Question

A motorboat is traveling with a speed of $ 3.0{\text{ }}m{\text{ }}se{c^{ - 1}}. $ If the force on it due to water flow is $ 500{\text{ }}N, $ the power of the boat is:
A. 150 KW
B. 15 KW
C. 1.5 KW
D. 150 W

Answer 1

Hint: Compute the territory of the stream by adding the normal profundities and partitioning by two, at that point duplicate the outcome by the width of the stream. Record this as the normal zone of the stream.

Complete answer:
 $ Power = Force \times velocity $
 $ {\text{power }} = {\text{ }}500 \times 3{\text{ }} = {\text{ }}1500{\text{ }}watt{\text{ }} = {\text{ }}1.5kW $
In ordinary use and kinematics, the speed (ordinarily alluded to as v) of an item is the greatness of the difference in its position; it is in this manner a scalar quantity. The normal speed of an article in a time frame is the distance gone by the article isolated by the span of the interval; the momentary speed is the restriction of the normal speed as the length of the time stretch methodologies zero.
Speed has the components of distance partitioned by time. The SI unit of speed is the meter every second, except the most widely recognized unit of speed in regular utilization is the kilometer every hour or, in the US and the UK, miles every hour. For air and marine travel, the bunch is usually utilized.
Hence the correct answer is an option (C).

Note:
When water from a flat line strikes a divider typically, power is applied to the divider. Assume the water hits the divider regularly with speed v and stops on striking the divider, the speed adjustment is then $ 0{\text{ }}-{\text{ }}v{\text{ }} = {\text{ }}-{\text{ }}v $ . From the second law, the power rises to the energy change every second of the water stream.