Okay.
Not 14 bro. 12.
Not 14 bro. 12.
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Hey are you there ?
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4)
10 ms = 1cm , total 8.5 waves in 6 cm, so if 10 ms = 1 cm so 6 cm = 60 ms and 60 ms = 0.006 s
So in 0.06 s there are 8.5 waves. so in 1 wave what is the value of time, that will be 0.00706 s
As you know f = 1 / T, f = 1 / 0.00706 = 141 that is 140 approx so B
10 ms = 1cm , total 8.5 waves in 6 cm, so if 10 ms = 1 cm so 6 cm = 60 ms and 60 ms = 0.006 s
So in 0.06 s there are 8.5 waves. so in 1 wave what is the value of time, that will be 0.00706 s
As you know f = 1 / T, f = 1 / 0.00706 = 141 that is 140 approx so B
Thought blocker
16)
The Efficiency = (Useful Output energy)/(Total Input Energy).
The Efficiency = (Useful Output energy)/(Total Input Energy).
Thought blocker
•Since the kinetic energy of the car doesn't change and remains at v, there is no change in the kinetic energy of the car.
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4)
lemme send u snap
lemme send u snap
Thought blocker
16)
•Since friction does not do any work (we can assume this, and even otherwise it is not useful output so it doesn't count) there is no energy lost to internal energy.
•Since friction does not do any work (we can assume this, and even otherwise it is not useful output so it doesn't count) there is no energy lost to internal energy.
Thought blocker
•However, potential energy does change - the car moves up a height of mgs * sin(α) (Since ΔP.E. = mgΔh, and sin(α) = Δh/s, Δh = s * sin(α) and ΔP.E. = mgs * sin(α)).
•Therefore, the useful energy output = mgs * sin(α)
•Therefore, the useful energy output = mgs * sin(α)
Thought blocker
•The work we put into the system is the only input to the system; no other influence is exerted by us, so the only input to the system is the work done by our force = F.
This work = F.s = Fs. <-- I guess you know this formula
This work = F.s = Fs. <-- I guess you know this formula
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So, (Useful Output Energy)/(Total Input Energy) = mgs * sin(α)/Fs = mgsin(α)/F
Hence your answer is D
Got it darling ?
Hence your answer is D
Got it darling ?
Thought blocker
12)
Its a simple concept of Elastic collision(E.C), we know that in E.C
the velocity of the object get exchanged, I mean, here, X is moving with V to right and and Y has V = 0 hence it is at rest (stop) , so when they collide there is E.C so velocity would be exchanged, i.e X would be at rest V = 0 (stop) and Y would be moving with velocity of V towards right. So D
Its a simple concept of Elastic collision(E.C), we know that in E.C
the velocity of the object get exchanged, I mean, here, X is moving with V to right and and Y has V = 0 hence it is at rest (stop) , so when they collide there is E.C so velocity would be exchanged, i.e X would be at rest V = 0 (stop) and Y would be moving with velocity of V towards right. So D
Thought blocker
Welcome, tell me if you got any more probs in this, I think your paper is still incomplete tell me if you have more doubts
Or if I am not here, ask in thread
tell me if you have more doubts Or if I am not here, ask in thread
For 14) Its clearly written in our book, do you use course book ?