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Student uses a digital ammeter to measure a current. Reading of the ammeter is found to fluctuate between 1.98A and 2.02A. Manufacturer of the ammeter states that any reading has a systematic uncertainty of ±1%. Which value of current should be quoted by the student?
A (2.00 ± 0.01) A
B (2.00 ± 0.02) A
C (2.00 ± 0.03) A
D (2.00 ± 0.04) A
Help please
A water cannon directs a jet of water towards a vertical wall. 300 kg of water hit the wall each minute. The water hits the wall horizontally with a velocity 20 m s–1. Assume the water falls vertically after hitting the wall.
What force does the water exert on the wall?A 100N B 200N C 3000N D 6000N
helpp!!!
Help needed....
View attachment 54762
Q.P http://maxpapers.com/wp-content/uploads/2012/11/9702_w12_qp_13.pdf
M.S http://maxpapers.com/wp-content/uploads/2012/11/9702_w12_ms_13.pdf
Can someone please explain Q 25?
And for Q 38, I get how gamma radiation will not be stopped by a few millimeters of lead, but inst alpha radiation absorbed by a few centimtres of air (1cm-3cm) ?
Thanks!!!! Can u please help me with these two ?Write down the values they've given you first.
Mass of water = 300 kg
each minute = every 60 seconds
final velocity = 20 ms^-1
They want the force, since we have mass, the formula we're going to use is F = ma.
We have the value of the mass, all we need is the acceleration.
a = v - u / t
Since the water hits every minute, that is 60 seconds between when it leaves the cannon and hits the wall.
a = v - u / 60
The water obviously starts from rest, before being fired out, so the initial velocity is 0.
You have the final velocity at which it hits the wall, which is 20 ms^-1/
a = 20 - 0 / 60
a = 20/60
a = 1/3 ms^-2, plug this value back into F = ma.
F = 300 kg * 1/3 ms^-2
F = 100 N, so your answer is A.
Hope that made sense!
6. Momentum is the area under graph. Convert time in to seconds.Thanks!!!! Can u please help me with these two ?
Ans for 6 is C while for the other one it is A
Which one? btw Thanks a lot..6. Momentum is the area under graph. Convert time in to seconds.
0.5*1120*0.005=2.8
Divide by mass 0.056 kg to get velocity 50.
Your previous question could've been solved by the same method.
33. d=0.01*10^-3 and sin θ for first order diffracted light is 0.05.
n λ = d sin θ
Read the chapter xD its mentioned in txt book.http://onlineexamhelp.com/wp-content/uploads/2015/02/9702_w14_qp_13.pdf
can someone tell me how is amplitude changing in Q28 ? :/
Haha really?? :ORead the chapter xD its mentioned in txt book.
I dont use coursebook, but yeah it should be given...Haha really?? :O
the coursebook ryt? xP
Stress= 1*10^9PLZ HELP !
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