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3 years ago

The horizontal component of the force acting on the crate is

Physics

1 answer:

Aleksandr [31]3 years ago

6 0

Answer: 19 n

Explanation:

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2 years ago

Please Help Me!

balandron [24]

The image distance when a boy holds a toy soldier in front of a concave mirror, with a focal length of 0.45 m. is -0.56 m.

<h3>What is image distance?</h3>

This is the distance between the image formed and the focus when an object is placed in front of a plane mirror.

To calculate the image distance, we use the formula below.

Formula:

1/f = 1/u+1/v........... Equation 1

Where:

f = Focal length of the mirror
v = Image distance
u = object distance

From the question,

Given:

f = 0.45 m
u = 0.25 m

Substitute these values into equation 1 and solve for the image distance

1/0.45 = 1/0.25 + 1/v
2.22 = 4+1/v
1/v = 2.22-4
1/v = -1.78
v = 1/(-1.78)
v = -0.56 m

Hence, The image distance is -0.56 m.

Learn more about image distance here: brainly.com/question/17273444

5 0

1 year ago

Unpolarized light with an intensity of 655 W / m2 is incident on a polarizer with an unknown axis. The light then passes through

Norma-Jean [14]

Answer:

1. $\theta=29.84^{0}$

2. $\theta=60.15^{0}$

Explanation:

Polarizes axis can create two possible angles with the vertical.

first we have to find the intensity of first polarizer

which is given as

$I=\frac{I_{0} }{2}$

$I= \frac{655\frac{W}{M^{2} } }{2}$

$I=327.5\frac{W}{m^{2} }$

For a smaller angle for the first polarizer:

According to Malus Law

$I_{2} =I_{1} Cos^{2}(90^{0} - \theta)$

$I_{2} =I_{1} sin^{2}\theta$

$\frac{I_{2} }{I_{1} }=Sin^{2}\theta$

taking square root on both sides

$\sqrt{\frac{163}{327.5} } = sin\theta$

$\theta=Sin^{-1}(0.4977)$

$\theta=29.84^{0}$

For a larger angle for the first polarizer:

According to Malus Law

$I_{2} =I_{1} cos^{2}\theta$

$\frac{I_{2} }{I_{1} }=Cos^{2}\theta$

taking square root on both sides

$\sqrt{\frac{163}{327.5} } = cos\theta$

$\theta=Cos^{-1}(0.4977)$

$\theta=60.15^{0}$

7 0

2 years ago