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hoa [83]
3 years ago
8

A worker does 25 J of work lifting a bucket, then sets the bucket back down in the same place. What is the total net work done o

n the bucket? a. –25 J c. 25 J b. 0 J d. 50 J
Physics
2 answers:
Sloan [31]3 years ago
6 0

Answer:

b. 0 J

Explanation:

Work is the cross product of force and displacement.  Since the bucket has 0 displacement, the net work is 0 J.

Nookie1986 [14]3 years ago
3 0

0 J this says that is there is no work done on the bucket

Option: b

<u>Explanation: </u>

Work is the scalar multiplication of force (F) and the displacement (d). In this particular question work done in lifting the bucket is 25 J. The worker lifts the bucket and puts the bucket down in the same place so displacement is zero. If the "displacement" is zero then there is no 'work' has been done by the worker. Since the bucket is kept back to the original position, there is no change in "potential energy". According to the "work-energy theorem," there is no network done. As the "displacement" is zero so the 'network done' is also zero.

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kifflom [539]

Answer:

1) d

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Explanation:

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1) Solve equation (ii) for acceleration a and plug the result in equation (i)

(iii) a = \frac{v -v_0}{t}

(iv) x = \frac{v-v_0}{2t}t^2+v_0t + x_0

Simplify equation (iv) and use the given values v = 0, x₀ = 0:

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3) Given v₀=0m/s, t₁=10s, t₂=1s and x₀=0. Looking for factor f = x(t₁)/x(t₂) using equation(i) to calculate x(t₁) and x(t₂):

f=\frac{x(t_1)}{x(t_2)}=\frac{\frac{1}{2}at_1^2 }{\frac{1}{2}at_2^2}=\frac{t_1^2}{t_2^2}=\frac{10^2}{1^2}=\frac{100}{1}

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If the average bond enthalpy for a C-H bond is 413 kJ/mol, When the C-H bond breaks in which energy will be required ,which will be an endothermic reaction.

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postnew [5]
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Answer:

Shown by explanation;

Explanation:

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