Answer:
The claim is not reasonable
Explanation:
From the question we are told that
The power consumed is
The height of the pool's free surface is
The rate is
Given that the rate is 53 liters in 1 second , let say the 1 liter is equivalent to 1 kg so 50 liter = 50 kg
So the workdone by the pump is mathematically represented as
=>
=>
So what this value is telling us is that 14700 J of work is done in one second
and generally power is workdone per time thus the power consumed to move water at the given rate is 14.7 kW which is far greater than the power stated in the question so the claim is not reasonable
Answer:
Explanation:
v = u +at
u = 0
a = 2.3 m /s²
t = 20 s
v = 2.3 x 20
= 46 m /s
Distance covered under acceleration of 2.3 m/s²
s = ut + 1/2 at²
= 0 + .5 x 2.3 x 20²
= 460 m
After that it moves under free fall ie g acts on it downwards .
v² = u² - 2gh , h is height moved by it under free fall
0 = 46² - 2 x 9.8 h
h = 107.96 m
Total height attained
= 460 + 107.96
= 567.96 m
b ) At its highest point ,it stops so its velocity = 0
c ) rocket's acceleration at its highest point = g = 9.8 downwards .
At highest point , it is undergoing free fall so its acceleration = g
Answer:
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The relative velocity of the athlete relative to the ground is 5.2 m/s
The given parameters;
constant velocity of the athlete, V = 5.2 m/s
let the velocity of the ground = Vg = 0
The relative velocity concept helps us to determine the velocity of a moving object relative to a stationary observer.
The athlete is the moving object in this question while the ground is stationary.
The relative velocity of the athlete relative to the ground is calculated as follows;
Thus, the relative velocity of the athlete relative to the ground is 5.2 m/s
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