Physics: Post your doubts here!

[Shimmery woods]

que 34 anyone ?

 

[Shimmery woods]

For q. 26:

equation for doppler effect:

[Speed of sound X Frequency of source]/[Speed of sound + or - speed of source]

so just put in the values and subtract when speed of sound - speed of source (approaching) from speed of sound + speed of source (Receding) ANS= C

for q. 25

for the length of wavelength: I used d = S X T

so wavelenght = 330 X 500 X 10^-6 = 0.165 m

so a translation of 0.165m means shift the graph by 1 lambda or 2 pi or 360 degrees

so basically it will be the same graph ANS= D

hope that helps

Thankyou!!

 

[Shimmery woods]

O


Only Ans A is correct as if wind blows, the aeroplane will move in the same direction as wind ( accelerate in that direction) and the velocity in wind will have a lower magnitude that without wind effect.

thankyou!

 

[Shimmery woods]

Areas under the graph (5-0.8)*20/2 + 0.8*20

thankyou

 

[Shimmery woods]

You must realise that this is a uniform block so the centre of mass is right in the center

Hope this helps

thankyou!

 

[Bishnu Dev]

que 34 anyone ?

Resistance of 1st Lamp = 960 (Power = 60W)
Resistance of 2nd Lamp = 40 (Power = 2.5W)

When Connected in series, PD across 1st lamp = (960/1000) * 250 = 240
PD across 2nd lamp = (40/1000) * 250 = 10
Since resistance is constant, Power Dissipated across 1st lamp = 60 W and Power Dissipated across 2nd lamp = 2.5 W.

Hence both lamps light normally.

 

[nigeller]

Answer is A. WHy?

 

[nigeller]

answer is D :/

 

[Holmes]

View attachment 62386

O/N 2016 (13)

hey anyone please

 

[Holmes]

Im trying but all I get is 60655...Pa <_> SOMEONE SOLVE THIS QUESTION!!

Attention please

 

[Bishnu Dev]

hey anyone please

The wavelength is 0.165m. After 0.165 further away, the same wave pattern will be repeated. Hence D is the appropriate option.

 

[janedoe]

Can anyone give some advice regarding estimation and approximation in the mcqs?
How to prepare for it?
There were questions on estimating resistance of copper and on estimating young modulus of metals.. how would you approximate such a value without calculation..?

 

[amina1300]

Attention please

I solved it and posted it too.

 

[selrey]

A giant squid of length 20.0m is vertical in seawater, with the top of the squid at a depth of
8.00m. The density of seawater is 1050kgm–3.
What is the difference in pressure between the top and the bottom of the squid?
A 82000Pa B 206000Pa C 288000Pa D 389000Pa
Anyone?

 

[GarryTheGhost]

A giant squid of length 20.0m is vertical in seawater, with the top of the squid at a depth of
8.00m. The density of seawater is 1050kgm–3.
What is the difference in pressure between the top and the bottom of the squid?
A 82000Pa B 206000Pa C 288000Pa D 389000Pa
Anyone?

use the pressure formula= density*g*depth
use depth as 20m

 

[GarryTheGhost]

how do we do this????

 

[dumbledore.]

Q19.
Ans=C

 

[GarryTheGhost]

how do we do this????

nvm i got it

 

[selrey]

use the pressure formula= density*g*depth
use depth as 20m

Ohh thankyou!

 

[GarryTheGhost]

Q19.
Ans=C

first find the power produced by the engine itself. so using force*velocity=power hence 30000/25=1200N
notice the car maintains 25m/s therefore forces are balanced and there is no net force.
we need to calculate the weight down the slope.
so 1400*g*sin(2)=479.3.....N
the car has to overcome this force to maintain a constant velocity and it does this by producing an extra output power of 479.3....N*25=12000watts
so total output power= 12000+30000=42000