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DerKrebs [107]
3 years ago
7

Two tiny conducting spheres are identical and carry charges of -18.4 µC and +53.0 µC. They are separated by a distance of 2.73 c

m. (a) What is the magnitude of the force that each sphere experiences, and is the force attractive or repulsive?
Physics
1 answer:
melomori [17]3 years ago
4 0

Answer:

-11776.36 N

This force is attractive since both charges are of opposite sign

Explanation:

Given that

q_1=-18.4\mu C=-18.4\times 10^{-6}C\ and\\\ q_2=53\mu C=53\times 10^{-6}C

Distance between the spheres = 2.73 cm =0.0273 m

where K is Coulomb's constant = 9.10^ 9 [N.m^2 /C^2]

According to coulombs law we know that force between two charges is given by

F = \frac{kq_1q_2}{r^2}

F = \frac{(9\times10^9)\times (-18.4 \times 10^-^6)(53  \times 10^-^6}{0.0273^2)} \\= -11776.36N

This force is attractive since both charges are of opposite sign

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A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help"
Digiron [165]

Answer: a) The cliff is 532.05m high

b) Her speed just before hitting the ground is 102.12 m/s

Explanation: To solve This, I'll use a sketch diagram, attached to this solution,

In 3seconds, the teacher heard the echo of her initial scream back. We can obtain the distance the teacher had fallen at the end of 3 seconds using the equations of motion,

Y1 = ut + 0.5g(t^2)

Since she's falling under the influence of gravity, her initial velocity, u = 0m/s, g = 9.8m/s2, t = 3s

Y1, distance she fell through in 3 seconds = 0.5×9.8(3^2) = 44.1m

Let the total height of the cliff be (44.1 + x); where is the remaining height of cliff that the teacher will fall through.

Using the equations of motion again, we can obtain distance travelled by the sound waves in 3s. sound waves travel with a constant speed of 340m/s, no acceleration,

Y2 = ut + 0.5g(t^2) where g = 0, u = 340m/s, t = 3seconds

Y2 = 340 × 3 = 1020m

But in 3 secs, the sound waves would have travelled through the total height of the cliff (44.1 + x) and back to the teacher's current height, x. That is, 1020 = 44.1 + x + x

x = 487.95m

So, total height of cliff = 44.1 + 487.95 = 532.05m

b) the speed of the teacher just before she hits the ground.

Using the equations of motion again,

(V^2) = (U^2) + 2gs

Where v is the final velocity to be calculated

U is the initial velocity = 0m/s

g is acceleration due to gravity = 9.8m/s2

S is the total height she fell through, that is, the height of the cliff = 532.05m

(V^2) = 0 + 2×9.8×532.05 = 10428.18

V = √(10428.18) = 102.12m/s

QED!

4 0
3 years ago
Suppose that two point charges, each with a charge of +3.00 Coulomb are separated by a distance of 3.00 meters. Determine the ma
Sveta_85 [38]

Answer:

9\cdot 10^9 N

Explanation:

The magnitude of the electrical force between the two point charges is

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q_1 =q_2 = +3.0 C is the magnitude of each charge

r = 3.00 m is the separation between the two charges

Substituting the numbers into the formula, we find

F=(9\cdot 10^9 Nm^2C^{-2})\frac{(+3 C)^2}{(3.00 m)^2}=9\cdot 10^9 N

3 0
3 years ago
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