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2 years ago

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What force must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2000-kg car (a large car) resti

ng on the slave cylinder? the master cylinder has a 2.00-cm diameter and the slave has a 24.0-cm diameter?

1 answer:

Nataliya [291]2 years ago

6 0

Archimedes principle states that

F1 / A1 = F2 / A2

F2 = (A2 / A1) * F1

Also, formula for the force is F = mg. Formula for the area of the cylinder is A = πr^2, therefore we get

F2 = (πr2^2 / πr1^2) * mg

Since the diameter of the cylinders are 2 cm and 24 cm, r1 = 12 and r2 = 1.

Substituting the values to the derived equation, we get

F2 = (π 1^2 / π 12^2) * 2400 * 9.8

F2 = 163.3333 N

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Answer: A) $6.38(10)^{6} m$

Explanation:

The equation for the moment of inertia $I$ of a sphere is:

$I=\frac{2}{5}mr^{2}$ (1)

Where:

$I=9.74(10)^{37}kg m^{2}$ is the moment of inertia of the planet (assumed with the shape of a sphere)

$m=5.98(10)^{24}kg$ is the mass of the planet

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$r=\sqrt{\frac{5I}{2m}}$ (2)

Solving:

$r=\sqrt{\frac{5(9.74(10)^{37}kg m^{2})}{2(5.98(10)^{24}kg)}}$ (3)

Finally:

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1 year ago

A 6.60-kg block slides with an initial speed of 1.56 m/s up a ramp inclined at an angle of 28.4° with the horizontal. The coeffi

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Answer:

The distance travel by block before coming to rest is 0.122 m

Explanation:

Given:

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Final speed of block $v_{f} = 0$ $\frac{m}{s}$

Coefficient of kinetic friction $\mu _{k} = 0.62$

Ramp inclined at angle $\theta =$ 28.4°

Using conservation of energy,

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$\mu _{k} mgd \cos 28.4 = \Delta K - \Delta U$

Where $\Delta U = mg d\sin 28.4$

$\mu _{k} mgd \cos 28.4 = \frac{1}{2}mv_{i} ^{2} - mgd\sin 28.4$

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$d = 0.122$ m

Therefore, the distance travel by block before coming to rest is 0.122 m

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2 years ago

A nonuniform, 80.0-g, meterstick balances when the support is placed at the 51.0-cm mark. At what location on the meterstick sho

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Given

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2 years ago