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vladimir1956 [14]

3 years ago

5

A disk rotates freely on a vertical axis with an angular velocity of 50 rpm . An identical disk rotates above it in the same dir

ection about the same axis, but without touching the lower disk, at 20 rpm . The upper disk then drops onto the lower disk. After a short time, because of friction, they rotate together.The final angular velocity of the disks is1. 49 rpm2. 21 rpm3. 77 rpm4. 35 rpm5. 42 rpm

1 answer:

True [87]3 years ago

5 0

Answer:

Final angular velocity is 35rpm

Explanation:

Angular velocity is given by the equation:

I1w1i + I2w2i = I1w1f -I2w2f

But the two disks are identical, so Ii =I2

wf can be calculated using

wf = w1i - w2i/2

Given: w1i =50rpm w2i= 30rpm

wf= (50 + 20) / 2

wf= 70/2 = 35rpm

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A wheel with radius 36 cm is rotating at a rate of 19 rev/s.(a) What is the angular speed in radians per second? rad/s(b) In a t

Sedaia [141]

(a) 119.3 rad/s

The angular speed of the wheel is

$\omega= 19 rev/s$

we need to convert it into radiands per second. We know that

$1 rev = 2 \pi rad$

Therefore, we just need to multiply the angular speed of the wheel by this factor, to get the angular speed in rad/s:

$\omega = 19 rev/s \cdot (2\pi rad/rev))=119.3 rad/s$

(b) 596.5 rad

The angular displacement of the wheel in a time interval t is given by

$\theta= \omega t$

where

$\omega=119.3 rad$

and

t = 5 s is the time interval

Substituting numbers into the equation, we find

$\theta=(119.3 rad/s)(5 s)=596.5 rad$

(c) 127.3 rad/s

At t=10 s, the angular speed begins to increase with an angular acceleration of

$\alpha = 1.6 rad/s^2$

So the final angular speed will be given by

$\omega_f = \omega_i + \alpha \Delta t$

where

$\omega_i = 119.3 rad/s$ is the initial angular speed

$\alpha = 1.6 rad/s^2$ is the angular acceleration

$\Delta t = 15 s - 10 s = 5 s$ is the time interval

Solving the equation,

$\omega_f = (119.3 rad/s) + (1.6 rad/s^2)(5 s)=127.3 rad/s$

(d) 616.5 rad

The angle through which the wheel has rotated during this time interval is given by

$\theta = \omega_i \Delta t + \frac{1}{2} \alpha (\Delta t)^2$

Substituting the numbers into the equation, we find

$\theta = (119.3 rad/s)(5 s) + \frac{1}{2} (1.6 rad/s^2) (5 s)^2=616.5 rad$

(e) 222 m

The instantaneous speed of the center of the wheel is given by

$v_{CM} = \omega R$ (1)

where

$\omega$ is the average angular velocity of the wheel during the time t=10 s and t=15 s, and it is given by

$\omega=\frac{\omega_i + \omega_f}{2}=\frac{127.3 rad/s+119.3 rad/s}{2}=123.3 rad/s$

and

R = 36 cm = 0.36 m is the radius of the wheel

Substituting into (1),

$v_{CM}=(123.3 rad/s)(0.36 m)=44.4 m/s$

And so the displacement of the center of the wheel will be

$d=v_{CM} t = (44.4 m/s)(5 s)=222 m$

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3 years ago

What is one way we can use agriscience to affect photosynthesis and increase plant growth?

pochemuha

Answer:

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During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between

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Answer:

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Explanation:

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T_1 sin14 + T_2sinA = mg

T_2sinA = mg - T_1sin12.5 ....................... (a)

Along horizontal axis, the equation can be written as

T_2×cosA = T_1×cos12.5 ......................... (b)

(a)/(b) given us

Tan A = (mg - T_1sin12.5) / T_1 cos12.5

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3 years ago

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Which of the following is a major ocean on Earth?

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The Indian Ocean.
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A 15-lb block B starts from rest and slides on the 25-lb wedge A, which is supported by a horizontal surface. Neglecting frictio

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The angle of the wedge is 30°.

Answer:

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Explanation:

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initial velocity u= 0

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acceleration, a = g sin 30° = 32 ft/s²× sin 30° = 16 ft/s²

Sliding displacement, s = 3ft

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$v^2-u^2 = 2as$

substitute the values and solve for v

$v^2-0 = 2\times 16 \times 3 =96 ft^2/s^2\\v = 9.8 ft/s$

b) Use conservation of momentum:

Initial momentum of the system = 0

final momentum = (15) ( 9.8)+ (25)(v')

v' = 5.88 ft/s

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3 years ago