Answer:
see explanation
Explanation:
You are missing the chart with the rates and time to do this, however, I wll do it with a similar exercise here, and you only need to replace the procedure with your data:
See the attached table.
From the left we have:
r = 1/2 (50 + 48 + 46 + 44 + 42 + 40) = 135 L/min
From the right we have:
r = 1/2 (48 + 46 +44 + 42 + 40 + 38) = 129 L/min.
And this should be the correct answer. Watch your chart and replace if it's neccesary.
Answer:
<h3>Because one Coulomb of charge is an abnormally large quantity of charge, the units of microCoulombs (µC) or nanoCoulombs (nC) are more commonly used as the unit of measurement of charge. To illustrate the magnitude of 1 Coulomb, an object would need an excess of 6.25 x 1018 electrons to have a total charge of -1 C.</h3>
Explanation:
<h3><em><u>mark as brainliast</u></em></h3><h3><em><u>indian </u></em><em><u>genius </u></em><em><u>s</u></em><em><u>a</u></em><em><u>r</u></em><em><u>thak</u></em></h3>
Explanation:
Acceleration is the change in speed over a given time period
Answer:
(a) v = 5.42m/s
(b) vo = 4.64m/s
(c) a = 2874.28m/s^2
(d) Δy = 5.11*10^-3m
Explanation:
(a) The velocity of the ball before it hits the floor is given by:
(1)
g: gravitational acceleration = 9.8m/s^2
h: height where the ball falls down = 1.50m

The speed of the ball is 5.42m/s
(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.
You use the following formula:
(2)
vo: velocity of the ball where it starts its motion upward
You solve for vo and replace the values of the parameters:

The velocity of the ball is 4.64m/s
(c) The acceleration is given by:


The acceleration of the ball is 2874.28/s^2
(d) The compression of the ball is:

THe compression of the ball when it strikes the floor is 5.11*10^-3m