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3 years ago

Which of these experiments would make use of qualitative data

Physics

2 answers:

Anna71 [15]3 years ago

8 0

Qualitative data does not have anything to do with numbers so i believe that it is C

7nadin3 [17]3 years ago

5 0

It has to due with numbers so I would say the last one!

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Is the state of the air in an isolated room completely specified by the temperature and the pressure? Explain

anyanavicka [17]

I think these two variables are sufficient to completely specify the state.

In an isolated room with air only ,the volume is fixed.Mass ,density and its specific volume can be easily known.

Other thermodynamic properties like entropy, enthalpy etc are also fixed at a given temperature & pressure.

4 0

3 years ago

The heating system in Jeff and Tony's new home takes cool air and passes it through the heat source using a fan or blower. What

-BARSIC- [3]

If it takes cool air and passes it through the heat source using a fan or blower, the type of heat transfer is convection heat transfer.

<h3>What is heat transfer by convection?</h3>

Convection is the transfer of heat from one place to another due to the movement of fluid.

During heat transfer by convection, cold air replaces hot risen air.

Thus, if it takes cool air and passes it through the heat source using a fan or blower, the type of heat transfer is convection heat transfer.

Learn more about convection here: brainly.com/question/893656

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8 0

1 year ago

*WILL MARK BRAINLIEST FOR RIGHT ANSWER* How much current must be applied across a 60 Ω light bulb filament in order for it to co

sp2606 [1]

Answer: The current must be equal to $\frac{\sqrt{33} }{6}$ amps, or ~0.9574 amps.

Explanation:

You can find the current in amperes using ohms and watts from this formula:

$I = \sqrt{\frac{P}{R} }$

Where P represents power in watts, R represents resistance in ohms, and I represents current in amperes.

You can then substitute 60 and 55 into the equation to find I:

$I = \sqrt{\frac{55}{60} } \\I = \frac{\sqrt{55} }{\sqrt{60} }$

Then, simplify the denominator:

$I = \frac{\sqrt{55} }{2\sqrt{15} }$

Rationalize the denominator:

$I = \frac{\sqrt{55} }{2\sqrt{15} } * \frac{\sqrt{15} }{\sqrt{15} } = \frac{\sqrt{825} }{30}$

Simplify the numerator by finding its factors:

$I = \frac{5\sqrt{33} }{30} = \frac{\sqrt{33} }{6}$

The current must be equal to $\frac{\sqrt{33} }{6}$ amps, or ~0.9574 amps.

8 0

3 years ago

A small, closed chamber of gas is heated. When the gas in the chamber expands,it does 5 J of work on a piston. The gas has an in

Anestetic [448]

Given:

Work done, W = 5 J

Initial energy = 8J

Final energy = 30J

Let's determine if the work done have a positive or nrgative value.

Appy the equation for the first lae of thermodynamics:

$\Delta U=Q-W$

Where:

U is the change in internal energy

Q is the added heat

W is the work done

To find the work done here, we have:

Rewrite the formula for W

$W=Q-\Delta U$

Where:

ΔU = 30J - 8J = 22J

Q = 5J

Thus, we have:

$\begin{gathered} W=5J-22J \\ \\ W=-17J \end{gathered}$

Therefore, the work done here is -17J.

This means the work done in this scenario has a negative value.

ANSWER:

The work done in this scenario has a negative value

3 0

1 year ago

What is the smallest sub atomic particle that orbit around the nucleus?

JulijaS [17]

Electron,a negative charged particles revolving around the nucleus.

5 0

3 years ago