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lutik1710 [3]
3 years ago
9

What does the electric field strength tell about the electric firld?

Physics
1 answer:
Solnce55 [7]3 years ago
8 0

Answer:

Explanation:

The region around a charged particle where another charged particle experiences a force of attraction or repulsion is called electric field.

The strength of electric field is defined as the force experienced by the unit positive test charge.

E = F / q

Electric field strength is a vector quantity and it is measured in newton per coulomb.

Where, F is the force of attraction or repulsion between the two charges and q is the test charge on which the electric field strength is to be calculated.

The strength of electric field is more if the field is strong. It means more be the electric field strength at a point more be the electric field.

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A machine carries a 100kg cargo to a boat at a rate of 10m/s2. The distance between the ground to the boat is 50ft. If the machi
klasskru [66]

Answer:

50.8 watt

Explanation:

we know that P=W÷t

W=F.S           S-->distance=50 ft= 15.24 m

F=ma

=100×10=1000 N

SO W= 1000×15.24

        =15240 J

NOW

P=W÷t          t=5 mints = 5×60=300 sec

P=15240÷300

P=50.8 watt

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3 years ago
What is a supernova?
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A supernova is a star  that suddenly increases greatly in brightness because of a catastrophic explosion that  ejects most of its mass.
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3 years ago
Given the distance between the crest of one wave and the crest of the next wave, you can determine the?
Sonja [21]
Answer: wavelength !!
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3 0
2 years ago
high school physics, no need detail explain, just give the answer, but you have to make sure thank you
Goshia [24]

Answer:many questions add point

6 0
3 years ago
Light propagate faster through medium “a” than medium “b”
dangina [55]

1) Medium "b" has more optical density

2) Light must hit the interface between the two mediums perpendicularly

Explanation:

1)

Refraction occurs when light propagates from a medium into a second medium.

The optical density of a medium is given by its index of refraction, which is defined as:

n=\frac{c}{v}

where

c is the speed of light in a vacuum

v is the speed of light in a medium

Higher index of refraction means higher optical density, and light propagater slower into a medium with higher optical density.

In this problem, light propagates faster through medium "a" than medium "b": this means that medium "a" has lower refractive index of medium "b", and so "b" has more optical density.

2)

We can answer this part by referring to Snell's law, which gives the relationship between the direction of the incident ray and of the refracted ray when light passes through the interface between two media:

n_1 sin \theta_1 = n_2 sin \theta_2

where

n_1, n_2 are the index of refraction of the two mediums

\theta_1, \theta_2 are the angle of incidence and of refraction (the angle that light makes with the normal to the surface in medium 1 and medium 2)

Here we want the direction of propagation of the light ray not to change: this means that it must be

sin \theta_1 = sin \theta_2 (1)

However, here we have two mediums "a" and "b" with different index of refraction, so

n_1\neq n_2

Therefore the only angle that can satisfy eq.(1) is

\theta_1 = \theta_2 = 0

So, the light must hit the surface perpendicular to the interface between the two mediums.

Learn more about refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

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3 years ago
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