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weqwewe [10]
2 years ago
14

What happens to the magnetic force as the distance between two magnetic poles decreases?

Physics
2 answers:
iragen [17]2 years ago
8 0
The magnetic force would also decrease. The pull wouldn’t be as strong between the two polls.
iren [92.7K]2 years ago
6 0
It becomes less magnetic. I also know this because I’m smart like that.
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Which state of matter would be described as a highly energized charge particles with moving extremely fast
Oksanka [162]
The correct answer is plasma
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3 years ago
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A car traveling at 14 m/s accelerates at 3.5 m/s² for 5 seconds. How much distance does it travel during that time?​
Rainbow [258]

Answer: 113.75

Explanation:

You know

acceleration = a = 3.5 m/s²

time = t = 5 seconds

initial velocity = u = 14 m/s

Unknown is distance = s = ?

Use equation: s = ut + \frac{1}{2} at²

Substitute all the known values inside the equation:

s = (14*5) + 0.5 * 3.5 * 5²

s = 70 + 43.75 = 113.75 m

The car travels 113.75 metres.

3 0
3 years ago
A billiard ball moving at 0.5 m/s strikes another identical billiard ball, which is at rest, in an elastic head-on collision on
NikAS [45]

Answer: assuming that the billiard balls are of identical weight the impacted billiard ball will move forward at around 0.5m/s (not considering energy conservation). The ball impacting the 2nd one would stop because most of its Kinetic energy would have been transferred into the not moving ball.

Explanation: hope this helps!

4 0
3 years ago
A pendulum that has a period of 2.67000 s and that is located where the acceleration due to gravity is 9.77 m/s2 is moved to a l
zalisa [80]

Answer:

Explanation:

Expression for time period of pendulum is given as follows

T=2\pi\sqrt{\frac{l}{g} }

where l is length of pendulum and g is acceleration due to gravity .

Putting the given values for first place

2.67=2\pi\sqrt{\frac{l}{9.77} }

Putting the values for second place

T=2\pi\sqrt{\frac{l}{9.81} }

Dividing these two equation

\frac{T}{2.67} =\sqrt{\frac{9.77}{9.81} }

T = 2.66455 s.

5 0
3 years ago
A spotlight on the ground shines on a wall 12 meters away. If a man 2 meters high walks away from the spotlight towards the wall
nordsb [41]

Answer: The length of the shadow on the wall is decreasing by 0.6m/s

Explanation:

the specified moment in the problem, the man is standing at point D with his head at point E.

At that moment, his shadow on the wall is y=BC.

The two right triangles ΔABC and ΔADE are similar triangles. As such, their corresponding sides have equal ratios:

ADAB=DEBC

8/12=2/y,∴y=3 meters

If we consider the distance of the man from the building as x then the distance from the spotlight to the man is 12−x.

(12−x) /12=2/y

1− (1 /12x )=2 × 1/y

Let's take derivatives of both sides:

−1 / 12dx = −2 × 1 / y^2 dy

Let's divide both sides by dt:

−1/12⋅dx/dt=−2/y^2⋅dy/dt

At the specified moment:

dxdt=1.6 m/s

y=3

Let's plug them in:

−1/121.6) = - 2/9 × dy/dt

dy/dt = 1.6/12 ÷ 2/9

dy/dt = 1.6/12 × 9/2

dy/dt = 14.4/24 = 0.6m/s

5 0
3 years ago
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