Question: Initially, the car travels along a straight road with a speed of 35 m/s. If the brakes are applied and the speed of the car is reduced to 13 m/s in 17 s, determine the constant deceleration of the car.
Answer:
1.29 m/s²
Explanation:
From the question,
a = (v-u)/t............................ Equation 1
Where a = deceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.
Given: v = 13 m/s, u = 35 m/s, t = 17 s.
a = (13-35)/17
a = -22/17
a = -1.29 m/s²
Hence the deceleration of the car is 1.29 m/s²
Answer:
the pencil i have on my desk is hard
Explanation:
i dont have 5 but here is one
Which questions do you need help? I could help you
Answer:The kinetic energy will decrease.
Explanation:As potential energy increases kinetic energy will increase, as kinetic or potential energy decreases, the kinetic or potential energy will decrease
Answer:
Answer: 18.3 km/s
Explanation:
If a satellite in Molniya orbit has an apogee at 48.000 km as measured from the center of Earth, and a velocity of 3.7 km/s. Its velocity in at perigee would be 18.3 km/s.
