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Evgesh-ka [11]
2 years ago
14

A 2kg mass is moving at 3m/s. What is its kinetic energy?

Physics
1 answer:
Illusion [34]2 years ago
3 0
<h2><u>KINETIC ENERGY</u></h2>

<h3>Problem:</h3>

» A 2kg mass is moving at 3m/s. What is its kinetic energy?

<h3>Answer:</h3>
  • \color{hotpink} \bold{9 \: J} \\

— — — — — — — — — —

<h3>Formula:</h3>

To calculate the velocity of a kinetic energy, we can use formula

  • \underline{ \boxed{  \tt KE =  \frac{1}{2} m{v}^{2}  } }

where,

  • v is the velocity in m/s
  • KE is the kinetic energy in J (joules)
  • m is the mass in kg

— — —

Based on the problem, the givens are:

  • KE (Kinetic energy) = ? (unknown)
  • m (mass) = 2 kg
  • v (velocity) = 3 m/s

<h3>Solution:</h3>

To get the velocity, substitute the givens in the formula above then solve.

\:   \: \tt KE =  \frac{1}{2} m{v}^{2}   \\ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \: \tt \:   KE =  \frac{1}{2} \times  2 \times {(3)}^{2}  \\ \tt  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  KE =  \frac{1}{2}  \times(2\times 9) \\ \tt  KE =  \underline{ \boxed{ \blue{ \tt9 \: J}}}

Therefore, the kinetic energy is 9 Joules.

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The magnitude of the force acting on the object lying on a flat surface without moving is 10 N.

The given parameters;

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Apply Newton's second law of motion to determine the magnitude of the force on the object.

Due to the position of the object, the magnitude of the force acting on it is calculated as;

\Sigma F_{net} = F\sin(\theta ) + F cos(\theta)\\\\\Sigma F_{net} = 10 sin(0) + 10cos(0)\\\\\Sigma F_{net} = 10 \ N

Therefore, the magnitude of the force acting on the object is 10 N.

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Answer:

65.625 m/s

Explanation:

F=20\times 10^9 Hz

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v=\frac{\Delta F}{2F}\times c\\\Rightarrow v=\frac{4750}{2\times 20\times 10^9}\times 3\times 10^8\\\Rightarrow v=35.625\ m/s

Relative speed = 35.625 m/s

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The average speed of the given car is 2.22 s and 3.13 s for 0.25 m and 0.50 m distance respectively.

<h3>How to calculate the Average speed?</h3>

The average speed can be calculated by adding the speed of each trial divided by the number of trials,

For 0.25 m the average speed will be:

S_{avg} = \dfrac{2.24 + 2.21 + 2.23}{ 3}\\\\S_{avg} = 2.22

For the 0.50 m, the average speed will:

S_{avg} = \dfrac {3.16 + 3.08 + 3.15} {3 }\\\\S_{avg}  = 3.13\rm \  s

Therefore, the average speed of the given car is 2.22 s and 3.13 s for 0.25 m and 0.50 m distance respectively.

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