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borishaifa [10]
2 years ago
9

A negative charge of -3.0 x 10^-5 C is placed 0.35 m from a positive charge of 9.0 x 10^-5 C. What is the force between the two

charges
Physics
1 answer:
Ilia_Sergeevich [38]2 years ago
8 0

Answer:

the force between the two charges is 198.37 N

Explanation:

Given;

charge of the first particle, Q₁ = -3 x 10⁻⁵ C

charge of the second particle, Q₂ = 9.0 x 10⁻⁵ C

distance between the two charges, r = 0.35 m

The force between the two charges is calculated from Coulomb's law;

F = \frac{kQ_1Q_2}{r^2}

where k is Coulomb's constant = 9 x 10⁹ Nm²/C²

F = \frac{(9\times 10^9)(3.0\times 10^{-5})(9\times10^{-5} )}{0.35^2} \\\\F = 198.37 \ N

Therefore, the force between the two charges is 198.37 N

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When Missourians carve their initials into the bark of a tree, the damage leaves the tree open to ____________________.
sasho [114]

Answer:

In fact, carving letters into a tree probably won't hurt it. ... In general, the tree will compartmentalize the wound and it will heal over. The initials that remain visible are essentially scar tissue, permanent scar tissue.

Explanation:

Unfortunately, when carving into the trunk of a tree the blade of a knife often penetrates the outer bark and cuts into the inner bark. ... In cases that the phloem is damaged all the way around the trunk (in a ring for example), the tree will slowly and eventually starve to death.

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luke_raines19

5 0
3 years ago
A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind
Marat540 [252]

Answer:

Part a)

F_v = 4.28 N

Part B)

L = 1.02 m

Part C)

v = 1.25 m/s

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

Tcos\theta = mg

T sin\theta = F_v

\frac{F_v}{mg} = tan\theta

F_v = mg tan\theta

F_v = 1.2\times 9.81 (tan20)

F_v = 4.28 N

Part B)

Here we can use energy theorem to find the distance that it will move

-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2

(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2

(-3.5 + 2.54)L = - 0.98

L = 1.02 m

Part C)

At terminal speed condition we know that

F_v = mg

bv^2 = mg

2.5 v^2 = 3.9

v = 1.25 m/s

7 0
3 years ago
Two balls move away from each other, both traveling at 7 m/s. One has a mass of 2 kg and the other has a mass of 3 kg
zavuch27 [327]
A 300-kg bear grasping a vertical tree slides down at constant velocity. The friction force between the
tree and the bear is
5 0
3 years ago
A 5.4 g lead bullet moving at 261 m/s strikes a steel plate and stops. If all its kinetic energy is converted to thermal energy
scZoUnD [109]

Answer:

Change in temperature ∆(tita) is 266.097°C

Explanation:

Ok kinectic energy = 1/2MV²

5.4 grams =( 5.4/1000) kilogram

Kinectic energy =( 1/2 )*(5.4/1000)*261²

Kinectic energy = 183.9267 joules

If kinetic energy = thermal energy

183.9267 joules = mc∆(tita)

Where ∆(tita) = change in temperature

And c = 128 J/kg

∆(tita) = 183.9267/((5.4/1000)*128)

∆(tita) = 266.097

∆(tita) = 266.097°C

8 0
3 years ago
While a balloon is being filled, if the temperature of the air in the balloon decreases, what happens to its volume?
Ugo [173]
Why is it always balloons?
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4 0
3 years ago
Read 2 more answers
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