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Alchen [17]
3 years ago
8

Please help answer #14!!!!!!!

Physics
1 answer:
jeka943 years ago
7 0

When a heavy football player and a light one run into each other, does the lighter player really exert as much force on the heavy player s the heavy player exerts on the light one. Yes. The interaction between the two players, the force each exerts on the other have equal strength.

Hope this helps you!!!

Please mark this as the brainliest one!!!

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Asap pls hurry will mark brainiest
kifflom [539]
#1. A. Waxing crescent.
#2. 1.
#3. C.
#4. C.
6 0
3 years ago
What is the magnitude of g at a height above Earth's surface where free-fall acceleration equals 6.5m/s^2?
prohojiy [21]

You've given the answer, right there in your question.

The "magnitude of gravity" is described in terms of the acceleration
due to it, and you just told us what that is.

We can also notice that the figure you gave is about 0.66 of the
acceleration due to gravity on the Earth's surface. That tells us that
the distance from the Earth's center at that height is about 

                     (1 / √0.66) = 1.23 times

the Earth's radius, so the height is about  910 miles above the surface.


7 0
3 years ago
A car travels around a level, circular track that is 750m across. What coefficient of friction is required to ensure the car can
Crank

The coefficient of friction must be 0.196

Explanation:

For a car moving on a circular track, the frictional force provides the centripetal force needed to keep the car in circular motion. Therefore, we can write:

\mu mg = m\frac{v^2}{r}

where the term on the left is the frictional force acting between the tires of the car and the road, while the term on the right is the centripetal force. The various terms are:

\mu is the coefficient of friction between the tires and the road

m is the mass of the car

g=9.8 m/s^2 is the acceleration of gravity

v is the speed of the car

r is the radius of the curve

In this problem,

r = 750 m is the radius

v=85 mph \cdot \frac{1609}{3600}=38.0 m/s is the speed

And solving for \mu, we find the coefficient of friction required to keep the car in circular motion:

\mu = \frac{v^2}{rg}=\frac{38.0^2}{(750)(9.8)}=0.196

Learn more about circular motion:

brainly.com/question/2562955  

brainly.com/question/6372960  

#LearnwithBrainly

8 0
3 years ago
A person walks 9.0 km directly east and then turns left and heads directly north for 12.0 km. What is his displacement from the
Readme [11.4K]

Explanation:

Given that,

A person walks 9.0 km directly east and then turns left and heads directly north for 12.0 km.

We need to find his displacement from the starting position.

We know that,

Displacement = shortest path covered

D=\sqrt{9^2+12^2} \\\\D=15\ km

For direction,

\theta=\tan^{-1}(\dfrac{d_y}{d_x})\\\\\theta=\tan^{-1}(\dfrac{12}{9})\\\\= $$53.13^{\circ}

Hence, this is the required solution.

5 0
2 years ago
Select all that apply.
Nana76 [90]

Answer: It is both B and D

Select all that apply.

At night, thermal energy moves _____.

from space to the atmosphere

from the land to the atmosphere

from the atmosphere to the land

from the atmosphere to space

3 0
3 years ago
Read 2 more answers
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