Radiation zone outward through the middle layer of the sun’s interior, the radiation zone. the radiation zone is a region of very tightly packed gas, where energy is transferred mainly in the form of electromagnetic radiation.
The answer is Absorption.
Gases at pressure are released by rockets as they travel towards space. According to Newton's third law, the combustion chamber's exhaust gases push the rocket with an accelerating force known as the thrust.
<h3>Explain exactly Newton's Third Law:</h3>
According to Newton's third law, if an object A pulls on an object B, then object B must exert an equal-sized and opposite-direction force on the first thing directed in the opposite direction. This law illustrates a symmetry in nature whereby forces always occur in pairs and whereby no body can exert a force without also being subjected to one.
<h3>What are Newton's 3rd law examples?</h3>
Action and response are always equal but always move in the opposite direction, according to Newton's third law of motion. A human walking on the ground, a hammer driving a nail, a magnet attracting a paper clip, and a horse pulling a cart are all examples of Newton's third rule of motion.
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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
Answer:
In order to prevent the aliased frequency from we need to sample at with a sampling frequency that is minimum 2 times the highest frequency where information exit i.e 2 × 750 = 1500 Hz
Explanation:
The explanation is shown on the first and second uploaded image
