Answer:
Distance, S = 440 meters.
Explanation:
Given the following data;
Initial velocity, u = 17m/s
Time, t = 20 seconds
Final velocity, v = 27m/s
To find the distance;
First of all, we would determine the acceleration of the truck.
Acceleration = (v-u)/t
Substituting the given values into the equation, we have;
Acceleration = (27 - 17)/20
Acceleration = 10/20
Acceleration = 0.5m/s²
Now, we would use the second equation of motion to find the distance traveled.
S = ut + ½at²
S = 17*20 + ½*0.5*20²
S = 340 + 0.25*400
S = 340 + 100
S = 440m
The equations of motion can be used to obtain the distance covered as 440 m.
We have to use of the equations that are used for uniformly accelerated motion in solving the problem. The chosen equation must be;
v^2 = u^2 + 2as and v = u + at
v = final velocity
u = initial velocity
a = acceleration
s = distance
To obtain the acceleration;
27 = 17 + 20(a)
27 - 17 = 20a
a = 0.5 ms-2
Now, to obtain the distance;
v^2 = u^2 + 2as
v^2 - u^2/as = s
s = (27)^2 - (17)^2/2(0.5)
s = 440 m
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A bottle of water at a room temperature of 21.0 C is placed into a refrigerator
with an air temperature of 4.5C. The thermal energy will move — *
A. from the cooler air to lower the temperature of the water to 4.5 C
B. in both directions until the temperature is equal in the water and the air
C. from the water to the air until the water temperature is zero degrees Celsius
O D. from the water to the air until the temperature is equal in both
Answer:
B. in both directions until the temperature is equal in the water and the air
Explanation:
When a warm body is in contact with a cool body , there is exchange of heat energy in both sides until there is attainment of equilibrium temperature . At this temperature both the body attains equal temperature . Initially rate of heat radiated by warm body is more than that from cool body , but after attainment of equilibrium , the rate becomes equal to each other . This is called dynamic equilibrium .
Hence option B is correct .
Express the speed of the electron in the Bohr model in terms of the fundamental constants (me, e, h, e0), the nuclear charge Z, and the quantum number n. Evaluate the speed of an electron in the ground states of He1 ion and U911. Compare these speeds with the speed of light c. As the speed of an object approaches the speed of light, relativistic effects become important. In which kinds of atoms do you expect relativistic effects to be greatest
Answer:
a) v = 4.37 10⁶ m / s, speed is much less than c
b) v = 2.01 10⁸ m / s, this value is 67% of the speed of light, , for which relativistic corrections should be used
Explanation:
The bohr model for the hydrogen atom and dendroids is a classical model with a quantization of the angular momentum
let's start by using Newton's second law with the electric force
F = m a
Coulomb's law electric force
F = [tex]k \frac{q_1q_2}{r^2}[/tex]
in this case in an atom the number of protons is equal to the atomic number and there is only one electron
q₁ = Ze
q₂ = e
acceleration is centripetal
a = v² / r
we substitute
[tex]k \frac{Z e^2}{r^2} = m \frac{v^2}{r}[/tex]
v² = [tex]k \frac{Ze^2}{m r}[/tex]
quantization is imposed without justification in this model,
L = p x r = n [tex]\hbar[/tex]
\hbar= h /2π
if we consider circular orbits, the speed and position are perpendicular
m v r = n \hbar
r = [tex]\frac{n \hbar}{m v}[/tex]
we substitute
v² = [tex]k \frac{Z e^2}{m} \frac{m v}{n \hbar}[/tex]
v = [tex]k \frac{Z e^2 }{ n \hbar}[/tex]
let's apply this equation
\hbar= h / 2π
\hbar= 6.626 10-34 / 2π
\hbar= 1.05456 10⁻³⁴ J s
a) He1 ion, the atomic number of helium is 2
v = [tex]\frac{9 \ 10^9 \ 2 ( 1.6 \ 10^{-19})^2 }{n \ 1.0546 \ 10^{-34}}[/tex]
v =4.3695 10⁶ / n m / s
the ground state occurs for N = 1
v = 4.37 10⁶ m / s
the relationship of this value to the speed of light is
v / c = 4.37 10⁶/3 10⁸
v / c = 1.46 10⁻²
speed is much less than c
b) the uranium ion with atomic number Z = 92
v = [tex]\frac{9 \ 10^9 \ 92 ( 1.6 \ 10^{-19})^2 }{n \ 1.054 \ 10^{-34} }[/tex]
v = 2.01 10⁸ m / s
v/c = [tex]\frac{2.01 \ 10^8 }{3 \ 10^8}[/tex]
v/c = 0.67
this value is 67% of the speed of light, for atoms with a higher atomic number the effects are increasingly important, for which relativistic corrections should be used
A runner starts from rest and stops in 12 seconds. He covers
100m distance. Using this information you can clain the
maximum absolute value of his acceleration was not less than:
a 0.69 m/s 2
b 1.39 m/s 2
c 2.78 m/s 2
d 3.47 m/s 2
Answer:
b 1.39 m/s²
Explanation:
Given the following data;
Time = 12 seconds
Distance, S = 100 m
Since it's starting from rest, the initial velocity is equal to 0m/s.
