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DerKrebs [107]
3 years ago
13

6d A boy pushes his little brother on a sled. The sled accelerates from rest to (4 m/s). If the combines of his brother and the

sled is (40.0 kg) and (20 W) of power is developed, how long time does the boy push the sled?​
Physics
1 answer:
Elanso [62]3 years ago
5 0

Answer:

16 s

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 40 Kg

Velocity (v) = 4 m/s

Power (P) = 20 W

Time (t) =?

Next, we shall determine the energy. This can be obtained as follow:

Mass (m) = 40 Kg

Velocity (v) = 4 m/s

Energy (E) =?

E = ½mv²

E = ½ × 40 × 4²

E = 20 × 16

E = 320 J

Finally, we shall determine the time. This can be obtained as follow:

Power (P) = 20 W

Energy (E) = 320 J

Time (t) =?

E = Pt

320 = 20 × t

Divide both side by 20

t = 320 / 20

t = 16 s

Thus, the time is 16 s

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Explanation:

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Citrus2011 [14]

Answer:

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part (b) s = 46.60 m

Explanation:

Given,

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part (a)

River is flowing due to east and the boat is moving in the north, therefore both the velocities are perpendicular to each other and,

Hence the resultant velocity i,e, the velocity of the boat relative to the shore is in the North east direction. velocities are the vector quantities, Hence the resultant velocity is the vector addition of these two velocities and the angle between both the velocities are 90^o

Let 'v' be the velocity of the boat relative to the shore.

\therefore v\ =\ \sqrt{v_r^2\ +\ v_b^2}\\\Rightarrow v\ =\ \sqrt{1.60^2\ +\ 10.3^2}\\\Rightarrow v\ =\ 10.42\ m/s.

Let \theta be the angle of the velocity of the boat relative to the shore with the horizontal axis.

Direction of the velocity of the boat relative to the shore.\therefore Tan\theta\ =\ \dfrac{v_b}{v_r}\\\Rightarrow Tan\theta\ =\ \dfrac{10.3}{1.60}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{10.3}{1.60}\ \right )\\\Rightarrow \theta\ =\ 81.17^o

part (b)

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Let 't' be the total time taken by the boat to cross the width of the river.\therefore w\ =\ v_bt\\\Rightarrow t\ =\ \dfrac{w}{v_b}\\\Rightarrow t\ =\ \dfrac{300}{10.3}\\\Rightarrow t\ =\ 29.12 s

Therefore the total distance traveled in the direction of downstream by the boat is equal to the product of the total time taken and the velocity of the river\therefore s\ =\ u_rt\\\Rightarrow s\ =\ 1.60\times 29.12\\\Rightarrow s\ =\ 46.60\ m

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