In the process of changing a flat tire, a motorist uses a hydraulic jack. She begins by applying a force of 48 N to the input piston, which has a radius r1. As a result, the output plunger, which has a radius r2, applies a force to the car. The ratio r2/r1 has a value of 9.0. Ignore the height difference between the input piston and output plunger and determine the force that the output plunger applies to the car.

Answers

Answer 1
Answer:

Answer:

Force that the output plunger applies to the car; F2 = 3888N

Explanation:

For a hydraulic device, the relationship between the force and the area using Pascal's principle is;

F1/A1 = F2/A2

Where;

F1 is force applied to the input piston

F2 is force that the output plunger applies to the car

A1 is Area of input piston

A2 is area of larger piston

We are given;

R2/R1 = 9

So,R2 = 9R1

F1 = 48N

Area of input piston;

A1 = π(R1)²

Area of output piston;

A2 = π(9R1)²

Since, (F1/A1) = (F2/A2)

Thus;

F1/(π(R1)²) = F2/(π(9R1)²)

If we simplify, π(R1)² will cancel out to give;

F1 = F2/9²

Thus;

F2 = 9² x F1

Plugging in 48N for F1, we have;

F2 = 9² x 48

F2 = 81 x 48

F2 = 3888N

Answer 2
Answer:

Final answer:

Using the principle of Pascal's law and ignoring the height difference, the output force is found by the formula F2 = F1*(r2/r1)^2. Given F1 is 48N and r2/r1 is 9.0, the output force F2 equates to 3888N.

Explanation:

In the case of a hydraulic jack, the principle of Pascal's law is applied. According to this law, pressure applied at one point of the fluid is transmitted equally in all directions. Therefore, if we ignore the height difference between pistons, the pressure exerted on both pistons would be the same.

Pressure is equal to the force divided by the area, where area equals π times the radius squared (π*r^2). So, the pressure at the input piston (P1) is the force at the input piston (F1) divided by its area (A1): P1 = F1/A1, where A1 = π*(r1)^2.

For the output plunger(P2 = F2/A2), where F2 = force at the output plunger and A2 = π*(r2)^2. By equating the pressures (P1=P2) and simplifying, we find that F2 = F1*(r2/r1)^2, where r2/r1 is given as 9.0. So, the output force F2 would be 48N*(9.0)^2 = 3888N.

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An electron in a vacuum chamber is fired with a speed of 7400 km/s toward a large, uniformly charged plate 75 cm away. The electron reaches a closest distance of 15 cm before being repelled. What is the plate's surface charge density?

We can calculate the force that the atmospheric pressure produces on a surface. Consider a living room that has a 4.0m×5.0m floor and a ceiling 3.0m high. What is the total force on the floor due to the air above the surface if the air pressure is 1.00 atm?

Answers

Answer:

Force, F=2.02* 10^6\ N

Explanation:

It is given that,

Length of the room, l = 4 m

breadth of the room, b = 5 m

Height of the room, h = 3 m

Atmospheric pressure, P=1\ atm=101325\ Pa

We know that the force acting per unit area is called pressure exerted. Its formula is given by :

P=(F)/(A)

F=P* l* b

F=101325* 4* 5

F=2.02* 10^6\ N

So, the total force on the floor due to the air above the surface is 2.02* 10^6\ N. Hence, this is the required solution.

A 50n brick is suspended by a light string from a 30kg pulley which may be considered a solid disk with radius 2.0m. the brick is released from rest and falls to the floor below as the pulley rotates. it takes 4 seconds for the brick to hit the floor. i) what is the tension in newtons in the string well the brick is falling? ii) what is the magnitude of the angular momentum in kg*m^2/s of the pulley at the instant the brick hits the floor?

Answers

Brick is held at a position which is at height 2 m from the floor

Now it is released from rest and hit the floor after t = 4 s

Now the acceleration of the brick is given by

y = v_i* t + 0.5 at^2

2 = 0 + 0.5 * a * 4^2

a = 0.25 m/s^2

a)

Now in order to find the tension in the string

we can use Newton's law

F_(net) = ma

mg - T = ma

50 - T = (50)/(9.8)*0.25

T = 48.72 N

part b)

Now for the pulley

moment of inertia= (1)/(2)mr^2

m = 30 kg

R = 2 m

I = (1)/(2)*30*2^2

I = 60 kg m^2

Now the angular speed just before brick collide with the floor

w = (v)/(r)[\tex]</p><p>here we have</p><p>[tex]v = v_i + a* t

v = 0 + 0.25 * 4

v = 1 m/s

Now we will have

L = angular momentum = I w = I*(v)/(R)

L = 60 *(1)/(2)

L = 30 kg m^2/s

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Answers

Answer:

3 min 55 sec is the solidification time if the cylinder height is doubled

7min 40 sec if the diameter is doubled

Explanation:

see the attachment

g During a collision with a wall, the velocity of a 0.4 KgKg ball changes from 25 m/sm/s towards the vall to 12 m/sm/s away from the wall. If the time the ball was in contact with the call was 0.5 secsec , what was the magnitude of the avarage force applied to the ball

Answers

Answer:

F = 10.8N

Explanation:

Given the mass m = 0.4kg, v1 = 25m/s, v2 = 12m/s and t =0.5s

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F×t = m(v2 – v1)

F = m(v2 – v1)/t = 0.4(25 – 12)/0.5 = 10.8N

Why does an astronaut in a spacecraft orbiting Earthexperience
a feeling of weightlessness?

Answers

Answer:

Astronaut in spacecraft while orbiting earth experience weightlessness because there is no gravity of earth or moon is acting on the body of an astronaut.                      

while on earth, we experience weight because the gravity of earth is acting on our body which is pulling us downward.

Both spacecraft and the astronauts both are in a free-fall condition.

Help me with my physics, please

Answers

The right answer would be

-20t+ 80