To find the acceleration, we would use the second equation of motion;
[tex] S = ut + \frac {1}{2}at^{2}[/tex]
Where;
S represents the displacement or height measured in meters.
u represents the initial velocity measured in meters per seconds.
t represents the time measured in seconds.
a represents acceleration measured in meters per seconds square.
Substituting into the equation, we have;
100 = 0(12) + ½*a*12²
100 = 0 + 72
100 = 72a
Acceleration, a = 100/72
Acceleration, a = 1.389 ≈ 1.39 m/s²
The membrane that surrounds a certain type of living cell has a surface area of 4.7 x 10-9 m2 and a thickness of 1.3 x 10-8 m. Assume that the membrane behaves like a parallel plate capacitor and has a dielectric constant of 4.7. (a) The potential on the outer surface of the membrane is 79.5 mV greater than that on the inside surface. How much charge resides on the outer surface
Answer:
Q = 1.2*10⁻¹² C
Explanation:
For any capacitor, by definition the capacitance C is equal to the relationship between the charge on one of the conductors and the potential difference between them, as follows:[tex]C = \frac{Q}{V} (1)[/tex]
For the special case of a parallel plate capacitor, just by application of Gauss' law to a rectangular surface half out of the outer surface, and half inside it, it can be showed that the value of the capacitance C is a parameter defined only by geometric constants, as follows:[tex]C = \frac{\epsilon_{0}*\epsilon _{r} * A}{d} (2)[/tex]
So, due to the left sides in (1) and (2) are equal each other, right sides must be equal too.Replacing ε₀, εr (dielectric constant), A, d and V by their values, we can solve for Q, as follows:[tex]Q =\frac{\epsilon_{0} * \epsilon_{r} *A* V}{d} = \frac{(8.85*(4.7)^{2}*79.5)e-24 (F/m*m2*V)}{1.3e-8m} = 1.2e-12 C = 1.2 pC (3)[/tex]
Imagine two cases: Block N is pushed by a hand, which exerts a constant force F_o. AND moves a distance d_ 0. In case 1, it takes a time T to move this distance. In a case 2, it takes time 2T to move this distance. The work done by the hand on N in case 1 is ____________ the work done by the hand in case 2.
a. greater than
b. less than
c, equal to
Answer:
C: Equal to
Explanation:
In calculating workdone, it is pertinent to know that it doesn't depend on time. The only relationship between work and time is when we want to calcite power where workdone/time taken = power.
Now, even if it took 2T time to love the same distance, it just means lesser force was used but still the workdone doesn't change.
Thus, the workdone in the first case will be equal to the workdone in the second case.
what do you call these sound waves whose frequency is above 20000 hertz
Answer:
Untrasound
Explanation:
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A 0.1 kg arrow with an initial velocity of 30 m/s hits a 4.0 kg melon initially at rest on a friction-less surface. The arrow emerges out the other side of the melon with a speed of 20 m/s. What is the speed of the melon? Why would we normally not expect to see the melon move with the is speed after being hit by the arrow?
