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2 years ago

8

A car with a mass of 1500 kg is pulled by a rope that is horizontal to the ground. The tension in the rope is 2000 N and a frict

ion force of 350 N opposes the car's motion. What is the magnitude of the car's acceleration?

1 answer:

Tanzania [10]2 years ago

7 0

Answer:

Explanation:

Assuming the ground is level as well.

F = ma

a = F/m

a = (2000 - 350) / 1500

a = 1.1 m/s²

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You throw a ball downward from a window at a speed of 2.5 m/s. how fast will it be moving when it hits the sidewalk 2.1 m below?

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List out all the variables that you do know;

acceleration=-9.8 ms⁻¹ (this remains constant on Earth)
Final velocity=?
Displacement (s)= -2.1 m
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v²=u²+2as
v²=(2.5)²+2(-9.8)(-2.1)
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v=6.88549 ≈ 6.9 ms⁻¹

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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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2 years ago