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JulsSmile [24]
2 years ago
15

Why must the test charge used to determine the electric field be very, very small?.

Physics
1 answer:
vekshin12 years ago
8 0

The test charge used to determine the electric field is very, very small so that its presence does not affect the electrical field because of the supply price. The electrical rate that produces the electric area is known as a source charge.

An electric-powered area is a physical area that surrounds electrically charged debris and exerts pressure on all differently charged particles inside the area, both attracting or repelling them. It also refers back to the physical field for a machine of charged particles.

Electric-powered discipline is described as the electrical force in keeping with unit fee. The path of the sector is taken to be the route of the force it might exert on a fine check price. the electric discipline is radially outward from a nice rate and radially in toward a bad point rate.

An electric discipline is a vicinity of space around an electrically charged particle or object in which an electric charge might sense force. An electric field is a vector amount and may be visualized as arrows going in the direction of or away from fees.

Learn more about  electric here brainly.com/question/24317284

#SPJ4

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Expain why a spinning top will not remain in motion forever
tankabanditka [31]
I think the force of spinning it can only hold for so long and because of gravitys pressure.(i hope i helped) :)
6 0
3 years ago
En una báscula hidráulica colocamos una persona de 75 kg sobre un émbolo y un camión de 7200 kg sobre una plataforma de 5 m de l
TEA [102]

...........................

4 0
4 years ago
What is the mass of a school bus if it can accelerate from rest to 15.5 m/s over 8.25
alexandr1967 [171]

The mass of a school bus if it can accelerate from rest to 15.5 m/s over 8.25s with 7,500 N of force is 3989.4kg.

HOW TO CALCULATE MASS:

  • The mass of an object can be calculated by dividing the force applied to the object by its acceleration.

  • According to this question, a bus can accelerate from rest to 15.5 m/s over 8.25s. The acceleration can be calculated as follows:

  • a = (v - u)/t

  • a = 15.5 - 0/8.25

  • a = 15.5/8.25

  • a = 1.88m/s²

  • The mass of the bus = 7500N ÷ 1.88m/s²

  • The mass of the bus = 3989.4kg

  • Therefore, the mass of a school bus if it can accelerate from rest to 15.5 m/s over 8.25s with 7,500 N of force is 3989.4kg.

Learn more about mass at: brainly.com/question/20259048?referrer=searchResults

3 0
3 years ago
A policeman in a stationary car measures the speed of approaching cars by means of an ultrasonic device that emits a sound with
Ymorist [56]

Answer:

the frequency of the beats is 8.7687 kHz

Explanation:  

Given the data in the question;

The frequency for stationary source and moving observer is;

f' = f( 1 +( v_observer/v_sound))

we know that, speed of sound in dry air =  343 m/s

so we substitute

f' = 41.2 kHz( 1 + (33.0 m/s / 343 m/s) ) = 41.2kHz( 1 + 0.0962) = 41.2kHz(1.0962)

f' = 45.1634 kHz

Now the frequency for stationary observer and moving source with frequency f' will be

f" = f'( 1 / (1 - ( v_observer/v_sound)))

45.1634 kHz( 343 / 343 - 33)

we substitute

f" = 45.1634 kHz( 1 / (1 - (33.0 m/s / 343 m/s)))

f" = 45.1634 kHz( 1 / (1 - 0.0962))

f" = 45.1634 kHz( 1 / 0.9038 )

f" = 45.1634 kHz( 1.1064 )

f" = 49.9687 kHz

Now the beat frequency will be;

f_beat = f' - f

we substitute

f_beat = 49.9687 kHz - 41.2 kHz

f_beat = 8.7687 kHz

Therefore, the frequency of the beats is 8.7687 kHz

7 0
3 years ago
Two speakers that are 12.0 m apart produce in-phase sound waves of frequency 245 Hz in a room where the speed of sound is 340 m/
Cerrena [4.2K]

Answer:

a) Please see below as the answer is self-explanatory

b) 0.35 m

c) 0.7 m

Explanation:

a) As there is no difference in the path (because she is exactly at the midpoint between both speakers) that the waves do in order to reach to her ear, she listens both waves with the same intensity.

The same effect would be experimented, if the difference between paths, were an even multiple of the half of the wavelength.

b) In order to have a destructive interference, the difference in path (in the worst case) must be equal to an odd multiple of the half wavelength.

In any wave, there is a fixed relationship between speed, frequency and wavelength, as follows:

λ = v / f

Replacing with our givens for frequency (245 Hz) and v (340 m/s), we have:

λ = 340 m/s / 245 Hz = 1.4 m

The condition for destructive interference, is as follows:

Δd = d₂-d₁ = (2n + 1) *λ/2

For n=0, we have the shortest distance:

Δd = λ/2 = 0.7 m

So, in order to have such a path difference, she must walk exactly the half of this distance from the center, i.e., 0.35 m.

c) With the same reasoning as above, as the condition for constructive interference, as we have already mentioned, is that the path difference be at least one full wavelength, we can say:

Δd = d₂-d₁ = λ

So, in order to have such a path difference, she must walk exactly the half of this distance from the center, i.e., 0.7 m.

4 0
3 years ago
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