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bekas [8.4K]
3 years ago
10

A 5 Kg bowling ball is thrown at a stationary 1.6 Kg bowling pin at 5 m/s. If the final velocity of the ball is 2.5 m/s. The fin

al velicoty of the bowling pin is _____m/s. ​
Physics
1 answer:
Sergio039 [100]3 years ago
7 0

Answer: hope this helps

Explanation:

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Explanation:

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The electric potential 1.34 m from a charge is 580 V. What is the value of the charge? Include the sign, + or -. (The answer is
blagie [28]

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(a) An object that has a small mass and an object that has a large mass have the same momentum. Which object has the largest kin
Iteru [2.4K]

Answer:

A) The smaller object; B) The larger object

Explanation:

A) Lets say the small object is 2 kg and the large one is 6 kg. Lets say they also have 30 kg*m/s of momentum each. The small object would have 15 m/s velocity and the large would have 5 m/s.

Now for kinetic energy(.5*m*v²), the small object is .5*2*15², which is 225 J

The large object is .5*6*5², which is 75 J, so the smaller object has more Kinetic energy. Since velocity is squared, it is more important than how large mass is.

B) Same masses as part A. Lets say the kinetic energy is 45 J for both of them. For the small object, 45=.5*2*v²

.5*2 is 1, so 45/1 is 45. Take the square root and we get v= 6.71 m/s

For the large object, 45=.5*6*v²

.5*6 is 3, so 45/3 is 15. Take the square root and we get v=3.87 m/s

Now we plug the velocities into p=mv

For the small object, p=2*6.71, which gives us p=13.42 kg*m/s

For the large object, p=6*3.87, which gives us p=23.22 kg*m/s

The larger object has the larger momentum.

Hope this helps

5 0
3 years ago
Compute the size of the charge necessary for two spheres separated by 1m to be attached with the force of 1N. How many electrons
yarga [219]

Answer:

q\approx 6.6\cdot 10^{13}~electrons

Explanation:

<u>Coulomb's Law</u>

The force between two charged particles of charges q1 and q2 separated by a distance d is given by the Coulomb's Law formula:

\displaystyle F=k\frac{q_1q_2}{d^2}

Where:

k=9\cdot 10^9\ N.m^2/c^2

q1, q2 = the objects' charge

d= The distance between the objects

We know both charges are identical, i.e. q1=q2=q. This reduces the formula to:

\displaystyle F=k\frac{q^2}{d^2}

Since we know the force F=1 N and the distance d=1 m, let's find the common charge of the spheres solving for q:

\displaystyle q=\sqrt{\frac{F}{k}}\cdot d

Substituting values:

\displaystyle q=\sqrt{\frac{1}{9\cdot 10^9}}\cdot 1

q = 1.05\cdot 10^{-5}~c

This charge corresponds to a number of electrons given by the elementary charge of the electron:

q_e=1.6 \cdot 10^{-19}~c

Thus, the charge of any of the spheres is:

\displaystyle q = \frac{1.05\cdot 10^{-5}~c}{1.6 \cdot 10^{-19}~c}

\mathbf{q\approx 6.6\cdot 10^{13}~electrons}

5 0
2 years ago
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