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mestny [16]
3 years ago
13

Two conductors having net charges of +16.0 HC and -16.0 HC have a potential difference of 16.0 V between them. (a) Determine the

capacitance of the system (b) What is the potential difference between the two conductors if the charges on each are increased to +256.0 C and-256.0 uC
Physics
1 answer:
Alex_Xolod [135]3 years ago
5 0

Answer:

(a) Capacitance is 1\mu C

(b) Potential difference is 256 V

Solution:

As per the question:

Charges on the two conductors, q is  + 16\muC and - 16\muC

The potential difference between the charges, V = 16.0 V

Now,

(a) The capacitance of the system is given by:

q = CV

C = \frac{q}{V} = \frac{16\times 10^{- 6}}{16} = 1\mu F

(Since, 1\mu C = 10^{- 6} C)

Now,

(b) When the charge, 'q' is increased to + 256\mu C and - 256\mu C

q = CV

V = \frac{q}{C} = \frac{256\times 10^{- 6}}{1\times 10^{- 6}} = 256\ V

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Angelina_Jolie [31]

Answer:

a) 3.43 m/s

Explanation:

Due to the law of conservation of momentum, the total momentum of the bullet - rifle system must be conserved.

The total momentum before the bullet is shot is zero, because they are both at rest, so:

p_i = 0

Instead the total momentum of the system after the shot is:

p_f = mv+MV

where:

m = 0.006 kg is the mass of the bullet

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v = 800 m/s is the velocity of the bullet

V is the recoil velocity of the rifle

The total momentum is conserved, therefore we can write:

p_i = p_f

Which means:

0=mv+MV

Solving for V, we can find the recoil velocity of the rifle:

V=-\frac{mv}{M}=-\frac{(0.006)(800)}{1.4}=-3.43 m/s

where the negative sign indicates that the velocity is opposite to direction of the bullet: so the recoil speed is

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5 0
3 years ago
The mass of a radioactive substance follows a continuous exponential decay model, with a decay rate parameter of 1% per day. A s
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Answer:

2,38kg

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Given that a sample decay 1% per day, that means that after first day you have 99% of mass.

m_{(1)} =m_{0} e^{-k(1)}, but m_{(1)}=\frac{99m_{0} }{100}, so we have \frac{99m_{0} }{100}=m_{0}e^{-k}, then k=-ln(\frac{99}{100})=0.01

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3 years ago
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Answer:

1)

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7 0
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