Physics: Post your doubts here!

[TheDealer]

Is it that hard?

 

[rz123]

Is it that hard?

did u really try?
work done = force into s , force component along the slope needs to be calculated. the angle which weight makes with the normal reaction line is 30 degree, equal to the angle given ORIGINALLY in the diagram. the weight component we require in these kind of questions is like this , 1 into 10^3 N into sin 30 degree. multiply with the distance given 5 cos 30 degree. we need to use horizontal distance i.e along the slope so dats why its 5 cos 30, multiply these two quantities it will come close to B..

 

[USMAN Sheikh]

salam hurriely need help phy 9702_w10_qp_22 kindly answer my question no 4(b)

and

9702_w10_qp_21 6(b) pls be quick need help before mocks!!!

 

[unique840]

salam hurriely need help phy 9702_w10_qp_22 kindly answer my question no 4(b)

and

9702_w10_qp_21 6(b) pls be quick need help before mocks!!!

first find out the resistance of 1 resistor. (230^2) / 1000 = 52.9 ohms

i) in parallel, the total resistance is 1/R = 1/52.9 + 1/52.9 = 26.45 ohms
power lost = V^2 / R
= (230^2) / 26.45 ohms
= 2000 W
= 2kW

ii) in series, the total resistance is R1 + R2 = 52.9 + 52.9 = 105.8 ohms
power = (230^2) / 105.8 = 5oo W
= 0.5 kW

iii) 1 in series and 2 in parallel so total resistance will be 26.45 + 52.9 = 79.35 ohms
power = (230^2) / 79.35 = 666.67 W
= 0.667 kW

 

[rz123]

salam hurriely need help phy 9702_w10_qp_22 kindly answer my question no 4(b)

and

9702_w10_qp_21 6(b) pls be quick need help before mocks!!!

q.4 b) qp 22 a) draw tangnt frm origin. stop height where tangent get separtd frm curve. the region below shows us Es.
b) area from the point where u left the tagent till .9mm will give u the answer close to 6. 1/2 into .34 into 38

c) dre is a change in the arrangement of atoms and molecules have changed their shape.

 

[rz123]

worksheet 6, question number 1. how its D? or more specifically can u tell me the different situations when different or same masses are approaching towards each other? any link

 

[fawaid]

can someone help me with this question

 

[princesszahra]

worksheet 6, question number 1. how its D? or more specifically can u tell me the different situations when different or same masses are approaching towards each other? any link

i'll answer ur question, check ur physics notes for other cases!

 

[rz123]

i'll answer ur question, check ur physics notes for other cases!

mujhe paki pakai kheer chaiye thi yahan. i forgot those cases in the notes..thanku

 

[princesszahra]

correct ans is
6kms-1 <6mhr-1 <6cms-1 < 6ms-1

 

[princesszahra]

mujhe paki pakai kheer chaiye thi yahan. i forgot those cases in the notes..thanku

hehe np!!!!!!!!!!!!!!!!!

 

[princesszahra]

ill upload the calculation if u want! tell me! @rvel

 

[Unicorn]

Can someone please explain to me how the graph would be D for question 5? Because that is what the mark scheme says but wouldn't the object be acceleratiog according to this graph?

 

[princesszahra]

where;s the question????:O

 

[Unicorn]

where;s the question????:O

whoops

 

[rz123]

correct ans is
6kms-1 <6mhr-1 <6cms-1 < 6ms-1

oooww which one r u solving princess of Mars, i asked for the momentum question. i m done with it. lazy of me to not c the notes. it says dat relative speed of approach equals to relative speed of separation so its D. hope ur doubt is clear.

 

[rz123]

Can someone please explain to me how the graph would be D for question 5? Because that is what the mark scheme says but wouldn't the object be acceleratiog according to this graph?

terminal velocity is when drag equals to the pull of gravity hence the object will travel with a uniform velocity, the discplacmeent time graph is showing us the velocity, i.e the gradient. velocity remains same after a point so it will eventually be a starlight line D option, acceleration will be zero as weight and air resistance balance each other.

bring more problems of AS.

 

[Unicorn]

terminal velocity is when drag equals to the pull of gravity hence the object will travel with a uniform velocity, the discplacmeent time graph is showing us the velocity, i.e the gradient. velocity remains same after a point so it will eventually be a starlight line D option, acceleration will be zero as weight and air resistance balance each other.

bring more problems of AS.

Thanks

 

[rz123]

Thanks

I need more problems!! Cumoon bring it people!

 

[Unicorn]

I need more problems!! Cumoon bring it people!

ok same paper question 15