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Physics: Post your doubts here!

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but
this one is quite a lot.i need like a summary showing the waves in order
do you want to know the frequencies and wavelengths of GXUVIMR? gamma rays to radio waves i mean and do u also want to know the practical use of these all waves?
 
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path difference = 28 cm
frequency is increased from 1.0 kHz to 4.0 kHz
that means (33o/1000)m to (330/4000)m
= 0.33 m to 0.0825 m
in cm,
= 33 cm to 8.25 cm
Now, lets check wavelengths for minima (n + ½)λ
0.5λ = 28 --> λ = 28/0.5 = 56 cm
1.5λ = 28 --> λ = 28/1.5 = 18.7 cm
2.5λ = 28 --> λ = 28/2.5 = 11.2 cm
3.5λ = 28 --> λ = 28/3.5 = 8 cm

In the range, 33 cm to 8.25 cm
There are only two wavelenghts, which are:
18.7 cm and 11.2 cm

Therefore two minima
:)
 
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3 b) you see the mass strikes the horizontal ground when it reaches 4.2 m/s, it remain contact with the ground till -3.6 m/s. so the time between is the time taken into consideration while calculating the momentum.

45/1000 ( v-u) ( -3.6 - 4.2) = -0.35Ns , but ms says +.35 i dnt knw why.

c) Force = change in momentum / time when the mass was in contact with the ground which is 0.57-0.43

so 0.35Ns/ 0.14 = 2.5 N

i can't understand why is 4.2 m/s taken as the final velocity.
 
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8) the displacement time graph shows us the velocity. its gradient telling us velocity so we can see after indicating their is a change in the velocity. hence D

15 ) i can't figure how its B? i think it shud be A ''16kj'' as the K.E will be 4 times higher then the previous. as force plus distance increases by twice. only one situation come to my mind if it has to be B option, that is maybe we are adding the possible friction present in the trolley at the second case. bcz it is NOT specified when the force is 2F and distance 2s dat it is still frictionless trolley
 
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Jaf

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- Question 11
Driving force = m x a = 2 x 9.81 = 19.62
Driving force - frictional force = m x a
19.62 - 6 = (8+2) x a
a = 1.362. Approximately = 1.4 m/s^2
Now the main doubt here is why are we doing 8+2. It would help if you've studied P4 mathematics, but if you haven't, then this is why:
T = tension, F = frictional force
Forces on box: T-F = 8a
Forces on 2kg mass = 2g - T = 2a (Note: we're taking -T here because in the previous equation we took T as positive when it was away from the box, now T is towards the box and away from the 2kg mass)
Substitute the first equation in the second one. 2x9.8 - 8a - F = 2a
10a = 13.6
a = 13.6/10 = 1.36 m/s^2. Approximately = 1.4 m/s^2

- Question 28 was already explained above.

- Question 32
All the wires are connected in parallel.
So 1/R = (1/10)X6 + 1/100
Therefore, 1/R = 61/100
R = 100/61 = 1.6 ohms.
 

Jaf

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Question 8 was already answered above.
Question 15:
F = ma1
1/2 m u^2 = 4
1/2 m v^2 = 8
That leaves us with a1 = F/m
u^2 = 8/m
v^2 = 16/m
We put these in the formula: v^2 = u^2 + 2a(1)s
That gives us: 16/m = 8/m + 2(F/m)s
8/m = (2Fs)/m
Fs = 4
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Now, in the next part 2F = ma2 (a2 is not 2 multiplied by a but rather a2 is a different acceleration [2 is in subscript])
a2= 2F/m
u^2 = 8/m (since initial KE remains same, so does u^2)
1/2mv^2 = E (E is the final KE)
v^2 = (E x 2)/m
v^2 = u^2 + 2a(2)s
2E/m = 8/m + 2(2F/m)2s
(2E-8)/m = (8Fs)/m
2E - 8 = 8Fs. But we know Fs = 4 from the previous working.
So 2E = 8x4 + 8 = 40
E = 40/2 = 20 J :D
[I know, I know not worth the 1 mark]
 
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