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Ksivusya [100]
3 years ago
11

Please help! What's the answer to this question, A 5.6 nC electric charge is placed in an Electric Field and experiences a force

of 7.4 μN. What is the magnitude of the Electric Field at that location?
Physics
1 answer:
zvonat [6]3 years ago
3 0

Answer:

1321 N/C

Explanation:

From the question,

Electric Field (E) = Electric Force(F)/Electric Charge(q)

E = F/q............ Equation 1

Given: F = 7.4 μN = 7.4×10⁻⁶ N, q = 5.6 nC = 5.6×10⁻⁹ C

Substitute these value into equation 1

E = ( 7.4×10⁻⁶)/(5.6×10⁻⁹)

E = 1.321×10³ N/C

E = 1321 N/C

Hence the magnitude of the electric field at that location is 1321 N/C

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Consider the medium of air as defined by the use of radio frequency. What made some of the early standards so slow compared to t
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Answer:

Explanation:

Earliest standards were dependent on a single frequency/channel to both send and receive. This shared medium creates the same problem as half-duplex coax cable. Because receivers had to wait for the signal before sending a response, this reduced the overall bandwidth.

Other factors affect wireless signal propagation, too, including RF interference, antenna choice, and obstacles such as walls, trees, and even weather (precipitation, for example).

3 0
4 years ago
Suppose that you connect the terminals of two batteries of different emfs positive to positive and negative to negative (opposin
gulaghasi [49]

Answer:

Answer is explained in the explanation section below.

Explanation:

This question is very basic and easy. The answer to this question is:

Answer: If both batteries are connected we would get less amount of charge as compared to connected a single battery.

Reasoning:

If both batteries are connected in a manner of positive terminal to positive terminal and negative terminal to negative terminal then a capacitor is added to charge it from the batteries then, total electromotive force (emf) would decrease.

As a result, the capacitor added would get less amount of charge stored. But capacitor added will get more amount of charge stored when a single battery is connected.

7 0
3 years ago
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GenaCL600 [577]
D. 65.1 is the answer
8 0
3 years ago
A 28-kg beginning roller skater is standing in front of a wall. By pushing against the wall, she is propelled backward with a ve
Scorpion4ik [409]

Answer:

33.6 Ns backward.

Explanation:

Impulse: This can be defined as the product of force and time. The S.I unit of impulse is Ns.

From Newton's second law of motion,

Impulse = change in momentum

I = mΔv................................. Equation 1

Where I = impulse, m = mass of the skater, Δv = change in velocity = final velocity - initial velocity.

Given: m = 28 kg, t = 0.8 s, Δv = -1.2-0 = -1.2 m/s (Note: the initial velocity of the skater = 0 m/s)

Substituting into equation 1

I = 28(-1.2)

I = -33.6 Ns

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3 0
4 years ago
In this example we will analyze the forces acting on your body as you move in an elevator. Specifically, we will consider the ca
dexar [7]

Answer:

The reading of the scale during the acceleration is 446.94 N

Explanation:

Given;

the reading of the scale when the elevator is at rest = your weight, w = 600 N

downward acceleration the elevator, a = 2.5 m/s²

The reading of the scale can be found by applying Newton's second law of motion;

the reading of the scale  = net force acting on your body

R = mg + m(-a)

The negative sign indicates downward acceleration

R = m(g - a)

where;

R is the reading of the scale which is your apparent weight

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g is acceleration due to gravity, = 9.8 m/s²

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R = m(g-a)

R = 61.225(9.8 - 2.5)

R = 446.94 N

Therefore, the reading of the scale during the acceleration is 446.94 N

4 0
3 years ago
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