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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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Two speakers both emit sound of frequency 320 Hz, and are in phase. A receiver sits 2.3 m from one speaker, and 2.9 m from the o
satela [25.4K]

Answer:

Option B

Explanation:

The phase difference is found by subtracting the 2.3m for the receiver from the other speaker which is 2.9m hence

Phase difference= 2.9-2.3= 0.6

3 0
2 years ago
Calculate the change in the kinetic energy (KE) of the bottle when the mass is increased. Use the formula KE = mv2, where m is t
Aliun [14]

kinetic energy is given as

KE = (0.5) m v²

given that : v = speed of the bottle in each case =  4 m/s

when m = 0.125 kg

KE = (0.5) m v² =  (0.5) (0.125) (4)² = 1 J

when m = 0.250 kg

KE = (0.5) m v² =  (0.5) (0.250) (4)² = 2 J

when m = 0.375 kg

KE = (0.5) m v² =  (0.5) (0.375) (4)² = 3 J

when m = 0.0.500 kg

KE = (0.5) m v² =  (0.5) (0.500) (4)² = 4 J

6 0
2 years ago
Read 3 more answers
A p-type Si sample is used in the Haynes-Shockley experiment. The length of the sample is 2 cm, and two probes are separated by
Airida [17]

Answer:

Mobility of the minority carriers, \mu_{n} =1184.21 cm^{2} /V-sec

Diffusion coefficient for minority carriers,D_{n} = 29.20 cm^2 /s

Verified from Einstein relation as  \frac{D_{n} }{\mu_{n} }  = 25 mV

Explanation:

Length of sample, l_{s} = 2 cm

Separation between the two probes, L = 1.8 cm

Drift time, t_{d} = 0.608 ms

Applied voltage, V = 5 V

Mobility of the minority carriers ( electrons), \mu_{n} = \frac{V_{d} }{E}

Where the drift velocity, V_{d} = \frac{L}{t_{d} }

V_{d} = \frac{1.8}{0.608 * 10^{-3} } \\V_{d} = 2960.53 cm/s

and the Electric field strength, E = \frac{V}{l_{s} }

E = 5/2

E = 2.5 V/cm

Mobility of the minority carriers:

\mu_{n} = 2960.53/2.5\\\mu_{n} =1184.21 cm^{2} /V-sec

The electron diffusion coefficient, D_{n} = \frac{(\triangle x)^{2} }{16 t_{d} }

\triangle x = (\triangle t )V_{d}, where Δt = separation of pulse seen in an oscilloscope in time( it should be in micro second range)

\triangle x = \frac{(\triangle t) L}{t_{d} } \\\triangle x = \frac{180*10^{-6} * 1.8}{0.608*10^{-3}  }\\\triangle x =0.533 cm

D_{n} = \frac{0.533^{2} }{16 * 0.608 * 10^{-3} }\\D_{n} = 29.20 cm^2 /s

For the Einstein equation to be satisfied, \frac{D_{n} }{\mu_{n} } = \frac{KT}{q} = 0.025 V

\frac{D_{n} }{\mu_{n} } = \frac{29.20}{1184.21} \\\frac{D_{n} }{\mu_{n} } = 0.025 = 25 mV

Verified.

4 0
2 years ago
Whennes
rodikova [14]

Answer:

See the explanation below.

Explanation:

12) When an object is falling, how does the objects velocity change? what formula do you use?

The speed of a falling object is increased by a value of 9.81 meters per second per second. That is if we throw any body regardless of mass from a considerable height, its speed in the first second will be 9.81[ m/ s] , in the next second will be equal to 19.62 [m/s] in the next will be equal to 29.43 [m/ s].

The formula is:

v=v_{0}+g*t

where:

vo = initial velocity = 0

g = gravity = 9.81[m/s^2]

t = time [s]

13)

what is a falling speed at 6s, 9s, 112s?

v = 0 + (9.81*6) = 58.86[m/s]

v = 0 + (9.81*9) = 88.29 [m/s]

v = 0 + (9*112) = 1098.72 [m/s]

14)

If an object is falling at 65 [m/s]. How long has it been falling ? what is the formula that you use?

The formula is the same:

v=v_{o}+g*t

65 = 0 + 9.81*t

t = 65/9.81

t = 6.62[s]

15)

What formula is used to determine the distance an object is falling ?

y = y_{o}+v_{o}*t + 0.5*9.81*t^{2}

where:

y = distance [m]

yo = initial distance, in most of the cases and depending of the reference point it will be eqaul to zero

vo = initial velocity, if it is free fall, then = 0

t = time [s]

g = gravity = 9.81[m/s^2]

This equation will be reduce to:

y =   0.5*g*t^{2}

16)

using the times given in problem 13. Determine the distance fallen for each.

y = 0.5*9,81*(6)^2 = 176.58 [m]

y = 0.5*9,81*(9)^2 = 397.3 [m]

y = 0.5*9,81*(112)^2 = 61528.3 [m]

17)

If an object has fallen a distance of 87.3 [m]. How long was it falling?

87.3 = 0.5*9.81*t^2

t=\sqrt{\frac{87.3}{0.5*9.81} }\\ t=4.21[s]

4 0
2 years ago
An initially neutral glass rod is rubbed with silk. It becomes positively charged by_____.a) electrons are created on the rod.
Dimas [21]

Answer:

e) electrons are transferred from the rod to the silk.

Explanation:

An initially neutral glass rod contains equal number of electron and proton.

If the rod becomes positively charged  after being rubbed with silk, then the rod must have lost some its electron to the silk since electrons are more mobile than protons, leaving the rod with excess positive charge (protons), and the silk will be negatively charged (excess electron).

Thus, the rod becomes positively charged by transfer of electrons from rod to the silk.

e) electrons are transferred from the rod to the silk.

3 0
2 years ago
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