Physics: Post your doubts here!

[Tkp]

cn any 1 explain me oct nov 03 p1-7,8,16,28,31,34

 

[littlecloud11]

cn any 1 explain me oct nov 03 p1-7,8,16,28,31,34

7) Speed is the gradient of a displacement time graph. When the gradient is 0 there is no displacement and hence the car is at rest. For the speed while it's moving calculate the gradient between 50 and 100 sec. That is, (70-30)/ (100-50) =.8m/s. So the answer is B.

8) Initially, when the object is dropped it accelerates at 'g' because there is no air resistance. As the object drops further it's velocity increases, so the air resistance acting on it increases and as a result it's acceleration decreases from 'g' to 0 gradually. This statement coincides with graph A, so that's the answer.

16) The vertical component of tension has to be equal to the sum of the vertical component of W and H as the object is in equilibrium. Hence, T has to be the greatest force. The only answer which lists T as the greatest is C.

28) Use the formula dSinθ =nλ , where d is the separation of the rods, θ is the angle of diffraction, n is the order and λ the wavelength.
Find the angle for the second order.
.08Sinθ = 2* .03,
θ= 48.5 degree
Then find the difference in angle between the first and second order. 48.5-22= 26.6
so the answer is B.

31) Calculate the total resistance in the parallel combination.
1/R = 1/6 + 1/3
Therefore R = 2
So the voltage will be divided equally between the resistor of 2.0 Ω in series and the resistors parallel combination. This means the voltage across the parallel combination is 6V.
Now just use V=IR to calculate the current across the 6.0 Ω resistor.
6= I*6
So, I= 1A
The answer is A.

34)Use the ratio of resistance* potential of supply to calculate the potential across the 3.0Ω in X and Y.
For X potential difference across 3.0Ω resistor is (3/3+.5)* 1.5 =1.28 V
For Y potential difference across 3.0Ω resistor is (3/3+2) *1.5 = .9 V
So greater potential across 3.0Ω in X.
P= v^2/R, R is constant (3.0Ω), as the potential is greater in X the power dissipated will also be greater in X.
The answer is B.

 

[Ahmedraza73]

U THINK IF SOMEONE WOULD SOLVE ALL 40QUESTIONS FOR U THEN STOP DREAMING JUST POST THE QUESTIONS THAT U THINK U CANT SOLVE

Agreed

 

[Tkp]

7) Speed is the gradient of a displacement time graph. When the gradient is 0 there is no displacement and hence the car is at rest. For the speed while it's moving calculate the gradient between 50 and 100 sec. That is, (70-30)/ (100-50) =.8m/s. So the answer is B.

8) Initially, when the object is dropped it accelerates at 'g' because there is no air resistance. As the object drops further it's velocity increases, so the air resistance acting on it increases and as a result it's acceleration decreases from 'g' to 0 gradually. This statement coincides with graph A, so that's the answer.

16) The vertical component of tension has to be equal to the sum of the vertical component of W and H as the object is in equilibrium. Hence, T has to be the greatest force. The only answer which lists T as the greatest is C.

28) Use the formula dSinθ =nλ , where d is the separation of the rods, θ is the angle of diffraction, n is the order and λ the wavelength.
Find the angle for the second order.
.08Sinθ = 2* .03,
θ= 48.5 degree
Then find the difference in angle between the first and second order. 48.5-22= 26.6
so the answer is B.

31) Calculate the total resistance in the parallel combination.
1/R = 1/6 + 1/3
Therefore R = 2
So the voltage will be divided equally between the resistor of 2.0 Ω in series and the resistors parallel combination. This means the voltage across the parallel combination is 6V.
Now just use V=IR to calculate the current across the 6.0 Ω resistor.
6= I*6
So, I= 1A
The answer is A.

34)Use the ratio of resistance* potential of supply to calculate the potential across the 3.0Ω in X and Y.
For X potential difference across 3.0Ω resistor is (3/3+.5)* 1.5 =1.28 V
For Y potential difference across 3.0Ω resistor is (3/3+2) *1.5 = .9 V
So greater potential across 3.0Ω in X.
P= v^2/R, R is constant (3.0Ω), as the potential is greater in X the power dissipated will also be greater in X.
The answer is B.

thnks.i dint get no. 16.its confusing.u made my confusions clear.thanks a lot

 

[Tkp]

mj 04 p1-13,16,25,31

 

[hela]

can you plaese solve q9 summer 11/ 2011

 

[hela]

can you plaese solve q9

 

[hela]

can you solve q9

 

[hela]

n someone provide me notes about momentum please urgent

 

[hela]

can you provide me with notes about momentum
please urgent

 

[hela]

can you please solve these question thanks a lot

 

