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Ierofanga [76]
2 years ago
6

A mechanic uses a hydraulic car jack to lift the front end of a car to change the oil. The jack used exerts 8,915 N of force fro

m the larger piston. To pump the jack, 444 N of force is exerted on the small piston, which has an area of 3.14 cm2. What is the area of the large piston?
Physics
1 answer:
Vaselesa [24]2 years ago
3 0

Answer:

63.05 cm²

Explanation:

We use Pascals law to find the answer.

The law says that in an incomprehensible , non-viscous fluid the pressure applied will transmit through out the fluid without a change.

So, Pressure on larger piston = pressure on smaller piston.

      \frac{444}{3.14}  = \frac{8915}{A}

                                     A ≅ 63.05 cm²

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loris [4]

Answer:

The intensity I₂ of the light beam emerging from the second polarizer is zero.

Explanation:

Given:

Intensity of first polarizer = Io/2

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5 0
2 years ago
(HELP!!! 30 pts if answered right. )What formula gives the strength of an electric field, E, at a distance from a known source c
umka2103 [35]

Answer:

E=\frac{k\,Q}{d^2}

Explanation:

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Since the electric field E is derived from the Coulomb Force per unit charge using a positive test charge, the field's units will be in units of Newtons/Coulomb, and be the formula for the Coulomb electric force between to charges (Q1 and Q2),

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8 0
2 years ago
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

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To Find:

Current, I =?

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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}

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R = Resistance

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A = Area of cross section = πr²

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

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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}}

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V= I\times R

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

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2 years ago
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Zigmanuir [339]

We solve this using special relativity. Special relativity actually places the relativistic mass to be the rest mass factored by a constant "gamma". The gamma is equal to 1/sqrt (1 - (v/c)^2). <span>

We want a ratio of 3000000 to 1, or 3 million to 1. 

</span>

<span>Therefore:
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<span>v = 3 x 10^8 m/s</span></span></span>

8 0
2 years ago
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