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Kruka [31]
3 years ago
14

Does dropping a magnet down a copper tube produce a current in the tube? explain your answer.

Physics
1 answer:
Ratling [72]3 years ago
3 0

Answer:

The current generated in the pipe by the falling magnet is induced where the magnetic field is changing, which is only close to the magnet. Try slitting the pipe lengthwise along one side.

Explanation:

I neep your brainlist answer

plzzzzzzz Mark my answer in brainlist

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Which one of the following activities is considered part of health care operations? A. Administering a stress test B. Reviewing
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<span>B. Reviewing the competency of health care workers</span> is considered part of health care operations
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4 years ago
Why is it safe to watch an eclipse of the Moon but not an eclipse of the Sun
gavmur [86]
A solar eclipse occurs when the moon crosses in front of the Sun, blocking some or all of its rays. A lunar eclipse happens when the moon is directly behind the earth, blocking the moon from receiving light. The only light comes from the light on earth's reflected shadow.

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8 0
4 years ago
In a race, Usain bolt accelerates at 1.99 m/s^2 for the first 60.0 m, then decelerates at -0.266 m/s^2 for the final 40.0 m. How
GalinKa [24]
7 miles long i think bc i dont know bc i am bad at math and math is good
6 0
3 years ago
A World-class sprinter can reach a top speed of about 11.5 m/s in the first 18.0 m of a race. What is the average acceleration o
irina [24]

Answer

a = 3.674 m / s ^ 2


t = 3.13 s

Using the kinematic equations for the movement we have:


h(t) = P_{0} + Vot + \frac{1}{2}at ^ 2 (1)


V_{f} = V_{0} + at (2)


Where:


P_{0} = initial position


V_{0}} = initial velocity


a = acceleration


t = time in seconds


V_{f} = final speed


We know:


P_{0}=0


V_{0}= 0

h = 18 m


V_{f} = 11.5\frac{m}{s}

  So:

 From (2) we have that: t =\frac{V_{f}}{a}


t =\frac{11.5}{a}

From (1) we have to:


h (t) = 0.5at ^ 2\\h = 18 = 0.5at ^ 2

Then we clear "a" to find the acceleration.


\frac{36}{t^2} = a\\a = \frac{36}{(\frac{11.5}{a})^2} \\\\a =\frac{11.5^2}{36}\\a = 3.674 m / s ^2

Then, the time it takes to reach this speed is:


t =\frac{V_{f}}{a}\\t =\frac{11.5}{3.674}\\t = 3.13 s

4 0
4 years ago
Read 2 more answers
A charge q of magnitude 6.4 × 10^-19 coulombs moves from point A to point B in an electric field of 6.5 × 10^4 newtons/coulomb.
lana [24]

In this problem we have the electric field intensity E:

E = 6.5 × 10^4 newtons/coulomb

We have the magnitude of the load:

q = 6.4 × 10 ^{-19} coulombs

We also have the distance d that the load moved in a direction parallel to the field 1.2 × 10^{-2} meters.

We know that the electric potential energy (PE) is:

PE = qEd

So:

PE = (6.4 × 10^{-19})(6.5 × 10^4)(1.2 × 10^{-2})

PE = 5.0 x 10^{-16} joules

None of the options shown is correct.

6 0
3 years ago
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