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3 years ago

A person on a merry-go-round makes a complete revolution in about 3.17 s.

Physics

1 answer:

GREYUIT [131]3 years ago

8 0

Answer:

second one

Explanation:

HOPE THIS helps

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A 5 kilograms bowling ball is dropped out a window. It hits the ground, and bounces upward. The velocity change of the ball is n

ioda

Answer:

13.5

Explanation:

Mass: 5kg

Initial Velocity: -15

Final Velocity: 12

Force: 10

We can use the equation: Vf = Vi + at

We need to find acceleration, and we can use the equation, F=ma,

We have mass and the force so it would look like this, 10=5a, and 5 times 2 would equal 10, so acceleration would be 2.

Now we have all the variables to find time.

Back to Vf = Vi + at, plug the numbers in, 12 = -15 + 2(t)

Plugging them in into desmos gives 13.5 for time.

4 0

2 years ago

A solenoid field is:

Bumek [7]

increased with an increased current flow

7 0

3 years ago

When electrons in an atom in an excited state fall to lower energy levels, energy is?

dusya [7]

Radiated away as electromagnetic radiation.

5 0

3 years ago

Justin depends 600watts if power pushing a car 10m in 5sec how much force did he exert

Natasha_Volkova [10]

Force exerted by Justin=300 N

Here power= 600 W

distance traveled=10 m

time=5 s

power is given by P= Work done/ time

600=work/5

so work= 60x5=3000J

now work done= force* distance

3000=F *10

F= 3000/10

F=300 N

5 0

4 years ago

The mass of a star is 1.210×1031 kg and it performs one rotation in 20.30 days. Find its new period (in days) if the diameter su

balandron [24]

The new period will be 2.486 days.

<h3>What is the period?</h3>

The period is found as the ratio of the angular displacement and the angular velocity. Its unit is the second and is denoted by t. The value of time needed to complete the rotation is the total period.

Given data;

Mass of a star,m= 1.210×10³¹ kg

The time period for one rotation of the star, T = 20.30 days

D' = 0.350 D

R' = 0.350 R

From the law of conservation of angular momentum;

$\rm I \omega = I' \omega' \\\\ \frac{2}{5} MR^2 \times \frac{2 \pi }{T}=\frac{2}{5} MR'^2 \\\\ R^2 \times \frac{1}{T}= R'^2 \times \frac{1}{T} \\\\ T' = \frac{R'^2T}{R^2} \\\\ T' = \frac{(0.350 R)^2 \times 26.1 }{R^2} \\\\T' = 0.1225 \times 20.30 \\\\ T'= 2.486 \ days$

Hence, the new period will be 2.486 days.

To learn more about the period, refer to the link;

brainly.com/question/569003

#SPJ1

8 0

2 years ago