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zlopas [31]
3 years ago
12

What is the voltage across six 1.5-V batteries when they are connected (a) in series, (b) in parallel, (c) three in parallel wit

h one another and this combination wired in series with the remaining three?
Physics
1 answer:
Katyanochek1 [597]3 years ago
6 0

Answer:

(a) 9.0 V (b) 1.5 V (c) 6.0 V

Explanation:

Here we use the simple concept of Kirchhoff's voltage law that potential in series will be added and potential in parallel will be same always. By this concept we answer all 3 parts of this question.

(a) When six 1.5 V batteries are is series means their potential is added.

So, Voltage across six batteries is

\triangle V = 1.5 V +1.5 V +1.5 V +1.5 V +1.5 V +1.5 V

\triangle V = 9.0 V

So, Net voltage is 9.0 Volts.

(b)When six 1.5 V batteries are in parallel then the net voltage across them will be voltage of either battery.

So, Voltage is

\triangle V = 1.5 V

Net voltage is 1.5 Volts.

(c) Now 3 are in series and 3 are in parallel,

the Net voltage of parallel is 1.5 V , now it behaves like <u>four </u>1.5 V are in series so the net voltage become

\triangle V = 1.5 V +  1.5 V +1.5 V +1.5 V

\triangle V = 6.0 V

Net voltage is 6.0 Volts.

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Answer:

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8 0
3 years ago
A stuntwoman is going to attempt a jump across a canyon that is 77 m wide. The ramp on the far side of the canyon is 25 m lower
liq [111]

initial speed of the stuntman is given as

v = 28 m/s

angle of inclination is given as

\theta = 15 degree

now the components of the velocity is given as

v_x = 28 cos15 = 27.04 m/s

v_y = 28 sin15 = 7.25 m/s

here it is given that the ramp on the far side of the canyon is 25 m lower than the ramp from which she will leave.

So the displacement in vertical direction is given as

\delta y = -25 m

\delta y = v_y * t + \frac{1}{2} at^2

-25 = 7.25 * t - \frac{1}{2}*9.8* t^2

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t = 3.12 s

Now in the above interval of time the horizontal distance moved by it is given by

d_x = v_x * t

d_x = 27.04 * 3.12 = 84.4 m

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5 0
3 years ago
What is the frequency of light with a wavelength of 7.9 x 10^-9 m? ( the speed of light is 3.00 x 10^8)
nlexa [21]
Using the formula v=f times lambada
then v=the speed of light.
and f=what’s we’re looking for
and lambada=the wavelength.

so then you sub what you have (v and lambada) in the formula.
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3 0
3 years ago
A satellite is in orbit 36000km above the surface of the earth. Its angular velocity is 7.27*10^-5 rad/s. What is the velocity o
bagirrra123 [75]

Answer:

v = 3.08 km/s

Explanation:

Given that,

The angular velocity of the satellite = \omega=7.27\times 10^{-5} rad/s

A satellite is in orbit 36000km above the surface of the earth.

The radius of the earth is 6400 km

Let v is the velocity of the satellite. It can be calculated as :

v=r\omega\\\\v=(36000\times 10^3+6400\times 10^3)\times 7.27\times 10^{-5}\\\\v=3082.48\ m/s\\\\v=3.08\ km/s

So, the velocity of the satellite is 3.08 km/s.

6 0
3 years ago
Anybody got any answers???
dexar [7]

Answer:

b? to a?

Explanation:

6 0
3 years ago
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