Answer:
Speed of the melon = 0.25 m/s
we would normally don't see the melon moving due to friction with the resting surface.
Explanation:
We use conservation of momentum:
Pi = Pf
with Pi = 0.1 kg * 30 m/s = 3 kg m/s
and Pf = 0.1 kg * 20 m/s + 4.0 kg * V = 2 kg m/s + 4 * V
Then using the equality above, we solve for V (velocity of the melon)
3 kg m/s = 2 kg m/s + 4 V
1 kg m/s = 4 kg * V
Then V = 1 / 4 M/s = 0.25 m/s
So we would normally don't see the melon moving due to friction with the resting surface.
in a football game, the kicker kicks a football a horizontal distance of 43 yards if the ball lands 3.9 seconds later, what is the balls horizontal velocity
Answer:
10s
Explanation:
Horizontal velocity is the velocity of an object in an horizontal direction
The ball's horizontal velocity is approximately 33.078 ft./s
Reason:
The known parameter are;
The horizontal distance the footballer kicks the ball, d = 43 yards
The time after which the ball lands, Δt = 3.9 seconds
Required:
To find the velocity of the ball
Solution:
[tex]Velocity = \dfrac{Distance}{Time} = \dfrac{d}{\Delta t}[/tex]
Therefore;
[tex]Horizontal \ velocity \ of \ the \ ball, \ v_x= \dfrac{43 \ yard}{3.9 \ seconds} \approx 11.026 \ yd/s[/tex]
The ball's horizontal velocity, vₓ ≈ 11.026 yd/s
1 yard = 3 feet
[tex]11.026 \ yard = 11.026 \ yard \times \dfrac{3 \ feet}{yard} = 22.078 \ feet[/tex]
The ball's horizontal velocity, vₓ ≈ 33.078 ft./s
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A 0.11 kg bullet traveling at speed hits a 18.3 kg block of wood and stays in the wood. The block with the bullet imbedded in it moves forward with a velocity of 8.8 m/s. What was the velocity (speed) of the bullet immediately before it hit the block (in m/s)?
Explanation:
The energy of the system before the collision must equal the energy after the collision.
After the collision the bullet and the block have a total mass of 18.41 kg and they move at a speed of 8.8 m/s. The kinetic energy after the collision is
[tex]\frac{18.41 kg (8.8 m/s)^2}{2} = 713 J[/tex]
Before the collision only the bullet has kinetic energy.
So we can now determine the speed of the bullet using
[tex]\frac{0.11kg (v^2)}{2} = 713 J\\v = 114 m/s[/tex]
The "problem of perception" is best characterized as?
Answer:
making sense of a 3-d world from 2-d data
Explanation:
A dog runs at 35 m/s at 45 degrees N of E. What are its x and y components (all answers
are in m/s)?
Answer:
its x and y component is 24.749m/s
Explanation:
Given
Speed of the dog = 35m/s
x component of the speed = xcos theta
y component of the speed = ycos theta
Given theta =45 degrees
x-component = 35cos45
x-component = 35(0.7071)
x-component = 24.749m/s
y-component = 35sin45
y-component = 35(0.7071)
y-component = 24.749m/s
Hence its x and y component is 24.749m/s
A student swings a 0.5kg rubber ball attached to a string over her head in a horizontal, circular
path. The string is 1.5 meters long and in 60 seconds the ball makes 120 complete circles.
What is the velocity of the ball?
What is the ball’s centripetal acceleration?
What is the ball's centripetal force?
Answer:
The balls velocity is 1 divided by 3
The velocity of the ball is 18.85 m/s.
The ball’s centripetal acceleration is 236.87 m/s².
The ball's centripetal force is 118.44 Newton.
What is centripetal acceleration?Centripetal acceleration is a characteristic of an object's motion along a circular path. Centripetal acceleration applies to any item travelling in a circle with an acceleration vector pointing in the direction of the circle's center.
Given parameters:
length of the string: l = 1.5 meters.
