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Physics: Post your doubts here!

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Which operation involves the greatest mean power?
A a car moving against a resistive force of 0.4kN at a constant speed of 20ms–1
B a crane lifting a weight of 3kN at a speed of 2ms–1
C a crane lifting a weight of 5kN at a speed of 1ms–1
D a weight being pulled across a horizontal surface at a speed of 6 m s–1 against a frictional
force of 1.5kN
 
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Which operation involves the greatest mean power?
A a car moving against a resistive force of 0.4kN at a constant speed of 20ms–1
B a crane lifting a weight of 3kN at a speed of 2ms–1
C a crane lifting a weight of 5kN at a speed of 1ms–1
D a weight being pulled across a horizontal surface at a speed of 6 m s–1 against a frictional
force of 1.5kN
Power is same as force times velocity.
A) 8Kpa
B) 6Kpa
C) 5Kpa
D) 9Kpa = Answer.
An electric power cable consists of six copper wires c surrounding a steel core s.

1.0 km of one of the copper wires has a resistance of 10 Ω and 1.0 km of the steel core has a
resistance of 100Ω.
What is the approximate resistance of a 1.0km length of the power cable?

Question 32, October 2008.
32)
All the wires are connected in parallel.
So 1/R = (1/10)X6 + 1/100
Therefore, 1/R = 61/100
R = 100/61 = 1.6 ohms.
 
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A particle has a charge of 4.8 × 10–19 C. The particle remains at rest between a pair of horizontal,
parallel plates having a separation of 15 mm. The potential difference between the plates is
660V.
What is the weight of the particle?
A 2.1 × 10−14N
B 2.1 × 10−15N
C 2.1 × 10−17N
D 1.1 × 10−23N

Question 30, October 2008
 
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A particle has a charge of 4.8 × 10–19 C. The particle remains at rest between a pair of horizontal,
parallel plates having a separation of 15 mm. The potential difference between the plates is
660V.
What is the weight of the particle?
A 2.1 × 10−14N
B 2.1 × 10−15N
C 2.1 × 10−17N
D 1.1 × 10−23N

Question 30, October 2008
This question is based on derivation of equations.
We know that F = mg = W
We also know that W = EQ Where E = v/d
So we got an expression related to the question we have here that is W = v/d * Q
Substituting the values you have in question final answer will be on your calculator display as "2.1 x 10^-14" that is A.
 
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Sorry, but I`m still not getting something. Maybe I'm having a confusion but I don't know what it is.

I'm thinking this way.

F = ke. e = F/k. For smaller k, shouldn't e be greater. So, why is C wrong.

Also, Young modulus E = stress / strain = (F/A) / (e/L) = FL / Ae. So, e = FL / AE. If E is smaller, e would be bigger. So, why D wrong?

Also, can you show me mathematically (using the formulae) how B is the answer. And what am I having wrong. The above logic shows that both C and D are good?
For smaller k, e WILL be greater but if you look at the question it says 'Which change will not have the desired effect?'
and e is extension which in this case is the compression. you haven't seen the not in the question. the e shouldn't be greater. If it is, it does not support the question they asked. both C and D are bad in that case.
 
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Hello.
For 1a)
You need a textbook and a fresh mind to understand whats written in your book. :)
4c)
Capacitor linearise the voltage signal. the signal given in the ques is already linearised once. if it is linearised again with the second capacitor the signal's ripple is further reduced but the time period and freq remains the same
 
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Zepudee
5B : This I saw somewhere, IDK but I had doubt too. Anyways here he showed this :¬
e-jpg.7238

7A : At t=0, the mass of iron will be zero grams. After 2.6 hours, mass of iron will be 0.7 micrograms. At t=5.2 h, mass of iron will be (0.7+0.35) micrograms. When m grams of manganese decays, it produces n grams of iron because both of them have same Ar.
 
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Hello.
For 1a)
You need a textbook and a fresh mind to understand whats written in your book. :)
4c)
Capacitor linearise the voltage signal. the signal given in the ques is already linearised once. if it is linearised again with the second capacitor the signal's ripple is further reduced but the time period and freq remains the same
for the diagram q4(c) im not sure if my diagram is right or wrong, just needed some clarification :D thanks thought blocker
 
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5B : This I saw somewhere, IDK but I had doubt too. Anyways here he showed this :¬
e-jpg.7238

7A : At t=0, the mass of iron will be zero grams. After 2.6 hours, mass of iron will be 0.7 micrograms. At t=5.2 h, mass of iron will be (0.7+0.35) micrograms. When m grams of manganese decays, it produces n grams of iron because both of them have same Ar.

Thanks ! :3
 
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