# A2Watch

#1
2 years ago
#2

For question 7:

The time period of a pendulum is given by T=2pi(L/g)0.5
If T=1.47 for T=2pi(L/g)0.5 ,

then for a value of g for mars, we'll call gmars, the time period is given by T=2pi(L/gmars)0.5

we are told that gmars = 0.37g. So substituting this into:
T=2pi(L/gmars)0.5 gives T=2pi(L/0.37g)0.5. Factoring out 0.37 from the bracket gives 1/(0.37)0.5 x 2pi(L/g)0.5

From here, you just times T=2pi(L/g)0.5 by 1/(0.37)0.5 which is the same as 1.47x 1/(0.37)0.5 which gives 2.33.

For question 17)

The question is easier than you probably think so I'll give you a clue, how many electrons would be able to "fit in" to the capacitor considering the equation Q=CV to calculate charge of the capacitor?
Consider e=1.6x10-19C

Hope this helps!
0
#3
(Original post by Nick-F007)
For question 7:

The time period of a pendulum is given by T=2pi(L/g)0.5
If T=1.47 for T=2pi(L/g)0.5 ,

then for a value of g for mars, we'll call gmars, the time period is given by T=2pi(L/gmars)0.5

we are told that gmars = 0.37g. So substituting this into:
T=2pi(L/gmars)0.5 gives T=2pi(L/0.37g)0.5. Factoring out 0.37 from the bracket gives 1/(0.37)0.5 x 2pi(L/g)0.5

From here, you just times T=2pi(L/g)0.5 by 1/(0.37)0.5 which is the same as 1.47x 1/(0.37)0.5 which gives 2.33.

For question 17)

The question is easier than you probably think so I'll give you a clue, how many electrons would be able to "fit in" to the capacitor considering the equation Q=CV to calculate charge of the capacitor?
Consider e=1.6x10-19C

Hope this helps!
Ahh okay this did help thanks
#4
(Original post by Nick-F007)
For question 7:

The time period of a pendulum is given by T=2pi(L/g)0.5
If T=1.47 for T=2pi(L/g)0.5 ,

then for a value of g for mars, we'll call gmars, the time period is given by T=2pi(L/gmars)0.5

we are told that gmars = 0.37g. So substituting this into:
T=2pi(L/gmars)0.5 gives T=2pi(L/0.37g)0.5. Factoring out 0.37 from the bracket gives 1/(0.37)0.5 x 2pi(L/g)0.5

From here, you just times T=2pi(L/g)0.5 by 1/(0.37)0.5 which is the same as 1.47x 1/(0.37)0.5 which gives 2.33.

For question 17)

The question is easier than you probably think so I'll give you a clue, how many electrons would be able to "fit in" to the capacitor considering the equation Q=CV to calculate charge of the capacitor?
Consider e=1.6x10-19C

Hope this helps!
Okay i need help on q7 again, whats l?
#5
http://filestore.aqa.org.uk/subjects...1-QP-JUN15.PDF
Help me with q15 & 20
2 years ago
#6
For question 15, draw a free-body diagram to "see" what are the forces acting on the charge. Remember that weight is always pointing downward.

Then make use of Newton's second law to solve for the magnitude of the electric field strength. You need to make use of the charge-to-mass ratio.

For question 20, take a look at this video to see if you can explain what is going on.

When the switch is closed, the current in the wire would give rise to a magnetic field and the interaction of the two magnetic fields results in a magnetic force acting on the wire. According to the Newton's third law, a magnetic force would act on the magnet due to the wire, which is the reading of 161 g. (I did not state the direction of the force - I hope you can figure it out or else reply to ask for help).

Remember the question ask when the current is reversed and doubled......

I did not give the full solutions but I hope the hints are enough for you work out the answer.
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