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Effectus [21]
3 years ago
15

An experiment is designed to test the relationship between the initial height of a basketball before it is dropped to the height

of its rebound bounce. The height of the rebound bounce is measured using a scale positioned behind the ball. In the above experiment, which condition would not be controlled?
A). the starting height of the ball
B). the surface the ball bounces on
C). the type of ball
D). the method to measure the rebound height of the ball

Physics
1 answer:
Virty [35]3 years ago
3 0

Answer:

A. The starting height of the ball

Explanation:

When we talk about controlled variables, we refer to the variable that should be kept the same throughout the experiment. The reason why we do this, is to limit anything else that is not being tested, that may affect the results of the experiment.

In the scenario given, the experiment is to see the relationship between the <u>initial height of a basketball</u> and the <u>height of its rebound bounce. </u>

So you the starting height of the ball should vary, meaning it is NOT controlled.

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A tank contains 350 liters of fluid in which 10 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu
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Answer:

A(t) = -340e^{-t/70} + 350

Explanation:

Since fluid is pumping in and out at the same rate (5L/min), the total fluid volume in the tank stays constant at 350L. Only the amount of salt and its concentration changed overtime.

Let A(t) be the amount of salt (g) at time t and C(t) (g/L) be the concentration at time t

A(0) = 10 g

Brine with concentration of 1g/L is pouring in at the rate of 5L/min so the salt income rate is 5 g/min

The well-mixed solution is pouring out at the rate of 5L/min at concentration C(t) so the salt outcome rate is 5C g/min

But the concentration is total amount of salt over 350L constant volume

C = A / 350

Therefore our rate of change for salt A' is

A' = 5 - 5A/350 = 5 - A/70

This is a first-order linear ordinary differential equation and it has the form of y' = a + by. The solution of this is

y = ce^{bt} + \frac{a}{b}

So A = ce^{\frac{-t}{70}} + \frac{5}{1/70} = ce^{-t/70} + 350

with A(0) = 10

c + 350 = 10

c = 10 - 350 = -340

A(t) = -340e^{-t/70} + 350

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