Work is equal to the applied multiplied by the moved.

Humberto builds two circuits using identical components.

vlada-n [284]

Humberto should expect to see that all bulbs in circuit 1 will shine more dimly than the original bulbs, while all bulbs in circuit 2 will have the same brightness as the original bulbs.

B) All bulbs in circuit 1 will shine more dimly than the original bulbs, while all bulbs in circuit 2 will have the same brightness as the original bulbs.

6 0

4 years ago

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A 240 kg motorcycle moves with a velocity of 8 m/s. What is its kinetic energy?

QveST [7]

1/2 x 240 x 64 = 120 X 64 = 7680 J

6 0

3 years ago

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A student drops a water balloon from tall building trying to hit her roommate on the ground below (and misses). After dropping t

Fudgin [204]

Answer:

H = 70.4 m

Explanation:

Here first of all the balloon will drop to the the floor under free fall motion and then the sound of splash will come up at constant speed

so we will say that if the total height is H

so the time to fall is given as

$H = \frac{1}{2}gt^2$

now we have

$t_1 = \sqrt{\frac{2H}{g}}$

also we know that the time to come the sound upwards is

$t_2 = \frac{H}{v}$

so we will have

$t_1 + t_2 = T$

$\sqrt{\frac{2H}{g}} + \frac{H}{v} = T$

now plug in all data

$\sqrt{\frac{2H}{9.81}} + \frac{H}{331} = 4$

by solving above equation we have

H = 70.4 m

4 0

4 years ago

What is the transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup

Fantom [35]

Answer:

3) Transmitted intensity of light if unpolarized light passes through a single polarizing filter = 40 W/m²

- Transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described = 7.5 W/m²

Explanation:

Complete Question

3) What is the transmitted intensity of light if unpolarized light passes through a single polarizing filter and the initial intensity is 80 W/m²?

- What is the transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described in Question 3 (the setup)? Show all work in your answer.

The image of this setup attached to this question as obtained from online is attached to this solution.

Solution

3) When unpolarized light passes through a single polarizer, the intensity of the light is cut in half.

Hence, if the initial intensity of unpolarized light is I₀ = 80 W/m²

The intensity of the light rays thay pass through the first single polarizer = I₁ = (I₀/2) = (80/2) = 40 W/m²

- According to Malus' law, the intensity of transmitted light through a polarizer is related to the intensity of the incident light and the angle at which the polarizer is placed with respect to the major axis of the polarizer before the current polarizer of concern.

I₂ = I₁ cos² θ

where

I₂ = intensity of light that passes through the second polarizer = ?

I₁ = Intensity of light from the first polarizer that is incident upon the second polarizer = 40 W/m²

θ = angle between the major axis of the first and second polarizer = 30°

I₂ = 40 (cos² 30°) = 40 (0.8660)² = 30 W/m²

In the same vein, the intensity of light that passes through the third/additional polarizer is related to the intensity of light that passes through the second polarizer and is incident upon this third/additional polarizer through

I₃ = I₂ cos² θ

I₃ = intensity of light that passes through the third/additional polarizer = ?

I₂ = Intensity of light from the second polarizer that is incident upon the third/additional polarizer = 30 W/m²

θ = angle between the major axis of the second and third/additional polarizer = 60° (although, it is 90° with respect to the first polarizer, it is the angle it makes with the major axis of the second polarizer, 60°, that matters)

I₃ = 30 (cos² 60°) = 30 (0.5)² = 7.5 W/m²

Hope this Helps!!!