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Physics: Post your doubts here!

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It's kg/s. And khud try ker k nikala hai :p Though search it online, zarur hoga.. Kg/m^3 *m/s = kg/m^2 s.This means, one kg per second per m^2 of area. Just multiply with the number of m^2 (area) you have and you get kg/s on that specified area.
(Y) nice try!! is ka matlab!! kuch samaj na ayee values ko multiply kar do..:D just kidning!!! Weldone (Y)
 
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It's kg/s. And khud try ker k nikala hai :p Though search it online, zarur hoga.. Kg/m^3 *m/s = kg/m^2 s.This means, one kg per second per m^2 of area. Just multiply with the number of m^2 (area) you have and you get kg/s on that specified area.
btw aaisa question like har baar ata hai :p
 
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Sorry another mistake. Eveything had to be in one decimal place. I don't know what was i thinking when i was in lab. image.jpg
 
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Help me with my superposition doubt please.

What changes is observed when the following is increased?
1. slit separation, a
2. slit width, w
3. angle of diffracted light from zero order, t
4. wavelength, l
5. perpendicular distance between the slit and screen, D

thanks
 
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A wind turbine has blades that sweep an area of 2000 m
2. It converts the power available in the

wind to electrical power with an efficiency of 50%.
What is the electrical power generated if the wind speed is 10 m s
–1? (The density of air is

1.3 kg m
–3.)

A
130 kW B 650 kW C 1300 kW D 2600 kW
can anyone explain this question
 
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A wind turbine has blades that sweep an area of 2000 m
2. It converts the power available in the

wind to electrical power with an efficiency of 50%.
What is the electrical power generated if the wind speed is 10 m s
–1? (The density of air is

1.3 kg m
–3.)

A
130 kW B 650 kW C 1300 kW D 2600 kW
can anyone explain this question
mass flow rwte intercepted by wind turbine blades is 2000*10*1.3=26000 kg/sec
energy=1/2* 26000 * 100 Joules/sec or watts.
due to 50% efficiency,power produced is half of that. Divide by 1000 o express in KW
Yahoo answer.
 
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The diagram shows a wheel of circumference 0.30 m. A rope is fastened at one end to a force
meter. The rope passes over the wheel and supports a freely hanging load of 100 N. The wheel is
driven by an electric motor at a constant rate of 50 revolutions per second.
When the wheel is turning at this rate, the force meter reads 20 N.
load 100N
force
meter
25

20
15
10
5
0
50rev s
–1

wheel of
circumference 0.30m
N
What is the output power of the motor?
A
0.3 kW B 1.2 kW C 1.8 kW D 3.8 kW
any one can explain this question
 
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A light wave of amplitude
A is incident normally on a surface of area S. The power per unit area

reaching the surface is
P.

The amplitude of the light wave is increased to 2
A. The light is then focussed on to a smaller

area
. 3

1
S

What is the power per unit area on this smaller area?
A
36P

B
18P

C
12P

D
6P

can any one explain this
 
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The diagram shows a wheel of circumference 0.30 m. A rope is fastened at one end to a force
meter. The rope passes over the wheel and supports a freely hanging load of 100 N. The wheel is
driven by an electric motor at a constant rate of 50 revolutions per second.
When the wheel is turning at this rate, the force meter reads 20 N.
load 100N
force
meter
25

20
15
10
5
0
50rev s
–1

wheel of
circumference 0.30m
N
What is the output power of the motor?
A
0.3 kW B 1.2 kW C 1.8 kW D 3.8 kW
any one can explain this question
It wpuld be better if you would just tell me the question number and year of paper.
 
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