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Elanso [62]
3 years ago
13

A student moves a box across the floor by exerting 56.7 N of force and doing 195 J of

Physics
1 answer:
Paha777 [63]3 years ago
8 0

Answer:

A. 3.4 m

Explanation:

Given the following data;

Force = 56.7N

Workdone = 195J

To find the distance

Workdone is given by the formula;

Workdone = force * distance

Making "distance" the subject of formula, we have;

Distance = \frac {workdone}{force}

Substituting into the equation, we have;

Distance = \frac {195}{56.7}

Distance = 3.4 meters.

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Answer:

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Explanation:

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3 years ago
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A 3.50 cm tall object is held 24.8 cm from a lens of focal length 16.0 cm. What is the image height?
Mandarinka [93]

Answer:

<h2>6.36 cm</h2>

Explanation:

Using the formula to first get the image distance

1/f = 1/u+1/v

f = focal length of the lens

u = object distance

v = image distance

Given f = 16.0 cm, u = 24.8 cm

1/v = 1/16 - 1/24.8

1/v = 0.0625-0.04032

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v = 1/0.02218

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To get the image height, we will us the magnification formula.

Mag = v/u = Hi/H

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Hi = (45.09*3.50)/24.8

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6 0
3 years ago
(a) What is the minimum width of a single slit (in multiples of λ ) that will produce a first minimum for a wavelength λ ? (b) W
Kitty [74]

Answer:

The minimum value of width for first minima is λ

The  minimum value of width for 50 minima is 50λ

The  minimum value of width for 1000 minima is 1000λ

Explanation:

Given that,

Wavelength = λ

For D to be small,

We need to calculate the minimum width

Using formula of minimum width

D\sin\theta=n\lambda

D=\dfrac{n\lambda}{\sin\theta}

Where, D = width of slit

\lambda = wavelength

Put the value into the formula

D=\dfrac{n\lambda}{\sin\theta}

Here, \sin\theta should be maximum.

So. maximum value of \sin\theta is 1

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D=\dfrac{1\times\lambda}{1}

D=\lambda

(b). If the minimum number  is 50

Then, the width is

D=\dfrac{50\times\lambda}{1}

D=50\lambda

(c). If the minimum number  is 1000

Then, the width is

D=\dfrac{1000\times\lambda}{1}

D=1000\lambda

Hence, The minimum value of width for first minima is λ

The  minimum value of width for 50 minima is 50λ

The  minimum value of width for 1000 minima is 1000λ

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3 years ago
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Physics is defined as:
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Answer:

B. Developed out of efforts of man trying to explain our physical environment.

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This is because physics, as we all know of, is considered to be <em>crucial to understanding the world around us</em>, making it the most basic and fundamental of science.

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