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stepan [7]
3 years ago
14

Simple pendulum, is show in several states In case A the mass is travelling back down to the bottom and is in between the bottom

and its max height. In case B the mass is at its max height and has a speed of zero. In case C the ball is at its lowest position which of the following statements about the magnitude of the mass's angular acceleration is true? Grade Sun O
It is maximum in case A O
It is maximum in case B O
It cannot be determined O
It is zero in all cases. O
It is maximum in case C
It is equal in all cases, but is non-zero
Physics
1 answer:
uysha [10]3 years ago
3 0

The angular acceleration is maximum in case C

Explanation:

The motion of the bob in the simple pendulum is a portion of a circular motion. In a circular motion, the angular acceleration is related to the tangential acceleration by

a=\alpha r

where

a is the tangential acceleration

\alpha is the angular acceleration

r is the radius (the length of the pendulum)

Moreover for a pendulum, the tangential acceleration is given by the component of the acceleration of gravity in the tangential direction, so

a=g sin \theta

where

g is the acceleration of gravity

\theta is the angle of the pendulum with the vertical

So we have

\alpha = \frac{g sin \theta}{r}

Since g and r are constant, we see that the angular acceleration is proportional to the angle: the larger the angle, the larger the angular acceleration, and the smaller the angle, the smaller the angular acceleration.

Therefore, the angular acceleration is maximum when the pendulum is at its maximum height, and zero when it is at the lowest position. So, the correct statements are

It is maximum in case C

Learn more about periodic motion:

brainly.com/question/5438962

#LearnwithBrainly

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A 0.270 m radius, 510-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a
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Answer:

0.35701 T

Explanation:

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B_f = Final magnetic field

\phi = Magnetic flux

t = Time taken = 4.17 ms

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Induced emf is given by

\epsilon=-N\frac{d\phi}{dt}\\\Rightarrow \epsilon=-N\frac{B_fAcos90-B_iAcos0}{dt}\\\Rightarrow \epsilon=N\frac{B_iA}{dt}\\\Rightarrow B_i=\frac{\epsilon dt}{NA}\\\Rightarrow B_i=\frac{10000 \times 4.17\times 10^{-3}}{510\times \pi 0.27^2}\\\Rightarrow B_i=0.35701\ T

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4 0
4 years ago
A 5.30 g bullet moving at 963 m/s strikes a 610 g wooden block at rest on a frictionless surface. The bullet emerges, traveling
Tems11 [23]

Answer:

a) V_{wf} = 4.67m/s

b) V = 8.29 m/s

Explanation:

Givens:

The bullet is 5.30g moving at 963m/s and its speed reduced to 426m/s. The wooden block is 610g.

a) From conservation of linear momentum

Pi = Pf

m_{b}V{b_{i} }  + V_{wi}  = m_{w} V_{wf} + m_{b}V_{bf}

where m_{b},V{b_{i} are the mass and the initial velocity of the bullet, m_{w} and V_{wi} are the mass and the initial velocity of the wooden block, and V_{wf} and V_{bf} are the final velocities of the wooden block and the bullet

The wooden block is initial at rest (V_{wi} = 0) this yields

m_{b}V{b_{i} }  = m_{w} V_{wf} + m_{b}V_{bf}

By solving for V_{wf} adn substitute the givens

V_{wf} = \frac{m_{b}V_{bi} - m_{b} V_{bf}    }{m_{W} }

= \frac{5.3(g)(963(m/s)-426(m/s) }{610(g)}

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b) The center of mass speed is defined as

V = \frac{m_{b} }{m_{b}+m_{w} } V_{bi}

substituting:

V = \frac{5.3(g)}{5.3(g)+610(g)} X 963(m/s)

V = 8.29 m/s

7 0
3 years ago
Could somebody double check my work???
irina1246 [14]

Answer:

I think it's correct. Hope it helps.

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