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melisa1 [442]
3 years ago
7

Two spheres have identical charges and are 75 cm apart. the force between them is +0.30 n. what is the magnitude of the charge o

n each sphere? (let x = charge) what can you tell about the charge signs on the spheres? ______________
Physics
1 answer:
ella [17]3 years ago
5 0

Answer:

4.33\cdot 10^{-6}C, charges are both positive or both negative

Explanation:

The electrostatic force between the two spheres is given by

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

where

k is the Coulomb's constant

q1 and q2 are the charges on the two spheres

r is the distance between the centres of the two spheres

In this problem, we have

F=+0.30 N is the force

r=75 cm=0.75 m is the distance between the spheres

q_1 =q_2 =q because the two spheres have identical charge

Solving the formula for q, we find

q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(+0.30 N)(0.75 m)^2}{9\cdot 10^9}}=4.33\cdot 10^{-6}C

And the two charges have the same sign (so, both positive or both negative), since the sign of the force is positive (+0.30 N), so it is a repulsive force.

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Answer:

4452.5 J.

Explanation:

The diver have both kinetic and potential energy.

Ek = 1/2mv² ................. Equation 1

Where Ek = Kinetic Energy of the diver, m = mass of the diver, v = velocity of the diver.

Given: m = 65 kg, v = 6.4 m/s.

Substitute into equation 1

Ek = 1/2(65)(6.4²)

Ek = 1331.2 J.

Also,

Ep = mgh ............................ Equation 2

Where Ep =  Potential energy of the diver when its above the water, h = height of the diver above the water, g = acceleration due to gravity.

Given: m = 65 kg, h = 4.9 m, g = 9.8 m/s²

Substitute into equation 2.

Ep = 65(4.9)(9.8)

Ep = 3121.3 J.

Note: When she hits the water, the potential energy is converted to kinetic energy.

E = Ek+Ep

Where E = Kinetic energy of the diver when she hits the water.

E = 1331.2+3121.3

E = 4452.5 J.

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3 years ago
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Influenced by the gravitational pull of a distant star, the velocity of an asteroid changes from from +19.3 km/s to −18.8 km/s o
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initial velocity = 19.3 km/s

final velocity = - 18.8 km/s

now in order to find the change in velocity

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acceleration is given by formula

a = \frac{\Delta v}{\Delta t}

given that

\Delta v =- 3.81 * 10^4 m/s

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4 0
3 years ago
Calculate the force which will produce an extension of 0.30mm in a steel wire with a length of 4.0m and a cross section area of
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Given data:

* The extension of the steel wire is 0.3 mm.

* The length of the wire is 4 m.

* The area of cross section of wire is,

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Y=2.1\times10^{11}\text{ Pa}

Solution:

The young modulus of the steel in terms of the force and extension is,

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Thus, the force which produce the extension of 0.3 mm of the steel wire is 31.5 N.

7 0
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