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grandymaker [24]
3 years ago
13

A tennis ball covers a distance of 12 meters in 0.4 seconds what is the velocity of the tennis ball?

Physics
2 answers:
Usimov [2.4K]3 years ago
7 0
V=d/t
V=12/0.4
V=30 m/s
Ksenya-84 [330]3 years ago
4 0

Answer:

30m/s

Explanation:

Given parameters:

Distance  = 12m

Time taken  = 0.4s

Unknown:

Velocity of the tennis ball = ?

Solution:

Velocity of a body is the displacement per unit of time.

  Velocity  = \frac{displacement}{time}  

So;

      Velocity  = \frac{12}{0.4}   = 30m/s

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kiruha [24]

Answer:

Option B

Explanation:

Gravitational force is a force that attracts two bodies (with a mass) towards each other. If an object has a higher mass, the gravitational pull will be greater.

According to Newton’s inverse square law:

<em>"The gravitational force is inversely proportional to the square of the distance between two bodies."</em>

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6 0
3 years ago
A large box slides across a frictionless surface with a velocity of 12 m/s and a mass of 4
denpristay [2]
Answer= 8m/s

Because total Momentum before= total momentum after

Momentum before (p=mu)
p=(4)(12)= 48
p=2(0)=0
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Momentum after (p=mu)
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Mb=Ma
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3 0
3 years ago
A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum
s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle=q=-2.74\times 10^{-6} C

Kinetic energy of particle=K_E=6.65\times 10^{-10} J

Initial time=t_1=6.36 s

Final potential difference=V_2=0.351 V

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

qV=K.E

Using the formula

2.74\times 10^{-6}V_1=6.65\times 10^{-10} J

V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V

Initial voltage=V_1=2.43\times 10^{-4} V

\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2

Using the formula

\frac{V_1}{V_2}=(\frac{6.36}{t})^2

\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}

t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}

t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}

t=241.7 s

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0
3 years ago
I will mark brainliest!
Dvinal [7]
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alina1380 [7]

Answer:

7.35 J

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5 0
3 years ago
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