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Gre4nikov [31]
2 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]2 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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The rule can be written as follows:
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E = 6.41 x 10^-16 joule

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3 0
2 years ago
For the meter stick shown in figure 10-4, the force F1 10.0 N acts at 10.0 cm. What is the magnitude of torque due to F1 about a
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Turner's treadmill runs with a velocity of -1.3 m/s and speeds up at regular intervals during a half-hour workout. after 25 min,
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Given : Initial velocity = -1.3 m/s

            Final Velocity = -6.5 m/s.

            Time = 25 minutes.

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Solution: We are given units in meter/second (m/s).

So, we need to convert time 25 minutes in seconds.

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25 minutes = 60*25 = 1500 seconds.

Formula for average acceleration is given by,

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We are not given intial time, so we can take initial time =0.

Plugging values in the above formula.

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3 0
2 years ago
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Calculate a pendulum's frequency of oscillation (in Hz) if the pendulum completes one cycle in 0.5 s.
Marina86 [1]
Time taken to complete one oscillation for a pendulum is Time Period, T = 0.5 s 
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3 0
2 years ago
A square loop of wire with initial side length 10 cm is placed in a magnetic field of strength 1 T. The field is parallel to the
Fofino [41]

Answer:

2 x 10⁻³ volts

Explanation:

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Induced emf is given as

E = B \frac{dA}{dt}

E = (1) (20 x 10⁻⁴ )

E = 2 x 10⁻³ volts

E = 2 mV

7 0
2 years ago
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