Time interval = 60 seconds.
Total number of complete rotation = 120.
Hence, the velocity of the ball = 120×2π×1.5/60 m/s
= 18.85 m/s.
The ball’s centripetal acceleration = (velocity)²/ radius
= (18.85)²/1.5 m/s²
= 236.87 m/s²
The ball's centripetal force = mass × centripetal acceleration
= 0.5 × 236.87 Newton
= 118.44 Newton
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what is air resistance means explain it with free falling body
how do i get the answer for keplers law 3
A police officer is parked by the side of the road, when a speeding car travelling at 50 mi/hrpasses. The police car immediately pursues it, accelerating at a rate of 10 mi/hr per second.The road is fairly busy, so the officer will not go faster than a top speed of 70 mi/hr. How longwill it take before the officer catches up to the speeding car, and how far will it have travelled inorder to do so
Answer:
a) time taken to catch up with speeding car is 12.25 secs
b) the police car will travel 273.8 m to catch up with the speeding car
Explanation:
Given that;
speed of car [tex]V_{c}[/tex] = 50 mi/hr = 22.352 m/s
acceleration of police car = 10 mi/hr = 4.47 m/s²
[tex]V_{f}[/tex] = 70 mi/hr = 31.29 m/s
Now time taken to reach maximum speed is t₁
so
[tex]V_{f}[/tex] = [tex]V_{i}[/tex] + at₁
we substitute
31.29 = 0 + 4.47t₁
t₁ = 31.29 / 4.47
t₁ = 7 sec
now
d₁ = 0 + 1/2 × at₁²
d₁ = 0 + 1/2 × 0 + 4.47×(7)²
d₁ = 109.5 m
so distance travelled by the speeding car in time t₁ will be
[tex]d_{c}[/tex] = [tex]V_{c}[/tex] × t₁
we substitute
[tex]d_{c}[/tex] = 22.352 × 7
[tex]d_{c}[/tex] = 156.46 m
now distance between polive car and speeding car
Δd = [tex]d_{c}[/tex] - d₁
Δd = 156.46 - 109.5
Δd = 46.96 m
time taken to cover Δd will be
t₂ = Δd / ( [tex]V_{f}[/tex] - [tex]V_{c}[/tex] )
t₂ = 46.96 / ( 31.29 - 22.352 )
t₂ = 46.96 / 8.938
t₂ = 5.25 sec
distance travelled by the police in time t₂ will be
d₂ = [tex]V_{f}[/tex] × t₂
d₂ = 31.29 × 5.25
d₂ = 164.3 m
a) How long will it take before the officer catches up to the speeding car;
time taken to catch up with speeding car;
t = t₁ + t₂
t = 7 + 5.25
t = 12.25 secs
Therefore, time taken to catch up with speeding car is 12.25 secs
b) how far will it have travelled in order to do so;
distance = d₁ + d₂
distance = 109.5 + 164.3
distance = 273.8 m
Therefore, the police car will travel 273.8 m to catch up with the speeding car
A car is traveling 100 km/hr. How many hours will it take to cover a distance of 850 km?
Your answer:
.118 hours
8.5 hours
7.5 hours
23 hours
Answer:
8.5 hours
Explanation:
Arrange the objects in order from greatst to least of potential energy assume that gravity is constant
Answer:
Water > Box of books > Stone > Ball
Explanation:
We'll begin by calculating the potential energy of each object. This can be obtained as follow:
For stone:
Mass (m) = 15 Kg
Acceleration due to gravity (g) = 10 m/s²
Height (h) = 3 m
Potential energy (PE) =?
PE = mgh
PE = 15 × 10 × 3
PE = 450 J
For water:
Mass (m) = 10 Kg
Acceleration due to gravity (g) = 10 m/s²
Height (h) = 9 m
Potential energy (PE) =?
PE = mgh
PE = 10 × 10 × 9
PE = 900 J
For ball:
Mass (m) = 1 Kg
Acceleration due to gravity (g) = 10 m/s²
Height (h) = 20 m
Potential energy (PE) =?
