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hichkok12 [17]
2 years ago
14

If speed quadruples from 11 m/s to 44 m/s, what happens to the kinetic energy?

Physics
1 answer:
Ad libitum [116K]2 years ago
5 0

Answer:

16 fold increase.

Explanation:

As kinetic energy is a function of velocity squared, if the velocity quadruples, kinetic energy increases by a factor of 4² = 16

You need to <u>do</u> it in less than an hour because it is <u>due</u> soon.

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If the atomic number of an atom is 4 and the mass number is 9, then how many electrons does it have
miskamm [114]
The atomic number of an atom says how many protons it has. This number cant change, since the atomic number is what gives elements their identities (in the periodic table, at least).
The mass number, on the other hand, says how many protons AND neutrons the atom has (so, the sum of P+ and N0). So, electrons have nothing to do with this number.
Atoms are neutrally charged, which means there has to be an equal number of positive and negative particles. The negative particles of an atom are its electrons, and since our atom has 4 protons, it must also have 4 electrons.
5 0
2 years ago
What would happen if you didn't have chemical energy in your body? Choose the best answer.
eimsori [14]

Answer:B

Explanation:

Because chemical energy fuels your body with energy so without it u wont have enough energy to move.

8 0
2 years ago
A 5.00 kg crate is on a 21.0° hill.
mojhsa [17]

Answer: 4575N

Explanation:

For y component, W = mgcosø

W = 500×9.8cos21

W = 4574.54N

Find the diagram in the attached file

8 0
3 years ago
A bowling ball of mass 5.8 kg moves in a straight line at 4.34 m/s How fast must a Ping-Pong ball of mass 2.214 g move in a stra
lilavasa [31]

Answer: 11369.46 m/s

Explanation:

We have the following data:

m_{1}=5.8 kg is the mass of the bowling ball

V_{1}=4.34 m/s is the velocity of the bowling ball

m_{2}=2.214 g \frac{1 kg}{1000 g}=0.002214 kg is the mass of the ping-pong ball

V_{2} is the velocity of the ping-pong ball

Now, the momentum p_{1} of the bowling ball is:

p_{1}=m_{1}V_{1} (1)

p_{1}=(5.8 kg)(4.34 m/s)  

p_{1}=25.172 kg m/s (2)

And the momentum p_{2} of the ping-pong ball is:

p_{2}=m_{2}V_{2} (3)

If the momentum of the bowling ball is equal to the momentum of the ping-pong ball:

p_{1}=p_{2} (4)

m_{1}V_{1}=m_{2}V_{2} (5)

Isolating V_{2}:

V_{2}=\frac{m_{1}V_{1}}{m_{2}} (6)

V_{2}=\frac{25.172 kg m/s}{0.002214 kg} (7)

Finally:

V_{2}=11369.46 m/s

6 0
2 years ago
PLEASE HELP!!
leva [86]
I think the answer is B
8 0
2 years ago
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