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Bingel [31]
3 years ago
15

A+10 u charge and a -10 4C (1 HC - 106 C), at a distance of 0.3 m,

Physics
1 answer:
Marina CMI [18]3 years ago
4 0

Answer:

B. Attract each other with a force of 10 newtons.

Explanation:

Statement is incorrectly written. <em>The correct form is: A </em>+10\,\mu C<em> charge and a </em>-10\,\mu C<em> at a distance of 0.3 meters. </em>

The two particles have charges opposite to each other, so they attract each other due to electrostatic force, described by Coulomb's Law, whose formula is described below:

F = \frac{\kappa \cdot |q_{A}|\cdot |q_{B}|}{r^{2}} (1)

Where:

F - Electrostatic force, in newtons.

\kappa - Electrostatic constant, in newton-square meters per square coulomb.

|q_{A}|,|q_{B}| - Magnitudes of electric charges, in coulombs.

r - Distance between charges, in meters.

If we know that \kappa  = 8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}, |q_{A}| = |q_{B}| = 10\times 10^{-6}\,C and r = 0.3\,m, then the magnitude of the electrostatic force is:

F = \frac{\kappa \cdot |q_{A}|\cdot |q_{B}|}{r^{2}}

F = 9.987\,N

In consequence, correct answer is B.

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Answer is B.
Because velocity is vector quantity, so magnitud and direction are needed to define it.
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3 0
3 years ago
Boxes A and B are in contact on a horizontal, frictionless surface. Box A has mass 21.0 kg and box B has mass 8.0 kg. A horizont
vfiekz [6]

Answer:

2.75 N

Explanation:

Given that,

Box A has a mass 21.0 kg and box B has a mass 8.0 kg.

A horizontal force of 100N is exerted on box A.

Let a be the acceleration of the system. Using second law of motion,

F=(m_A+m_B)a\\\\a=\dfrac{F}{(m_A+m_B)}\\\\a=\dfrac{10}{(21+8)}\\\\a=0.344\ m/s^2

Now applying Newton's second law to box B. So,

F_A=m_Ba\\\\=8\times 0.344\\\\=2.75\ N

So, 2.75 N is the force that box A exerts on box B.

8 0
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A pulley consists of a wrapped around a
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Of a rope wrapped around a pulley chain
6 0
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(ANSWER NOW PLS) A diver with a mass of 90kg is at a height of 10m, and he has not jumped off of the board yet (v=0m/s) what's h
Korolek [52]

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3 0
3 years ago
You have a spring-loaded air rifle. When it is loaded, the spring is compressed 0.3 m and has a spring constant of 150 N/m. In j
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The potential energy of the spring is 6.75 J

The elastic potential energy stored in the spring is given by the equation:

E= \frac{1}{2} kx^2

where;

k is the spring constant

x is the compression/stretching of the string

In this problem, we have the spring as follows:

k = 150 N/m is the spring constant

x = 0.3 m is the compression

Substituting in the equation, we get

E=\frac{1}{2} (150) (0.3)^2

E=6.75J

Therefore. the elastic potential energy stored in the spring is 6.75J .

Learn more about potential energy here:

brainly.com/question/10770261

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6 0
2 years ago
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