The air pressure in a car is 210kPa at a temperature of 26°C. What is the temperature of the air in the tyre of the pressure of 220kPa? [assume that the volume of the air in the tyre is constant]

313.24K

Explanation:

Given that:

Initial air pressure P1 = 210kPa

Initial Temperature T1 = 26°C

(Since, the SI unit of temperature is Kelvin, convert temperature in celsius to kelvin

26°C + 273 = 299K)

Final temperature T2 = ?

Final pressure P2 = 220kPa

Since, pressure and temperature are involved while volume is constant, apply the formula for pressure law

P1/T1 = P2/T2

210kPa/299K = 220kPa/T2

Cross multiply

210kPa x T2 = 220kPa x 299K

210 x T2 = 65780

T2 = (65780/210)

T2 = 313.24K

Thus, the temperature of the air in the tyre is 313.24K

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//