1) 4.8
2) 5.6
3)0.8
.............………
Bigger change in velocity because the object is lighter than the object with more mass so it would move further (sorry it’s not a great explanation)
Answer:
one-third of its weight on Earth's surface
Explanation:
Weight of an object is = W = m*g
Gravity on Earth = g₁ = 9.8 m/s
Gravity on Mars = g₂ =
g₁
Weight of probe on earth = w₁ = m * g₁
Weight of probe on Mars = w₂ = m * g₂ -------- ( 1 )
As g₂ = g₁/3 --------- ( 2 )
Put equation (2) in equation (1)
so
Weight of probe on Mars = w₂ = m * g₁ /3
Weight of probe on Mars =
m * g₁ =
w₁
⇒Weight of probe on Mars =
Weight of probe on earth
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