Given:
mass: 100 kg
height: 500 m
1 kJ = 1000 J
gravity = 9.8 m/s²
velocity before impact: v = √2gh ; v = √2 * 9.8 m/s² * 500 m ; v = 98.99494 m/s
KE = 1/2 m v²
KE = 1/2 * 100 kg * (98.99494 m/s)²
KE = 490,000 J
Pls. see attachment.
Answer:
The fall in temperature of the liquid is 8.6 +/- 0.1 ⁰C
Explanation:
Given;
initial temperature of the liquid, t₁ = 76.3 +/- 0.4⁰C
final temperature of the liquid, t₂ = 67.7 +/- 0.3⁰C
The change in temperature of the liquid is calculated as;
Δt = t₂ - t₁
Δt = (67.7 - 76.3) +/- (0.3 - 0.4)
Δt = (-8.6) +/- (-0.1)
Δt = 8.6 +/- 0.1 ⁰C
Therefore, the fall in temperature of the liquid is 8.6 +/- 0.1 ⁰C
A roller coaster car starts from the rest from a first summit, descends a vertical distance of 45 meters and then climbs a second summit, reaching the top with a speed of 15m / s. How high is the second summit? Do not consider friction
You would weigh 58.967 kilograms