Physics: Post your doubts here!

[69ermalik007]

Help needed....

 

[69ermalik007]

 

[princess Anu]

answer?

View attachment 54764

 

[princess Anu]

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!!!

 

[princess Anu]

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

 

[Xylferion]

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

The readings are between 1.98 and 2.02, however the important part in the question is that these values are to +- 1%.

So it's 1.98 +- 1% and 2.02 +- 1%.

Since the readings " fluctuate " betweem 1.98 and 2.02, find the average value. This is 1.98 + 2.02 / 2 = 2.00, as stated in the answers.

Now write out the percentage uncertainties as actual uncertainties.

1/100 * 1.98 = 0.02
1/100 * 2.02 = 0.02

When finding the average, we added the two values. So we do the same with the actual uncertainties, giving us 0.04.
This leads to 2.00 +- 0.04, so the answer is D.

Hope that helped!

 

[Xylferion]

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!!!

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!

 

[Xylferion]

Help needed....
View attachment 54762

http://physics-ref.blogspot.ae/2015/05/physics-9702-doubts-help-page-154.html ---> He explains it well here.

 

[Xylferion]

View attachment 54764

Ducility is a measure of the ability of a solid to deform under the action of stress. If you think about it, Youngs modulus is the ratio of stress / strain. A high youngs modulus means it will be harder to deform the solid. A low youngs modulus means that it would be easier to deform the solid. So clearly, ductility is dependent on youngs modulus.

If you consider the others,

The elastic limit applies to anything under the influence of Hooke's law, that is the force and its equivalent extension. So this can be ruled out, since youngs modulus plays no role here.

Again, the spring constant depends on the load and the extension of the spring from its original length. k = F/x
Youngs modulus has no business here either.

The ultimate tensile stress is basically the braking point of an object. By that I mean, it's the maximum value of stress you can apply to an object, before it brakes.

Youngs modulus depends on stress, but stress does not depend on youngs modulus. We only get youngs modulus by seeing the amount of stress applied, and how much the object strains. So since we're looking for a quantity that depends upon youngs modulus, we cannot use this.

Therefore the only possible answer is A, hope that helps!

 

[Xylferion]

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) ?

Alpha is stopped by paper and Beta by about 5 mm of aluminum, neither of them are stopped by air.

 

[princess Anu]

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!

Thanks!!!! Can u please help me with these two ?
Ans for 6 is C while for the other one it is A

 

[Wolfgangs]

Do you have an idea about the minimum marks needed to score A*?

 

[princess Anu]

 

[fantastic girl]

http://onlineexamhelp.com/wp-content/uploads/2015/02/9702_w14_qp_13.pdf
can someone tell me how is amplitude changing in Q28 ? :/

 

[Ahmed Aqdam]

Thanks!!!! Can u please help me with these two ?
Ans for 6 is C while for the other one it is A

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

 

[princess Anu]

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

Which one? btw Thanks a lot..
ans for the attached one is 1200.. can u help me with this one too

 

[The Sarcastic Retard]

http://onlineexamhelp.com/wp-content/uploads/2015/02/9702_w14_qp_13.pdf
can someone tell me how is amplitude changing in Q28 ? :/

Read the chapter xD its mentioned in txt book.

 

[fantastic girl]

Read the chapter xD its mentioned in txt book.

Haha really?? :O
the coursebook ryt? xP

 

[The Sarcastic Retard]

Haha really?? :O
the coursebook ryt? xP

I dont use coursebook, but yeah it should be given...

 

[eliyeap]

PLZ HELP !

Stress= 1*10^9
Area= 1/4*3.142*(1*10^-3)^2 =7.855*10^-7
Force= 785.5N
The question asks what the range includes this value, so ans is C