A1500 kg car accelerates uniformly from rest to reach 10.0 m sin 5.00 s. determine (i) the average power delivered by the engine.

1.50 × 10⁴ W

Explanation:

Given data

Mass (m): 1500 kg

Initial speed (vi): 0 m/s (rest)

Final speed (vf): 10.0 m/s

Time (t): 5.00 s

First, we will calculate the acceleration (a) of the car.

a = vf - vi/ t = (10.0 m/s - 0 m/s) / 5.00 s = 2.00 m/s²

Next, we find the associated force (F) using Newton's second law of motion.

F = m . a = 1500 kg . 2.00 m/s² = 3.00 × 10³ N

Then, we find the distance (d) traveled in this time.

d = 1/2 . a . t² = 1/2 . 2.00 m/s² . (5.00 s)² = 25.0 m

The work (w) exerted by the engine is:

w = F . d = 3.00 × 10³ N . 25.0 m = 7.50 × 10⁴ J

Finally, the average power (P) delivered by the engine is:

P = w / t = 7.50 × 10⁴ J / 5.00 s = 1.50 × 10⁴ W

hi,

what occurs during a solar eclipse?

there are small tides across earth.

the moon is between earth and the sun.

p.s. im not sure if the first answer is correct because during a solar ecliphse there are spring tides and i dont know if they are the same.