Physics: Post your doubts here!

[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.

 

[Kanekii]

How to solve this

 

[Kanekii]

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)

Why is Angle 35?

 

[hellodjfos;s'ff]

How to solve this

P=E/t, KE=1/2(m)(v^2), D=m/V
m=D*V
KE=1/2(D*V)(v^2)
The volume is broken up into area and length
P=1/2(D*A*l)(v^2)/t
l/t=v, so:
P=1/2(D*A*v^3)
This gives an input power of 1300kW. Since the efficiency is 50%, the output power is 650kW, hence the answer B.

 

[MShaheerUddin]

Hellp ? :/

 

[hellodjfos;s'ff]

Why is Angle 35?

Because they gave the angle between the two first order maximum. The angle in the equation dsin(theta)=n(lamda) is the angle between the central maxima and the order, so to get this angle you divide 70 by 2 which is 35.

 

[A*****]

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

That's good but I'm asking myself...that won't be showing off