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Physics: Post your doubts here!

Student12

Student12

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Question 28http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_w11_qp_11.pdf
 
A

amal sharkawi

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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
 
A

amal sharkawi

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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??
 
Phoenix Blood

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amal sharkawi said:
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
Click to expand...
Just find the mean, i.e, the average of the two speeds.
So (400+600)/2
=500km/h
 
Thought blocker

Thought blocker

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Kamihus said:
Suchal Riaz sagar65265 Thought blocker
Click to expand...
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.
 
Thought blocker

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Student12 said:
Question 27 http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf

Question 29
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_11.pdf
Click to expand...
27)
find d which is 1/n -->1/500 * 10^ (-3) = 2 x 10^-6.
then d sin90=n x (600 x 10^-9) = 3 then he asked for the images so it is 3 orders for one side which is 45 degrees so for the 90 degrees it is 3 x 2= 6 + the normal ray = 7 so D

29)
For a stationary wave in P there should be a node at the closed end and an antinode at the open end.For a wave to be formed in Q there should be an antinode at both ends.
length of P=35 cm
wavelength of given wave = 20 cm
so 35/20 = 1.75 lambda can fit in P. if you draw a wave 1.75 lambda long starting with a node you will find that it ends with an antinode so this means that a stationary wave would be formed in P
for Q 50/20 = 2.5 lambda would fit in it.
start by drawing an antinode at one open end and continue until you draw 2.5 lambda .u will find that an antinode also forms at the other end.This satisfies the condition for a stationary wave .hence you will get a stationary wave in Q as well
 
Princess Raven

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Thought blocker said:
s12_s12
17)
At the bottom height is zero so mgh = zero
now going upwards..
2 block has height = 1 so mg1h
3 block = mg2h
4 block = mg3h
Sum up P.E ans is 6mgh

29)
c = f * λ
We know f = 1 / t
so c = 1 / t * λ
t = λ / c
now it took 3 wavefronts to reach XY to P
so t = 3λ / c

30)
Its a standing wave, so the 33cm is the distance between two nodes. The wavelength thus becomes .66m. F = 330/0.66 = 500. T = 1/500 = 0.002s = 2ms. B is the right answer

s07_1
Suchal Riaz
me gtg :)
Click to expand...
Thank u :) just one question y do u do distance between 2 nodes I mean isnt the normal way just 33 cm?
 
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