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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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The frequency of a wave is defined as the number of cycles completed per second while the period refers to the time taken to complete a cycle. The frequency is the inverse of period.

So;

Period(T) = 54 minutes or 3240 seconds

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A car travels initially at 24 m/s, until it enters the highway. If the car accelerates at 4 m/s^2 for a 96 meters, what is the c
marishachu [46]
  • initial velocity=u=24m/s
  • Acceleration=a=4m/s^2
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Using 3rd equation of kinematics

\boxed{\Large{\sf v^2-u^2=2as}}

\\ \Large\sf\longmapsto v^2=u^2+2as

\\ \Large\sf\longmapsto v^2=24^2+2(4)(96)

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\\ \Large\sf\longmapsto v^2=1344

\\ \Large\sf\longmapsto v=\sqrt{1344}

\\ \Large\sf\longmapsto v=36.6m/s

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3 years ago
A 1.50-V battery supplies 0.414 W of power to a small flashlight. If the battery moves 4.93 1020 electrons between its terminals
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Answer:

2.86×10⁻¹⁸ seconds

Explanation:

Applying,

P = VI................ Equation 1

Where P = Power, V = Voltage, I = Current.

make I the subject of the equation

I = P/V................ Equation 2

From the question,

Given: P = 0.414 W, V = 1.50 V

Substitute into equation 2

I = 0.414/1.50

I = 0.276 A

Also,

Q = It............... Equation 3

Where Q = amount of charge, t = time

make t the subject of the equation

t = Q/I.................. Equation 4

From the question,

4.931020 electrons has a charge of (4.931020×1.6020×10⁻¹⁹) coulombs

Q = 7.899×10⁻¹⁹ C

Substitute these value into equation 4

t =  7.899×10⁻¹⁹/0.276

t = 2.86×10⁻¹⁸ seconds

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Your answer is A

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