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jarptica [38.1K]
3 years ago
7

Modern oil tankers weigh over a half-million tons and have lengths of up to one-fourth miles. Such massive ships require a dista

nce of 5.3 km (about 3.3 mi) and a time of 22 min to come to a stop from a top speed of 29 km/h. What is the magnitude of such a ship’s average acceleration in m/s^2 in coming to a stop? What is the magnitude of the ship’s average velocity in m/s?
Chemistry
1 answer:
uranmaximum [27]3 years ago
4 0

Answer:

Acceleration = (change in speed) / (time for the change)

Change in speed= (0 - 26 km/hr) = -26 km/hr

(-26 km/hr) x (1,000 m/km) x (1 hr / 3,600 sec) = -7.222 m/sec

Average acceleration = (-7.222 m/s) / (22 min x 60sec/min) = -0.00547 m/sec²

Average speed during the stopping maneuver =

              (1/2) (start speed + end speed) = 13 km/hr = 3.6111 m/sec

Explanation:

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Given question is incomplete. The complete question is as follows.

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When a chemical equation contains same number of atoms on both reactant and product side then this equation is known as balanced equation.

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