Answer:
a) During the reaction time, the car travels 21 m
b) After applying the brake, the car travels 48 m before coming to stop
Explanation:
The equation for the position of a straight movement with variable speed is as follows:
x = x0 + v0 t + 1/2 a t²
where
x: position at time t
v0: initial speed
a: acceleration
t: time
When the speed is constant (as before applying the brake), the equation would be:
x = x0 + v t
a)Before applying the brake, the car travels at constant speed. In 0.80 s the car will travel:
x = 0m + 26 m/s * 0.80 s = <u>21 m </u>
b) After applying the brake, the car has an acceleration of -7.0 m/s². Using the equation for velocity, we can calculate how much time it takes the car to stop (v = 0):
v = v0 + a* t
0 = 26 m/s + (-7.0 m/s²) * t
-26 m/s / - 7.0 m/s² = t
t = 3.7 s
With this time, we can calculate how far the car traveled during the deacceleration.
x = x0 +v0 t + 1/2 a t²
x = 0m + 26 m/s * 3.7 s - 1/2 * 7.0m/s² * (3.7 s)² = <u>48 m</u>
Because the atoms and molecules all have different properties
Answer:
The net change in the internal energy of the gas in the piston is -343J
Explanation:
Because heat and workdone are the only means of energy transfer between the system and the surrounding, change in internal energy is given by;
∆E = q + w
q = heat transfer
w = workdone
Because heat is lost by the system, the heat transfer is negative
q = -413J
Because work is done on the system, workdone is positive
w = +70J
∆E = -413J + 70J
∆E = -343J
Answer:
(a) Distance traveled = 75.3846 m
(b) Velocity of car at that instant will be 14 m/sec
Explanation:
We have given acceleration of the car 
Initial velocity of the cart u = 0 m/sec
(a) According to second equation of motion we know that 
So distance traveled by car 
As the truck is moving with constant speed
So distance traveled by truck 
As the truck overtakes the car
So 
So distance traveled 
(b) From second equation of motion we know that v = u+at
So v = 0+1.3×10.769 = 14 m /sec
