In part a, you determined that the heat added to the coffee is what you got. if it takes 4.2 j of energy to raise the temperature of 1.0 ml of coffee by 1.0∘c, then it should take twice as much energy to raise the temperature of 2.0 ml of coffee by 1.0∘c, or to raise the temperature of 1.0 ml of coffee by 2.0∘c. multiply carefully to find the total amount of energy needed to raise the temperature of 200 ml of coffee by the required number of degrees.

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j

the speed of the cart after 3 seconds of low fan speed is ⇒ 54.0 cm/s.

the speed of the cart after 5 seconds of medium fan speed is ⇒ 120.0 cm/s.

the speed of the cart after 2 seconds of high fan speed is ⇒ 64.0 cm/s. 

just did the lab.