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Virty [35]
3 years ago
5

14. Three identical light bulbs are connected in series, then are disconnected and arranged in parallel. For each of the scenari

os below indicate what changes (if any) take place (1 pt. ea.):
a. Total resistance of the circuit
b. Total current of the circuit
c. Power dissipated by the circuit
Physics
1 answer:
Inessa [10]3 years ago
8 0
In order to make things easier to describe and explain, let's call
the resistance of each bulb 'R', and the battery voltage 'V'.

a).  In series, the total resistance is 3R.
In parallel, the total resistance is R/3.
Changing from series to parallel, the total resistance of the circuit
decreases to 1/9 of its original value.

b).  In series, the total current is  V / (3R) .
In parallel, the total current is  3V / R .
Changing from series to parallel, the total current in the circuit
increases to 9 times its original value.

c).  In series, the power dissipated by the circuit is 

                                   (V) · V/3R  =  V² / 3R .

In parallel, the power dissipated by the circuit is

                                   (V) · 3V/R  =  3V² / R .

Changing from series to parallel, the power dissipated by
the circuit (also the power delivered by the battery) increases
to 9 times its original value.

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A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical ener
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Complete question:

A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical energy of the spring–load system is 2.00 J. Find

(a) the force constant of the spring and (b) the amplitude of the motion.

Answer:

(a) the force constant of the spring = 47 N/m

(b) the amplitude of the motion = 0.292 m

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Given;

mass of the spring, m = 200g = 0.2 kg

period of oscillation, T = 0.410 s

total mechanical energy of the spring, E = 2 J

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\omega = \frac{2\pi}{T} \\\\\omega = \frac{2\pi}{0.41} \\\\\omega = 15.33 \ rad/s

(a) the force constant of the spring

\omega = \sqrt{\frac{k}{m} } \\\\\omega^2 = \frac{k}{m} \\\\k = m \omega^2\\\\k = (0.2)(15.33)^2\\\\k = 47 \ N/m

(b) the amplitude of the motion

E = ¹/₂kA²

2E = kA²

A² = 2E/k

A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2\times 2}{47} }\\\\A = 0.292 \ m

7 0
3 years ago
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Hence the window is 37.33m above the ground

7 0
3 years ago
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