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Kisachek [45]
3 years ago
6

A box exerts 10,000 Pa of pressure on the ground. If the box weighs 1000 N, how much area is in contact with the ground?

Physics
1 answer:
lara [203]3 years ago
4 0

Pressure = (total force) / (Area)

10,000 Pa = (1,000 N) / (Area)

Multiply each side by (Area) :

(10,000 Pa) x (Area) = 1,000 N

Divide each side by (10,000 Pa) :

Area = (1,000 N) / (10,000 Pa)

<em>Area = 0.1 m² </em>

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What are the unit for acceleration
Flura [38]
<h3>Answer</h3>

m/s^2 (meter per sec square)

Explanation:

acc = change in velocity/time

= distance/time

----------------

time

= m/s

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s

=m/s^2

7 0
3 years ago
I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi
Masteriza [31]

Answer:

<h3>a.</h3>
  • After it has traveled through 1 cm : I(1 \ cm) = 0.5488 I_0
  • After it has traveled through 2 cm : I(2 \ cm) = 0.3012 I_0
<h3>b.</h3>
  • After it has traveled through 1 cm : od( 1\ cm) =  0.2606
  • After it has traveled through 2 cm :  od( 2\ cm) =  0.5211

Explanation:

<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient \mu the formula is:

I(x) = I_0 e^{-\mu x}

where I is the intensity of the beam, I_0 is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = 0.5488 \ I_0

After travelling 2 cm:

I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}

I(2 \ cm) = I_0 e^{- 1.2}

I(2 \ cm) = I_0 e^{- 1.2}

I(2 \ cm) = 0.3012 \ I_0

<h2>b</h2>

The optical density od is given by:

od(x) = - log_{10} ( \frac{I(x)}{I_0} ).

So, after travelling 1 cm:

od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )

od( 1\ cm) = - log_{10} ( 0.5488 )

od( 1\ cm) = - (  - 0.2606)

od( 1\ cm) =  0.2606

After travelling 2 cm:

od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )

od( 2\ cm) = - log_{10} ( 0.3012 )

od( 2\ cm) = - (  - 0.5211)

od( 2\ cm) =  0.5211

3 0
3 years ago
A hill that has a 36.1% grade is one that rises 36.1 m vertically for every 100.0 ml of distance in the horizontal direction. At
GaryK [48]

Answer:

angle is 19.85°

Explanation:

given data

grade = 36.1 %

vertically distance y = 36.1 m

horizontal distance x = 100 m

to find out

angle

solution

we know that according to question here

it make a triangle shape that contain 100 m base and height 36.1 m

and make angle with base is θ

so by triangle we know here

tanθ = y/x   ..........1

so we now put x and y and find θ

tanθ = 36.1 / 100

tanθ = 0.361

θ = 19.85°

so angle is 19.85°

7 0
3 years ago
If you could live any where except for where you are now where would you live
laila [671]

Answer:

I would live in the Atlantic ocean on a lux liner

Explanation:

:)

4 0
3 years ago
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Zanzabum
The runner's displacement and average velocity are both zero. (I hope this answer is on the list of choices that you neglected to share.)
4 0
3 years ago
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