let the mass of Venus is M then mass of Saturn is 100 M
similarly if the radius of Venus is R then the radius of Saturn is 10 R
now the force of gravity on a man of mass "m" at the surface of Venus is given by

now similarly the gravitational force on the man if he is at the surface of Saturn


so here if we divide the two forces

so here we can say
F1 = F2
so on both planets the gravitational force will be same
a content creator because if i was a rapper i probably wouldn't make good songs lol
Answer:
29.4855 grams of chlorophyll
Explanation:
From Raoult's law
Mole fraction of solvent = vapor pressure of solution ÷ vapor pressure of solvent = 457.45 mmHg ÷ 463.57 mmHg = 0.987
Mass of solvent (diethyl ether) = 187.4 g
MW of diethyl ether (C2H5OC2H5) = 74 g/mol
Number of moles of solvent = mass/MW = 187.4/74 = 2.532 mol
Let the moles of solute (chlorophyll) be y
Total moles of solution = moles of solute + moles of solvent = (y + 2.532) mol
Mole fraction of solvent = moles of solvent/total moles of solution
0.987 = 2.532/(y + 2.532)
y + 2.532 = 2.532/0.987
y + 2.532 = 2.565
y = 2.565 - 2.532 = 0.033
Moles of solute (chlorophyll) = 0.033 mol
Mass of chlorophyll = moles of chlorophyll × MW = 0.033 × 893.5 = 29.4855 grams
I belive it would power I’m not 100% sure. If it’s not power then force
Answer:
μ = 0.692
Explanation:
In order to solve this problem, we must make a free body diagram and include the respective forces acting on the body. Similarly, deduce the respective equations according to the conditions of the problem and the directions of the forces.
Attached is an image with the respective forces:
A summation of forces on the Y-axis is performed equal to zero, in order to determine the normal force N. this summation is equal to zero since there is no movement on the Y-axis.
Since the body moves at a constant speed, there is no acceleration so the sum of forces on the X-axis must be equal to zero.
The frictional force is defined as the product of the coefficient of friction by the normal force. In this way, we can calculate the coefficient of friction.
The process of solving this problem can be seen in the attached image.