[snowbrood]

can you please solve these question thanks a lot

for first question
momentum is conserved no matter whether the collision is elastic or inelastic
both bodies have same mass "m"
total momentum before collision
mv+m(0) since the second body is at rest
since total momentum before collision equals total momentum after collision
i.e mv
let V2 be the velocity of these two bodies
we know v is the velocity of first mass
from momentum of conversation
mv=2mV2
V2=v/2
k.e
1/2*2m*V2^2
m(v/4)
so A is the correct answer

for second question
take velocity in the right as positive and velocity in left as negative
(note:question asks about change in momentum not total momentum)
change in momentum=mv-mu
=m(v-u)
=0.1(20-(-30) (as 100g equal 0.1kg)
=0.1*50
=5
so B is the correct answer

for third question
let mass of one body equal m
take velocity in the right as positive and velocity in left as negative u can take the other way round
mv+mu
m(v+u)
m(60+(-40)
20m
since total momentum before collision equals total momentum after collision
20m=2mu
u=10
so A is the correct answer

for the fourth question
using area under graph
it is a trapezium
area of trapezium
=1/2*(base1+base2) *h
=1/2*(6+12)*2
=18m
so C is the correct answer

for the fifth question
difference between elastic and inelastic collision is that k.e is conserved in elastic collision
meaning total k.e before collision equals total k.e after collision
(1/2*2m*u^2)+(1/2*mu^2)
we get 1.5mu^2 as k.e before collision it must equal the k.e after collision only option that satisfies this condiiton is A so correct option is A
there is really no need to denote one velocity positive and the other as negative since squarring the number would always result in a positive answer

for the sixth question

Kinetic energy of 1st trolley = ½mv² = ½ x 2 x 2² =4J
Momentum of 1st trolley = mv = 2 x 2 = 4kgm/s to the left
From The conservation of momentum, the 2nd trolley's momentum = 4kgm/s to the right.
Since the 2nd trolley's mass =1kg, its velocity is 4m/s.
Kinetic energy of 2nd trolley = ½ x 1 x 4² =8J
Total kinetic energy = 4+8=12J. Since this energy has been supplied by the spring, the answer is D.

 

[littlecloud11]

thnks.i dint get no. 16.its confusing.u made my confusions clear.thanks a lot

Ok, Let me try again. For an object to be in equilibrium all the forces as to be balanced, right?
The Tension and H are acting at angles on the plank so they have to be resolved into their vertical and horizontal components when you're calculating the resultant force.
You can form the equation-
vertical component of T = vertical component of H + W
As component of T equals that of H and W, it has to be greater then both forces. Hence T is the largest force. Only answer C lists T as the largest.

Hope this helps!

 

[Tkp]

Ok, Let me try again. For an object to be in equilibrium all the forces as to be balanced, right?
The Tension and H are acting at angles on the plank so they have to be resolved into their vertical and horizontal components when you're calculating the resultant force.
You can form the equation-
vertical component of T = vertical component of H + W
As component of T equals that of H and W, it has to be greater then both forces. Hence T is the largest force. Only answer C lists T as the largest.

Hope this helps!

OH BUddy ur great.nw i got it.ur really genius

 

[Tkp]

hey can u solve my mj o4 prblm.pls if u have time

 

[snowbrood]

hey can u solve my mj o4 prblm.pls if u have time

which question of summer 2004?

 

[hela]

for first question
momentum is conserved no matter whether the collision is elastic or inelastic
both bodies have same mass "m"
total momentum before collision
mv+m(0) since the second body is at rest
since total momentum before collision equals total momentum after collision
i.e mv
let V2 be the velocity of these two bodies
we know v is the velocity of first mass
from momentum of conversation
mv=2mV2
V2=v/2
k.e
1/2*2m*V2^2
m(v/4)
so A is the correct answer

for second question
take velocity in the right as positive and velocity in left as negative
(note:question asks about change in momentum not total momentum)
change in momentum=mv-mu
=m(v-u)
=0.1(20-(-30) (as 100g equal 0.1kg)
=0.1*50
=5
so B is the correct answer

for third question
let mass of one body equal m
take velocity in the right as positive and velocity in left as negative u can take the other way round
mv+mu
m(v+u)
m(60+(-40)
20m
since total momentum before collision equals total momentum after collision
20m=2mu
u=10
so A is the correct answer

for the fourth question
using area under graph
it is a trapezium
area of trapezium
=1/2*(base1+base2) *h
=1/2*(6+12)*2
=18m
so C is the correct answer

for the fifth question
difference between elastic and inelastic collision is that k.e is conserved in elastic collision
meaning total k.e before collision equals total k.e after collision
(1/2*2m*u^2)+(1/2*mu^2)
we get 1.5mu^2 as k.e before collision it must equal the k.e after collision only option that satisfies this condiiton is A so correct option is A
there is really no need to denote one velocity positive and the other as negative since squarring the number would always result in a positive answer

for the sixth question

Kinetic energy of 1st trolley = ½mv² = ½ x 2 x 2² =4J
Momentum of 1st trolley = mv = 2 x 2 = 4kgm/s to the left
From The conservation of momentum, the 2nd trolley's momentum = 4kgm/s to the right.
Since the 2nd trolley's mass =1kg, its velocity is 4m/s.
Kinetic energy of 2nd trolley = ½ x 1 x 4² =8J
Total kinetic energy = 4+8=12J. Since this energy has been supplied by the spring, the answer is D.

thanks a lot it was very helpfull

would you plaese solve the remaining question 7 ,8,9,10.
thanks again

 

[hela]

please solve these 3 question

 

[Tkp]

which question of summer 2004?

13,16,25,31

 

[Abu mota]

help q8

s=1/2at^2
L is directly proportional to t^2
so if u halve t u will quarter L so the answer is 0.25L which is B