Answer:
The latent heat of vaporization of water is 2.4 kJ/g
Explanation:
The given readings are;
The first (mass) balance reading (of the water) in grams, m₁ = 581 g
The second (mass) balance reading (of the water) in grams, m₂ = 526 g
The first joulemeter reading in kilojoules (kJ), Q₁ = 195 kJ
The second joulemeter reading in kilojoules (kJ), Q₂ = 327 kJ
The latent heat of vaporization = The heat required to evaporate a given mass water at constant temperature
Based on the measurements, we have;
The latent heat of vaporization = ΔQ/Δm
∴ The latent heat of vaporization of water = (327 kJ - 195 kJ)/(581 g - 526 g) = 2.4 kJ/g
The latent heat of vaporization of water = 2.4 kJ/g
264Ns
Explanation:
Given parameters:
Time of the push = 3s
Force of push = 88N
Unknown:
Impulse = ?
Solution:
Impulse is defined as the change in momentum of a body when force acts on it.
Impulse = Force x time
Inputting the parameters:
Impulse = 88 x 3 = 264Ns
Learn more:
Momentum brainly.com/question/9484203
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Answer:
Cruising at 35,000 feet in an airliner, straight toward the east,
at 500 miles per hour
Explanation:
Answer:
Explanation:
Time dilation formula is
T = T₀ / √ 1-v²/c²
T₀ is time elapsed in moving reference , T time elapsed in stationary reference.
Here T₀ = 1 second
T = 1/√ 1-0.9² = 1/.4358 = 2.3 second
So 2.3 second will pass for each second on moving reference.
The conservation of energy always holds true even when not clearly observable in machines that are less than 100% efficient. More often than not a machine will suffer energy losses (e.g. consider for a cooling fan: friction between the rotating blades, drag resistance in the air the fan is pushing around, resistance in the wire, and heat radiating/conducting away from the circuitry).
