Ask question

Ask question

All categories

2 years ago

15

An object with a charge of 0.9 x 10^-5 C is separated from a second object with a chare of 2.5 x 10^-4 C by a distance of 0.5 m.

Calculate the force acting between the charges.

1 answer:

meriva2 years ago

4 0

Answer: Force = 81 N

Explanation:

from Columbs law,

F = k(q1*q2)/r²

k = 9 x 10^9 Nm²/C²

F = (9 x 10^9)x (0.9x10^-5 x 2.5x10^-4)/(0.5)²

F = 81 Newtons

You might be interested in

A plane flies from base camp to lake a, 200 km away in the direction 20.0° north of east. after dropping off supplies it flies t

Liono4ka [1.6K]

Distance of lake a is 200 km at 20 degree north of east

distance between lake a and b is 230 km at 30 degree west of north

now the distance between base and lake b is given as

$d = d_1 + d_2$

given that

$d_1 = 200 cos20 i + 200 sin20 j$

$d_1 = 187.94 i + 68.4 j$

$d_2 = -230 sin30 i + 230 cos30 j$

$d_2 = -115 i + 199.2 j$

now the total distance is

$d = (187.94 - 115)i + (199.2 + 68.4)j$

$d = 72.94 i + 267.6 j$

now the magnitude of the distance is given as

$d = \sqrt{72.94^2 + 267.6^2}$

$d = 277.4$

also the direction is given as

$\theta = tan^{-1}\frac{267.6}{72.94}$

$\theta = 74.7 degree$

<em>so it is 277.4 km at 74.7 degree North of East</em>

8 0

2 years ago

Three point charges, two positive and one negative, each having a magnitude of 20 C are placed at the vertices of an equilateral

Daniel [21]

The resultant force on the positive charge is mathematically given as

X=40N

<h3>What is the magnitude of the electrostatic force on the negative charge?</h3>

Question Parameters:

Three-point charges, two positive and one negative, each having a magnitude of 20

Generally, the -ve charge is mathematically given as

$Q+=\sqrt{x^2+x^2+2x.xcos120}\\\\Q+=\sqrt{2x^2+2x*(1/2)}$

Q+=X

Therefore

$x=\frac{Kq1q2}{r2}\\\\x=\frac{9*10^9*20*10^{-6}*20*10^{-6}}{(30*10^-2)^2}$

X=40N

For more information on Force

brainly.com/question/26115859

5 0

1 year ago

Calculate the amount of heat liberated (in kj) from 411 g of mercury when it cools from 88.0°c to 12.0°c.

Katen [24]

You should note that the melting point of mercury is -38.83°C, while the boiling point is at 356.7°C. Then, that means that there is no latent heat involved here. We only compute for the sensible heat.

ΔH = mCpΔT
The Cp of mercury is 0.14 J/g·°C
Thus,
ΔH = (411 g)(0.14 J/g·°C)(88 - 12°C)
<em>ΔH = 4,373.04 J</em>

5 0

3 years ago

Consider a small car of mass 1200 kg and a large sport utility vehicle (SUV) of mass 4000 kg. The SUV is traveling at the speed

Karolina [17]

Answer:

63.9 m/s

Explanation:

Parameters given:

Mass of small car, m = 1200 kg

Mass of SUV, M = 4000 kg

Speed of SUV, V = 35 m/s

Their kinetic energy of the small car is equal to the kinetic energy of the SUV, hence:

0.5 * m * v² = 0.5 * M * V²

=> 0.5 * 1200 * v² = 0.5 * 4000 * 35²

600 * v² = 2450000

v² = 2450000/600

v² = 4083.3

=> v = 63.9 m/s

The speed of the small car is 63.9 m/s.

6 0

2 years ago

If spring has a spring constant of 500 N/m and is stretched .50 meters,how much energy is stored in the spring ((show work for f

Free_Kalibri [48]

Answer:

The energy stored is: 62.5 Joules

Explanation:

Given

$k = 500N/m$ --- spring constant

$x = 0.5m$ --- stretch

Required

The amount of energy

This is calculated as:

$U = \frac{1}{2} kx^2$

$U = \frac{1}{2} * 500N/m * (0.5m)^2$

$U = 250N/m * (0.5m)^2$

$U = 62.5\ J$

7 0

2 years ago