2E-4 is 2*10-4
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Physics: Post your doubts here!
- Thread starter XPFMember
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Thankyou so much
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6: Area under graph in last 2 seconds=Area of trapezium
0.5(6+12)*2=18m
8: Elastic collision so they will move in opposite directions. Only A and B but in B momentum is not conserved. Thought blocker can explain you this one better.
21: Convert diameter to radius then pi*r^2 to get area. force/area to get stress and stress/young modulus to get strain. Multiply by 100 to get the percentage.
25: 7 maxima are on one side, 7 are on the other and 1 is the zero order maxima so total 15.
34: pd across R1 is ratio of R1 and total resistance. Ratio is same as ratio pd divided by total pd or emf.
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http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_11.pdf
Q32 How is torque anti-clockwise?
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_11.pdf
Q14 Ans is A.
Q32 How is torque anti-clockwise?
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_11.pdf
Q14 Ans is A.
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I also Have doubt in this question 11,12,13.. Sorry..but Thank you fr ur help..
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- Messages
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Question 28http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_w11_qp_11.pdf
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- Messages
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An aeroplane travels at an average speed of 600 km h–1 on an outward flight and at 400 km h–1 on
the return flight over the same distance.
What is the average speed of the whole flight?
A 111 m s–1 B 167 m s–1 C 480 km h–1 D 500 km h–1
the return flight over the same distance.
What is the average speed of the whole flight?
A 111 m s–1 B 167 m s–1 C 480 km h–1 D 500 km h–1
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An aeroplane travels at an average speed of 600 km h–1 on an outward flight and at 400 km h–1 on
the return flight over the same distance.
What is the average speed of the whole flight?
A 111 m s–1 B 167 m s–1 C 480 km h–1 D 500 km h–1
can any one answer this question??
the return flight over the same distance.
What is the average speed of the whole flight?
A 111 m s–1 B 167 m s–1 C 480 km h–1 D 500 km h–1
can any one answer this question??
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32)
The arrows point downwards which means the charge at the top is positive and the bottom is negative. So, the +Q will be attracted downwards and the -Q will be attracted upwards, spinning the rod anti-clockwise. The resultant force is zero because F = E/Q and the Q (both charges) are equal but opposite. So resultant force is zero. If the charges weren't the same in magnitude, then we would have a resultant force.
14)
sagar had explained v.well..
Suppose we take the upward direction to be positive, and the downward direction to be negative, we can again write the momentum equations and use them to find the final velocity of the system - the forces between the clay and the lead pellet are huge compared to other external forces, so even though there are external forces acting on the system during the collision, we can assume the momentum stays approximately constant.
So, the initial momentum is the momentum of the bullet alone, which is equal to mv = (5.0/1000) * 200 = 1.0 kg ms^-1 . The clay block is stationary, so it does not contribute any momentum to the system initially.
When the bullet collides with the clay, the bullet gets stuck in the clay and they both move off with the same velocity, which we'll call v(f). The mass of the lead pellet+ the clay block = (95/1000 + 5/1000) = (100/1000) = 0.1 kg.
Their final velocity = v(f)
Therefore, the final momentum of the system is 0.1 * v(f)
Since this is equal to the initial momentum, we can write (0.1 kg) * v(f) = (1.0 kg ms^-1) and so v(f) = 10 ms^-1.
This is the speed with which they rise after the collision. While the system moves upwards, gravity accelerates them downwards at a constant rate of -9.81 ms^-2. When they reach the highest point above the original position, their velocity is 0. Therefore, we can use the formula v^2 = u^2 + 2as to give us "s", which is the maximum height. So,
(0)^2 = (10)^2 +2(-9.81)s
19.62 * s = 100
s = 5.09 m = 5.1 meters = A.
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Power = Energy / time.
Power = ( 0.5 * m * v^2 ) / t
Power = ( 0.5 *(rho * volume)* v^2) / t
Power = ( 0.5 *(rho * area * length)* v^2) / t
Power = (0.5 *(rho * area) * v^3)
Power = 1300kW