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elena55 [62]
3 years ago
10

Gaussian surfaces A and B enclose the same positive charge+Q. The area of Gaussian surface A is three times larger than that of

Gaussian surface B. The flux of electric field through Gaussian surface A is A) nine times larger than the flux of electric field through Gaussian surface B. B) three times smaller than the flux of electric field through Gaussian surface B. C) unrelated to the flux of electric field through Gaussian surface B. D) equal to the flux of electric field through Gaussian surface B. E) three times larger than the flux of electric field through Gaussian surface B
Physics
1 answer:
GalinKa [24]3 years ago
4 0

Answer:

D) equal to the flux of electric field through the Gaussian surface B.

Explanation:

Flux through S(A) = Flux through S (B ) =  Charge inside/ ∈₀

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If a car's speed triples, how does the momentum and kinetic energy of the
Harrizon [31]

Answer: When the car speed triples, momentum also triples but Kinetic energy increases 9 times or by 9 fold.

Explanation:

The momentum of a car (an object) is

p= mv

where

m is =the mass of the object( in this case car)

v is its= velocity

While the kinetic energy is is given by the formulae

K=1/2mv²

To determine how momentum and kinetic energy of the  car changes when the speed of the object triples, We have that the new velocity,

v¹= 3v

So that  the momentum  change becomes

p¹=mv¹=m (3v)= 3mv

mv=p

therefore p¹= 3p

we can see that the momentum also triples.

And the kinetic energy change  becomes

K¹=1/2m(v¹)²= 1/2m (3v)²

= 1/2m9v²= 1/2 x m x 9 x v²=9 x1/2mv²

1/2mv²=K

K¹= Kinetic energy = 9k

but Kinetic energy increases 9 times

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3 years ago
What if is a forecasting game, below are actions on your document and all you have to do is predict what will happen after the a
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2 years ago
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What happens to its surface temperature as a star changes color from red to blue
MAVERICK [17]
The surface temperature of the star increases when it changes its colour changes from red to blue.
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5 0
3 years ago
A tuning fork vibrating at 508 Hz falls from rest and accelerates at 9.80 m/s^2. How far below the point of release is the tunin
JulijaS [17]

Answer:

Explanation:

given,

tuning fork vibration = 508 Hz

accelerates = 9.80 m/s²

speed of sound = 343 m/s

observed frequency = 490 Hz

f_s = f(\dfrac{v}{v-(-v_s)})

f_s = f(\dfrac{v}{v+v_s})

v_s = v[\dfrac{f_s}{f_o}-1]

      = 343[\dfrac{508}{490}-1]

      v_s=12.6 m/s

distance the tunning fork has fallen

y_1=\dfrac{v^2}{2a_y}

     =\dfrac{12.6^2}{2\times 9.8}

     =8.1 m

now, time required for the observed will be

t = \dfrac{8.1}{343} = 0.023 s

now, for the distance calculation

y_2 = u\ t + \dfrac{1}{2}at^2

  = 12.6\times 0.023 +\dfrac{1}{2}\times 9.8 \times 0.023^2

  =0.293 m

total distance

 = 8.1 + 0.293 = 8.392 m

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2 years ago
_h2o2(aq)-_h2o(l) +o2(g)
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Answer:

V02=9.78L

Explanation:

6 0
2 years ago
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