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bogdanovich [222]
3 years ago
6

A piece of aluminum is dropped vertically downward between thepoles of an electromagnet. Does the magnetic field affect theveloc

ity of the aluminum?a. yesb. noExplain.
Physics
1 answer:
sashaice [31]3 years ago
8 0

When aluminum begins to fall and move towards the field, the currents found in it begin to have an effect on Aluminum. Said magnetic field what it will do is generate a slowing force on the plate that acts with the current of the electroiman and will slow down its fall. Similarly, when the plate leaves the magnetic field, a current is induced and, once again, there is an upward force to reduce the speed of the plate.

Therefore the correct answer is A. Yes.

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Why does the pattern of colors repeat in a thin soap film? Please use 2 content related sentences. (ref: p.522-530)
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Becuz when you wash up in the tub you want layers of soap so you don’t stink
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What is the typical treatment for Hodgkin's disease?
Evgesh-ka [11]

Answer:

Pretty sure its D.

radiation, chemotherapy, and bone marrow transplant

Explanation:

7 0
3 years ago
Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc
kumpel [21]

Answer:\frac{D_A}{v_B-v_A}

Explanation:

Given

car A had a head start of D_A

and it starts at x=0 and t=0

Car B has to travel a distance of D_A and d_a

where d_a is the distance travel by car A in time t

distance travel by car A is

d_a=v_A\times t

For car B with  speed v_B

d_B=D_A+d_a

v_B\times t=D_A+v_A\times t

t=\frac{D_A}{v_B-v_A}

7 0
4 years ago
A student that jumps a vertical height of 50 cm during the hang time activity.
muminat

The hang time of the student is 0.64 seconds, and he must leave the ground with a speed of 3.13 m/s

Why?

To solve the problem, we must consider the vertical height reached by the student as max height.

We can use the following equations to solve the problem:

<u>Initial speed calculations:</u>

v_{f}^{2}=v_{o}^{2}+2*a*d

At max height, the speed tends to zero.

So, calculating, we have:

<u>v_{f}^{2}=v_{o}^{2}+2*a*d\\\\0=v_{o}^{2}+2*(-9.81\frac{m}{s^{2}})*0.5m\\\\v_{o}^{2}=9.81\frac{m^{2} }{s^{2}}\\\\v_{o}=\sqrt{9.81\frac{m^{2} }{s^{2}}}=3.13\frac{m}{s}</u>

<u>Hang time calculations:</u>

We must remember that the total hang time is equal to the time going up plus the time going down, and both of them are equal,so, calculating the time going down, we have have:

y-yo=vo.t+\frac{1}{2}*9.81\frac{m}{s^{2}}*t^{2} \\\\0.5m-0=0*t+4.95*t^{2}\\\\t^{2}=\frac{0.5}{4.95}\\\\t=\sqrt{0.101}=0.318s

Then, for the total hang time, we have:

TotalHangTime=2*0.318seconds=0.64seconds

Have a nice day!

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4 years ago
The time between high and low tide when the current changes direction is called ____.
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<span>slack water is the ansewer to that question</span>
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