Physics: Post your doubts here!

[dwightedknight]

This equation is not correct I guess, there's no solution. Is there a question about it?

Oh there is not solution. I got confused.

 

[dwightedknight]

View attachment 64102

Why can't it be C??

It's a slight inaccuracy from CIE I believe. Instead of "infinite", you should interpret as "very large".

 

[Mush999]

View attachment 64102

Why can't it be C??

That's a galvenometer and it a very sensitive device, upto 0.001A
Ita resistance is almost negligible!

The correct answer is B!

 

[A*****]

It's a slight inaccuracy from CIE I believe. Instead of "infinite", you should interpret as "very large".

There is no problem with the word 'infinite' but I just wanted to ask that why this option is not correct although it is possible

 

[A*****]

That's a galvenometer and it a very sensitive device, upto 0.001A
Ita resistance is almost negligible!

The correct answer is B!

Ohh right!

 

[A*****]

How to do this? I did it just by comparing their units but I want a proper method if anyone could help plz

 

[Kanekii]

Is it D?
If so then:
The equation used will be F=ma, Density=M/V and v=s/t
so M=Density*(A*s/t)
M=kgs-1
F=M*v
F=kgms-2
F=Density*A*v^2

 

[hellodjfos;s'ff]

View attachment 64102

Why can't it be C??

Can't be C, because if the resistance was infinite, the ammeter would show an error sign. If the pd is zero between the two points, then the ammeter would show zero since no current would flow between those point.

 

[A*****]

Is it D?
If so then:
The equation used will be F=ma, Density=M/V and v=s/t
so M=Density*(A*s/t)
M=kgs-1
F=M*v
F=kgms-2
F=Density*A*v^2

Thankss!!

 

[hellodjfos;s'ff]

How to do this? I did it just by comparing their units but I want a proper method if anyone could help plz

View attachment 64103

I think there are about 3 methods to attempt this one.
P=pgh, P=F/A
F=P*A
F=pgh*A
g*h becomes v^2(their units multiply to form the units of v^2).
F=p(v^2)(A) which is D

2nd method:
F=mv/t, p=m/V, v=l/t
p=m/V becomes m=p*V
Replace m=p*V into F=mv/t
F=p*V*v/t
V can be broken up into A and l(area and length)
F=p*A*l*v/t
l/t=v, so:
F=p*A*v^2

 

[Ayesha_m01]

 

[hellodjfos;s'ff]

Intensity is directly proportional to amplitude squared.
The equation becomes I=A^2/x^2 which can be simplified to I=A/x
At P: I=8/1 so intensity at P is 8 micro meter
At Q: I=8/2 so intensity at Q is 4 micro meter(D)

 

[Ayesha_m01]

Intensity is directly proportional to amplitude squared.
The equation becomes I=A^2/x^2 which can be simplified to I=A/x
At P: I=8/1 so intensity at P is 8 micro meter
At Q: I=8/2 so intensity at Q is 4 micro meter(D)

may god bless you

 

[Hisham Khan]

Help?

 

[Kanekii]

Help?

P=Density*g*h
P=Po*90/100
0.1Po= Density*g*h
h=0.1Po/density*g (multiply by 10)
h=Po/10*density*g

 

[Kanekii]

help

 

[A*****]

View attachment 64107 help

Is it C?

 

[Kanekii]

Is it C?

Yes and why though?

 

[hellodjfos;s'ff]

Help?

Ok this is a difficult question to understand. So the concept of this drink is that the child decreases the pressure on the top of the straw to 90% of the atmospheric pressure. This means there is a pressure difference between the air in the straw and the atmosphere of 10%. So the atmosphere applies a force on the surface of the liquid which pushes the liquid up the straw(due to the pressure difference). This is just so that you understand what happens, now onto the formulas:
P=F/A and W=mg(the weight of the light being pulled up the straw is equal to the force being applied by the atmosphere)
F=P*A
mg=0.1P*A(you use 0.1 here because that is the pressure difference)
P=m/V which makes m=P*V ----> m=P*A*l(V is broken into area and length)
P*A*l*g=0.1P*A
Rearrange the equation to make l the subject of the formula.
You will get l=0.1P/10pg which is l=P/10pg

 

[hellodjfos;s'ff]

View attachment 64107 help

Only one field exists here which is electric. The electric field's direction is towards the right so the positive terminal is at the left and the negative terminal is at the right side. Since the electron is approaching the negative terminal, there would be a repulsion force. This would case the electron to decelerate to rest, so the speed of the electron would decrease. It would be a linear path since the electric field is acting horizontally, so there is no force acting vertically, hence the electron will move in a straight line.