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Norma-Jean [14]
4 years ago
11

If a ball leaves the ground with a vertical velocity of 5.46 m/s, how long does it take

Physics
1 answer:
kifflom [539]4 years ago
8 0

Answer:

0.557 s

Explanation:

Given:

v₀ = 5.46 m/s

v = 0 m/s

a = -9.8 m/s²

Find: t

v = at + v₀

0 m/s = (-9.8 m/s²) t + 5.46 m/s

t = 0.557 s

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A man tries to push a 200 kg Car that moves at a acceleration 0.50 m/s2. The man is able to displace the car 10 m. How much work
yawa3891 [41]

The work done by the man pushing the car over the given distance is 1000J.

Given the data in the question;

  • Mass of car; m = 200kg
  • Acceleration of the car; a = 0.5m/s^2
  • Distance covered by the car; d = 10m

Work done; W = \ ?

<h3>Work done</h3>

Work done is simply defined as the energy transfer that takes place when an object is either pushed or pulled over a certain distance by an external force. It is expressed as;

Work\ done = f * d

Where f is force applied and d is distance travelled.

To determine the work done by the man, we first solve for the force applied F.

From Newton's Second Law; Force \ F = m * a

We substitute our given values into the expression

F = m * a \\\\F = 200kg * 0.5m/s^2\\\\F = 100kg.m/s^2

Next we substitute our values into the expression of work done above.

Work \ done = f * d\\\\Work \ done = 100kg.m/s^2 * 10m\\\\Work \ done = 1000kgm^2/s^2\\\\Work \ done = 1000J

Therefore, the work done by the man pushing the car over the given distance is 1000J.

Learn more about work done: brainly.com/question/26115962

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2 years ago
If you measured all the energy related to motion and all the stored energy in the particles of a substance, which would you be m
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<span>If you measured all the energy related to motion and all the stored energy in the particles of a substance, you would be measuring thermal energy. The correct option among all the options that are given in the question is the last option or option "C". I hope that the answer has come to your great help.</span>
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g calculate the effectiveness radiation dosage in sieverts for a 79 kg person who is exposed 6.8x10^9
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Answer:

The answer is "\bold{dosage = 0.031 rem}"

Explanation:

please find the complete question in the attached file.

Given value:

m = 79\  kg  \\\\n = 3.4 \times  10^9 \\\\E = 5.5  \times  10^{-13} \\\\ RBE = 15

\to E = n E\\

        = 3.4  \times  10^9  \times  5.5  \times  10^{-13} \\\\      = 1.87 \times  10^{-3}

\to E(absorbed) = 1.87  \times 10^{-3}  \times  0.87 = 1.63  \times  10^{-3}

calculating the radiation absorbed per kg:

= \frac{1.63  \times  10^{-3}}{79}  \\\\ = 2.06  \times  10^{-5} \\\\ = 0.00206 \  rad

\to Dosage = 0.00206  \times  15 \\

                 = 0.031 \ \ rem

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3 years ago
Which of the following statements is false?
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