Two cars collide head on while each is traveling at 90 km/hkm/h. Suppose all their kinetic energy is transformed into the thermal energy of the wrecks. What is the temperature increase of each car?

Complete question:

Two cars collide head on while each is traveling at 90 km/h. Suppose all their kinetic energy is transformed into the thermal energy of the wrecks. What is the temperature increase of each car?

you may assume that the specific heat capacity of each car is that of

iron..

The temperature increase of each car is 0.68 k

Explanation:

Given;

velocity of each car, v = 90 km/h

velocity of each car (in m/s) = (90 x 10)/36 = 25 m/s

Kinetic energy of each car, KE = ¹/₂ mv²

Total Kinetic energy of the two cars = 2 x ¹/₂ mv² = mv²

Thermal energy = mcΔT

Total thermal energy of the the two cars = 2 x mcΔT = 2mcΔT

Suppose all their kinetic energy is transformed into the thermal energy of the wrecks, then; 2mcΔT = mv²

divide both sides by m,

2cΔT = v²

where;

c is the specific heat of iron = 460.548 J/kg.K

ΔT is the temperature increase of each car

v is the velocity of each car = 25 m/s

Substitute these values and calculate ΔT

ΔT = v²/2c

ΔT  = (25)²/(2 x 460.548)

ΔT  = 0.679 k = 0.68 k

Therefore, the temperature increase of each car is 0.68 k

the force required to move the cart is divided by three or 33.33 n.

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge