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Zanzabum
3 years ago
9

Suddenly a worker picks up the bag of gravel. Use energy conservation to find the speed of the bucket after it has descended 2.3

0 m from rest. (You can check your answer by solving this problem using Newton's laws.)
Physics
1 answer:
fiasKO [112]3 years ago
8 0

Explanation:

A worker picks up the bag of gravel. We need to find the speed of the bucket after it has descended 2.30 m from rest. It is case of conservation of energy. So,

\dfrac{1}{2}mv^2=mgh\\\\v=\sqrt{2gh}

h = 2.3 m

v=\sqrt{2\times 9.8\times 2.3} \\\\v=6.71\ m/s

So, the speed of the bucket after it has descended 2.30 m from rest is 6.71 m/s.

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If you travel 450 meters in 40 seconds, what is your average speed in meters per
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Answer:

The answer is a 11.25m/s

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Point charges q1 = 14 µC and q2 = −60 µC are fixed at r1 = (5.0î − 4.0ĵ) m and r2 = (9.0î + 7.5ĵ) m. What is the force (in N) of
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Answer:

The force on q₁ due to q₂ is (0.00973i + 0.02798j) N

Explanation:

F₂₁ = \frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|}

Where;

F₂₁ is the vector force on q₁ due to q₂

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|r₂₁| is the magnitude of the unit vector

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|q₂| is the absolute charge on point charge two

r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)

|r₂₁| = \sqrt{(4^2)+(11.5^2)} = \sqrt{148.25}

(|r₂₁|)² = 148.25

F_2_1=\frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|} = \frac{8.99X10^9(14X10^{-6})(60X10^{-6})}{148.25}.\frac{(4i + 11.5j)}{\sqrt{148.25} }

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