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3 years ago

Another one! Your kind of gonna need it.

Physics

2 answers:

kifflom [539]3 years ago

7 0

Answer:

the right answer is option C.

rusak2 [61]3 years ago

5 0

Cartilage attaches muscle and bone

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Your friend tells you “I no the moon does not rotate because we always see the same side.” do you agree or disagree with your fr

nydimaria [60]

Answer:

no the moon does not rotate it only goes in circle just like the sun so I disagree with your friend

3 0

3 years ago

Worth 50 point!!!

stira [4]

Answer:

fact one

Explanation:

fact two is going slower and a longer distance than fact one so fact one will get there first.

hope this helps

4 0

3 years ago

Alex pushes on a 2.0 kg book, resulting in a net force of 6.0 N on the book.

Yakvenalex [24]

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

From the question we have

$a = \frac{6}{2} \\$

We have the final answer as

Hope this helps you

4 0

3 years ago

A ball is thrown straight downward with a speed of 0.50 meter per second from a height of 4.0 meters. What is the speed of the b

Bond [772]

Assuming that reaching a height 0 doesn’t stop the ball, and that it accelerates at 9.8 m/s^2, the ball would be traveling at 0.5 + 0.7*9.8 = 7.36 m/s downwards.

3 0

2 years ago

Read 2 more answers

The Coulomb force between two charges q1 and q2 at separation r in the air is 10N. If half of the separation is filled with medi

adoni [48]

Answer:

The value of new coulomb force is 1.43 N.

Explanation:

Given;

Coulomb's force in vacuum (air), $F_v$ = 10 N

dielectric constant, K = 7

The Coulomb's force between two charges separated by a distance r in a vacuum is given as;

$F_v = \frac{1}{4\pi \epsilon_0} \frac{q_1q_2}{r^2}$

The Coulomb's force between two charges separated by a distance r in a medium with dielectric constant is given as

$F_m = \frac{1}{4\pi K\epsilon_0} \frac{q_1q_2}{r^2}$

Take the ratio of the two forces;

$\frac{F_v}{F_m} = \frac{1}{4\pi \epsilon_0} \frac{q_1q_2}{r^2} \ \times \ \frac{4\pi K\epsilon_0 r^2}{q_1q_2} = K\\\\\frac{F_v}{F_m} = K\\\\\frac{10}{F_m} = 7\\\\F_m = \frac{10}{7} \\\\F_m = 1.43 \ N$

Therefore, the value of new coulomb force is 1.43 N.

5 0

2 years ago