Answer:
true! : )
(i underlined the place where the answer is the other information is just as important but if you do not want to read it you do not have to)
Explanation:
Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces. So as two objects are separated from each other, the force of gravitational attraction between them also decreases. the greater the mass, the greater the gravitational pull. <u>gravitational pull decreases with an increase in the distance between two objects.</u> Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces. So as two objects are separated from each other, the force of gravitational attraction between them also decreases.
Answer:
mass = 0.18 [kg]
Explanation:
This is a classic problem where we can apply the definition of density which is equal to mass over volume.
![density = \frac{mass}{volume} \\\\where:\\volume = 1 [m^3]\\density = 0.18[kg/m^3]](https://tex.z-dn.net/?f=density%20%3D%20%5Cfrac%7Bmass%7D%7Bvolume%7D%20%5C%5C%5C%5Cwhere%3A%5C%5Cvolume%20%3D%201%20%5Bm%5E3%5D%5C%5Cdensity%20%3D%200.18%5Bkg%2Fm%5E3%5D)
mass = 0.18*1
mass = 0.18 [kg]
Answer:
The change in temperature of 576.9°C will produce an elongation of 9 inches per feet in steel.
Explanation:
The formula for linear expansion of a material is:
ΔL = αLΔT
where, ΔL = change in length
L = Original length
ΔT = Change in temperature
α = coefficient of linear expansion
For steel, α = 13 x 10^-6 /°C
L = 100 ft
ΔL = (9 in)(1 ft/12 in) = 0.75 ft
Therefore,
0.75 ft = (13 x 10^-6 /°C)(100 ft)ΔT
<u>ΔT = 576.9°C</u>
<u></u>
For the swimmer to go straight the sin component of his velocity should cancel the river velocity.
So, we have 
θ
This angle θ is the angle shall the swimmer point upstream from the shore.
0.32 = 0.85 sin[/tex]θ
θ = 
The velocity of swimmer across the river is given by cos componet of his velocity
v = 0.85 cosθ = 0.85 cos 22 = 0.79 m/s
