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3 years ago

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You are to design a rotating cylindrical axle to lift 800-N buckets of cement from the ground to a rooftop 78.0 m above the grou

nd. The buckets will be attached to a hook on the free end of a cable that wraps around the rim of the axle; as the axle turns, the buckets will rise. (a) What should the diameter of the axle be in order to raise the buckets at a stead

1 answer:

Ronch [10]3 years ago

8 0

Answer:

The diameter of the axle is 5.08 cm.

Explanation:

Given that,

Force = 800 N

Distance = 78.0 m

Suppose we need to find the diameter of the axle be in order to raise the buckets at a steady 2.00 cm/s when it is turning at 7.5 rpm.

We need to calculate the radius of axle

Using formula of linear velocity

$v = r\omega$

$r=\dfrac{v}{\omega}$

Where, v =velocity

r = radius

$\omega$ =angular velocity

Put the value into the formula

$r=\dfrac{2.00}{7.5\times\dfrac{2\pi}{60}}$

$r=2.54\ cm$

We need to calculate the diameter of axle

Using formula of diameter

$d=2r$

$d=2\times2.54$

$d=5.08\ cm$

Hence, The diameter of the axle is 5.08 cm.

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Answer:

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Explanation:

Use Newton's Second Law to determine the acceleration that each object has.

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The force applied in both cases is 50 N, but the mass for object C and object D is different.

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Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{1} V_{o} + m_{2} U_{o}$ (2)

$p_{f}=(m_{1} + m_{2}) V_{f}$ (3)

$m_{1}=110 kg$ is the mass of the first football player

$V{o}=-7 m/s$ is the velocity of the first football player (to the south)

$m_{2}=75 kg$ is the mass of the second football player

$U_{o}=4 m/s$ is the velocity of the second football player (to the north)

$V_{f}$ is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

$m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}$ (5)

$V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg}$ (6)

$V_{f}=-2.54 m/s$ (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy $\Delta K$ is defined as:

$\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}$ (8)

Simplifying:

$\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2})$ (9)

$\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2})$ (10)

Finally:

$\Delta K=-2351.25 J$ (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

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