Answer:
The tea will generally stop cooling when it reaches room temperature ️
Answer
When a body moves in projectile motion it has two components. One of the components is a horizontal component of the velocity and another is a vertical component.
The velocity along the horizontal component does not change because there is no acceleration long horizontal component.
Whereas velocity along vertical direction keeps on changing because the acceleration due to gravity is acting on the object. At a maximum height of the projectile velocity is equal to zero.
A selective breeding. Put your best animals together to get better offspring.
<span>Foods cook faster when placed in a pressure cooker.
This is because in the pressure cooker, the pressure on the surface
of the water is greater than the normal atmospheric pressure. Under
those conditions, the boiling temperature of water is increased.
So the food can be heated to a greater temperature than the normal
boiling point of water. That means that the food gets hotter, and it
cooks through and through faster.</span>
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