PE = mgh
PE = 1 × 10 × 20
PE = 200 J
For box of books:
Mass (m) = 25 Kg
Acceleration due to gravity (g) = 10 m/s²
Height (h) = 2 m
Potential energy (PE) =?
PE = mgh
PE = 25 × 10 × 2
PE = 500 J
Summary:
Object >>>>>>>> Potential energy
Stone >>>>>>>>> 450 J
Water >>>>>>>>> 900 J
Ball >>>>>>>>>>> 200 J
Box of books >>> 500 J
Arranging from greatest to least, we have:
Object >>>>>>>> Potential energy
Water >>>>>>>>> 900 J
Box of books >>> 500 J
Stone >>>>>>>>> 450 J
Ball >>>>>>>>>>> 200 J
Water > Box of books > Stone > Ball
A movie stunt double is supposed to run across the top of a train (in the opposite direction that the train is moving) and just barely jump off before reaching a tunnel, but after reaching the end of the train (starting from the front). If the train is moving at 150 km/hr, is 2 km long and the tunnel is 20 km away from the end (where the stunt double is going to jump from), how fast (in km/hr) will the stunt double need to run
Answer:
the required speed/velocity of the stunt double is 13.633 km/h
Explanation:
Given the data in the question;
velocity of train V = 150 km/h
distance = length of train + distance between the tunnel and the end
= 2 km + 20 km = 22 km
first we calculate time t taken by the train to reach the tunnel;
t = distance / velocity
we substitute
t = 22 km / 250 km/h
t = 0.1467 hr
so the velocity of the of the stunt double will be;
velocity = distance / time
we substitute
velocity = 2 km / 0.1467 hr
velocity = 13.633 km/h
Therefore, the required speed/velocity of the stunt double is 13.633 km/h
g A boat is anchored 2000 ft from shore anddirects its searchlight towards an automobile travelingdown the straight road. At the particular moment whenthe distance rfrom the searchlight to the automobile is3000 ft, the automobile has speed 80 ft/s and increasesits speed at a rate of 15 ft/s2 down the road.Find the required angular velocity and angularacceleration of the boat's searchlight to track the automobile at this instant.(Answers: 0.018 rad/s CCW, 0.004 rad/s2 CCW)
Answer:
Explanation:
The distance of searchlight will act as radius R and velocity of car may be supposed to be tangential velocity v . We are required to calculate angular velocity ω .
v = 80 ft /s
R = 3000 ft
ω = v / R
= 80 / 3000 = .027 rad / s
For angular acceleration the formula is
angular acceleration α = a / R
a is linear acceleration = 15 ft / s²
α = 15 / 3000 = .005 rad / s².
You are sitting in your car at rest at a traffic light with a bicyclist at rest next to you in the adjoining bicycle lane. As soon as the traffic light turns green, your car speeds up from rest to 49.0 mi/h with constant acceleration 8.00 mi/h/s and thereafter moves with a constant speed of 49.0 mi/h. At the same time, the cyclist speeds up from rest to 21.0 mi/h with constant acceleration 14.00 mi/h/s and thereafter moves with a constant speed of 21.0 mi/h.
A. For what time interval (in s) after the light turned green is the bicycle ahead of your car?
B. By what maximum distance does the bicycle lead the car?
Answer:
A. 2.63 s B. 12.38 m
Explanation:
A. For what time interval (in s) after the light turned green is the bicycle ahead of your car?
The time interval at which the bicycle is ahead of the car is the time it takes for the car to reach the bicycle's speed of 21.0 mi/h.
So, using v = u + at where u = initial speed of car = 0 mi/h, v = final speed of car = 21.0 mi/h, a = acceleration of car = 8.00 mi/h/s and t = time taken for acceleration.
So, v = u + at
t = (v - u)/a
substituting the values of the variables into the equation, we have
t = (21.0 mi/h - 0 mi/h)/8.00 mi/h/s
= 21.0 mi/h ÷ 8.00 mi/h/s
= 2.63 s
B. By what maximum distance does the bicycle lead the car?
To find this distance, we find the distance moved by both the car in this time of t = 2.63 s
So, using s = ut + 1/2at² where u = initial speed of car = 0 mi/h = 0 m/s, t = time = 2.63 s, a = acceleration of car = 8.00 mi/h/s = 8.00 × 1609 m/3600 s = 3.58 m/s/s = 3.58 m/s² and s = distance moved by car.
