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elena-14-01-66 [18.8K]
2 years ago
6

How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is

Physics
1 answer:
inn [45]2 years ago
5 0
Two small spheres spaced 20.0 cm apart have equal charge.

How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 4.57 x 10-21 N?

Number of Electrons = ?
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A 60 kg gorilla named Anthony Falcon is standing on his skateboard. This is on planet Erf,
djyliett [7]
F= ma
F= (600/-10) -10
F= 580n

At least I think that’s the answer
4 0
2 years ago
Ok, what determines the color of light?
gladu [14]

Explanation:

C . frequency

is the correct answer I think .

7 0
3 years ago
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You place an ice cube of mass 7.50×10−3kg and temperature 0.00∘C on top of a copper cube of mass 0.220 kg. All of the ice melts,
g100num [7]

Answer:

T = 29.66 ^o C

Explanation:

As we know that here final equilibrium temperature is 0 degree C

so we can use energy conservation here

heat given by the cube = heat absorbed by the ice

so we have

ms \Delta T = m' L

so here we have

0.220(385)(T - 0) = (7.50 \times 10^{-3})(3.35 \times 10^5)

T = \frac{2512.5}{0.220 \times 385}

T = 29.66 ^o C

5 0
3 years ago
Consider the space between a point charge and the surface of a neutral spherical conducting shell. If the charge sits at the cen
Furkat [3]

Answer:

True

Explanation:

If a thin, spherical, conducting shell carries a negative charge, We expect the excess electrons to mutually repel one another, and, thereby, become uniformly distributed over the surface of the shell. The electric field-lines produced outside such a charge distribution point towards the surface of the conductor, and end on the excess electrons. Moreover, the field-lines are normal to the surface of the conductor. This must be the case, otherwise the electric field would have a component parallel to the conducting surface. Since the excess electrons are free to move through the conductor, any parallel component of the field would cause a redistribution of the charges on the shell. This process will only cease when the parallel component has been reduced to zero over the whole surface of the shell

According to Gauss law

∅ = EA =-Q/∈₀

Where ∅  is the electric flux through the gaussian surface and E is the electric field strength

If the gaussian surface encloses no charge, since all of the charge lies on the shell, so it follows from Gauss' law, and symmetry, that the electric field inside the shell is zero. In fact, the electric field inside any closed hollow conductor is zero

8 0
2 years ago
During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between
Whitepunk [10]

Answer:

a. 12.12°

b. 412.04 N

Explanation:

Along vertical axis, the equation can be written as

T_1 sin14 + T_2sinA = mg

T_2sinA = mg - T_1sin12.5           ....................... (a)

Along horizontal axis, the equation can be written as

T_2×cosA = T_1×cos12.5    ......................... (b)

(a)/(b) given us

Tan A = (mg - T_1sin12.5) / T_1 cos12.5

 = (176 - 413sin12.5) / 413×cos12.5

A = 12.12 °

(b) T2 cosA = T1 cos12.5

T2 = 413cos12.5/cos12.12

= 412.04 N

4 0
3 years ago
Read 2 more answers
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