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3 years ago

How are force and motion related

1 answer:

5 0

Force is used to put things in motion you can’t have motion without force!
Even if it just pushing a piece of paper you still use force to put the paper in motion!

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A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N

Dmitriy789 [7]

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

$r=\sqrt{(r_{1})^2+(r_{2})^2}$

Put the value into the formula

$r=\sqrt{(9.8)^2+(2.1)^2}$

$r=10.02\ cm$

We need to calculate the force

Using formula of force

$F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}$

Force F₁₂,

$F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}$

$F_{12}=13.33\ N$

$F_{21}=-13.33\ N$

Force F₂₃,

$F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}$

We need to calculate the value of third charge

$F_{net}=F_{21}+F_{23}$

$4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}$

$q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}$

$q_{3}=7.99\times10^{-7}\ C$

$q_{3}=0.8\times10^{-6}\ C$

Hence, The value of third charge is 0.8μC.

4 0

3 years ago

What item can be arranged to transform electrical energy to mechanical energy

meriva

Answer:

battery iron wire

Explanation:

7 0

3 years ago

The tundra is located _______.

Leona [35]

Answer:

d: both north and south

Explanation:

8 0

3 years ago

Read 2 more answers

A ____ exerted by an object on another is a force

Hitman42 [59]

Answer:

A gravitational force exerted by an object on another is a force.

Explanation:

hope it helps

5 0

3 years ago

Read 2 more answers

A classroom that normally contains 40 people is to be airconditioned with window air-conditioning units of 5-kW cooling capacity

oksano4ka [1.4K]

Answer:

About 2 units.

Explanation:

We assume that there is no heat dissipating instrument in the room

Total cooling load of the room is defined from the given below equation

$Q_{cooling}$ = $Q_{light}$ + $Q_{people}$ + $Q_{heat gain}$

where

$Q_{light}$ = 10*100 W =1 KW

$Q_{people}$ = 40*360 KJ/h= 4 KW

$Q_{heat gain}$ =15000KJ/h= 4.17 KW

$Q_{cooling}$ = 1+4+4.17=9.17 KW

the number of air conditioner unit is = $\frac{9.17}{5}$ =1.83

which is approximately 2 units

7 0

3 years ago