So, substituting the values of the variables into the equation, we have
s = ut + 1/2at²
s = 0 m/s × 2.63 s + 1/2 × 3.58 m/s² × (2.63 s)²
s = 0 m + 1/2 × 3.58 m/s² × (2.63 s)²
s = 1.79 m/s² × 6.9169 s²
s = 12.38 m
which is also the maximum distance with which the bicycle leads the car.
Grace drives her car 168 km in 2 hours. What is her average speed in kilometers per hour?
Answer:
84kliometers
Explanation:
divide one hundred and sixty eight kilo meters by two hours
2. Am 80.0 kg astronaut is training for accelerations that he will experience upon re-entry to earth’s gravity from space. He is placed in a centrifuge (r = 25.0 m) and spun at a constant angular velocity of 10.0 rpm (revolutions per minute).
a. Find the linear velocity of the centrifuge in m/s. Show your work
b. Find the magnitude and direction of the centripetal acceleration when he is spinning at this constant velocity.
c. How many g’s is the astronaut experiencing? (at constant velocity)
d. Find the linear deceleration and torque required to bring the centrifuge (5000.0 kg) to a stop over a 5 minute time period.
Answer:
a) v = 26.2 m / s, b) acceleration is radial, a = 27.4 m / s², c) a = 2.8 g and
d) a = - 8.73 10⁻² m / s², τ = 1.09 10⁴ N m
Explanation:
a) For this exercise we can use the relationships between rotational and linear motion
v = w r
let's reduce the magnitudes to the SI system
w = 10 rpm (2pi rad / 1 rev) 1 min / 60s) = 1,047 rad / s
r = 25.0 m
let's calculate
v = 1.047 25.0
v = 26.2 m / s
b) When the body is rotating at constant speed, the relationship must be perpendicular to the speed, therefore the direction of acceleration is radial, that is, towards the center of the circle and its magnitude is
a = v² / r
a = 26.2²/25
a = 27.4 m / s²
c) Let's look for the relationship between the centripetal acceleration and the acceleration due to gravity
a / g = 27.4 / 9.8
a / g = 2.8
a = 2.8 g
d) let's find the deceleration and torque to stop the centripette in 5 min
t = 5 min (60 s / 1min) = 300 s
let's use the rotational kinematics relations
w = w₀ + α t
initial angular velocity is wo = 1,047 rad / s and the final as is stop do w = 0
α = - w₀ / t
α = - 1,047 / 300
α = -3.49 10⁻³ rad / s²
angular and linear are related
a = α r
a = -3.49 10⁻³ 25
a = - 8.73 10⁻² m / s²
the negative sign indicates that the acceleration is stopping the movement
torque is
τ = F r
The force can be found with Newton's second law
F = m a
we substitute
τ = m a r
τ = 5000.0 8.73 10⁻² 25
τ = 1.09 10⁴ N m
1. State the law of conservation of energy and what it means for you as a human considering how energy works.
2. Explain how different forms of energy are related.
PLEASE I NEED HELP!! I NEED IT NOW!! AND PLEASE DO IT IN YOUR OWN WORDS!! THANK YOU!
Answer: 1. The law of consevation of energy sates that energy can neither be created nor destroyed. It can only be transformed or transfered from one form to another. The law of conservation of energy is found everywhere for example, Water falls from the sky, converting potential energy to kinetic energy.
2. Different forms of energy are related because energy cannot be created or destroyed. they can all be transformed into from one form to another.
Explanation:
3.
A student swings a ball attached to the end of a string 0.5m in length in
a vertical circle. The speed of the ball is 2m/s at the highest point and
6m/s at its lowest point: Find the acceleration of the ball at (ii) its highest
point and (ii) its lowest point.
