Physics: Post your doubts here!

[dumbledore.]

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

 

[Daniel Jeng]

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.

 

[dumbledore.]

34. C

 

[Daniel Jeng]

34. C

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.

 

[Heba Nazir]

this might be a very easy question,but i dont seem to get it.
the answer is A.
please helppp!

 

[A*****]

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

 

[dumbledore.]

View attachment 62752
this might be a very easy question,but i dont seem to get it.
the answer is A.
please helppp!

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

 

[Heba Nazir]

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!

 

[Deeksha Tamang]

9702/12/m/j/16

The answer is C. 52000 Pa . How can we get it?

 

[auxy12]

9702/13/m/j/17

The answer is A... can anyone tell me how to get it?

 

[sushil bista]

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

 

[sushil bista]

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

 

[MShaheerUddin]

Can anybody help me with part b) iii)
Which force acts on A?? :S

 

[ftfayek]

Can you post singapore-cambridge physics as level questionbpapers from 2005-15?

 

[ftfayek]

You should be more specific otherwise we cannot help..

SINGAPORE-CAMBRIDGE AS LEVEL PHYSICS 2005-2015 QUESTION PAPER ?

 

[Hamnah Zahoor]

can anyone help me with Question 7 .....have exam tomorrow!

 

[MShaheerUddin]

can anyone help me with Question 7 .....have exam tomorrow!

Initial speed u=50 ms–1
Final speed v=100 ms–1
Time t=20 s
Find deceleration a using v = u + at.
10=50 + a(20)
a=-2 ms–2.

Find marker distance using s=ut + ½ at2.
s=(50)(20) + ½ (-2)(202)
s=600 m


Answer: D

 

[saraawan]

can someone explain how will we do this question
A car of mass 1400kg is travelling on a straight, horizontal road at a constant speed of 25ms–1. The output power from the car’s engine is 30kW. The car then travels up a slope at 2° to the horizontal, maintaining the same constant speed.
What is the output power of the car’s engine when travelling up the slope?
A 12kW B 31kW C 42kW D 65kW
ans = C

 

[MShaheerUddin]

can someone explain how will we do this question
A car of mass 1400kg is travelling on a straight, horizontal road at a constant speed of 25ms–1. The output power from the car’s engine is 30kW. The car then travels up a slope at 2° to the horizontal, maintaining the same constant speed.
What is the output power of the car’s engine when travelling up the slope?
A 12kW B 31kW C 42kW D 65kW
ans = C

When Car is travelling at horizontal road it's speed is 25 m/s .. The car maintains this speed at slope WHICH MEANS THE CAR ALSO MAINTAINS 30kW OUTPUT POWER AT THE SLOPE (with new power at slope which is calculated below) AND SPEED OF 25 m/s.

At Slope the horizontal component of weight is forward driving force which is [mgSin(2)]=[1400x9.8Sin2]=478.8 N
Power is Force x Velocity ----> 478.8 x 25 = 11970.52 W
Since Car maintains the Output power of 30K at slope also so Output power of the car's engine is 11970.52 + 30kW Which is equals to 41970.5 W
Round it off to get 42kW of Power. (C ANS)

 

[MShaheerUddin]

Need help..