A2 Physics | Post your doubts here

[shiningstar]

CAN SOMEONE PLEASE ANSWER ME ?

It said at a height o.5 above Earth's surface which means that r = R of Earth + 0.5 R(Height ABOVE Earth's Surface) = 1.5R
Hope you get it

 

[namename]

Assuming that the Earth is a uniform sphere of radius 6.4 x 106 m and mass 6.0 x 1024 kg, find the gravitational field strength g at a point:

(b) at height 0.50 times the radius of above the Earth's surface.​

g = GM / r2 = (6.67 × 10-11)(6.0 x 1024) / ( (1.5 × 6.4 x 106)2 = 4.34ms-2​

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this example I found in the revision section of this site ermmm could someone explain me why do we multiply the radius of earth with 1.5 but not 0.5 mmmm confused ? anyone ? ​

Like me if u understood.
It is 0.5 times the radius ABOVEE THE earth's SURFACE. so it is the RADIUS + 0.5RADIUS = 1.5 radius

 

[farrukh]

What is meant by frequency response?

 

[shiningstar]

What is meant by frequency response?

This is the quantity that us proportional to the width of the resonance curve. As the damping inceases the frequency response also increases and the resonance curve becomes broader.

 

[RGBM211]

thank you guyz

 

[namename]

Any1?

Given

Question

 

[namename]

Any1?

GivenView attachment 9393

QuestionView attachment 9394

And this (The same question)

 

[SkyPilotage]

And thisView attachment 9396 (The same question)

which year is this?

 

[namename]

which year is this?

No idea.. my teachers just sent me those cut-sections

 

[RGBM211]

Can someone elxplain this

Example 1:
A ship is at rest on the Earth's equator. Assuming the earth to be a perfect sphere of radius R and the acceleration due to gravity at the poles is go, express its apparent weight, N, of a body of mass m in terms of m, go, R and T (the period of the earth's rotation about its axis, which is one day).

At the North Pole, the gravitational attraction is F = GMEm / R2 = mgo
At the equator,
Normal Reaction Force on ship by Earth = Gravitational attraction - centripetal force
N = mgo – mRω2= mgo – mR (2π / T)2​

 

[shiningstar]

Can someone elxplain this

Example 1:
A ship is at rest on the Earth's equator. Assuming the earth to be a perfect sphere of radius R and the acceleration due to gravity at the poles is go, express its apparent weight, N, of a body of mass m in terms of m, go, R and T (the period of the earth's rotation about its axis, which is one day).

At the North Pole, the gravitational attraction is F = GMEm / R2 = mgo​

At the equator,​

Normal Reaction Force on ship by Earth = Gravitational attraction - centripetal force​

N = mgo – mRω2= mgo – mR (2π / T)2​

At the poles there is no centripetal force as it rotates about Earth's axis so the radius of rotation is zero. so ther is only gravitaional force: F = GMEm / R2 = mgo
At the Equator, there are both Centripetal and Gravitational forces. But since the Gravitational force is greater the resulta nt is the Normal reaction with is outwards.
The Centripetal force is less as the Earth travels with very small speed and since Fc= (mv*2)/r , Fc will be small.
Hope this will help.

 

[angelgirl:)]

o/n 2009 ppr 42....Q1c...?

plz ppl explain tne answer...

 

[user]

..

Aoa wr wb!
the answer to that question from the redspot..: [and this is supposed to be correct cuz a teacher has confirmed it...]

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Now tell me if you can understand something from this..
i dont get it....

 

[leadingguy]

please mention the syllabus point which states, learning bandwidth of op-amp.

no I am sorry i just wrote it in a hurry bandwidth of op-amp is nt in sylabus

 

[angelgirl:)]

no I am sorry i just wrote it in a hurry bandwidth of op-amp is nt in sylabus

yaaaaaaaa

 

[Mobeen]

plz ppl explain tne answer...

the geostationary satellites dont cover the polar regions .. but for GPS we need a satelite in the polar regions so thats why geostationary are not used ..

 

[Mobeen]

Any1?

GivenView attachment 9393

QuestionView attachment 9394

do you have the answers?

 

[user]

do you have the answers?

it's this paper: http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w04_qp_6.pdf
Q:15

 

[histephenson007]

Oh right! I think I get it. Also, I know this may sound stupid, but for the same question, they said:
- water vapour volume = 2.96 × 10^–2 m3
- volume occupied by liquid water at 100 °C is 1.87 × 10^–5 m3

So, how would the change in volume (dV) = 2.96 × 10^–2 - 1.87 × 10^–5 ?

I do not quite get ur question here. The dV is positive. And, because W = -P (dV). The total work done is negative.

 

[histephenson007]

Hey histephenson007...could you please explain me the answer to MAY/JUNE 2009 Q2 PART a?

Nah, I don't have any reason for that one. I just know that if anything like that comes up in the test, I will answer that the half-life is too large