of 1.64 μm and density 0.851 g/cm3 is suspended in the lower chamber when a downward-pointing electric field of 1.92 * 105 N/C is applied.
What is the weight of the drop?
Find the charge on the drop, in terms of e.
How many excess or deficit electrons does it have?
(a) The weight of the drop is 1.54 x 10⁻²⁵ N,
(b) The the charge on the drop, in terms of e is 5 x 10⁻¹²e and
(c) The excess electrons is 5 x 10⁻¹² electron.
Weight of the dropThe weight of the drop is calculated as follows;
Volume of the drop; V = ⁴/₃πr³
V = ⁴/₃π(1.64 x 10⁻⁶)³ = 1.845 x 10⁻¹⁷ m³
mass = density x volume
mass = 0.851 g/cm³ x 1.845 x 10⁻²³ cm³ = 1.572 x 10⁻²³ g = 1.57 x 10⁻²⁶ kg.
Weight = 1.57 x 10⁻²⁶ kg x 9.8 m/s² = 1.54 x 10⁻²⁵ N.
Charge on the dropq = F/E
q = (1.54 x 10⁻²⁵ N)/(1.92 x 10⁵)
q = 8.01 x 10⁻³¹ C
1.6 x 10⁻¹⁹ C = 1e
8.01 x 10⁻³¹ C = ?
= 5 x 10⁻¹²e
Excess electron on the drop1.6 x 10⁻¹⁹ C ------- 1 electron
8.01 x 10⁻³¹ C ------- ?
= 5 x 10⁻¹² electron
Thus, the weight of the drop is 1.54 x 10⁻²⁵ N, the the charge on the drop, in terms of e is 5 x 10⁻¹²e and the excess electrons is 5 x 10⁻¹² electron.
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Aster walks first 70 m in the direction 37° north of east, and then walks 82 m in the
direction 20° south of east, and finally walks 28 m in the direction 30° west of north.(2pt)
a) How far and at what angle is the Aster's final position from her initial position?
b) In what direction would she has to head to return to her initial position?
The Aster's final position from her initial position is 64 m
The angle is 300° and She has to head in West north direction to return to her initial position
What is Displacement ?The displacement is the distance travelled in a specific direction. Displacement is a vector quantity.
Given that Aster walks first 70 m in the direction 37° north of east, and then walks 82 m in the direction 20° south of east, and finally walks 28 m in the direction 30° west of north.
This can be solved by using bearing method. Cosine formula will be the best to solve for the distance D.
[tex]D^{2}[/tex] = [tex]70^{2}[/tex] + [tex]82^{2}[/tex] - 70 x 82 x Cos (37 + 20)
[tex]D^{2}[/tex] = 4900 + 6724 - 5740Cos57
[tex]D^{2}[/tex] = 11624 - 3126.23
[tex]D^{2}[/tex] = 8497.8
D = [tex]\sqrt{8497.8}[/tex]
D = 92.2 m
a) The Aster's final position from her initial position is 92.2 - 28 = 64 m
The angle = 270° + 30° = 300°
b) She has to head in West north direction to return to her initial position
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