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Dima020 [189]
3 years ago
6

A bat hits a baseball with a force of 500N. The magnitude of the force that the baseball exerts on the bat is...

Physics
1 answer:
g100num [7]3 years ago
5 0

Answer:

45

Explanation:

because

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Write a hypothesis for Part II of the lab, which is about the relationship described by F = ma. In the lab, you will use a toy c
Temka [501]

Answer:

F=ma is the relationship where, F is force, m is mass and a is acceleration.

Newton's second law states that  the unbalanced force applied to the object accelerates the object which is directly proportional to the force and inversely to the mass.

If we apply force to a toy car then It will accelerate.

This is how Newton's second law of motion is verified.

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2 years ago
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Please help me with the equations for this! Three uniform spheres are fixed at the positions shown in the diagram. ( there is a
lora16 [44]
The change in gravitational potential energy due to change in position must be the change in it's kinetic energy as the system is isolated! so find out the potential energies of the two different points!

<span>PE=−[G<span>M1</span><span>M2</span>]÷R

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3 years ago
An automobile with an initial speed of 5.13 m/s accelerates uniformly at the rate of 3.0 m/s2. Find the final speed of the car a
Digiron [165]

Answer:

Final velocity v=19.83 m/sec

Explanation:

We have given initial velocity u =5.13 m/sexc

Acceleration of automobile a=3m/sec^2

Time t =4.9 sec

We have to find the final velocity v

According to first law of motion v = u+at ,here v is the final velocity , a is acceleration and t is time

So v=5.13+3\times 4.9=19.83m/sec

So the final velocity is 19.83 m/sec

7 0
2 years ago
What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of −25.0 nc? g?
Daniel [21]

To calculate the force between two negative charges, we use the formula which is given by the Coulomb`s Law as

F=\frac{kq_{1}q_{2}  }{r^{2} }

Here, q_{1} andq_{2} are the charges on the pith balls, r is the separation between the charges and k is constant and its value is  8.99\times 10^9 N m^2/C^2.

Given  q_{1} =q_{2} =-25 nC=-25\times 10^{-9}C and  r=8 cm=8\times10^{-2} m.

Substituting these values in above formula we get,

F= 8.99\times 10^9 N m^2/C^2\frac{(-25\times 10^{-9}C)(-25\times 10^{-9}C)}{(8\times10^{-2} m)^2} \\\\\ F= 877929.7\times 10^{-9}  N\\\\F=8.8\times10^{-4}N

Thus, the repulsive force between two pith balls is 8.8\times10^{-4}N.

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3 years ago
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If a one-pound weight were one foot from the pivot or shaft of a lever, how many pound-feet of force would result? A. 4 B. 1 C.
mixas84 [53]

The force is still 1 pound.

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2 years ago
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