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navik [9.2K]
3 years ago
13

A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with oil with a d

ensity of 800 kg/m3 and capped at both ends with tight-fitting pistons. The wider arm of the U-tube has a radius of 18.0 cm and the narrower arm has a radius of 5.00 cm. The car rests on the piston on the wider arm of the U-tube. The pistons are initially at the same level. What is the force that must be applied to the smaller piston in order to lift the car after it has been raised 1.20 m
Physics
1 answer:
const2013 [10]3 years ago
3 0

Answer:

The answer is below

Explanation:

Pascal's law states that pressure is exerted in all parts of a static fluid equally.

The area of the narrower arm with a radius of 0.05 m (5 cm) is given as:

Area of narrow arm = π(0.05)²

The area of the wider arm with a radius of 0.18 m (18 cm) is given as:

Area of narrow arm = π(0.18)²

The ratio if the wider arm area to the narrow arm area = π(0.18)² / π(0.05)² = 12.96

To move the 12000 N car, the amount of force needed = 12000/12.96 = 926 N

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A closely wound, circular coil with radius 2.50 cmcm has 740 turns. Part A What must the current in the coil be if the magnetic
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Answer:

The current in the coil is 4.086 A

Explanation:

Given;

radius of the circular coil, R = 2.5 cm = 0.025 m

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magnetic field at the center of the coil, B = 0.076 T

The magnetic field at the center of the coil is given by;

B = \frac{N\mu_o I}{2R}

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The current in the circular coil is given by

B = \frac{N\mu_o I}{2R} \\\\I = \frac{2BR}{N\mu_o} \\\\I =\frac{2*0.076*0.025}{740*4\pi*10^{-7}} \\\\I = 4.086 \ A

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An ideal monatomic gas at temperature T is held in a container. If the gas is compressed isothermally, that is at constant tempe
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Answer:

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d) ΔQ = W

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The gas is compressed isothermally, then, there is no change in the internal energy and you have

ΔU = 0 J

b) The work is done by the gas, not over the gas.

The work is given by the following formula:

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T: constant temperature

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Vi: initial volume

Vf < Vi, then W < 0 and the work is done on the gas

c) The gas has been compressed. Thus, its temperature increases and heat has been transferred to the gas.

The amount of heat is equal to the work done W

d)

\Delta U = \Delta Q-W\\\\0=\Delta Q-W\\\\\Delta Q=W=nRTln(\frac{V_f}{V_i})

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