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Marina CMI [18]
3 years ago
6

5

Physics
1 answer:
mina [271]3 years ago
5 0

Answer:

B West

Explanation:

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A basketball of mass 0.608 kg is dropped from rest from a height of 1.37 m. It rebounds to a height of 0.626 m.
kenny6666 [7]

Answer:

a)|\Delta E|=4.58\: J  

b)F=61.90\: N

Explanation:

a)

We can use conservation of energy between these heights.

\Delta E=mgh_{2}-mgh_{1}=mg(h_{2}-h_{1})  

\Delta E=0.608*9.81(0.6026-1.37)

Therefore, the lost energy is:

|\Delta E|=4.58\: J  

b)

The force acting along the distance create a work, these work is equal to the potential energy.

W=\Delta E

F*d=mgh

Let's solve it for F.

F=\frac{mgh}{d}

F=\frac{0.608*9.81*1.37}{0.132}

Therefore, the force is:

F=61.90\: N

I hope is helps you!

6 0
3 years ago
A painter on a ladder, painting the ceiling of a room comments. “It is hotter up here by the ceiling than it is down on the floo
ira [324]

Answer:

1. Hot air is less dense and has moved upward allowing cool air to move downward which is less dens

(Convection current)

2. The ceiling has transferred heat to him by radiation.

Explanation:

During the day when air is heated as a result of the ceiling transferring heat to it, it becomes less dens and it gains energy, which make it lighter in weight than cooler air, this hot air moves upward allowing cooler air to move downward.

4 0
3 years ago
A proton moves perpendicularly to a uniform magnetic field b with a speed of 3.7 × 107 m/s and experiences an acceleration of 5
svlad2 [7]
The magnetic force experienced by the proton is given by
F=qvB \sin \theta
where q is the proton charge, v its velocity, B the magnitude of the magnetic field and \theta the angle between the direction of v and B. Since the proton moves perpendicularly to the magnetic field, this angle is 90 degrees, so \sin \theta=1 and we can ignore it in the formula.

For Netwon's second law, the force is also equal to the proton mass times its acceleration:
F=ma

So we have
ma=qvB
from which we can find the magnitude of the field:
B= \frac{ma}{qv}= \frac{(1.67 \cdot 10^{-27}kg)(5\cdot 10^{13}m/s^2)}{(1-6 \cdot 10^{-19}C)(3.7 \cdot 10^7 m/s)}=0.014 T
4 0
3 years ago
Consider a constant density gas flowing steadily over an airfoil. Far upstream the velocity is V0. Halfway along the top surface
Mashutka [201]

Answer:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Explanation:

5 0
3 years ago
What’s the answer to this
ladessa [460]
Choices  1,  2,  and 4 . . . . . Yes

Choices  3  and 5 . . . . . No
6 0
3 years ago
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