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3 years ago

14

A remote-controlled car runs on a 9V battery. If a 1.5A current is running through the circuit, what resistance is the car prov

iding?

1 answer:

Aleonysh [2.5K]3 years ago

4 0

Ohm's law:

V = I * R, so R = V/I = 9/1.5 = 6 Ohm

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An 8.0-ohm resistor and a 6.0-ohm resistor are connected in series with a battery. The potential difference across the 6.0-ohm r

ipn [44]

Answer:

a) 14 Ω

b) 2.0 A

c) 28 V

Explanation:

a) The total resistance of resistors in series is the sum:

R = R₁ + R₂

R = 8.0 Ω + 6.0 Ω

R = 14 Ω

b) The current in the 6.0 Ω resistor can be found with Ohm's law:

V = IR

12 V = I (6.0 Ω)

I = 2.0 A

c) Since the resistors are in series, they have the same current. So the total voltage is:

V = IR

V = (2.0 A) (14 Ω)

V = 28 V

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2 years ago

For a two-level system, the weight of a given energy distribution can be expressed in terms of the number of systems, N, and the

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Answer:

W = N!/(n0! * n1!)

Explanation:

Let n0 = number of particles in the lowest energy state

n1 = number of particles in the excited energy state.

Using this, we can say that N = n0 + n1

From this we can then express the weight, W of the close system by finding the factorials of each particles

W = N!/(n0! * n1!)

Hence, the weight W is expressed as W = N!/(n0! * n1!)

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2 years ago

Which circuit would have the most electrical power?

aliya0001 [1]

the answer is c because it has the most volts

a

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1 year ago

A) A spaceship passes you at a speed of 0.800c. You measure its length to be 31.2 m .How long would it be when at rest?

rosijanka [135]

Answer:

a

$l_o =52 \ m$

b

$l = 37.13 \ LY$

Explanation:

From the question we are told that

The speed of the spaceship is $v = 0.800c$

Here c is the speed of light with value $c = 3.0*10^{8} \ m/s$

The length is $l = 31.2 \ m$

The distance of the star for earth is $d = 145 \ light \ years$

The speed is $v_s = 2.90 *10^{8}$

Generally the from the length contraction equation we have that

$l = l_o \sqrt{1 -[\frac{v}{c } ]}$

Now the when at rest the length is $l_o$

So

$l_o =\frac{l}{\sqrt{ 1 - \frac{v^2}{c^2 } } }$

$l_o =\frac{ 31.2 }{ \sqrt{1 - \frac{(0.800c ) ^2}{c^2} } }$

$l_o=52 \ m$

Considering b

Applying above equation

$l =l_o \sqrt{1 - [\frac{v}{c } ]}$

Here $l_o =145 \ LY(light \ years )$

So

$l=145 * \sqrt{1 - \frac{v_s^2}{c^2 } }$

$l =145 * \sqrt{ 1 - \frac{2.9 *10^{8}}{3.0*10^{8}} }$

$l = 37.13 \ LY$

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2 years ago

Which statement is TRUE? A) In a series circuit, the resistors do not share the current. B) In a series circuit, as light bulbs

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The answer is B, actually did the research

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2 years ago

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