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Mila [183]
4 years ago
12

Explain how variable and controlled parameters are used in scientific experiments.

Physics
2 answers:
adell [148]4 years ago
7 0

Answer:

Ok, suppose you have an experiment in which you want to see how the variable X affects the variable Y.

In this experiment, you can change the variable X and see how the other changes, and you may understand in this way how those two variables are related. Suppose that the variable X is the weight of a car and Y is the speed of the car (X is the independent variable, and Y is the dependent variable)

Now, we can change the weight and see how the speed changes in consequence. Here we also have the "controlled parameters" that are other variables that we have fixed during the experiment, in this case, we can define things like the shape of the car, the hardware, etc as controlled parameters (we even can do the same experiment again with little changes in those parameters, and in this way find the optimal combination).

aksik [14]4 years ago
5 0
Im not sure about controlled parameters, but i can explain how variables work. There are different types of variables. An independent one is the thing you change every time you do the experiment. Dependent variables are what you keep the same every time you do the experiment. (i hope this somewhat helped)
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Answer:

The Hubble space telescope.

Explanation:

Hubble is a telescope that observers the sky 24/7 non-stop, which means that for every day of the year it would have made a significant discovery, which of course includes your birthday. Furthermore, you can actually go to NASA website and find out what discovery was made on your birthday! This shows both the vastness of the universe <em>(it really has to be huge for a telescope to have a discovery for each day of the year!) </em> and the ceaseless work of the telescope!

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erastova [34]

Answer:

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3 years ago
The Sex Equity in Education Act requires educational institutions to distribute a sexual harassment policy to:
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3 years ago
A projectile enters a resisting medium at x = 0 with an initial velocity v0 = 910 ft/s and travels 5 in. before coming to rest.
evablogger [386]

Answer:

a = - 1.987 × 10⁶ ft/s²

t = 6.84 × 10⁻⁴ s

Explanation:

v₀ = 910 ft/s

x = 5 in.

relation v = v₀ - k x

v = 0 as body comes to rest

0 = 900 - 5k/12

k = 2184 s⁻¹

acceleration

\frac{\mathrm{d} v}{\mathrm{d} t} = -k\frac{\mathrm{d} x}{\mathrm{d} t}

where

(A) a = -k × v

 at v= 910 ft/s

     a = - 1.987 × 10⁶ ft/s²

(B)  at x = 3.9 in.

v = 910 - 3.9(2184)/12

v = 200.2 m/s

\frac{\mathrm{d} v}{\mathrm{d} t} = -k\frac{\mathrm{d} x}{\mathrm{d} t}

\frac{dv}{v} = -kdt

\int\limits^{200.2}_{900} {\frac{1}{v} }dv = -k\int\limits^t_0 dt

ln(200.2)-ln(900) = -kt

t = 6.84 × 10⁻⁴ s

3 0
3 years ago
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