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2 years ago

9

Differences between work against friction and work against gravity.

2 answers:

Goshia [24]2 years ago

4 0

Friction is the force that is resisting the motion of an object so it will always point in the opposite direction of that of movement. ... The force of gravity points downwards . So when you do work against gravity it means that the force acting on that object points in the upward direction .

oksian1 [2.3K]2 years ago

4 0

Explanation:

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Answer:

Part 1 $E = 2.42 * 10^{21} N/C$

Part 2 $E = 1.3 * 10^{13} N/C$

Part 3 $E = 0$

Explanation:

Given

Number of protons $= 92$

Radius of nucleus $r_n = 7.4 * 10^{-15} m$

Distance of the electrons $r_1 = 1.0 * 10^ {-10} m$

Part 1

Electric field produced by just outside its surface

$E = \frac{q}{4\pi*E_0* r_n^2 } \\E = \frac{9 * 10^ 9 * 92 * 1.6 * 10^{-19}}{(7.4* 10^{-15})^2} \\E = 2.42 * 10^{21} N/C$

Part 2

Electric field produced by just outside its surface

$E = \frac{q}{4\pi*E_0* r_n^2 } \\E = \frac{9 * 10^ 9 * 92 * 1.6 * 10^{-19}}{(1* 10^{-15})^2} \\E = 1.3 * 10^{13} N/C$

Part 3

The net electric field inside a uniform shell of negative charge is zero because the electric flux lines cancel out each other

hence, the solution is

Part 1 $E = 2.42 * 10^{21} N/C$

Part 2 $E = 1.3 * 10^{13} N/C$

Part 3 $E = 0$

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3 years ago

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3 years ago

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kogti [31]

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3 years ago

A scientist is testing the seismometer in his lab and has created an apparatus that mimics the motion of the earthquake felt in

otez555 [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

$a_{max} = 0.00246 \ m/s^2$

b

$k =722.2 \ N/m$

Explanation:

From the question we are told that

The amplitude is $A = 1.8 \ cm = 0.018 \ m$

The period is $T = 17 \ s$

The test weight is $W = 13 \ N$

Generally the radial acceleration is mathematically represented as

$a = w^2 r$

at maximum angular acceleration

$r = A$

So

$a_{max} = w^2 A$

Now $w$ is the angular velocity which is mathematically represented as

$w = \frac{2 * \pi }{T}$

Therefore

$a_{max} = [\frac{2 * \pi}{T} ]^2 * A$

substituting values

$a_{max} = [\frac{2 * 3.142}{17} ]^2 * 0.018$

$a_{max} = 0.00246 \ m/s^2$

Generally this test weight is mathematically represented as

$W = k * A$

Where k is the spring constant

Therefore

$k = \frac{W}{A}$

substituting values

$k = \frac{13}{0.018}$

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3 years ago

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Ahat [919]

Answer:

C

Explanation:

Some of the mass

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2 years ago

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