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iris [78.8K]
2 years ago
13

A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceler

ation α. The flywheel is assumed to be at rest at time t=0 in Parts A and B of this problem.
Part A

Find the time t1 it takes to accelerate the flywheel to ω1 if the angular acceleration is α.

Express your answer in terms of ω1 and α.

Part B

Find the angle θ1 through which the flywheel will have turned during the time it takes for it to accelerate from rest up to angular velocity ω1.

Express your answer in terms of some or all of the following: ω1, α, and t1.

Part C

Assume that the motor has accelerated the wheel up to an angular velocity ω1 with angular acceleration α in time t1. At this point, the motor is turned off and a brake is applied that decelerates the wheel with a constant angular acceleration of −5α. Find t2, the time it will take the wheel to stop after the brake is applied (that is, the time for the wheel to reach zero angular velocity).

Express your answer in terms of some or all of the following: ω1, α, and t1.
Physics
1 answer:
aleksandrvk [35]2 years ago
6 0

Answer:

Part a)

t_1 = \frac{\omega_1}{\alpha}

Part b)

\theta = \frac{1}{2}\alpha t_1^2

Part c)

t = \frac{t_1}{5}

Explanation:

Part a)

As we know that it is having constant torque so here the time taken by it to accelerate is given as

\omega_f = \omega_i + \alpha t

\omega_1 = 0 + \alpha t_1

t_1 = \frac{\omega_1}{\alpha}

Part b)

angular displacement is given as

\theta = \omega_i t_1 + \frac{1}{2}(\alpha) t_1^2

\theta = 0 + \frac{1}{2}\alpha t_1^2

\theta = \frac{1}{2}\alpha t_1^2

Part c)

As we know that the angular deceleration produced by the brakes is given as

\alpha_d = - 5\alpha

now we have

\omega_f = \omega _i + \alpha t

0 = \omega_1 - 5 \alpha_1 t

t = \frac{\omega_1}{5 \alpha}

As we know that

t_1 = \frac{\omega_1}{\alpha}

so we have

t = \frac{t_1}{5}

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A(n) 71.1 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 70.2 m away from the shuttle
denpristay [2]

Answer:

10.347 minutes.

Explanation:

According to F = ma, she exerts force on camera of the magnitude

F = 0.67Kg*12m/s^{2} = 8.04N, assuming it took her one second to accelerate camera to 12m/s, then by newtons third law, which says every action has equal and opposite reaction , the camera exerts the same amount of force on the astronaut which gives her acceleration of a = \frac{8.04N}{70.2Kg} = 0.1130801680m/s^2.

and velocity of V = 0.1130801680m/s.

at this velocity , the astronaut has to cover the distance of 70.2 meters, it will take her 620.7985075s = 10.347 min to get to the shuttle (using S = vt).

4 0
2 years ago
Consider the following:
pychu [463]

Answer:

They have different wavelengths.

They have different frequencies.

They propagate at different speeds through non-vacuum media depending on both their frequency and the material in which they travel.

Explanation:

The complete question is

Consider the following:

a) radio waves emitted by a weather radar system to detect raindrops and ice crystals in the atmosphere to study weather patterns;

b) microwaves used in communication satellite transmissions;

c) infrared waves that are perceived as heat when you turn on a burner on an electric stove;

d) the multicolor light in a rainbow;

e) the ultraviolet solar radiation that reaches the surface of the earth and causes unprotected skin to burn; and

f) X rays used in medicine for diagnostic imaging.

Which of the following statements correctly describe the various forms of EM radiation listed above?

check all that apply to the above

They have different wavelengths.

They have different frequencies.

They propagate at different speeds through a vacuum depending on their frequency.

They propagate at different speeds through non-vacuum media depending on both their frequency and the material in which they travel.

They require different media to propagate.

All the above phenomena are due the electromagnetic wave spectrum. Electromagnetic waves travel at a constant speed of 3 x 10^8 m/s in a vacuum. Within the spectrum, the different types of electromagnetic waves exists in different band range of frequencies and wavelengths unique to each of the waves, and the energy they carry. When these waves enter a non-vacuum medium, their speed change, depending on the nature of the material of the medium, and the frequency or the wavelength of the incoming wave.

5 0
2 years ago
A person driving a car suddenly applies the brakes. The car takes 4 s to come to rest while traveling 20 m at constant accelerat
Zinaida [17]

Answer:

Yes we can find the initial velocity of car without finding acceleration.

u = 10 m/s.

Explanation:

Given that

s=20 m

Car takes 4 s to come in rest.

We know that when acceleration is constant then we can apply motion equation

v=u+at        ----------1

s=ut+\dfrac{1}{2}at^2       ------2

From equation 1 and 2

s=ut+\dfrac{1}{2}t^2\left (\dfrac{v-u}{t} \right )

So we can say that

s= \left(\dfrac{v+u}{2}\right)t

Given that the velocity of car at final condition will be zero (v=0)

s= \left(\dfrac{0+u}{2}\right)t

s= \left(\dfrac{u}{2}\right)t

From the above equation we can find the initial velocity of car without finding the acceleration

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7 0
2 years ago
What is the length of the x-component of the vector plotted below?
deff fn [24]

Answer:

4

Explanation:

7 0
2 years ago
If a rock is thrown upward on the planet mars with a velocity of 11 m/s, its height (in meters) after t seconds is given by h =
Butoxors [25]
(a) 3.56 m/s 
(b) 11 - 3.72a 
(c) t = 5.9 s 
(d) -11 m/s  
For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule. 
y = 11t - 1.86t^2 
y' = 11 - 3.72t 
Now that you have the first derivative, it will give you the velocity as a function of t. 
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(b) Velocity after a seconds. 
y' = 11 - 3.72t 
y' = 11 - 3.72a  
So the answer is 11 - 3.72a  
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.  
(d) Plug in the value of t calculated for (c) into the velocity function, so: 
y' = 11 - 3.72a
 y' = 11 - 3.72*5.913978495
 y' = 11 - 22
 y' = -11 
 So the velocity is -11 m/s which makes sense since the total energy of the rock will remain constant, so it's coming down at the same speed as it was going up.
3 0
2 years ago
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