Physics: Post your doubts here!

[Shah34]

Plz solve

 

[mistique_bee]

 

[mistique_bee]

 

[Rizwan Javed]

View attachment 60242

The radio waves emitted travel a distance 'd' by going from the source to the reflector. And then when reflected they again travel a distance 'd' from the reflector the source. That means that the radio waves travel a total distance of '2d'.
Now first find the time period for travelling this distance. This can be found by counting the number of boxes between the two peaks shown on the cro. They are 4 in this case. Then multiply the number of boxes, with the time base setting.
4 * 0.20 = 0.8us = 0.80 * 10^-6 s <--- this is the time taken in moving from the source to the reflector, and from the the reflector to the source.

Radio waves travel with a speed of 3.0 * 10^8 m/s in a vacuum (as they are part of EM spectrum)
So use the formule for calculating distance :

s = v * t
s is 2d here. v is 3.0*10^8 and time is 0.80 * 10^-6

so substitute them,

2d = 3.0*10^8 * 0.80 * 10^-6
solve this to find d which is the the distance b/w the source and reflector.

 

[Rizwan Javed]

View attachment 60242

For (b) part you know the distance between source and reflector as calculated before. Also you know the speed of sound which is given. Calculate the time taken using :

t = 2s/v (the distance will be multiplied by 2 because it sound travels distance 'd' twice from source to reflector and from reflector to source.)

t = 2 * 120 / 300 = 0.8s

We're asked that the separation between the peaks must be same, so divide this time with the number of boxes you counted in the previous part (i.e. 4 boxes). This will give you the time base setting: 0.8/4 = 0.2s/cm

 

[mistique_bee]

For (b) part you know the distance between source and reflector as calculated before. Also you know the speed of sound which is given. Calculate the time taken using :

t = 2s/v (the distance will be multiplied by 2 because it sound travels distance 'd' twice from source to reflector and from reflector to source.)

t = 2 * 120 / 300 = 0.8s

We're asked that the separation between the peaks must be same, so divide this time with the number of boxes you counted in the previous part (i.e. 4 boxes). This will give you the time base setting: 0.8/4 = 0.2s/cm

thnku so much xD

 

[Rizwan Javed]

View attachment 60243

View attachment 60243

Can you post the whole question?

 

[mistique_bee]

Can you post the whole question?

 

[mistique_bee]

 

[Rizwan Javed]

View attachment 60245

The trace will be something like this^

 

[mistique_bee]

View attachment 60248
The trace will be something like this^

i dont get tht

 

[Rizwan Javed]

i dont get tht

I think i made a mistake It should have been like this^

First find the time periodf or the wave = 1/500 = 2ms
1 cm == 0.1 ms

draw a straight line across the grid as shown by the red horizontal line here. Then draw marks at 1 cm intervals. You can see that you can have at max 1ms on the grid. This means that half wave will be seen on cro.

The amplitude for the wave is given in the previous part, 5mV = 1 cm. So draw a point 1 cm above the straight line you drew. then join that point with smooth lines as shown.

Get it?

 

[Shah34]

Can someone plz solve my question????????

 

[mistique_bee]

Plz solve

For static equilibrium, the resultant force is zero. So the (upward) force in the spring is equal to the total (downward) weight of all the masses.
Spring force....T = 0.20 (10) + 0.1 (0.10) = 3N

When the thread is burnt, the weight of the 0.10kg mass no longer contributes any forces to the system, but at that (instant)... the spring force is still 3N.

So, the weight of the 0.20kg acts downwards while the spring force acts upwards. The resultant (upward) force on the 0.20kg mass is given by
Resultant force = ma = T – mg = 3 – 0.20(10) = 1N
Acceleration, a = 1 / 0.20 = 5ms-2

 

[Shah34]

O

For static equilibrium, the resultant force is zero. So the (upward) force in the spring is equal to the total (downward) weight of all the masses.
Spring force....T = 0.20 (10) + 0.1 (0.10) = 3N

When the thread is burnt, the weight of the 0.10kg mass no longer contributes any forces to the system, but at that (instant)... the spring force is still 3N.

So, the weight of the 0.20kg acts downwards while the spring force acts upwards. The resultant (upward) force on the 0.20kg mass is given by
Resultant force = ma = T – mg = 3 – 0.20(10) = 1N
Acceleration, a = 1 / 0.20 = 5ms-2

Okaaaay thanks

 

[mistique_bee]

part b(ii)

 

[mistique_bee]

 

[Rizwan Javed]

View attachment 60262
part b(ii)

Spring constant can be found by finding the gradient of the graph, since k = F/x

k = 4-0/(5-0)*10^-2
= 80 N/m

 

[Rizwan Javed]

View attachment 60263

Find the area under the graph upto F = 6.4N

work done = area under the graph
= 1/2 * 6.4 * 8 * 10^-2
= 0.256J

 

[mistique_bee]

Find the area under the graph upto F = 6.4N

work done = area under the graph
= 1/2 * 6.4 * 8 * 10^-2
= 0.256J

thnx alot