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Svetradugi [14.3K]
3 years ago
7

A portable television using a 12 volt, 3 Ah rechargeable battery can operate for a period of about 6 hours. What is the average

current drawn in amperes during this period
Physics
1 answer:
bezimeni [28]3 years ago
7 0

Answer:

0.5 A

Explanation:

Current: This is defined as the rate flow of electric charge round a circuit,

The S.I unit of current is Ampere (A). Mathematically,

Current (I) = C/t

I = C/t................  Equation 1

Where  C = capacity of the battery, t = time

Given: C= 3 Ah, t = 6 hours.

Substitute into equation 1

I = 3/6

I = 1/2

I = 0.5 A.

Hence the average current drawn = 0.5 A

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when a charge of 1 C has an electric PE of 1 J, it has an electric potential of 1 V. When a charge of 2 C has an electric PE of
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What determines the number of possible sublevels?
Talja [164]
<h2>Hello!</h2>

The answer is A.   the principal energy level

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The principal energy level or principal quantum number n, tell us about the position of an determined electron in the energy levels relative to the greater average distance of an electron from the nucleus, the larger the value of n and the higher its energy.

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For example, the number 1, has one orbital which is contained in a energy sublevel, its called s orbital, and it's just an orbital with 2 electrons.

Have a nice day!

7 0
3 years ago
Read 2 more answers
A light beam is traveling through an unknown substance. When it exits that substance and enters into air, the angle of reflectio
bogdanovich [222]

Answer:

0.79

Explanation:

Using Snell's law, we have that:

n(1) * sin θ1 = n(2) * sinθ2

Where n(1) = refractive index of air = 1.0003

θ1 = angle of incidence

n(2) = refractive index of second substance

θ2 = angle of refraction

The angle of reflection through the unknown substance is the same as the angle of incidence of air. This means that θ1 = 32°

=> 1.0003 * sin32 = n(2) * sin42

n(2) = (1.0003 * sin32) / sin42

n(2) = 0.79

3 0
4 years ago
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At t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is 6.00 i + 4.00 j m/s2 . It moves at constan
Jet001 [13]

The particle moves with constant speed in a circular path, so its acceleration vector always points toward the circle's center.

At time t_1, the acceleration vector has direction \theta_1 such that

\tan\theta_1=\dfrac{4.00}{6.00}\implies\theta_1=33.7^\circ

which indicates the particle is situated at a point on the lower left half of the circle, while at time t_2 the acceleration has direction \theta_2 such that

\tan\theta_2=\dfrac{-6.00}{4.00}\implies\theta_2=-56.3^\circ

which indicates the particle lies on the upper left half of the circle.

Notice that \theta_1-\theta_2=90^\circ. That is, the measure of the major arc between the particle's positions at t_1 and t_2 is 270 degrees, which means that t_2-t_1 is the time it takes for the particle to traverse 3/4 of the circular path, or 3/4 its period.

Recall that

\|\vec a_{\rm rad}\|=\dfrac{4\pi^2R}{T^2}

where R is the radius of the circle and T is the period. We have

t_2-t_1=(5.00-2.00)\,\mathrm s=3.00\,\mathrm s\implies T=\dfrac{3.00\,\rm s}{\frac34}=4.00\,\mathrm s

and the magnitude of the particle's acceleration toward the center of the circle is

\|\vec a_{\rm rad}\|=\sqrt{\left(6.00\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(4.00\dfrac{\rm m}{\mathrm s^2}\right)^2}=7.21\dfrac{\rm m}{\mathrm s^2}

So we find that the path has a radius R of

7.21\dfrac{\rm m}{\mathrm s^2}=\dfrac{4\pi^2R}{(4.00\,\mathrm s)^2}\implies\boxed{R=2.92\,\mathrm m}

8 0
3 years ago
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