Explanation:
Given that,
Mass of ball, m = 0.425 kg
Initial speed of the ball, u = 12 m/s
Initial speed of a person, u' = 0
Mass of a person, m' = 68 kg
(a) Let V is the combined speed of the person and the ball. Using conservation of momentum as :
(b) If the ball hits the person and bounces off his chest, so afterwards it is moving horizontally at 9.00 m/s in the opposite direction,. Let v' is the speed of the person after the collision. So,
v = -9 m/s
Hence, this is the required solution.
Answer:
=1419.19 meters.
Explanation:
The time it takes for the shell to drop to the tanker from the height, H =1/2gt²
610m=1/2×9.8×t²
t²=(610m×2)/9.8m/s²
t²=124.49s²
t=11.16 s
Therefore, it takes 11.16 seconds for a free fall from a height of 610m
Range= Initial velocity×time taken to hit the tanker.
R=v₁t
Lets change 300 mph to kph.
=300×1.60934 =482.802 kph
Relative velocity=482.802 kph-25 kph
=457.802 kph
Lets change 11.16 seconds to hours.
=11.16/(3600)
=0.0031 hours.
R=v₁t
=457.802 kph × 0.0031 hours.
=1.41918 km
=1.41919 km × 1000m/km
=1419.19 meters.
Multiply the power (1,800 watts) by time (1,200 seconds) to get 2,160,000 joules (or 2.16 MJ)
A table would be the most appropriate because that way you can compare the data.
Answer:
0.6983 m/s
Explanation:
k = spring constant of the spring = 0.4 N/m
L₀ = Initial length = 11 cm = 0.11 m
L = Final length = 27 cm = 0.27 m
x = stretch in the spring = L - L₀ = 0.27 - 0.11 = 0.16 m
m = mass of the mass attached = 0.021 kg
v = speed of the mass
Using conservation of energy
Kinetic energy of mass = Spring potential energy
(0.5) m v² = (0.5) k x²
m v² = k x²
(0.021) v² = (0.4) (0.16)²
v = 0.6983 m/s
