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3 years ago

13

A battery operating a large flashlight produces a current of 2 amperes in a lightbulb that has a resistance of 2.4 ohms. The vol

tage of the battery is:

1 answer:

NeX [460]3 years ago

3 0

Answer:

Voltage of the battery = 4.8 volt

Explanation:

It is given that the flashlight produces a current of 2 ampere

So current i = 2A

And the resistance R = 2.4 ohm

We have to find the voltage of the battery

According to ohm's law voltage of the battery is given by = current × resistance

So voltage = 2×2.4 =4.8 volt

The voltage of the battery depends on the current and resistance.

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The force on the wire is directed?

USPshnik [31]

Answer: A

Explanation:

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1 year ago

A 10 kg block slides on a frictionless surface at 10 m/s . It hits a rough patch 3 meters long with a coefficient of kinetic fri

Valentin [98]

12 MPH

I DIDNT do the math my brother did hes in colledge so good lucks guys

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2 years ago

A child with a mass of 20 kg sits at a distance of 2 m from the pivot point of a seesaw. where should a 16-kg child sit to balan

Maksim231197 [3]

In the above problem, we need to find mass of the second child, so that the Center of Mass remains at the origin( pivot).

CM= m1r1+m2r2/m1+m2
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4 0

2 years ago

An all-electric car (not a hybrid) is designed to run from a bank of 12.0 V batteries with total energy storage of 2.30 ✕ 107 J.

AnnyKZ [126]

Answer:

a) $I=733.33\ A$

b) $d=52272.7273\ m$

c) $d'=51948.0519\ m$

Explanation:

Given:

voltage of the battery, $V=12\ V$

energy storage capacity of the battery, $E=2.3\times 10^7\ J$

speed of the car, $v=20\ m.s^{-1}$

a)

power drawn by the car, $P=8.8\ kW$

<u>Now the Current delivered to the motor:</u>

we the relation between the power and electrical current,

$P=V.I$

$8800=12\times I$

$I=733.33\ A$

b)

<u>Distance travelled before battery is out of juice:</u>

we first find the time before the battery runs out,

$t=\frac{E}{P}$

$t=\frac{2.3\times 10^7}{8800}$

$t=2613.636\ s$

Now the distance:

$d=v.t$

$d=20\times 2613.636$

$d=52272.7273\ m$

c)

When the head light of 55 W power is kept on while moving then the power consumption of the car is:

$P'=P+55$

$P'=8800+55$

$P'=8855\ W$

<u>Now the time of operation of the car:</u>

$t'=\frac{E}{P'}$

$t'=\frac{2.3\times 10^7}{8855}$

$t'=2597.4026\ s$

<u>Now the distance travelled:</u>

$d'=v.t'$

$d'=20\times 2597.4025$

$d'=51948.0519\ m$

5 0

2 years ago

Two equal charges are 2m2m apart. If the charges and the distance are divided by two, how is the force between the charges affec

lara31 [8.8K]

Answer:

The force between the charges are not affected.

Explanation:

Given;

distance between two equal charges, R = 2m

The force between the charges is given by;

$F = \frac{kq^2}{R^2}\\\\F_1 = \frac{kq_1^2}{R_1^2}\\\\When\ the \ charges \ and \ the \ distance \ are \ divided \ by \ two \ (q_2 = \frac{q_1}{2}, \ R_2 = \frac{R_1}{2} )\\\\ F_2 = \frac{kq_2^2}{R_2^2}\\\\F_2 = \frac{k(q_1/2)^2}{(R_1/2)^2}\\\\F_2= \frac{4k*q_1^2}{4*R_1^2}\\\\F_2 = \frac{k*q_1^2}{R_1^2}\\\\F_2 = F_1$

Therefore, the force between the charges are not affected.

7 0

2 years ago