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hichkok12 [17]
3 years ago
10

Moving 2.0 coulombs of charge a distance of 6.0 meters from point A to point B within an electric field requires a 5.0-newton fo

rce. What is the electric potential difference between points A and B?
Physics
1 answer:
amm18123 years ago
5 0

Answer:

15 V

Explanation:

Electric potential is given as the product of Electric field and distance. Mathematically:

V = E * r

Where E = Electric field

r = distance = 6m

Electric field, E, is:

E = F / q

Where F = Electric force = 5N

q = Electric charge = 2C

Therefore, electric potential will be:

V = (F * r) / q

V = (5 * 6) / 2

V = 15V

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8a2-10ab+15b+10 Explaintion:

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The most common atom used in fission is ​
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Answer:

Uranium

Explanation:

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3 years ago
The magnitude of the net force versus time graph has a rectangular shape. Often in physics geometric properties of graphs have p
Blizzard [7]

Answer:

True

Explanation:

In this particular case, the area of the graph represents the impulse.

In fact, impulse is defined as the change in momentum of an object:

I=\Delta p

Moreover, impulse is also defined as the product between the magnitude of the force acting on an object and the duration of the collision:

I=F\Delta t

If we plot a graph of the force versus the time, if the force is constant then this graph will have a rectangular shape, and the area under the graph will simply be the product

F\cdot \Delta t

which corresponds to the definition of impulse.

8 0
3 years ago
A stretched string has a mass per unit length of 5.40 g/cm and a tension of 17.5 N. A sinusoidal wave on this string has an ampl
kondaur [170]

Answer:

Part a)

y_m = 0.157 mm

part b)

k = 101.8 rad/m

Part c)

\omega = 579.3 rad/s

Part d)

here since wave is moving in negative direction so the sign of \omega must be positive

Explanation:

As we know that the speed of wave in string is given by

v = \sqrt{\frac{T}{m/L}}

so we have

T = 17.5 N

m/L = 5.4 g/cm = 0.54 kg/m

now we have

v = \sqrt{\frac{17.5}{0.54}}

v = 5.69 m/s

now we have

Part a)

y_m = amplitude of wave

y_m = 0.157 mm

part b)

k = \frac{\omega}{v}

here we know that

\omega = 2\pi f

\omega = 2\pi(92.2) = 579.3 rad/s

so we  have

k = \frac{579.3}{5.69}

k = 101.8 rad/m

Part c)

\omega = 579.3 rad/s

Part d)

here since wave is moving in negative direction so the sign of \omega must be positive

4 0
3 years ago
15. A locomotive moved 18.0 m [W] in a time of 6.00 s and stopped. After stopping, the
mariarad [96]

Answer:

Distance = 30m

Displacement = 6m W

Explanation:

Given the following:

Movement 1 = 18m W

Movement 2 = 12m E

Diatance is a scalar quantity with only magnitude and no direction. That is, in Calculating the distance moved by the locomotive, the direction of travel or movement of the object is not considered. It only measures the total amount of movement made during the Time of motion.

Therefore, total distance traveled equals :

Movement 1 + movement 2

18m + 12m = 30m

B) Displacement also measures the movement made by an object. However, Displacement is a vector quantity and therefore, considers both magnitude and direction of travel of the object. Therefore, it measures the overall change in position of the object from its starting position.

Therefore, Displacement of the locomotive equals:

18m W - 12m E = 6m E

3 0
2 years ago
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