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3 years ago

11

Four 240 Ω light-bulbs are connected in series. What is the total resistance of the circuit? What is their resistance if they ar

e connected in parallel?

1 answer:

irina [24]3 years ago

5 0

Had to submit as image as it wouldn't let me paste these symbols in the answer box

hope this helps:)

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Electrical metallic tubing would be used in an electrical installation because it

maw [93]

Hey There!

Electrical metallic tubing would be used in an electrical installation because it <span>is less expensive.</span>

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4 years ago

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A 3kg book falls from a shelf. If it lands with a speed of 4.8 m/s, from what height did it fall? A. 28.8 m , B. 1.2 m , C. 0.6

Irina-Kira [14]

The correct answer is B. 1.2 m (apex)

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3 years ago

A 2290 kg car traveling at 10.5 m/s collides with a 2780 kg car that is initially at rest at the stoplight. The cars stick toget

avanturin [10]

Answer:

0.41

Explanation:

given,

mass of the car, m = 2290 Kg

initial speed = 10.5 m/s

mass of another car, M = 2780 Kg

distance moved = 2.80 m

coefficient of friction = ?

conservation of energy

m u = (M + m) V

2290 x 10.5 = (2290 + 2780) V

V = 4.74 m/s

using equation of motion

v² = u² + 2 a s

4.74² = 2 x a x 2.8

a = 4.02 m/s²

now using equation

a = μ g

4.02 = μ x 9.8

μ = 0.41

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3 years ago

In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the

MatroZZZ [7]

Answer:

The wavelength is $\lambda_2 = 534 *10^{-9} \ m$

Explanation:

From the question we are told that

The wavelength of the first light is $\lambda _ 1 = 587 \ nm$

The order of the first light that is being considered is $m_1 = 10$

The order of the second light that is being considered is $m_2 = 11$

Generally the distance between the fringes for the first light is mathematically represented as

$y_1 = \frac{ m_1 * \lambda_1 * D}{d}$

Here D is the distance from the screen

and d is the distance of separation of the slit.

For the second light the distance between the fringes is mathematically represented as

$y_2 = \frac{ m_2 * \lambda_2 * D}{d}$

Now given that both of the light are passed through the same double slit

$\frac{y_1}{y_2} = \frac{\frac{m_1 * \lambda_1 * D}{d} }{\frac{m_2 * \lambda_2 * D}{d} } = 1$

=> $\frac{ m_1 * \lambda _1 }{ m_2 * \lambda_2} = 1$

=> $\lambda_2 = \frac{m_1 * \lambda_1}{m_2}$

=> $\lambda_2 = \frac{10 * 587 *10^{-9}}{11}$

=> $\lambda_2 = 534 *10^{-9} \ m$

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3 years ago

Which is not one of the three forms of energy that travels to earth

Tju [1.3M]

Is there options for this??

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3 years ago