Answer:
C. Momentum is conserved but not kinetic energy.
Explanation:
This case represents an entirely inelastic collision, that is, a collision between the car and the truck that reduces total kinetic energy of the entire system, whereas linear momentum is conserved. Hence, correct answer is C.
From 50km/h to 0km/h in 0.5s we need next acceleration:
First we convert km/h in m/s:
50km/h = 50*1000/3600=13.8888 m/s
a = v/t = 13.88888/0.5 = 27.77777 m/s^2
Now we use Newton's law:
F=m*a
F=1700*27.7777 = 47222N
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
The anwser is 1.475124 horsepower.