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ELEN [110]
3 years ago
10

Electric potential difference is also known as

Physics
1 answer:
Sauron [17]3 years ago
8 0

Answer:

volt / voltage

Explanation:

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When an automobile moves with constant speed down a highway, most of the power developed by the engine is used to compensate for
uysha [10]

Answer:

F=5449 N

Explanation:

Work done is a product of force and displacement ie

Work done, W, = Force*Displacement

Power, P, is Work done/Time

P=\frac {W}{t}=\frac {FS}{t} where P is power, W is work done, F is force, S is displacement and t is time

In this case, F is the frictional force. Converting the power from hp to W, we multiply by 746 hence P=746*168=125328  W

Since displacement/time is velocity, then

P=FV where V is velocity in m/s

Making F the subject

F=\frac {P}{V}

F=\frac {125328}{23}=5449.043478  N

F=5449 N

7 0
3 years ago
A yoyo is hanging motionless from a string. Identify and describe the forces exerted on the string.
sashaice [31]
Potential and kinetic energy along with an external force
5 0
4 years ago
There are 5,280 feet in 1 mile and 12 inches in one foot. How many inches are in a mile?
ohaa [14]
63360
5280 feet*12 inches in every foot=63360
4 0
3 years ago
What will be the effect on the loudness of sound if the amptitude is doubled and halved?
kap26 [50]
<em>remember that
<u>loudness is directly proportional to the square of amplitude...
</u>
so when amplitude is doubled ,,loudness increases four times..
when it is halved ,,,loudness is halved,,,</em>
4 0
3 years ago
Tap on the photo. For each diagram, explain why the light behaves in the way that it does.
dem82 [27]

Answer:

Diagram 1, 3 and 4 can be explained with the phenomenon of refraction.

Refraction occurs when a ray of light crosses the interface between two mediums with different optical density: when this occurs, the ray of light is bent and its speed changes, according to Snell's law

n_1 sin \theta_1 = n_2 sin \theta_2

where n_1,n_2 are the refractive index of the 1st and 2nd medium

\theta_1, \theta_2 are the angle that the incident ray and the refracted ray makes with the normal to the interface

In diagram, 1, the ray of light arrives perpendicularly to the interface, so it is refracted through the medium but it doesn't change its direction (only its speed).

In diagram 3, the ray of light is refracted twice: at the 1st interface and at the 2nd interface. In the 1st case, it goes from a medium with lower refractive index to a medium with higher refractive index (n_1), this means that \theta_2, so the ray bends towards the normal. Vice-versa, in the 2nd case the ray goes from a medium with higher refractive index to a medium with lower refractive index (n_1>n_2), so it bends away from the normal (\theta_2>\theta_1).

In diagram 4, the ray of light is also refracted twice. The ray of light here acts exactly the same as in diagram 3, h

However, this time the 2nd interface is the opposite direction with respect to diagram 3, so in this case the ray of light at the 2nd interface bends in the opposite direction (still away from the normal).

Diagram 2 instead is an example of reflection, that occurs when a ray of light bounces off the interface between the two mediums, withouth entering the 2nd medium.

According to the law of reflection:

- The incoming ray, the reflected ray and the normal to the boundary are all in the same plane

- The angle of incidence is equal to the angle of reflection (both are measured relative to the normal to the boundary)

Therefore in this diagram, the ray of light hits the boundary at approx. 45 degrees from the normal, and then it is reflected back approximately at 45 degrees on the other side with respect to the normal.

3 0
4 years ago
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