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AfilCa [17]
3 years ago
5

How does the force exerted by two magnets change as the magnets are moved farther apart?

Physics
1 answer:
galben [10]3 years ago
8 0
<span>The magnetic force exerted by two magnets decreases as the magnets are moved farther apart.

</span><span>The magnetic force between two magnets is the force of repulsion or attraction between two magnets which leads to the calculation of a positive work.
</span><span>The magnetic of a magnet is strongest at its poles.</span>
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A car travels 92 miles in 2 hours. What is the car's AVERAGE SPEED?
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about 46 mph

Explanation:

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What is the form of energy that batteries store energy as
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The answer you're looking for is Electrical Energy.

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If a cell phone is dropped from a very tall building, how far has the phone fallen after 2.7 seconds, neglecting air resistance?
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The free fall of the phone is an uniformly accelerated motion toward the ground, with constant acceleration equal to
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6 0
3 years ago
A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 780 N on him. The
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Answer:

f = 556.4 N

Explanation:

Mass of the losing player with its all equipment is given as

M = 86 kg

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F = 780 N

also we know that acceleration of the losing player is given as

a = 2.6 m/s^2

now by Newton's 2nd law we will have

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7 0
3 years ago
Two trains on separate tracks move toward each other. Train 1 has a speed of 145 km/h; train 2, a speed of 72.0 km/h. Train 2 bl
tekilochka [14]

Answer:

Therefore,

The frequency heard by the engineer on train 1

f_{o}=603\ Hz

Explanation:

Given:

Two trains on separate tracks move toward each other

For Train 1 Velocity of the observer,

v_{o}=145\ km/h=145\times \dfrac{1000}{3600}=40.28\ m/s

For Train 2 Velocity of the Source,

v_{s}=90\ km/h=90\times \dfrac{1000}{3600}=25\ m/s

Frequency of Source,

f_{s}=500\ Hz

To Find:

Frequency of Observer,

f_{o}=?  (frequency heard by the engineer on train 1)

Solution:

Here we can use the Doppler effect equation to calculate both the velocity of the source v_{s} and observer v_{o}, the original frequency of the sound waves f_{s} and the observed frequency of the sound waves f_{o},

The Equation is

f_{o}=f_{s}(\dfrac{v+v_{o}}{v -v_{s}})

Where,

v = velocity of sound in air = 343 m/s

Substituting the values we get

f_{o}=500(\dfrac{343+40.28}{343 -25})=500\times 1.205=602.64\approx 603\ Hz

Therefore,

The frequency heard by the engineer on train 1

f_{o}=603\ Hz

7 0
3 years ago
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