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2(D) Use s=ut + 1/2 at^2. Sub in time from c(ii) and acceleration from c(i). Your answer should be s=12.71cm. But the length of the metal plates is only 12cm. Since s calculated is longer than metal plates, electron will emerge from plates.
For Q no. 4 the total range with which the values are fluctuating is +- 0.04 not +- 0.02 . So the range is +- 0.04. That's my reason . And for question no. 10 we know that in case of elastic collision u1 - u2 = v2 - v1 ( i.e Speed of approach = Speed of separation ) . Since the body are initially travelling in opposite direction so one velocity must be in negative direction. Since u1, v1 and v2 are in same direction so it is better to take u2 as negative ( But you can alter that and take u2 as positive and others as negative. You will end up with the same answer in either case. ) Now since u2 is negative so the above eqn. becomes (u1 + u2 = v2 - v1 ) So D not A. I guess you are clear now.Can anyone help clarify? For Q4, why is the answer D not B? For Q10, Why is it not A but D? Thanks!!
Yes for the question no. 2 of Paper 3 . You have to write the errors and ways to solve them and you usually don't have to draw graph in those questions.
Thank you soo muchFor Q no. 4 the total range with which the values are fluctuating is +- 0.04 not +- 0.02 . So the range is +- 0.04. That's my reason . And for question no. 10 we know that in case of elastic collision u1 - u2 = v2 - v1 ( i.e Speed of approach = Speed of separation ) . Since the body are initially travelling in opposite direction so one velocity must be in negative direction. Since u1, v1 and v2 are in same direction so it is better to take u2 as negative ( But you can alter that and take u2 as positive and others as negative. You will end up with the same answer in either case. ) Now since u2 is negative so the above eqn. becomes (u1 + u2 = v2 - v1 ) So D not A. I guess you are clear now.
For Q no. 4 the total range with which the values are fluctuating is +- 0.04 not +- 0.02 . So the range is +- 0.04. That's my reason . And for question no. 10 we know that in case of elastic collision u1 - u2 = v2 - v1 ( i.e Speed of approach = Speed of separation ) . Since the body are initially travelling in opposite direction so one velocity must be in negative direction. Since u1, v1 and v2 are in same direction so it is better to take u2 as negative ( But you can alter that and take u2 as positive and others as negative. You will end up with the same answer in either case. ) Now since u2 is negative so the above eqn. becomes (u1 + u2 = v2 - v1 ) So D not A. I guess you are clear now.
haha, no the motorist drives a vehicle powered by a motor,here, a carTobi will help you solve it!
Alright
u = 10 m/s
v = 0 m/s
s = 10 m
That's the first case
So...
You gotta find the retardation of the car first [Basically negative acceleration]
So....
Yeah
* = Multiply
V*V = U*U - 2*a*s
0 = 100 - 20a
-100 = -20a
a = 100/20 =( 5m/s*s)
That's the acceleration of your car! CAR FTW ! XD
Now the question is IF he were to travel at 30m/s,what'd be the distance
Therefore use the same equation again nacho
But in this case:
u = 30 m/s
v = 0m/s
s = ??? m
a = 5 m/s*s
V*V = U*U - 2*a*s
0 = 30*30 - 2*5*s
0 = 900 - 10s
- 900 = -10s
s = -900/-10 = 90m
90m OMG. That's the answer lol xD
btw doesn't a motorist drive a motorcycle instead of a car?How can a biker drive a car at the same time when he's driving a bike?:/
Its not for elastic collision. That's for inelastic collision. In case of elastic collision the kinetic energy of the body is conserved so we have to use the another equation that I mentioned earlier. I guess you are clear now.
For Q1, cube root 2.7 x 10 ^ 25 m ^-3 and you get 6.46 x 10^-9 m. Divide the answer by two again and you get the final answer. I don't know why divide by two though...
For Q2, use P1/T1 =P2/T2 to get the pressure when box is 200 degree celsius. Remember that temp should be in Kelvin. You should get 1.65 x 10^ 5 Pa. Since it says net force, find its net pressure. Outside the box there's atmospheric pressure of 1.01 x 10^5 Pa already. Inside the box, there's 1.65 x 10^5 Pa. So minus that and you get 6.5 x 10^4 Pa. That times the area which is 0.09m squared, you'll get the final answer
thanks!Its not for elastic collision. That's for inelastic collision. In case of elastic collision the kinetic energy of the body is conserved so we have to use the another equation that I mentioned earlier. I guess you are clear now.
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