1050kg sorry Ik this is not correct
Answer:
the velocity of car when it passes the truck is u = 16.33 m/s
Explanation:
given,
constant speed of truck = 28 m/s
acceleration of car = 1.2 m/s²
passes the truck in 545 m
speed of the car when it just pass the truck = ?
time taken by the truck to travel 545 m
time =
time =
time =19.46 s
velocity of the car when it crosses the truck
u = 16.33 m/s
the velocity of car when it passes the truck is u = 16.33 m/s
Here are the steps you would need to follow:
#1). Define what 'the 'X is, and how it's related to the ball.
#2). Be clear on how 'the X' is related to the 'known velocity'.
#3). Identify how the 'known velocity' is related to the action of the ball when it's launched.
With this information in front of you, you'll have a much better chance
of answering the question.
With none of it in front of me, I have no chance at all.
Answer:
823.46 kgm/s
Explanation:
At 9 m above the water before he jumps, Henri LaMothe has a potential energy change, mgh which equals his kinetic energy 1/2mv² just as he reaches the surface of the water.
So, mgh = 1/2mv²
From here, his velocity just as he reaches the surface of the water is
v = √2gh
h = 9 m and g = 9.8 m/s²
v = √(2 × 9 × 9.8) m/s
v = √176.4 m/s
v₁ = 13.28 m/s
So his velocity just as he reaches the surface of the water is 13.28 m/s.
Now he dives into 32 cm = 0.32 m of water and stops so his final velocity v₂ = 0.
So, if we take the upward direction as positive, his initial momentum at the surface of the water is p₁ = -mv₁. His final momentum is p₂ = mv₂.
His momentum change or impulse, J = p₂ - p₁ = mv₂ - (-mv₁) = mv₂ + mv₁. Since m = Henri LaMothe's mass = 62 kg,
J = (62 × 0 + 62 × 13.28) kgm/s = 0 + 823.46 kgm/s = 823.46 kgm/s
So the magnitude of the impulse J of the water on him is 823.46 kgm/s
