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3 years ago

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Two identical batteries, each with an emf of 18 V and an internal resistance of 1 :, are wired in parallel by connecting their p

ositive terminals together and connecting their negative terminals together. The combination is then wired across a 4-: resistor. The current in the 4-: resistor is:

1 answer:

bulgar [2K]3 years ago

8 0

Answer:

I = 4 A

Explanation:

When the batteries are connected in parallel, the voltage is the same as the voltage of each battery, but the internal resistance is less, in general we can assume that the internal resistance is also in parallel.

1 / $r_{eq}$ = 1 / r₁ + 1 / r₂ = 2 / r = 2/1

r_{eq} = 0.5 ohm

The current in the resistor is

V = I R + I r_{eq}

I = V / R + r_{eq}

I = 18 / (4 +0.5)

I = 4 A

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Solution:
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The wavelength of the standing wave is instead twice the length of the string:
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A proton orbits a long charged wire, making 1.80 ×106 revolutions per second. The radius of the orbit is 1.20 cm What is the wir

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Answer:

linear charge density = -9.495 × $10^{-34}$ C/m

Explanation:

given data

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radius = 1.20 cm

solution

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so

- q × E = m × w² × r ..................1

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and w = $\frac{2*\pi}{T}$

w = $\frac{d\theta }{dt}$

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He produced the first orderly arrangement of known elements, he used patterns to predict undiscovered elements

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Anastasy [175]

-- We're going to be talking about the satellite's speed.
"Velocity" would include its direction at any instant, and
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-- The mass of the satellite makes no difference.

Since the planet's radius is 3.95 x 10⁵m and the satellite is
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=    0.04 x 10⁵ m/sec

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3 years ago

1. An express train, traveling at 36 m/s, is accidentally sidetracked onto a local train track. The express engineer spots a loc

Colt1911 [192]

Answer:

(i) 12 seconds

(ii) 216 meters from the initial position

(iii) 132 meters from the initial position

(iv) No

Explanation:

Speed of express train =36 m/s

Speed of local train =11 m/s

The initial distance between the local train and passenger train =100 m.

Due to the application of breaks, the express train slows at the rare of $3.0 m/s^2.$

So, the acceleration of the express train, $a=-3 m/s^2$ .

(i) Let t be the time the express train takes to stop.

From the equation of motion,

v=u+at

where, v: final velocity, u: initial velocity, a: constant acceleration, t: time taken to change the speed from u to v.

In this case, v=0, u=36 m/s, $a=-3 m/s^2$

So, 0=36+(-3)t

$\Rightarrow t= 36/3=12$ seconds.

(ii) To compute the distance traveled, s, till the express train stops, using

$v^2=u^2+2as$

$\Rightarrow 0^2=36^2+2(-3)s$

$\Rightarrow s=\frac{36\times36}{6}$

$\Rightarrow s=216$ meters.

(iii) The local train is moving at a speed of 11 m/s

So, in 12 seconds, the distance, d, traveled by the local train

d= 11x12=132 meters [as distance= speed x time]

(iv) Let 0 be the reference position which is the initial position of the express train.

So, at the initial time, the position of the local train is at 100m.

After 12 seconds:

The position of the express train is at 216 m [using part (ii)]

and the position of the local train is at 100+132=232m [using part (iii)].

So, the local train is still ahead of the express train, hence the trains didn't collide.

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3 years ago