Physics: Post your doubts here!

[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.

 

[A*****]

Why not B?

 

[hellodjfos;s'ff]

Why not B?

View attachment 64111

Bro it is a VELOCITY-DISTANCE graph not a velocity-time graph. If it were a velocity-time graph, it would be B since the velocity is constant initially and then there is a constant deceleration(gradient is constant in that graph). But that's not the case. The answer would be A, because the car is decelerating so as its speed decreases, the rate at which it covers a distance decreases rapidly. This is shown by A, since when the velocity is decreasing, the distance the car covers is small(compared to other graphs).

 

[Hisham Khan]

Bro it is a VELOCITY-DISTANCE graph not a velocity-time graph. If it were a velocity-time graph, it would be B since the velocity is constant initially and then there is a constant deceleration(gradient is constant in that graph). But that's not the case. The answer would be A, because the car is decelerating so as its speed decreases, the rate at which it covers a distance decreases rapidly. This is shown by A, since when the velocity is decreasing, the distance the car covers is small(compared to other graphs).

Are you einstein?

 

[Kanekii]

 

[hellodjfos;s'ff]

View attachment 64112

The angle between the central maxima and first order is 35(which is the theta in the equation dsin(theta)=n(lamda) )
1x10^-6*sin35=1(lamda)
lamda=5.7x10^-7 which is 574nm(C)

 

[Hisham Khan]

The angle between the central maxima and first order is 35(which is the theta in the equation dsin(theta)=n(lamda) )
1x10^-6*sin35=1(lamda)
lamda=5.7x10^-7 which is 574nm(C)

This pls

 

[A*****]

The angle between the central maxima and first order is 35(which is the theta in the equation dsin(theta)=n(lamda) )
1x10^-6*sin35=1(lamda)
lamda=5.7x10^-7 which is 574nm(C)

If u don't mind, can u tell what were ur O level grades?

 

[Mr."S"]

How to solve this

 

[A*****]

View attachment 64113
This pls

First apply the equation to the part with distance x and then to the whole descent...then solve the two equations simultaneously to eliminate x

 

[Hisham Khan]

Nerds aren't that bad

First apply the equation to the part with distance x and then to the whole descent...then solve the two equations simultaneously to eliminate x

View attachment 64115

 

[A*****]

Nerds aren't that bad

I'm not a nerd

 

[Kanekii]

View attachment 64113
This pls

Two simultaneous equations will be formed by s=ut +0.5at^2
1- x=0.5a*t1^2
2- x+h=0.5*t2^2
0.5a*t1^2 + h = 0.5*t2^2
h=0.5a*t2^2 - 0.5*t1^2
h=0.5a*(t1-t2)^2
u=o as it starts from rest.

 

[hellodjfos;s'ff]

View attachment 64113
This pls

Firstly, the ball is released meaning it's initial velocity(u) is zero. Now, you form two equations, one for the motion of the ball from zero point to m1, then from zero point to m2. First equation:
u=o, s=x, a=finding, t=t1
s=ut+1/2(a)(t^2)
x=1/2(a)(t1^2) --- 1st equation

u=o, s=x+h, a=finding, t=t2
x+h=1/2(a)(t2^2)--- 2nd equation
1/2(a)(t1^2)+h=1/2(a)(t2^2)

Now just rearrange that final equation and you will get answer D.

 

[hellodjfos;s'ff]

If u don't mind, can u tell what were ur O level grades?

They're great but I don't like to show off.