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As physics p1 MCQS YEARLY ONLY.

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guys can anyone please tell me the solution for MCQ 11 year2016 paper11 may/june.
the mass of man n platform is 96 kg so its weight is
96 x 10 = 960N downwards

n the man n platform r at a constant height so that means the resultant force on them is 0.
this shows that the water is applying a force of 960N upwards as it moves downwards.

Now we use the formula F = ma
However since the mass is actually in kgs-1 we will use
F= m/t x a
a = (v - u)/t
substituting a into F = m/t x a we have
F = m x (v - u)

We will put the values in to find the change in speed of water
960 = 40 x (v - u)
Now the water must have started from rest so we put a value of 0 for u
960 = 40 x (v - o)
v = 24ms-1

i hope this helps
 
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the mass of man n platform is 96 kg so its weight is
96 x 10 = 960N downwards

n the man n platform r at a constant height so that means the resultant force on them is 0.
this shows that the water is applying a force of 960N upwards as it moves downwards.

Now we use the formula F = ma
However since the mass is actually in kgs-1 we will use
F= m/t x a
a = (v - u)/t
substituting a into F = m/t x a we have
F = m x (v - u)

We will put the values in to find the change in speed of water
960 = 40 x (v - u)
Now the water must have started from rest so we put a value of 0 for u
960 = 40 x (v - o)
v = 24ms-1

i hope this helps
thank you so much!
 
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que no 13 and 14 as well please
For question 13

the weight of the bus is equal to the two upward forces so W = 25000N
Take distance between 16000N and the weight to be "x" m and the distance between 9000N and the weight to be (1.5 - x) m
Use your idea of moment to find x. then add x and 0.6m to get the answer.

For question 14

This is quite a tricky question. There might be a better way to solve it but I couldn't think of one.

what you could do here is resolve the forces horizontally like this
1) 20170602_132823.jpg 2) 20170602_132830 (2).jpg

here we proved that F2 > F1 . So the answer could be either A or C.

Now I figured by intuition that A cannot be the answer since the sum of F1 and F2 is 10. So the answer must be C. Because when we resolve the forces vertically we are taking the components of F1 and F2 which means sum of F1 and F2 should be greater . To prove that I did this 20170602_132840 (2).jpg.

Now you may not understand this but what you have to remember is that
1) if alpha is bigger than beta then the force F1 will be smaller than F2 . that way you don't have to find an expression to prove, just use your intuition on it.
2) sum of F1 and F2 should be bigger than the weight here . ( to understand why, check the picture I uploaded)
 
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que 19 please!
Power = Force x velocity

On the horizontal road

30000 = F x 25
F = 1200N

On the slope

Force from the car's engine = 1200N
backward force = mg sin(theta) = 14000sin(2)

To go up the slope the minimum forward force = 1200 + 14000sin(2) to overcome the backward force

Therefore P = (1200 + 14000sin(2)) x 25 = 42214W = 42kW . Ans is C
 
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I don t understand that in some cases they use R=V/I saying resistance is directly propotional to voltage and in some cases I=V/R is used, saying Current is directly propotional to voltage. I don t understand which concept to use where. Help please
 
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Power = Force x velocity

On the horizontal road

30000 = F x 25
F = 1200N

On the slope

Force from the car's engine = 1200N
backward force = mg sin(theta) = 14000sin(2)

To go up the slope the minimum forward force = 1200 + 14000sin(2) to overcome the backward force

Therefore P = (1200 + 14000sin(2)) x 25 = 42214W = 42kW . Ans is C
thankyou!
 
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Can anyone give some advice regarding estimation and approximation in the mcqs?
How to prepare for it?
There were questions on estimating resistance of copper and on estimating young modulus of metals.. how would you approximate such a value without calculation..?
 
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What is the order of magnitude of the Young modulus for a metal such as copper?
A: 10^–11 Pa B: 10^–4 Pa C: 10^4 Pa D: 10^11 Pa
9702_w16_qp_11 Q.1
How do we even answer such questions?
 
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What is the order of magnitude of the Young modulus for a metal such as copper?
A: 10^–11 Pa B: 10^–4 Pa C: 10^4 Pa D: 10^11 Pa
9702_w16_qp_11 Q.1
How do we even answer such questions?
The Answer is always 10^11 no matter what
 
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