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goldfiish [28.3K]
3 years ago
10

Through which of the following can magnetic force pass

Physics
2 answers:
laiz [17]3 years ago
8 0
All of the above hope this helps :)
Stells [14]3 years ago
7 0
All of the above hope this helps
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If it takes you 10 seconds to move a chair 5 meters across the floor, using a force of 2 Newtons, how much power did you put out
Maru [420]

Answer:

power=work done÷time taken

2×5=10

10÷10=1

ans 1J per second

5 0
2 years ago
The universe is made up of
sasho [114]
The universe is made up of baryonic matter.(neutrons,electrons,protons)
5 0
3 years ago
Derive an expression for the gravitational potential energy U(r) of the object-earth system as a function of the object's distan
Drupady [299]

Answer:

U(r)=-\frac{Gm_Emr^2}{2R^3_E}

Explanation:

We are given that

Gravitational force=F_g=\frac{Gm_Emr}{R^3_E}

r=0,U(0)=0

We know that

Gravitational potential energy=-\int F_gdr

U(r)=-\int\frac{Gm_Emr}{R^3_E}dr

U(r)=-\frac{Gm_Em}{R^3_E}\times \frac{r^2}{2}+C

Substitute r=0 ,U(0)=0

0=0+C

C=0

Substitute the value

U(r)=-\frac{Gm_Emr^2}{2R^3_E}

4 0
3 years ago
a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?
julsineya [31]

Answer:

Approximately 7.0\; \rm m \cdot s^{-1}.

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant g = 9.81 \; \rm m \cdot s^{-2} near the surface of the earth.

For an object that is accelerating constantly,

v^2 - u^2 = 2\, a \, x,

where

  • u is the initial velocity of the object,
  • v is the final velocity of the object.
  • a is its acceleration, and
  • x is its displacement.

In this case, x is the same as the change in the ball's height: x = 2.5\; \rm m. By assumption, this ball was dropped with no initial velocity. As a result, u = 0. Since the ball is accelerating due to gravity, a = 9.81\; \rm m \cdot s^{-2}.

v^2 - u^2 = 2\, g \cdot h.

In this case, v would be the velocity of the ball just before it hits the ground. Solve for

v^2 = 2\, a\, x + u^2.

\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}.

3 0
3 years ago
Three light bulbs are connected to a battery in a series circuit. How will the bulbs behave if the circuit is closed?
stiv31 [10]

all three will glow A

6 0
3 years ago
Read 2 more answers
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