Answer:
I.72m/s²
II.8m/s²
Explanation:
acceleration equal velocity² divided by length
A 3.50 kg box that is several hundred meters above the surface of the earth is suspended from the end of a short vertical rope of negligible mass. A time-dependent upward force is applied to the upper end of the rope, and this results in a tension in the rope of T(t) = (36.0 N/s)t. The box is at rest at t = 0. The only forces on the box are the tension in the rope and gravity.
Required:
a. What is the velocity of the box at (i) t = 1.00 s and (ii) t = 3.00 s?
b. What is the maximum distance that the box descends below its initial position?
c. At what value of t does the box return to its initial position?
Answer:
a. i. -4.65 m/s ii. -13.95 m/s b. 5.89 m c. 2.85 s
Explanation:
a. What is the velocity of the box at (i) t = 1.00 s and (ii) t = 3.00 s?
We write the equation of the forces acting on the mass.
So, T - mg = ma where T = tension in vertical rope = (36.0 N/s)t, m = mass of box = 3.50 kg, g = acceleration due to gravity = 9.8 m/s² and a = acceleration of box = dv/dt where v = velocity of box and t = time.
So, T - mg = ma
T/m - g = a
dv/dt = T/m - g
dv/dt = (36.0 N/s)t/3.50 kg - 9.8 m/s²
dv/dt = (10.3 m/s²)t - 9.8 m/s²
dv = [(10.3 m/s²)t - 9.8 m/s²]dt
Integrating, we have
∫dv = ∫[(10.3 m/s³)t - 9.8 m/s²]dt
∫dv = ∫(10.3 m/s³)tdt - ∫(9.8 m/s²)dt
v = (10.3 m/s³)t²/2 - (9.8 m/s²)t + C
v = (5.15 m/s³)t² - (9.8 m/s²)t + C
when t = 0, v = 0 (since at t = 0, box is at rest)
So,
0 = (5.15 m/s³)(0)² - (9.8 m/s²)(0) + C
0 = 0 + 0 + C
C = 0
So, v = (5.15 m/s³)t² - (9.8 m/s²)t
i. What is the velocity of the box at t = 1.00 s,
v = (5.15 m/s³)(1.00 s)² - (9.8 m/s²)(1.00 s)
v = 5.15 m/s - 9.8 m/s
v = -4.65 m/s
ii. What is the velocity of the box at t = 3.00 s,
v = (5.15 m/s³)(3.00 s)² - (9.8 m/s²)(3.00 s)
v = 15.45 m/s - 29.4 m/s
v = -13.95 m/s
b. What is the maximum distance that the box descends below its initial position?
Since v = (5.15 m/s³)t² - (9.8 m/s²)t and dy/dt = v where y = vertical distance moved by mass and t = time, we need to find y.
dy/dt = (5.15 m/s³)t² - (9.8 m/s²)t
dy = [(5.15 m/s³)t² - (9.8 m/s²)t]dt
Integrating, we have
∫dy = ∫[(5.15 m/s³)t² - (9.8 m/s²)t]dt
∫dy = ∫(5.15 m/s³)t²dt - ∫(9.8 m/s²)tdt
∫dy = ∫(5.15 m/s³)t³/3dt - ∫(9.8 m/s²)t²/2dt
y = (1.72 m/s³)t³ - (4.9 m/s²)t² + C'
when t = 0, y = 0.
So,
0 = (1.72 m/s³)(0)³ - (4.9 m/s²)(0)² + C'
0 = 0 + 0 + C'
C' = 0
y = (1.72 m/s³)t³ - (4.9 m/s²)t²
The maximum distance is obtained at the time when v = dy/dt = 0.
