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3241004551 [841]

3 years ago

7

Two blocks of masses m1 and m2 are connected to each other on an Atwood machine (the blocks are connected by a string going over

a light pulley with no friction in the bearing). A person holds one of the blocks with her hand. When the system is released, the heavier block moves down with an acceleration of 3.5 m/s2 and the lighter object moves up with an acceleration of the same magnitude. Suppose that the lighter block has a mass of m1 = 100 g. Determine the mass m2.

1 answer:

ddd [48]3 years ago

8 0

Answer:211.11 gm

Explanation:

Given

mass of lighter block $m_1=100 gm$

mass of heavier block is $m_2$

acceleration of system $a=3.5 m/s^2$

From diagram

for lighter block

$T-m_1g=m_1a$

$T=m_1(g+a)$

For heavier Block

$m_2g-T=m_2a$

$T=m_2(g-a)$

Equating Tension T

$m_1(g+a)=m_2(g-a)$

$m_2=m_1\cdot \frac{g+a}{g-a}$

$m_2=m_1\cdot \frac{9.8+3.5}{9.8-3.5}$

$m_2=0.1\cdot 2.11$

$m_2=0.2111 kg$

$m_2=211.11 gm$

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Answer:

The ratio of the new force over the original force is 16

Explanation:

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So now, if the masses M1 and M2 are quadrupled and the distance stays the same, the new force becomes:

$F'_G=G\,\frac{4M_1\,\,4M_2}{D^2}=G\,\frac{16\,\,M_1\,\,M_2}{D^2}=16\,\,G\,\frac{M_1\,\,M_2}{D^2}= 16\,\,F_G$

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3 years ago

A Ping-Pong ball moving East at a speed of 4 m/s collides with a stationary bowling ball. The Ping-Pong ball bounces back to the

In-s [12.5K]

Answer:

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We can re-arrange the equation as follows

$p_p - p'_p = p_b'-p_b$

or

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so the magnitude of the change in momentum of the ping-pong ball is equal to the magnitude of the change in momentum of the bowling ball.

However, we also know that the magnitude of the impulse on an object is equal to the change of momentum of the object:

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4 years ago

What is a concave lens?

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3 years ago

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mrs_skeptik [129]

Answer:

$t=16.67s$

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From the acceleration equation, we can obtain the velocity equation:

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4 years ago

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Hope this helps :)

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3 years ago