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Physics: Post your doubts here!

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Torque of Couple = Force x Distance between two forces which is perpendicular to the line of action of force
Here we will take vertical component of force (8Sin60) so that distance is perpendicular to it.
8Sin60 x 0.6 = 4.2 N
 
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Q21. According to principle of conversation of momentum :
Total momentum before emission (P1) = Total momentum after (P2)

P1 = o (stationary)

The nucleus emits one proton, so the daughter nucleus will have a nucleon number of A- 1.
It's mass will be (A-1)m, where m is the mass of one nucleon.

Momentum of nucleus = (A-1)mu
Momentum of proton = -mv
(Opposite signs because of the opposite directions of the velocities).

P2 = (A-1)mu - mv

P1 = P2
0 = (A-1)mu - mv
mv = (A-1)mu
v = (A-1)u (B)
 
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Q21. According to principle of conversation of momentum :
Total momentum before emission (P1) = Total momentum after (P2)

P1 = o (stationary)

The nucleus emits one proton, so the daughter nucleus will have a nucleon number of A- 1.
It's mass will be (A-1)m, where m is the mass of one nucleon.

Momentum of nucleus = (A-1)mu
Momentum of proton = -mv
(Opposite signs because of the opposite directions of the velocities).

P2 = (A-1)mu - mv

P1 = P2
0 = (A-1)mu - mv
mv = (A-1)mu
v = (A-1)u (B)
You sir, deserve a f***ing medal....Thanks alot dude
 
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Q22. The gradient of displacement-time graph is velocity, so when the gradient is zero the velocity and thus KE are also zero.
The gradient is zero at time = 0, 0.5T, T, 1.5T and 2T
So the KE must be zero at those time coordinates too, only graph D shows this.

Q14. F is the force provided by each beam.
They act towards the hinge like this:
upload_2018-6-9_4-5-5.png

The system is in equilibrium, the vertical components = the weight and the horizontal components cancel out.
Vertical component of one beam = Fsin(50)
Total upward forces = 2Fsin(50)

2Fsin(50) = 4 x 10^4

solving for F gives F = 2.6 x 10^4 (A)

Q10.
upload_2018-6-9_4-15-31.png
ma = force in direction of motion - opposing force

for block M:
Ma = Mg- T
for block m:
ma = T - mg

Add equation 2 to 1:
Ma + ma = Mg- mg
Solving for a gives B as the answer.

Q37.
upload_2018-6-9_4-24-18.png
Use KCL (I flowing into = I flowing out of a junction) to find the currents in the resistors.

Use KVL (EMFs = PDs across a closed loop) to find R.

EMF = PD
12 = (4)(0.5) + (1)(R) + (1.5)(R)
10 = 2.5R
4 = R (B).

Q6. The sandbag was in motion in the balloon, meaning that it's initial velocity when dropped is 3m/s upwards.
It will have a parabolic motion while under acceleration of gravity.
The displacement of the bag during the 5 seconds will be the distance between its starting position and the ground, which is the height of the balloon when it was released.

S = ut + 0.5at^2
S = (-3)(5) + 0.5(9.81)(5)^2
S = 107.625 (B)

Note that the initial velocity and acceleration are in opposite directions.
 
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What do they mean by this?
You apply v=f(lamda) to find the frequency. You know the wavelength of the visible spectrum ranges from 400nm to 700nm, and that the velocity of electromagnetic waves is 3x10^8.
So 3x10^8=400x10^-9(f)
The f turns out to be 7.5x10^14 Hz. Hence, the order of frequency is 10^14 Hz(C).
 
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You apply v=f(lamda) to find the frequency. You know the wavelength of the visible spectrum ranges from 400nm to 700nm, and that the velocity of electromagnetic waves is 3x10^8.
So 3x10^8=400x10^-9(f)
The f turns out to be 7.5x10^14 Hz. Hence, the order of frequency is 10^14 Hz(C).
Thank you so much!!!!! for the three answers, gosh I really need to pay more attention when I do the papers I did not see the word "visible" until now.
 
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Q22. The gradient of displacement-time graph is velocity, so when the gradient is zero the velocity and thus KE are also zero.
The gradient is zero at time = 0, 0.5T, T, 1.5T and 2T
So the KE must be zero at those time coordinates too, only graph D shows this.

Q14. F is the force provided by each beam.
They act towards the hinge like this:
View attachment 64089

The system is in equilibrium, the vertical components = the weight and the horizontal components cancel out.
Vertical component of one beam = Fsin(50)
Total upward forces = 2Fsin(50)

2Fsin(50) = 4 x 10^4

solving for F gives F = 2.6 x 10^4 (A)

Q10.
View attachment 64090
ma = force in direction of motion - opposing force

for block M:
Ma = Mg- T
for block m:
ma = T - mg

Add equation 2 to 1:
Ma + ma = Mg- mg
Solving for a gives B as the answer.

Q37.
View attachment 64091
Use KCL (I flowing into = I flowing out of a junction) to find the currents in the resistors.

Use KVL (EMFs = PDs across a closed loop) to find R.

EMF = PD
12 = (4)(0.5) + (1)(R) + (1.5)(R)
10 = 2.5R
4 = R (B).

Q6. The sandbag was in motion in the balloon, meaning that it's initial velocity when dropped is 3m/s upwards.
It will have a parabolic motion while under acceleration of gravity.
The displacement of the bag during the 5 seconds will be the distance between its starting position and the ground, which is the height of the balloon when it was released.

S = ut + 0.5at^2
S = (-3)(5) + 0.5(9.81)(5)^2
S = 107.625 (B)

Note that the initial velocity and acceleration are in opposite directions.
Thanks alot mann!!
 
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38th question, I'm slightly confused. Can someone help?
 

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