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FinnZ [79.3K]
3 years ago
6

4. Which of the following usually occurs with a short circuit? A. Most of the current will flow through one part of the circuit.

B. All parts of the circuit will begin to carry higher amounts of current than normal. C. All parts of the circuit will begin to carry lower amounts of current than normal. D. The material within a fuse may begin to melt when the current ceases to exist.
Physics
2 answers:
Svetllana [295]3 years ago
6 0
<span>A. Most of the current will flow through one part of the circuit. </span>
ira [324]3 years ago
4 0
A is the correct answer :)
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A straight, nonconducting plastic wire 9.50 cm long carries a charge density of 130 nC/m distributed uniformly along its length.
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Answer:

A) E = 3.70*10^{4} N/C

B)  E = 2.281*10^3 N/C

Explanation:

given data:

charge density \lambda = 130*10^{-9} C/m

length of wire = 9.50 cm

a) at x  = 4.5 m above midpoint, electric field is calculated as

E = \frac{1}{ 2\pi \epsilon} * \frac{ \lambda}{x\sqrt{(x^2/a^2)+1}}

x = 4.5 cm

midpoint a = 4.5 cm = 0.0475 m

E =2{\frac{1}{ 8.99*10^9} * \frac{130*10^{-9} }{0.045\sqrt{(4.5^2/4.75^2)+1}}

E = 3.70*10^{4} N/C

B) when wire is in circle form

Q = \lambda * L

= 130*10^{-9} *9.5*10^{-2}

   = 1.235*10^{-8} C

Radius of circle

r = \frac{L}{2\pi}

r = \frac{9.5*10^{-2}}{2\pi}

r = 1.511*10^{-2} m

E = \frac{1}{ 2\pi \epsilon} * \frac{Qx}{(x^2+r^2)^{3/2}}

E =8.99*10^{9} * \frac{1.23*10^{-8}*4.5*10^{-2}}{((4.5*10^{-2})^2+(1.511*10^{-2})^2)^{3/2}}

E = 2.281*10^3 N/C

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Answer:

15.24 m/s in the downward direction

Explanation:

Given that the initial upward velocity of the ball is 24 m/s.

Assuming that the upward direction is positive.

As gravitational force acts in the downward direction and the direction of acceleration is the same as the direction of force, so the acceleration due to gravity will be negative.

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