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vivado [14]
2 years ago
8

Carla draws two circuit diagrams that connect the same components in different ways, as shown. which statement about the circuit

s is correct? the total resistance in circuit a is greater than that in circuit b. when a bulb in circuit a goes out, the other bulbs keep shining. the bulbs in circuit a shine brighter than those in circuit b. when a bulb in circuit b goes out, the other bulbs go out.
Physics
1 answer:
vagabundo [1.1K]2 years ago
5 0

Carla draws two circuit diagrams that connect the same components in different ways, as shown. The correct statement is "the total resistance in circuit a is greater than that in-circuit b.". Option A. This is further explained below.

<h3>What is a circuit?</h3>

Generally, a circuit is simply defined as a fixed schedule of activities or places for a certain activity, usually including athletics or public performance.

In conclusion, Carla creates two circuit schematics, as illustrated, that link the same components in various ways. "The overall resistance in circuit an is larger than that in b," is the correct statement.

Read more about  resistance

brainly.com/question/11431009

#SPJ4

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A horizontal clothesline is tied between 2 poles, 12 meters apart. When a mass of 4 kilograms is tied to the middle of the cloth
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Answer:

T=26.03 N

Explanation:

Given that

Distance between poles = 12 m

Mass of block m= 4 kg

Sag distance = 5 m

Lets take tension in the clothesline is T.

The component of tension in vertical direction will be T cosθ.

By force balancing

2 T cosθ = 40

here tan\theta =\dfrac{5}{6}

θ=39.80°

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3 years ago
Calculate the energy, wavelength, and frequency of the emitted photon when an electron moves from an energy level of -3.40 eV to
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Answer:

(a) The energy of the photon is 1.632 x 10^{-8} J.

(b) The wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency of the photon is 2.47 x 10^{25} Hz.

Explanation:

Let;

E_{1} = -13.60 ev

E_{2} = -3.40 ev

(a) Energy of the emitted photon can be determined as;

E_{2} - E_{1} = -3.40 - (-13.60)

           = -3.40 + 13.60

           = 10.20 eV

           = 10.20(1.6 x 10^{-9})

E_{2} - E_{1} = 1.632 x 10^{-8} Joules

The energy of the emitted photon is 10.20 eV (or 1.632 x 10^{-8} Joules).

(b) The wavelength, λ, can be determined as;

E = (hc)/ λ

where: E is the energy of the photon, h is the Planck's constant (6.6 x 10^{-34} Js), c is the speed of light (3 x 10^{8} m/s) and λ is the wavelength.

10.20(1.6 x 10^{-9}) = (6.6 x 10^{-34} * 3 x 10^{8})/ λ

λ = \frac{1.98*10^{-25} }{1.632*10^{-8} }

  = 1.213 x 10^{-17}

Wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency can be determined by;

E = hf

where f is the frequency of the photon.

1.632 x 10^{-8}  = 6.6 x 10^{-34} x f

f = \frac{1.632*10^{-8} }{6.6*10^{-34} }

 = 2.47 x 10^{25} Hz

Frequency of the emitted photon is 2.47 x 10^{25} Hz.

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