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Gre4nikov [31]
4 years ago
9

A 13,000-N vehicle is to be lifted by a 25-cm diameter hydraulic piston. What force needs to be applied to a 5.0 cm diameter pis

ton to accomplish this? Assume the pistons each have negligible weight.
Physics
1 answer:
MaRussiya [10]4 years ago
4 0

Answer:

Explanation:

We shall apply Pascal's Law in fluid mechanics

According to it , pressure is transmitted in liquid from one point to another without any change .

25 cm diameter = 12.5 x 10⁻² m radius

Area = 3.14 x (12.5 x 10⁻²)²

= 490.625 x 10⁻⁴ m²

Pressure by vehicle

Force / area

13000 / 490.625 x 10⁻⁴

= 26.497 x 10⁴ Pa

5 cm diameter = 2.5 x 10⁻² radius

area = 3.14 x (2.5 x 10⁻²)²

= 19.625 x 10⁻⁴ m²

If we assume required force F on this area

Pressure = F / 19.625 x 10⁻⁴ Pa

According to Pascal Law

F / 19.625 x 10⁻⁴  = 26.497 x 10⁴

F = 19.625 x 26.497

= 520 N

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Approximately 6.8 x 10⁻¹⁵

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CHECK ATTACHMENT FOR Step by step solution to the answer

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A 4.3-g object moving to the right at 22 cm/s makes an elastic head-on collision with a 8.6-g object that is initially at rest.
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7 0
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Read 2 more answers
You are a pirate working for dread pirate roberts. you are in charge of a cannon that exerts a force 20000 n on a cannon ball wh
Crazy boy [7]
Refer to the diagram shown below.

F = 2000 N, the force exerted on the cannonball
L = 2.41 m, the length of the barrel
V₀ = 83 m/s, the launch velocity
θ = 35°, the launch angle

Let m =  the mass of the cannonball.
Let a  =  the acceleration of the ball when fired.

The net force acting on the ball is
F - mg sinθ = 2000 - 9.8*m*sin(35°) =  2000 - 5.621*m N

Then, from Newton's Law of motion,
F = ma
2000 - 5.621*m = m*a             (1)

The launch velocity is V₀ = 8.3 m/s, therefore
V₀² = 2*a*L
(83 m/s)² = 2*(a m/s²)*(2.41 m)
6889 = 4.82*a
a = 6889/4.82 = 1429.25 m/s²      (2)

Insert (2) into (1)
2000 - 5.621*m = 1429.25*m
2000 = 1434.871*m
m = 1.394 kg

Answer: 1.394 kg



4 0
4 years ago
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