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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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How long must a 0.70-mm-diameter aluminum wire be to have a 0.40 a current when connected to the terminals of a 1.5 v flashlight

Natalka [10]

By using Ohm's law, we can find what should be the resistance of the wire, R:
$R= \frac{V}{I}= \frac{1.5 V}{0.40 A} =3.75 \Omega$

Then, let's find the cross-sectional area of the wire. Its radius is half the diameter,
$r=35 mm=0.35 \cdot 10^{-3} m$
So the area is
$A=\pi r^2 = \pi (0.35 \cdot 10^{-3} m)^2=3.85 \cdot 10^{-7} m^2$

And by using the resistivity of the Aluminum, $\rho=2.65 \cdot 10^{-8} \Omega m$ , we can use the relationship between resistance R and resistivity:
$R= \frac{\rho L}{A}$
to find L, the length of the wire:
$L= \frac{RA}{\rho}= \frac{(3.75 \Omega)(3,85 \cdot 10^{-7} m^2)}{2.65 \cdot 10^{-8} \Omega m}=54.48 m$

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3 years ago

How will decreasing the amplitude of the sound waves from a television affect its intensity?

Colt1911 [192]

A. The sound will decrease in volume

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3 years ago

A violin string is 45.0 cm long and has a mass of 0.242 g. When tightened on the neck of the violin, the distance between the pi

stiks02 [169]

Answer:

The tension is 75.22 Newtons

Explanation:

The velocity of a wave on a rope is:

$v=\sqrt{\frac{TL}{M}}$ (1)

With T the tension, L the length of the string and M its mass.

Another more general expression for the velocity of a wave is the product of the wavelength (λ) and the frequency (f) of the wave:

$v= \lambda f$ (2)

We can equate expression (1) and (2):

$\sqrt{\frac{TL}{M}}$ = $\lambda f$

Solving for T

$T= \frac{M(\lambda f)^2}{L}$ (3)

For this expression we already know M, f, and L. And indirectly we already know λ too. On a string fixed at its extremes we have standing waves ant the equation of the wavelength in function the number of the harmonic $N_{harmonic}$ is:

$\lambda_{harmonic}=\frac{2l}{N_{harmonic}}$

It's is important to note that in our case L the length of the string is different from l the distance between the pin and fret to produce a Concert A, so for the first harmonic:

$\lambda_{1}=\frac{2(0.425m)}{1}=0.85 m$