Answer: 361° C
Explanation:
Given
Initial pressure of the gas, P1 = 294 kPa
Final pressure of the gas, P2 = 500 kPa
Initial temperature of the gas, T1 = 100° C = 100 + 273 K = 373 K
Final temperature of the gas, T2 = ?
Let us assume that the gas is an ideal gas, then we use the equation below to solve
T2/T1 = P2/P1
T2 = T1 * (P2/P1)
T2 = (100 + 273) * (500 / 294)
T2 = 373 * (500 / 294)
T2 = 373 * 1.7
T2 = 634 K
T2 = 634 K - 273 K = 361° C
Answer:
The answer to your question is: F = 0.4375 N. The force will be 16 times lower than with the first conditions.
Explanation:
Data
F = 7 N
F = ? if the masses is quartered
Formula

Process
Normal conditions F = Km₁m₂/r² = 7
When masses quartered F = K(m₁/4)(m₂/4)/r² = ?
F = K(m₁m₂/16)/r²
F = K(m₁m₂/16r² = 7/16 = 0.4375 N
Answer:
The rate at which energy is transferred is called power and the amount of energy that is usefully transferred is called efficiency.