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aniked [119]
3 years ago
5

Does anyone know how to do this question??? Force = 7kN Pressure = 1mPa Area = ?

Physics
1 answer:
Andreas93 [3]3 years ago
5 0

Answer:

7000 m^2

Explanation:

Pressure=Force/Area

1=7000/Area

1(Area)=7000

Area=7000 m^2

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As waves get closer to a beach they _____.
krok68 [10]

Answer:

Increase in energy

Explanation:

4 0
2 years ago
A book is sitting on a desk. What best describes the normal force acting on the book?
hjlf

Answer:

umm im not sure

3 0
3 years ago
Find the sum 5.24 g, 43.261 g, and 7.3458 g. Write your answer with the correct amount of significant figures.
lapo4ka [179]

Answer:

This is how I do it:

  • 5.24 rounded to one significant figure is 5
  • 43.261 rounded to one significant figure is 40.
  • 7.3458 rounded to one significant figure is 7.
  • 5 + 40 + 7 is 52 g

Hope this helps you.

Explanation:

7 0
3 years ago
How would decreasing the volume of the reaction vessel affect each of the following equilibria?2NOBr(g)⇌2NO(g)+Br2(g)
mafiozo [28]

Answer:

The equilibrium position will shift towards the left hand side or reactants side

Explanation:

Decreasing the volume (increasing the pressure) of the system will shift the equilibrium position towards the lefthand side or reactants side. This is because, decreasing the volume (increasing the pressure) implies shifting the equilibrium position towards the side having the least number of moles.

There are two moles of reactants and a total of three moles of products(total). Hence decreasing the volume and increasing the pressure of the gas phase reaction will shift the equilibrium position towards the lefthand side.

5 0
3 years ago
The period of the earth around the sun is 1 year and its distance is 150 million km from the sun. An asteroid in a circular orbi
SOVA2 [1]

Answer:

5.024 years

Explanation:

T1 = 1 year

r1 = 150 million km

r2 = 440 million km

let the period of asteroid orbit is T2.

Use Kepler's third law

T² ∝ r³

So,

\left ( \frac{T_{2}}{T_{1}} \right )^2=\left ( \frac{r_{2}}{r_{1}} \right )^3

\left ( \frac{T_{2}}{1} \right )^2=\left ( \frac{440}{150} \right )^3

T2 = 5.024 years

Thus, the period of the asteroid's orbit is 5.024 years.

4 0
3 years ago
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