Answer:
Solon,
total mass (kg)= 100kg
height (h)= 25m
acceleration due to gravity = 9.8m/s²
so,
work done =m*g*h
= 100*9.8*25
= 24,500 joule
Answer:
A
Explanation:
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Answer:
The correct solution is:
(a) 
(b) 
(c) 
(d) 
Explanation:
The given values are:
Effective duration of the flash,
ζ = 0.25 s
Average power,


Average voltage,

Now,
(a)
⇒ 
On substituting the values, we get
⇒ 
⇒ 
(b)
⇒ 
then,
⇒ 
On substituting the values, we get
⇒ 
⇒ 
(c)
⇒ 
⇒ 
⇒ 
(d)
As we know,
⇒ 
⇒ 
⇒ 
⇒ 
The answer is C, a peer group.
The best and most correct answers among the choices provided by your question are he second and third choices.
<span>The velocity at any instant the average velocity during some time interval cannot be obtained from the graph alone.</span>
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