So,
dy/dt = (5.15 m/s³)t² - (9.8 m/s²)t = 0
(5.15 m/s³)t² - (9.8 m/s²)t = 0
t[(5.15 m/s³)t - (9.8 m/s²)] = 0
t = 0 or [(5.15 m/s³)t - (9.8 m/s²)] = 0
t = 0 or (5.15 m/s³)t = (9.8 m/s²)
t = 0 or t = (9.8 m/s²)/(5.15 m/s³)
t = 0 or t = 1.9 s
Substituting t = 1.9 s into y, we have
y = (1.72 m/s³)t³ - (4.9 m/s²)t²
y = (1.72 m/s³)(1.9 s)³ - (4.9 m/s²)(1.9 s)²
y = (1.72 m/s³)(6.859 s³) - (4.9 m/s²)(3.61 s²)
y = 11.798 m - 17.689 m
y = -5.891 m
y ≅ - 5.89 m
So, the maximum distance that the box descends below its initial position is 5.89 m
c. At what value of t does the box return to its initial position?
The box returns to its original position when y = 0. So
y = (1.72 m/s³)t³ - (4.9 m/s²)t²
0 = (1.72 m/s³)t³ - (4.9 m/s²)t²
(1.72 m/s³)t³ - (4.9 m/s²)t² = 0
t²[(1.72 m/s³)t - (4.9 m/s²)] = 0
t² = 0 or (1.72 m/s³)t - (4.9 m/s²) = 0
t = √0 or (1.72 m/s³)t = (4.9 m/s²)
t = 0 or t = (4.9 m/s²)/(1.72 m/s³)
t = 0 or t = 2.85 s
So, the box returns to its original position when t = 2.85 s
What is the difference between elastic PE and gravitational PE?
A car is sitting still. It accelerates to a constant speed then it decelerates again to zero speed. While the car is accelerating how do the directions of the angular acceleration and angular velocity of one of the wheels compare
Answer:
in the acceleration process the quantity α and w must increase
the deceleration process the alpha quantity must constant a direction opposite to the angular velocity
Explanation:
Acceleration and angular velocity are related to linear
v = w xr
a = αx r
The bold letters indicate vectors and the cross is a vector product, therefore if
we can see that the relationship between linear and angular variables is direct
therefore in the acceleration process the quantity α and w must increase as well as their linear counterparts
in the deceleration process the alpha quantity must constant as the linear acceleration and must have a direction opposite to the angular velocity
what occurred when the photosynthetic began to pump free oxygen into oceans?
when the photosynthetic began to pump free oxygen into oceans, the ocean had enough oxygen to support the life of non-photosynthetic organisms. So, non-photosynthetic organisms came into being.
6th grade science I mark as brainliest
Answer:
7 would be C, a cell.
Explanation:
Hi.
7 would be C, a cell.
A cell is the basic unit of structure and function in all living things.
If it is living, it is made of cells.
Hope this helps.
Answer:
7. Cell
8. Organelle
A 107 gram apple falls from a branch that is 2 meters above the ground. (a) How much time elapses before the apple hits the ground
Answer:
The time of motion is 0.64 s.
Explanation:
Given;
mass of the apple, m = 107 g
height of fall, h = 2 m
The velocity of the apple when it hits the ground is calculated from the law of conservation of energy;
[tex]P.E = K.E\\\\mgh = \frac{1}{2} mv^2\\\\gh = \frac{1}{2} v^2\\\\v^2 = 2gh\\\\v = \sqrt{2gh} \\\\v = \sqrt{2\times 9.8\times 2} \\\\v = 6.261 \ m / s[/tex]
The time of motion is calculated;
v = u + gt
6.261 = 0 + 9.8t
6.261 = 9.8t
t = 6.261 / 9.8
t = 0.64 s
Therefore, the time of motion is 0.64 s
The time taken for the apple to hit the ground is 0.64 s.
The time taken for the apple to hit the ground can be calculated using the formula below.
Formula:
s = ut+gt²/2............ Equation 1Where:
s = heightt = timeu = initial velocityg = acceleration due to gravity.
From the question,
Given:
s = 2 mu = 0 m/s (fall from a height)g = 9.8 m/s²Substitute these values into equation 1
2 = 0(t)+9.8(t²)/2Solve for t.
9.8t² = 4t² = 4/9.8t² = 0.4081t = √0.4081t = 0.64 s.Hence, The time taken for the apple to hit the ground is 0.64 s
Learn more about time taken here: https://brainly.com/question/4931057