Physics Exam Question A Level
Please can someone explain how to answer Q5, 6, 7 and 15? Any help would be greatly appreciated!
https://www.ocr.org.uk/Images/171727-unit-h556-01-modelling-physics-sample-assessment-materials.pdf
https://www.ocr.org.uk/Images/171727-unit-h556-01-modelling-physics-sample-assessment-materials.pdf
I will be posting one by one
for Q5,
would you agree that total force exerted is mg?
and area touching the surface is thin metal ring (the bottom bit of the tube), this can be approximated by multiplying circumference by the thickness of the metal, which is 2pie R times t,
and pressure is force per unit area so the answer is A
for Q5,
would you agree that total force exerted is mg?
and area touching the surface is thin metal ring (the bottom bit of the tube), this can be approximated by multiplying circumference by the thickness of the metal, which is 2pie R times t,
and pressure is force per unit area so the answer is A
Q6)
because these trucks are traveling at a constant speed, driving force (to be accurate driving force - resistance on locomotive) is 4000*6=24000N.
resistance force is decreased to 8000N at the instance when the coupling snaps. and now there are only 2 trucks and locomotive with total mass of 12*10^4 kg. resultant force is 16000N, from F=ma, a should be 0.044. correct answer is B
because these trucks are traveling at a constant speed, driving force (to be accurate driving force - resistance on locomotive) is 4000*6=24000N.
resistance force is decreased to 8000N at the instance when the coupling snaps. and now there are only 2 trucks and locomotive with total mass of 12*10^4 kg. resultant force is 16000N, from F=ma, a should be 0.044. correct answer is B
Q7)
first of all, you have to find the maximum height reached by the sandbag. this can be calculated using v^2=u^2+2as. In this case v=0, u=7 a=-9.8. So s is 2.5m which is 3.8m above the ground. it starts to fall freely from here. when it hits the ground, vertical speed is v=8.63 (using v^2=2as as u=0) and then combine it with horizontal speed which is 5ms^-1. the answer should be about 10ms^-1, C
first of all, you have to find the maximum height reached by the sandbag. this can be calculated using v^2=u^2+2as. In this case v=0, u=7 a=-9.8. So s is 2.5m which is 3.8m above the ground. it starts to fall freely from here. when it hits the ground, vertical speed is v=8.63 (using v^2=2as as u=0) and then combine it with horizontal speed which is 5ms^-1. the answer should be about 10ms^-1, C
Q15)
this is essentially asking the change in gravitational potential energy. which is given by the equation -GMm/r
so work done will be given by (GPE in the orbit) - (GPE on earth surface)
this should give u 6.0*10^9J but considering air resistance, more work should be done so I would choose B as a reasonable answer but I am not sure about this
this is essentially asking the change in gravitational potential energy. which is given by the equation -GMm/r
so work done will be given by (GPE in the orbit) - (GPE on earth surface)
this should give u 6.0*10^9J but considering air resistance, more work should be done so I would choose B as a reasonable answer but I am not sure about this
(edited 3 years ago)
Original post by scottishpenguin
Q15)
this is essentially asking the change in gravitational potential energy. which is given by the equation -GMm/r
so work done will be given by (GPE in the orbit) - (GPE on earth surface)
this should give u 6.0*10^9J but considering air resistance, more work should be done so I would choose B as a reasonable answer but I am not sure about this
this is essentially asking the change in gravitational potential energy. which is given by the equation -GMm/r
so work done will be given by (GPE in the orbit) - (GPE on earth surface)
this should give u 6.0*10^9J but considering air resistance, more work should be done so I would choose B as a reasonable answer but I am not sure about this
You are honestly a life saver! This all makes so much more sense thank you so much.
Just one quick question, for Q7 does that mean that if I drop the sandbag it initially travels upwards?
Sorry for how bad my physics is but thanks for your help, it does mean a lot!
Original post by maya_jai_singh
You are honestly a life saver! This all makes so much more sense thank you so much.
Just one quick question, for Q7 does that mean that if I drop the sandbag it initially travels upwards?
Sorry for how bad my physics is but thanks for your help, it does mean a lot!
Just one quick question, for Q7 does that mean that if I drop the sandbag it initially travels upwards?
Sorry for how bad my physics is but thanks for your help, it does mean a lot!
Yes and no.
According to the question, there is a cross-wind. When the sandbag is released from the balloon, it has an initial upward velocity of 7.0 m s–1 and an initial horizontal velocity of 5.0 m s–1.
When the sandbag is released, it travels upward and rightward simultaneously.
Quick Reply
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