A rapid release of stored up energy
Answer: 5.47m/s
Explanation:
Mass = 72.3kg
K.E = 1080.0J
V =?
K.E = 1 /2MV^2
V^2 = 2K.E /M = (2x1080)/72.3
V = sqrt [(2x1080)/72.3]
V = 5.47m/s
Answer:
My bad i didnt mean to put that carry on.
Explanation:
Answer: 18.65L
Explanation:
Given that,
Original volume of oxygen (V1) = 30.0L
Original temperature of oxygen (T1) = 200°C
[Convert temperature in Celsius to Kelvin by adding 273.
So, (200°C + 273 = 473K)]
New volume of oxygen V2 = ?
New temperature of oxygen T2 = 1°C
(1°C + 273 = 274K)
Since volume and temperature are given while pressure is held constant, apply the formula for Charle's law
V1/T1 = V2/T2
30.0L/473K = V2/294K
To get the value of V2, cross multiply
30.0L x 294K = 473K x V2
8820L•K = 473K•V2
Divide both sides by 473K
8820L•K / 473K = 473K•V2/473K
18.65L = V2
Thus, the new volume of oxygen is 18.65 liters.