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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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Answer:

Potential energy =mass* acceleration due to gravity * height

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hence potential energy of the puppy= weight * height

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3 years ago
When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin
dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) 1.21\times 10^{15}Hz

Explanation:

We have given Wavelength of the light  \lambda = 240 nm

According to plank's rule ,energy of light

E = h\nu = \frac{hc}{}\lambda

E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

E = KE_{max}+\Phi _{0}, here \Phi _0 is work function.

\Phi _{0}=E - KE_{max}= 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}

\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }

\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm

Part (C) In this part we have to find the cutoff frequency

\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz

5 0
3 years ago
What is the answer to this?
jeka57 [31]

Answer:

I think it is the forth one

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Students want to investigate the inverse relationship between the pressure and temperature of an ideal gas as predicted by the i
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Answer:

Option B, Fix the piston in place so the volume of the pas remains constant

Explanation:

As we know

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