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3 years ago

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A spool of thread, free to unwind, is on a horizontal surface (with friction) and pulled directly upward (+y) at all times. The

spool rolls without slipping on the floor. The spool has a mass M, moment of inertia I, and a radius of R. Assume the string is wrapped around the spool at a radius r (where r < R). spool rolling on the floor
Which equation below is a correct expression of Newton’s Second Law ( ΣF=ma ) in the x (horizontal) -direction for this situation. Assume all the forces below are magnitudes and that to the right is considered the positive direction.

a. F_T - Ff= Max
b. F_T+Ff= Max
c. Ff= Max

1 answer:

k0ka [10]3 years ago

6 0

Answer:

The Correct Answer is "D"

Explanation:

For Detail of this answer Please see the attached file.

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3 years ago

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Suppose a skydiver (mass = 80kg) is falling toward the Earth. Calculate the skydiver’s gravitational potential energy at a point

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Ep is gravitational potential energy
m is mass (kg)
g is gravitational field strength (N/kg)
h is height (m)

Ep= mgh
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3 years ago

if 130N centripetal force is needed to keep a 0.45kg ball that is attached to a string that is 0.7m long to complete 5 full rota

andrezito [222]

Answer:

centripetal acceleration of the ball is 6.9 m/s/s

tangential speed of the ball is 2.2 m/s

Explanation:

As we know that ball complete 5 rotations in 10 seconds

so frequency of rotation of ball is given as

$f = \frac{5}{10} = 0.5 Hz$

now we know that angular frequency is given as

$\omega = 2\pi f$

$\omega = 2(\pi)(0.5)$

$\omega = \pi rad/s$

Now centripetal acceleration is given as

$a_c = \omega^2 R$

$a_c = \pi^2 (0.7)$

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3 years ago

Jose was out drinking with his friends for nearly the whole night. The next morning he was confused and vomiting, and had a low

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Answer:

He has a hangover.

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2 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

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

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

$V_{wf}$ = 4.67m/s

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