Answer:
its terminal velocity is 19.70 m/s
the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance is 8.85 m/s
Explanation:
Firstly,
given that
m = 580g = 0.58kg
Area A = 0.11 * 0.22 = 0.0242m
g = 9.8
idensity constant p = 1.21 kg/m^3
the terminal velocity of the sphere Vt is ;
Vt = √ ( 2mg / pCA)
we substitute
Vt = √ ( (2*0.58*9.8) / (1.21*1*0.0242)
Vt = √ (11.368 / 0.029282)
Vt = √ ( 388.22)
Vt = 19.70 m/s
its terminal velocity is 19.70 m/s
What will be the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance?
The Velocity of the person is;
V2 = √ 2ax
V2 = √ ( 2 * 9.8 * 4 )
V2 = √ (78.4)
V2 = 8.85 m/s
the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance is 8.85 m/s
Answer:
naturally occurring
Explanation:
it has to come from the earth and not be man made
We know, x+y = 180
A.T.Q, x = 3y, y= x/3
substitute the value of y,
x+x/3 = 180
4x = 540
x = 540/4
x=135
I agree with Maria and disagree with Mike.
The earth has gravity and that pulls objects towards the center of the planet
Answer:
A
Explanation:
The greatest concentration of atomic mass is in the nucleus because it is made up of protons and neutrons. The electrons surrounding the nucleus don't have as much mass as protons or neutrons.
Hopefully this helps...