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Alexxandr [17]
3 years ago
8

Which situation would create a field like the one shown here?

Physics
1 answer:
kiruha [24]3 years ago
7 0
D. Interaction between a magnetic South Pole and a copper bar
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How many revolutions per minute would a 25 m -diameter ferris wheel need to make for the passengers to feel "weightless" at the
kozerog [31]
Refer to the diagram shown below.

Let ω =  the angular velocity (rad/s) of the wheel.

At the topmost point, the passenger, with mass m,  will feel weightless if the passenger's weight matches the centripetal force.
The passenger's weight is mg.
The centripetal force is mω²r.

Therefore
mω²r = mg
ω = √(g/r)

Because g = 9.8 m/s² and r = 25/2 = 12.5 m, therefore
ω = √(9.8/12.5) = 0.8854 rad/s

Because 1 revolution per second is 2π rad/s, therefore
\omega = (0.8854 \,  \frac{rad}{s} )*( \frac{1}{2 \pi} \,  \frac{rev}{rad} )*(60 \,  \frac{s}{min} ) = 8.455 \, rev/min

Answer: 8.455 rev/min

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3 years ago
Help for physical science u4 limiting reactants
Yuliya22 [10]
The reactants are on the left and the products are on the right of the equation
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4 years ago
How long does it take long a light from the sun to reach the earth if it travels a distance of 1.5 × 10^11m? (velocity of light
mojhsa [17]

Answer:

Therefore, light travelling at 3.0x10^8 meters per second takes 500 seconds (8 minutes, 20 seconds) to reach the Earth, which is 1.5x10^11 meters away from the sun

Explanation:

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