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soldi70 [24.7K]
2 years ago
13

Write the formula for kinetic energy that uses force in its equation. Calculate the kinetic energy generated when a 2500 N force

is applied to a car when moving it 200 m. Show work.
Physics
1 answer:
djyliett [7]2 years ago
8 0

Answer:

500000 J

Explanation:

ΔKE =Work = F*d = 2500 N * 200 m =500000 J

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solong [7]

Answer:

= 2 beats per seconds

Explanation:

  • From |f -f'| = modulus of the difference between the frequency given.
  • f = 510Hz and f' = 512Hz
  • Difference between the frequency will give us the number of beat per seconds.
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These also shows how to get the period of the tuning forks.

7 0
2 years ago
A car is moving at 10 m/s to the right. It accelerates for 10 s after which it is moving at 5 m/s to the left. What was the car'
NNADVOKAT [17]

Answer:

Acceleration, a=-1.5\ m/s^2

Explanation:

It is given that,

Initial velocity of the car, u = 10 m/s (in right)

Final velocity of the car, v = -5 m/s (in left)  

Time taken, t = 10 s

Let a is the acceleration of the car. It can be calculated using the equation of kinematics. The equation is as :

v=u+at

a=\dfrac{v-u}{t}

a=\dfrac{-5-10}{10}    

a=-1.5\ m/s^2

So, the acceleration of the car is -1.5\ m/s^2. Hence, this is the required solution.

6 0
3 years ago
The valence electrons of metals are weakly attracted to the parent nuclei, so the electrons break free and float. The moving ele
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The valence electrons of metals are weakly attracted to the parent nuclei, so the electrons break free and float. The moving electrons form a electron <u>negative</u> blanket that binds the atomic <u>positive</u> nuclei together, forming a metallic bond.

So the answers are <u>{ Negative }</u> and <u>{ Positive }.</u>  

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5 0
2 years ago
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If the charge on the negative plate of the capacitor is 121 nano-Coulomb, how many excess electrons are on that plate? Write you
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Answer:

n = 756.25 giga electrons

Explanation:

It is given that,

If the charge on the negative plate of the capacitor, Q=121\ nC=121\times 10^{-9}\ C

Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :

Q=ne

e is the charge on electron

n=\dfrac{Q}{e}

n=\dfrac{121\times 10^{-9}}{1.6\times 10^{-19}}

n=7.5625\times 10^{11}

or

n = 756.25 giga electrons

So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.

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3 years ago
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