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vagabundo [1.1K]
3 years ago
10

When light bulbs A and B are connected in series to a battery, A glows brightly and B glows dimly. You remove bulb B so that the

circuit is just the battery and bulb A and connect bulb B to an identical battery.With the two bulbs connected to identical batteries but in separate circuits, do they glow equally brightly now? a. Bulb A glows brighter b. The two bulbs glow equally brighty. c. Bulb B glows brighter.
Physics
2 answers:
torisob [31]3 years ago
6 0

Answer:

The bulb B glows brighter.

Explanation:

Given that,

A glows brightly and B glows dimly.

According to ohm's law,

Two light bulbs A and B are connected in series to a battery then the current will be same in both bulbs and the resistance is high of bulb A and low in bulb B.

If bulb A connect to a battery and bulb B connect to a same battery separately.

Then bulb B glows brighter because the resistance is high in bulb A so the current will be low.

The resistance is low in bulb B so the current will be high.

Hence, The bulb B glows brighter.

JulsSmile [24]3 years ago
5 0

Answer:

Explanation:

When the two bulbs A and B are connected in series, the current is same in both the bulbs. Power is given by

P = i²R

so, as the bulb A is glowing more than the B, so the resistance of A is more than B.

Now they are connected seperately by two indentical batteries. The potential difference is same in both the bulbs.

Power is given by

P = V²/R

as the resistance of B is smaller than the resistance of A, so the brightness of bulb B is more than A.

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A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ
vitfil [10]

Answer:

<em>0.85c </em>

Explanation:

Rest mass of Kaon M_{0K} = 494 MeV/c²

Rest mass of proton M_{0P}  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy E_{0K} = 494c² MeV

for the proton, rest energy E_{0P} = 938c² MeV

Recall that the rest energy, and the total energy are related by..

E = γE_{0}

which can be written in this case as

E_{K} = γE_{0K} ...... equ 1

where E = total energy of the kaon, and

E_{0} = rest energy of the kaon

γ = relativistic factor = \frac{1}{\sqrt{1 - \beta ^{2} } }

where \beta = \frac{v}{c}

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

E_{K} = E_{0P} ......equ 2

where E_{K} is the total energy of the kaon, and

E_{0P} is the rest energy of the proton.

From E_{K} = E_{0P} = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = \frac{1}{\sqrt{1 - \beta ^{2} } } = 1.89

1.89\sqrt{1 - \beta ^{2} } = 1

squaring both sides, we get

3.57( 1 - \beta^{2}) = 1

3.57 - 3.57\beta^{2} = 1

2.57 = 3.57\beta^{2}

\beta^{2} = 2.57/3.57 = 0.72

\beta = \sqrt{0.72} = 0.85

but, \beta = \frac{v}{c}

v/c = 0.85

v = <em>0.85c </em>

7 0
3 years ago
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
defon

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

Explanation:

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

The angular speed is calculated as follows;

\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

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