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Tju [1.3M]
2 years ago
11

If the magnitude of a charge is half as much as another charge, but the force experienced is the same, then the electric field s

trength of this charge is _____ the strength of the other charge.
one-fourth
one-half
two times
four times
Physics
1 answer:
Kazeer [188]2 years ago
8 0

Answer:

the electric field strength of this charge is two times the strength of the other charge

Explanation:

Using the relationship between electric field and the charge, which is inversely proportionality. Let the the magnitude of the first charge be Q and the respective electric field be E. It implies that;

E1/E2 = Q2/Q1

E2 = E1 x Q1/Q2

      = E x Q/ (Q/2)

       = 2E

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Answer:

4 A

Explanation:

We are given that

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We have to find the current flowing through each resistor.

We know that in parallel combination current flowing through different resistors are different and potential difference across each resistor is same.

Formula :

\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}

Using the formula

\frac{1}{R}=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}

R=\frac{4}{3}\Omega

V=IR

Substitute the values

V=12\times \frac{4}{3}=16 V

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A tug boat pulls a ship with a constant net horizontal force of 5.00•10*3 N and causes the ship to move through a harbor. How mu
Murljashka [212]

The work done on the ship is 1.5\cdot 10^7 J

Explanation:

The work done by a force on an object is given by:

W=Fd cos \theta

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d is the displacement

\theta is the angle between the direction of the force and of the displacement

In this problem, we have:

F=5.00\cdot 10^3 N (force acting on the ship)

d = 3.00 km = 3000 m (displacement of the ship)

\theta=0^{\circ} (because the force is horizontal, and the displacement is horizontal as well)

Therefore, the work done on the ship is

W=(5.00\cdot 10^3)(3000)(cos 0^{\circ})=1.5\cdot 10^7 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

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2 years ago
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