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Physics: Post your doubts here!

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Metallic9896

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Dynamo said:
View attachment 62530
Anyone please help?
Click to expand...

They are asking for a hypothetical relationship, a "possible relationship" as they say. You know that Power depends on v and F, and F depends on v. So power depends on v twice as much as F depends on v. Thus, in a potential relationshiip, the dependence of Power on v must be one exponent higher than the dependence of F on v. A is impossible 'cause this shows the reverse. B and C are impossible cause here both are dependent on v to the same extent. This leaves only D as plausible answer where Power shows dependence on v to the cube while F shows dependence on v to the square. I'm not so good at maths hopefully someone else can explain if I didn't make sense.
 
Rizwan Javed

Rizwan Javed

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Dynamo said:
View attachment 62530
Anyone please help?
Click to expand...
Metallic9896 said:
They are asking for a hypothetical relationship, a "possible relationship" as they say. You know that Power depends on v and F, and F depends on v. So power depends on v twice as much as F depends on v. Thus, in a potential relationshiip, the dependence of Power on v must be one exponent higher than the dependence of F on v. A is impossible 'cause this shows the reverse. B and C are impossible cause here both are dependent on v to the same extent. This leaves only D as plausible answer where Power shows dependence on v to the cube while F shows dependence on v to the square. I'm not so good at maths hopefully someone else can explain if I didn't make sense.
Click to expand...
well if I were to do this, I would have done it like this:
P = Fv
Since the object is moving, it definitely possesses some Ek. So substitute it into the above equ.

Ek/t = Fv

Take 't' on the other side, you will get:

Ek = F*d 0.5*m*v^2 = F*d

take 'd' on other side and divide. you will get:

0.5m/d * v^2 = F

since speed is constant, resistive force = driving force, so (Fr for resistive force)

Fr = 0.5m/d * v^2

Mass of the object, and distance moved, let them to be constant. so Fr depends on V^2 like this: Fr = k v^2

=> Fr ∝ v^2

Now to find the relationship with power, we know: P = Fv

F= kv^2 <--- put this into above ^ equ

P = kv^2 * v
P = k v^3
P ∝ v^3

Hence the answer is D.
 
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Metallic9896

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ughkno said:
31 and 32 please
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Q31:

V = IR (Voltage = Current / Resistance)

R = pL/A (Resistance = Resistivity x Length / Area)

Thus, V = IpL/A (Voltage = Current x Resistivity x Length / Area)

I = VA/pL (Current = Voltage x Area / Resistivity x Length)

V is the same, p is the same, L is the same. A is directly proportional to I (also proven by I = nAvq), so we can say I = kA, where k is a constant.

A itself = (pi x d squared)/4, so A is proportional to sqaure of d or we can say A = kd, where k is a constant.

Wire S has double the diameter of Wire T, so based on A = kd^2, if we double d, we multiply A by 4. So Wire S has 4 times the area as Wire T.

I = kA, and if A is 4 times in Wire S, then I in Wire S is also 4 times that of Wire T.

Ratio of current in Wire S to Wire T then must 4, D.

Q32:

Power does not depend on the sign of the current, only on the magnitude. Mean power = (Power when current = 2) + (Power when current = 1) / 2

Get power from I^2 x R

Mean power = (2 x 2 x 100) + (1 x 1 x 100) / 2 = (400) + (100) / 2 = 500/2 = 250 W.
 
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Rizwan Javed said:
well if I were to do this, I would have done it like this:
P = Fv
Since the object is moving, it definitely possesses some Ek. So substitute it into the above equ.

Ek/t = Fv

Take 't' on the other side, you will get:

Ek = F*d 0.5*m*v^2 = F*d

take 'd' on other side and divide. you will get:

0.5m/d * v^2 = F

since speed is constant, resistive force = driving force, so (Fr for resistive force)

Fr = 0.5m/d * v^2

Mass of the object, and distance moved, let them to be constant. so Fr depends on V^2 like this: Fr = k v^2

=> Fr ∝ v^2

Now to find the relationship with power, we know: P = Fv

F= kv^2 <--- put this into above ^ equ

P = kv^2 * v
P = k v^3
P ∝ v^3

Hence the answer is D.
Click to expand...

So beautiful. :')
 
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Rizwan Javed

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Your 69Mom said:
View attachment 62531

Help Please!
Click to expand...
The question states that density decreases LINEARLY with altitude. So you can consider it in terms of a graph like this (taking sea level to be altitude of zero!)
upload_2017-6-7_22-28-15.png
Finding the area under the graph will give you the value for 'ph' which can be used in the pressure equation (P = pgh)

So find the area = 0.5 * (1.22+0.74) * 5000 = 4900

now find the pressure change that occurs = pgh = 4900 * 9.81 = 48069
Note that this is the PRESSURE CHANGE, not the pressure at 5000m

so you can calculate the pressure at 5000m altitude as:

change = pressure(sea level) - pressure(5000m)
pressure(5000) = pressure(sea level) - change
pressure(5000) = 10^5 - 48069 = 51931 ~ 52000 Pa
 
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