Answer:
77.88 lbm/ft³
Explanation:
Given,
Specific gravity, SG = 1.25
Density of water, ρ = 62.30 lbm/ft³
density of the fluid =
= S.G x ρ_{water}
= 62.30 x 1.25
= 77.88 lbm/ft³
Density of the fluid is equal to 77.88 lbm/ft³
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The answer to this question is <span>13,537</span>
Answer:
Explanation:
The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:
Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.
Also, we know that the centripetal force of an object describing a circular motion is given by:
Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.
Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:
Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So and (Since ). Then, we get:
In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).