Answer:
Astronaut in spacecraft while orbiting earth experience weightlessness because there is no gravity of earth or moon is acting on the body of an astronaut.
while on earth, we experience weight because the gravity of earth is acting on our body which is pulling us downward.
Both spacecraft and the astronauts both are in a free-fall condition.
I would say 5.6 to the 4 and I’ll I did was add it to both side
You can compare the velocity of the car, 60 mph, with the velocity that a mass would acquire when falls from certain height.
First, convert 60 mph to m/s:
60 miles/h * 1.60 km/mile * 1000 m/km * 1h/3600s = 26.67 m/s
Second, calculate from what height a body in free fall reachs 26.67 m/s velocity when hits the floor.
free fall => Vf^2 = 2g*H => H = Vf^2 / (2g)
H = (26.67m/s)^2 / (2*9.8 m/s) = 36.2 m
If you consider that the height between the floors of a building is approximately 3.6 m, you get 36.2 m / 3.6 m/floor = 10 floors.
Then, you conclude that the force of impact is the same as driving you vehicle off a 10 story building.
Answer:
As collision is elastic,thus we can use conservation of momentum equation
mA=0.2 kg
(vB)1=0 m/s.......................as it is on rest before collision
(vA)1=4 m/s
(vA)2=-1 m/s
(vB)2=2 m/s
using equation
(mA*vA+mB*vB)1= (mA*vA+mB*vB)2
Where 1 and 2 represents before and after collision
(0.2*4)+(mB*0)=(0.2*-1)+(mB*2)
0.8=-0.2+(2mB)
mass of object B=mB=0.3 Kg
Answer:
The work done by the applied force is 259.22 J.
Explanation:
The work done by the applied force is given by:

Where:
F: is the applied horizontal force = 108.915 N
d: is the distance = 2.38 m
Hence, the work is:

Therefore, the work done by the applied force is 259.22 J.
I hope it helps you!