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Sunny_sXe [5.5K]
2 years ago
6

Sam moves a box with a force of 400N a distance of 5 meters. How long did it take him to move the box if 20 Watts of power was u

sed?
Physics
1 answer:
kipiarov [429]2 years ago
5 0

Answer:

<em><u>Work</u></em><em><u> </u></em><em><u>done</u></em><em><u> </u></em><em><u>by</u></em><em><u> </u></em><em><u>sam</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>force</u></em><em><u> </u></em><em><u>×</u></em><em><u> </u></em><em><u>distance</u></em>

<em><u>=</u></em><em><u> </u></em><em><u>4</u></em><em><u>0</u></em><em><u>0</u></em><em><u>×</u></em><em><u>5</u></em>

<em><u>=</u></em><em><u> </u></em><em><u>2</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>J</u></em>

<em><u>Time</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>work</u></em><em><u>/</u></em><em><u>power</u></em>

<em><u>=</u></em><em><u> </u></em><em><u>2</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>/</u></em><em><u>2</u></em><em><u>0</u></em>

<em><u>=</u></em><em><u> </u></em><em><u>1</u></em><em><u>0</u></em><em><u>0</u></em><em><u>s</u></em>

Explanation:

<h2>HOPE IT WILL HELP YOU✌✌✌✌✌</h2>
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A rock is dropped from a garage roof from rest. the roof is 6.0 m from the ground. determine the velocity of the rock as it hits
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From rest, a rock is dropped from a garage roof. The roof is 6.0 meters above ground level. The rock will reach the earth at a speed of 10.849 meters per second.

<h3>What is velocity?</h3>

The change of displacement with respect to time is defined as the velocity.  Velocity is a vector quantity.

it is a time-based component. Velocity at any angle is resolved to get its component of x and y-direction.

Given data:

V(Final velocity)=? (m/sec)

h(height)= 6.0 m

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g(gravitational acceleration)=9.81 m/s²

Newton's third equation of motion:

\rm v_y^2 = u_y^2+ 2gh \\\\\rm v_y^2 = 0+ 2gh\\\\\  v_y= \sqrt{2\times 9.81 \ (m/s^2)\times 6.0 (m)} \\\\ v_y=10.849 \ m/sec

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2 years ago
Q 19.23: A proton is initially moving at 3.0 x 105 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude
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Answer:

The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

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initial velocity of proton, v_p_i = 3 x 10⁵ m/s

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electric field strength, E = 120 N/C

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Consider work-energy theorem;

W = ΔK.E

W =K.E_f - K.E_i

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K.E_f =EQd + \frac{1}{2}m_pv_p_i^2

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K.E_f = 120\times 1.6 \times 10^{-19} \times 3.5   \ + \ \frac{1}{2}(1.673\times 10^{-27})(3\times 10^5)^2 \\\\

K.E_f = 6.72\times 10^{-17} \ + \ 7.53 \times 10^{-17} \\\\K.E_f = 14.25 \times 10^{-17} J\\\\K.E_f = 1.425\times 10^{-16} \ J

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Answer: Hello! An objects speed is constant and has the units meters per second (m/s); thus, it does not change overtime. Acceleration is a rate of change where the speed does either increase or decrease overtime from its inital value; its units are meters per second second (m/s/s). I hope that helps!

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