Physics: Post your doubts here!

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So .. tomorrow we all have variant 33 practicals.. here's what I've heard
For Q1.. They have some glass jars,clay and masses..
For Q2.. a clamp, the tennis ball type thingy and a wooden board and maybe even a protector too
 
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So .. tomorrow we all have variant 33 practicals.. here's what I've heard
For Q1.. They have some glass jars,clay and masses..
For Q2.. a clamp, the tennis ball type thingy and a wooden board and maybe even a protector too
 
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So .. tomorrow we all have variant 33 practicals.. here's what I've heard
For Q1.. They have some glass jars,clay and masses..
For Q2.. a clamp, the tennis ball type thingy and a wooden board and maybe even a protector too
 
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So .. tomorrow we all have variant 33 practicals.. here's what I've heard
For Q1.. They have some glass jars,clay and masses..
For Q2.. a clamp, the tennis ball type thingy and a wooden board and maybe even a protector too
 
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So .. tomorrow we all have variant 33 practicals.. here's what I've heard
For Q1.. They have some glass jars,clay and masses..
For Q2.. a clamp, the tennis ball type thingy and a wooden board and maybe even a protector too
hmm not sure for Q1, but maybe Q2 is about ramps then and rolling the ball?
 
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So .. tomorrow we all have variant 33 practicals.. here's what I've heard
For Q1.. They have some glass jars,clay and masses..
For Q2.. a clamp, the tennis ball type thingy and a wooden board and maybe even a protector too
too bad i only saw this after my p3 ended. btw how did you know?
 
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Please can anybody explain me this
In first part we find U
then W
both were positive
but when we have to find total q why did we add them?
isn't U=q+w
shouldn't we subtract it like this q= U-W?Phy prob.PNG
 
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Please can anybody explain me this
In first part we find U
then W
both were positive
but when we have to find total q why did we add them?
isn't U=q+w
shouldn't we subtract it like this q= U-W?View attachment 64706
The equation depends on how you define the terms:

(generally these two are constant)
U is increase in internal energy
q is heat supplied to system

If w is work done on system, then:
U = q + w

If w is work done by the system, then:
U = q - w

Think of how the work changes the internal energy, if the system is doing the work then its using up some internal energy, if work is being done on it then its internal energy increases.
 
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Please can anybody explain me this
In first part we find U
then W
both were positive
but when we have to find total q why did we add them?
isn't U=q+w
shouldn't we subtract it like this q= U-W?View attachment 64706
You should try to send the whole question... I would use the definition of the 1st law of thermodynamics here. Since the gas is expanding and pushing onto the piston, first of all energy is being supplied to the system(positive q) and work is being done BY THE SYSTEM to push the piston(negative w). So the equation we use is U=q-w, isolate q and you get U+q=w(now just insert values and find answer).
 
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Use the equation F=BILsin(theta). The variables are F and theta in this case with B, I and L being constant so just take some values of theta and find the relevant values of F(using F=sin(theta) since the rest of the variables are constant as I stated). Allowing you to draw a graph for example, I'll take Theta=90, 45, 0 and find the values of F(relative to scale).
 
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Use the equation F=BILsin(theta). The variables are F and theta in this case with B, I and L being constant so just take some values of theta and find the relevant values of F(using F=sin(theta) since the rest of the variables are constant as I stated). Allowing you to draw a graph for example, I'll take Theta=90, 45, 0 and find the values of F(relative to scale).
Bro the marking scheme says that A line and a curve should be drawn .. but when I searched the similar solution it said to draw decreasing gradient curve only
 
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