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Physics: Post your doubts here!

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and
Microwaves are used to transmit television signals to and from a satellite.
Which statement about microwaves is correct?
A They have a longer wavelength than radio waves.
B They penetrate the atmosphere without significant loss of energy.
C They travel much faster than radio waves in a vacuum.
D They warm the satellite and stop it freezing.
why is it B
A is incorrect because microwaves have more energy than radio waves. This means that the microwaves have a higher frequency and therefore lower wavelength.
C is incorrect as all electromagnetic waves travel at the same speed in space i.e. 3x10^8 ms^-1.
And D is just absolutely ridiculous.
By way of elimination, B is the answer which makes the most sense.
 
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A is incorrect because microwaves have more energy than radio waves. This means that the microwaves have a higher frequency and therefore lower wavelength.
C is incorrect as all electromagnetic waves travel at the same speed in space i.e. 3x10^8 ms^-1.
And D is just absolutely ridiculous.
By way of elimination, B is the answer which makes the most sense.
Thought I'd just make D a little clearer: the main problem is the sun overheating the satellite, rather than freezing, ironically.
 
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How to do this??
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The diagram shows a junction in a circuit where three wires
P, Q and R meet. The currents in P and Q are 1 A and 3 A
respectively, in the directions shown.

How many coulombs of charge pass a given point in wire R in
5s?
A) 0.4 B) 0.8 C) 2 D) 10
 

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How to do this??
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The diagram shows a junction in a circuit where three wires
P, Q and R meet. The currents in P and Q are 1 A and 3 A
respectively, in the directions shown.

How many coulombs of charge pass a given point in wire R in
5s?
A) 0.4 B) 0.8 C) 2 D) 10
By Kirchhoff's law of the conservation of charge, the current in wire R must be 2A into the junction, as current in=current out.
To figure out the charge passing through the wire Q the equation Q=It can be used, so the charge passing through Q would be 2x5=10C.
Therefore the answer will be D, 10C.
 
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A loaded aeroplane has a total mass of 1.2x10^5 kg while climbing after take off. It climbs at an agle 23 degree to horizontal with a speed of 50 m/s. what is the rate at which it is gaining potential energy at this time?
ans: 2.3x10^7
 
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A loaded aeroplane has a total mass of 1.2x10^5 kg while climbing after take off. It climbs at an agle 23 degree to horizontal with a speed of 50 m/s. what is the rate at which it is gaining potential energy at this time?
ans: 2.3x10^7
Figure out the vertical component of the velocity first. The vertical component of the velocity is 50sin23. Now you know the rate at which the height of the plane increases.

Now just use GPE=mgh, to find the potential energy increase per unit time. This will be equal to 1.2x10^5x9.81x50sin23. That should give you the answer you entered there.
 
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First use Q=It to obtain the current through the circuit.

Then find the resistance of R. Use the equation of E=I^2Rt, by combining the equations P=VI=I^2R and E=VIt together.

Plug the values of the I, R and E (given in the question) into E=I(R+r) (the equation you written there), and you should be able to solve the problem. The answer should come out as 20 ohms like you have stated.
 
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View attachment 62752
this might be a very easy question,but i dont seem to get it.
the answer is A.
please helppp!
Suppose the diameter of the wider portion is 2r, so its radius will be r and area will be pi x r²
Since the dia of the narrow portion is half, so it will be r and its radius will be r/2=0.5r and area will be pi x (0.5r)²= pi x 0.25r²
Now the ratio of stress will be (T/pi x r²) / (T/ pi x 0.25 r²)
T and pi will be cancelled out and This will be simplified to (1/r²)/(1/0.25r²)= 0.25
Hope it helps
 
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STRESS= FORCE/AREA

Stress (wide) =F/[d^2]
Stress (narrow)= F/[(d/2)^2]
F and d^2 will cancel out each other by substituting them.

StressW/StressN= [1]/[4]
=0.25
Suppose the diameter of the wider portion is 2r, so its radius will be r and area will be pi x r²
Since the dia of the narrow portion is half, so it will be r and its radius will be r/2=0.5r and area will be pi x (0.5r)²= pi x 0.25r²
Now the ratio of stress will be (T/pi x r²) / (T/ pi x 0.25 r²)
T and pi will be cancelled out and This will be simplified to (1/r²)/(1/0.25r²)= 0.25
Hope it helps

Thank you! :D
 
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10 A firework rocket is fired vertically upwards. The fuel burns and produces a constant upwards force on the rocket. After 5 seconds there is no fuel left. Air resistance is negligible. What is the acceleration before and after 5 seconds? before 5 seconds after 5 seconds A constant constant B constant zero C increasing constant D increasing zero.9702/12May/June 2015 qn no 10
 
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10 A firework rocket is fired vertically upwards. The fuel burns and produces a constant upwards force on the rocket. After 5 seconds there is no fuel left. Air resistance is negligible. What is the acceleration before and after 5 seconds? before 5 seconds after 5 seconds A constant constant B constant zero C increasing constant D increasing zero.9702/12May/June 2015 qn no 10
 
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