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Bond [772]
1 year ago
14

A hydraulic lift is designed to lift cars up to 2000 kg in mass.If the lift under the car is 1.0 m by 1.2m and the area of the i

nput piston is 10 cm by 10 cm, how much force must be exerted on the input piston to lift a 2000-kg car? A) 136 N B 24x 10^4 N C) 160 N D) 196 N E) 17N
Physics
1 answer:
Lerok [7]1 year ago
6 0

Answer:

C) 160 N

Explanation:

m = mass of the car = 2000 kg

Weight of the car is given as

W = mg \\W = (2000) (9.8)\\W = 19600 N

F_{i} = Force exerted on the input piston

A_{i} = Area of input piston = 10 cm x 10 cm = 0.1 m x 0.1 m = 0.01 m²

A_{lift} = Area of lift = 1 m x 1.2 m = 1.2 m²

Using Pascal's law

\frac{F_{i}}{A_{i}} = \frac{W}{A_{lift}}\\\frac{F_{i}}{0.01} = \frac{19600}{1.2}\\F_{i} = 160 N

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A tin can whirled on the end of a string moves in a circle because
Ilya [14]

Answer:

There is an inward force acting on the can

Explanation:

This inward force is known as Centripetal force and it is responsible for making the can whirl on the end of a string in circle and it is also directed towards the center around which the can is moving.

8 0
2 years ago
A steel tank of weight 600 lb is to be accelerated straight upward at a rate of 1.5 ft/sec2. Knowing the magnitude of the force
VikaD [51]

Answer:

a) the values of the angle α is 45.5°

b) the required magnitude of the vertical force, F is 41 lb

Explanation:

Applying the free equilibrium equation along x-direction

from the diagram

we say

∑Fₓ = 0

Pcosα - 425cos30° = 0

525cosα - 368.06 = 0

cosα = 368.06/525

cosα = 0.701

α = cos⁻¹ (0.701)

α = 45.5°

Also Applying the force equation of motion along y-direction

∑Fₓ = ma

Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)

525sin45.5° + F + 212.5 - 600 = 27.95

374.46 + F + 212.5 - 600 = 27.95

F - 13.04 = 27.95

F = 27.95 + 13.04

F = 40.99 ≈ 41 lb

8 0
2 years ago
Two objects move toward each other because of gravitational attraction. As the objects get closer and closer, the force between
bearhunter [10]
<span>Two objects move toward each other because of gravitational attraction. As the objects get closer and closer, the force between them increases. </span>
5 0
1 year ago
Read 2 more answers
Calculate the current through a 10.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V
Kipish [7]

Answer:

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

Explanation:

Given:

Length = l = 10 meter

Radius = r = 0.321\ mm =0.321\times 10^{-3}\ meter

Resistivity=\rho=1.00\times 10^{-6}\ ohm\ meter

V = 12 Volt

To Find:

Current, I =?

Solution:

Resistance for 0.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V battery given as

R=\dfrac{\rho\times l}{A}

Where,

R = Resistance

l = length

A = Area of cross section = πr²

\rho=Resistivity=1.00\times 10^{-6}\ ohm\ meter

Substituting the values we get

R=\dfrac{1\times 10^{-6}\times 10}{3.14\times (0.321\times 10^{-3})^{2}}

R=\dfrac{1\times 10^{-5}}{3.23\times 10^{-7}}

R=\dfrac{1\times 10^{2}}{3.23}

R=30.95\ ohm

Now by Ohm's Law,

V= I\times R

Substituting the values we get

I=\dfrac{V}{R}=\dfrac{12}{30.95}=0.3876\ Ampere

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

4 0
1 year ago
The lightest duty and most widely used non flexible metal conduit is
Alenkinab [10]
Aluminium, or heavier version copper.
6 0
1 year ago
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