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3 years ago

1. (a) The battery on your car has a rating stated in ampere-minutes which permits you to

Physics

1 answer:

Alborosie3 years ago

6 0

Answer:

Energy is stored by a 50 ampere-minute 12

volt battery is approximately = 36,000 J = 36 kJ

Explanation:

Power in electrical circuits is given as

Power = IV

But power generally is defined as energy expended per unit time

Power = (Energy/time)

Energy = Power × Time

Energy = IV × Time

Energy = (I.t × V)

I.t = 50 Ampere-minute = 50 × 60 = 3000 Ampere-seconds

V = 12 V

Energy = 3,000 × 12 = 36,000 J = 36 kJ

Hope this Helps!!!

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while working out a man performed 2525j of work in 19seconds . what was his power A:132.9w. B:241.5w C 47.975w. D100.5w

Olenka [21]

A: 132.9w because 2525\19 is how much energy transferred per second which is also known as the power

7 0

4 years ago

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Calculate the RMS voltage of the following waveforms with 10 V peak-to-peak:

Deffense [45]

Answer:

a) $T=0.01s$

b) $T=0.001s$

c) $T=0.00001s$

Explanation:

From the question we are told that:

Given Frequencies

a. 100 Hz,

b. 1 kHz,

c. 100 kHz.

Generally the equation for Waveform Period is mathematically given by

$T=\frac{1}{f}$

Therefore

For

$T=100 Hz$

$T=\frac{1}{100}$

$T=0.01s$

For

$F=1kHz$

$T=\frac{1}{1000}$

$T=0.001s$

For

$F=100kHz$

$T=\frac{1}{100*100}$

$T=0.00001s$

6 0

3 years ago

What factors affect potential energy

jok3333 [9.3K]

Mass ,gravity and height

6 0

3 years ago

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A uniform-density 7 kg disk of radius 0.21 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is

GREYUIT [131]

Answer:

$\omega = 22.67 rad/s$

Explanation:

Here we can use energy conservation

As per energy conservation conditions we know that work done by external source is converted into kinetic energy of the disc

Now we have

$W = \frac{1}{2}I\omega^2$

now we know that work done is product of force and displacement

so here we have

$W = F.d$

$W = (44 N)(0.9 m) = 39.6 J$

now for moment of inertia of the disc we will have

$I = \frac{1}{2}mR^2$

$I = \frac{1}{2}(7 kg)(0.21^2)$

$I = 0.154 kg m^2$

now from above equation we will have

$39.6 = \frac{1}{2}(0.154)\omega^2$

$\omega = 22.67 rad/s$

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3 years ago

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A 0.12 kg body undergoes simple harmonic motion of amplitude 8.5 cm and period 0.20 s. (a) What is themagnitude of the maximum f

Neporo4naja [7]

Answer:

a)F=698.83 N

b)K=8221.56 N/m

Explanation:

Given that

mass ,m = 0.12 kg

Amplitude ,A= 8.5 cm

time period ,T = 0.2 s

We know that

$T=\dfrac{2\pi}{\omega}$

${\omega}=\dfrac{2\pi }{0.2}\ rad/s$

${\omega}=31.41\ rad/s$

We know that

${\omega}^2=m\ K$

K=Spring constant

$K=\dfrac{\omega^2}{m}$

$K=\dfrac{31.41^2}{0.12}\ N/m$

K=8221.56 N/m

The maximum force F

F= K A

F= 8221.56 x 0.085 N

F=698.83 N

a)F=698.83 N

b)K=8221.56 N/m

3 0

4 years ago