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murzikaleks [220]

3 years ago

10

If a resistance is added in parallel to a circuit resistance is than?

2 answers:

irakobra [83]3 years ago

8 0

Answer: The total resistance in a parallel circuit is always less than any of the branch resistances. Adding more parallel resistances to the paths causes the total resistance in the circuit to decrease. As you add more and more branches to the circuit the total current will increase because Ohm's Law states that the lower the resistance, the higher the current.

Explanation:

Serggg [28]3 years ago

4 0

Answer:

Explanation: Adding more parallel resistances to the paths causes the total resistance in the circuit to decrease. As you add more and more branches to the circuit the total current will increase because Ohm's Law states that the lower the resistance, the higher the current.

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What is the total energy of a particle with a rest mass of 1 gram moving with half the speed of light? 1 eV = 1.6 x 10^-19 J. An

GREYUIT [131]

Answer:

6.5 x 10^32 eV

Explanation:

mass of particle, mo = 1 g = 0.001 kg

velocity of particle, v = half of velocity of light = c / 2

c = 3 x 10^8 m/s

Energy associated to the particle

E = γ mo c^2

$E=\frac{m_{0}c^}2}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$E=\frac{m_{0}c^}2}{\sqrt{1-\frac{c^{2}}{4c^{2}}}}$

$E=\frac{2m_{0}c^}2}{\sqrt{3}}$

$E=\frac{2\times0.001\times9\times10^{16}}{1.732}$

$E=1.04\times10^{14}J$

Convert Joule into eV

1 eV = 1.6 x 10^-19 J

So, $E=\frac{1.04\times10^{14}}{1.6\times10^{-19}}=6.5\times10^{32}eV$

4 0

2 years ago

An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p

igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Explanation:

Given:

Distance from the wire to the field point $r = 2.83 \times 10^{-2}$ m

Speed of electron $v = 35.5 \%c$

Current $I = 17.7$ A

For finding the acceleration,

First find the magnetic field due to wire,

$B = \frac{\mu _{o}I }{2\pi r }$

Where $\mu_{o} = 4\pi \times 10^{-7}$

$B = \frac{4\pi \times 10^{-7} \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }$

$B = 12.50 \times 10^{-5}$ T

The magnetic force exerted on the electron passing through straight wire,

$F = qvB$

$F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}$

$F = 21.3 \times 10^{-16}$ N

From the newton's second law

$F = ma$

Where $m =$ mass of electron $= 9.1 \times 10^{-31}$ kg

So acceleration is given by,

$a = \frac{F}{m}$

$a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }$

$a = 2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Therefore, the magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

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2 years ago

Explain why some places can use only certain kinds of energy to generate electricity.

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Answer:

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Answer:

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Explanation:

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v = -25 m/s

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(-25 m/s)² = (15 m/s)² + 2 (-10 m/s²) Δy

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Answer:

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Explanation:

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