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Bezzdna [24]
3 years ago
7

11 e) Can a conductor be given limitless charge? Obtain the equivalent resistance of several resistors if (a) they are in series

, (b) they are in parallel​
Physics
1 answer:
horsena [70]3 years ago
6 0

Answer:

(e) no

(a) Rs = R' + R'' + R'''

(b) 1/Rp = 1/R' + 1/R'' + 1/R'''

Explanation:

11 e)

Practically it is not possible to give limitless charge to a conductor. It depends to the number of valence electrons.

(a) When the three resistances R'. R'' and R''' is in series combination.

Let they are connected to the voltage V and the current in each resistance is I.

According to Ohm's law

Voltage across R', V' = I R'

Voltage across R'', V'' = I R''

Voltage across R''', V''' = I R'''

So, let the equivalent resistance is Rs.

I Rs = I R' + I R'' + I R'''

Rs = R' + R'' + R'''

(b)

When the three resistances R'. R'' and R''' is in parallel combination.

Let they are connected to the voltage V and the current in each resistance is I', I''. I'''.

Current in R', I' = V/R'

Current in R'', I'' = V/R''

Current in R''', I''' = V/R'''

The equivalent resistance is Rp.

V/Rp = V/R' + V/ R'' + V/R'''

1/Rp = 1/R' + 1/R'' + 1/R'''

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She must be launched with a speed of 74.2 m/s.

Explanation:

Hi there!

The equations of the horizontal component of the position vector and the vertical component of the velocity vector are the following:

x = v0 · t · cos θ

vy = v0 · sin θ + g · t

x = horizontal distance traveled at time t.

v0 = initial velocity.

t = time.

θ = launching angle.

vy = vertical component of the velocity vector at time t.

g = acceleration due to gravity (-9.8 m/s²).

To just cross the 520-m gap, the maximum height of the flight must be reached halfway of the gap at 260 m horizontally (see attached figure).

When she is at the maximum height, her vertical velocity is zero. So, when x = 260 m, vy = 0. Using both equations we can solve the system for v0:

x = v0 · t · cos θ

Solving for v0:

v0 = x/ (t · cos θ)

Replacing v0 in the second equation:

vy = v0 · sin θ + g · t

0 = x/(t·cos(56°)) · sin(56°) + g · t

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9.8 m/s² · t = 260 m · tan (56°) / t

t² = 260 m · tan (56°) / 9.8 m/s²

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v0 = x/ (t · cos θ)

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v0 = 74.2 m/s

She must be launched with a speed of 74.2 m/s.

5 0
3 years ago
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