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Luden [163]
3 years ago
8

Which factor indicates the amount of charge on the source charge?

Physics
2 answers:
ICE Princess25 [194]3 years ago
8 0

Answer:

B. the number of field lines on the source charge

Explanation:

As we know that electric flux is defined as the number of electric field lines passing through a given area.

So here  electric flux due to a point charge "q" is given by

\phi = \frac{q}{\epsilon_0}

so here we know that flux depends on the magnitude of charge and hence we can say that number of filed lines originating from a point charge will depends on the magnitude of the charge.

so here correct answer will be

B.   the number of field lines on the source charge

Elina [12.6K]3 years ago
4 0
B. The number of field lines on the source charge.
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Answer: reaction force = -558N

Explanation:

w = f = 558N

since action force and reaction force are equal in magnitude and opposite in direction,

reaction force = -(f)

reaction force = -558N

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A student makes the following statement
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Convert horsepower to Kilo Newton.
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3 years ago
A military helicopter on a training mission is flying horizontally at a speed of 90.0 m/s when it accidentally drops a bomb (for
Elena-2011 [213]

Answer:

1) 10.1 s  2) 909 m 3) 90.0 m/s 4) -99m/s 5) just over the bomb.

Explanation:

1)

  • In the vertical direction, as the bomb is dropped, its initial velocity is 0.
  • So, we can find the time required for the bomb to reach the earth, applying the following kinematic equation for displacement:

       \Delta y = \frac{1}{2}*a*t^{2} (1)

  • where Δy = -500 m (taking the upward direction as positive).
  • a=-g=-9.8 m/s²
  • Replacing these values in (1), and solving for t, we have:

       t =\sqrt{\frac{2*\Delta y}{-g}} = \sqrt{\frac{2*(-500m)}{-9.8m/s2}} = 10.1 s

  • The time required for the bomb to reach the earth is 10.1 s.

2)

  • In the horizontal direction, once released from the helicopter, no external influence acts on the bomb, so it will continue moving forward at the same speed. that it had, equal to the helicopter.
  • As the time must be the same for both movements, we can find the horizontal displacement just as the product of this speed times the time, as follows:

       x = v_{0x} * t = 90.0 m/s * 10.1 s = 909 m.

3)

  • The horizontal component of the bomb's velocity is the same that it had when left the helicopter. i.e. 90 m/s.

4)

  • In order to find the vertical component of the bomb's velocity just before it strikes the earth, we can apply the definition of acceleration, remembering that v₀ = 0, as follows:

        v_{f} = -g*t = -9.8 m/s2*10.1 s = -99 m/s

5)

  • If the helicopter keeps flying horizontally at the same speed, it will be always over the bomb, as both travel horizontally at the same speed.
  • So, when the bomb hits the ground, the helicopter will be exactly over it.

8 0
3 years ago
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