The answer is C.) mass is the matter of an object
200g*1 mole/ 18g=11.1 moles There are 11.1 moles of water.
The vertical weight carried by the builder at the rear end is F = 308.1 N
<h3>Calculations and Parameters</h3>
Given that:
The weight is carried up along the plane in rotational equilibrium condition
The torque equilibrium condition can be used to solve
We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person
This would lead to:
F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)
F(1cos20)= 197/2(3.10sin20 + 2 cos 20)
Fcos20= 289.55
F= 308.1N
Read more about vertical weight here:
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The acceleration due to gravity is g/4
The acceleration above the earth surface is given by the relation
g^'=gr^2/〖(h+r)〗^2
Since the satellite orbits the earth in a orbit of radius equal to earth radius, therefore
g^'=(gr^2)/〖(r+r)〗^2 =g/4
Thus the acceleration due to gravity on the satellite is g/4.
Answer:
The tension is 
Explanation:
The free body diagram of the question is shown on the first uploaded image From the question we are told that
The distance between the two poles is 
The mass tied between the two cloth line is 
The distance it sags is 
The objective of this solution is to obtain the magnitude of the tension on the ends of the clothesline
Now the sum of the forces on the y-axis is zero assuming that the whole system is at equilibrium
And this can be mathematically represented as

To obtain
we apply SOHCAHTOH Rule
So 
![\theta = tan^{-1} [\frac{opp}{adj} ]](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D%20%5B%5Cfrac%7Bopp%7D%7Badj%7D%20%5D)
![= tan^{-1} [\frac{1}{7}]](https://tex.z-dn.net/?f=%3D%20tan%5E%7B-1%7D%20%5B%5Cfrac%7B1%7D%7B7%7D%5D)





