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Anestetic [448]
3 years ago
9

A student heats a liquid on a burner. What happens to the portion of liquid that first begins to warm? It loses energy. It becom

es less dense. It sinks downward. Its molecules get closer together.
PLEASE HURRY
Physics
2 answers:
4vir4ik [10]3 years ago
6 0

Answer:

B

Explanation:

Because i said so

scoray [572]3 years ago
5 0

Answer:

they get closer together

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kenny6666 [7]

Answer:

c

Explanation:

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1) Momenum is
adelina 88 [10]

Answer:

momentum formula = Mass × Velocity

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3 years ago
Which of the following is typically NOT a method scientist use to determine astronomic distances
tangare [24]
I think the answer is A
5 0
3 years ago
What problem do refractor telescopes have that reflectors don't? Group of answer choices bad seeing light loss from secondary el
Paul [167]

chromatic aberration problem do refractor telescopes have that reflectors don't

<u>Explanation:</u>

Chromatic aberration is a phenom in which light rays crossing through a lens focus at various points, depending on their wavelength. Chromatic aberration is a dilemma in which lens or refracting, telescopes undergo from. The various image distances for the respective colors affect various image sizes for them.

This involves the creation of disturbing color fringes in the image. Chromatic aberration can be pretty well adjusted by the use of an achromatic doublet. Here, a positive biconvex lens is coupled with a negative lens placed backward with greater dispersion. Thus partly compensates for the chromatic aberration.

8 0
3 years ago
A cube has a drag coefficient of 0.8. What would be the terminal velocity of a sugar cube 1 cm on a side in air ( = 1.2 kg/mº)?
anzhelika [568]

0.495 m/s

Explanation

the formula for the terminal velocity is given by:

\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ \text{where} \\  \end{gathered}

m is the mass

g is 9.81 m/s²

ρ is density

A is area

C is the drag coefficient

then

Step 1

Let's find the mass

\begin{gathered} \sigma=\frac{m}{v} \\ m=\sigma\cdot v \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(0.01m)^3 \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(1\cdot10^{-6}) \\ \text{mass}=2\cdot10^{-3}\operatorname{kg} \\ \text{mass}=0.002\text{ kg } \\ \text{Area}=(0.01\text{ m}\cdot0.01m)=0.0001m^2 \end{gathered}

now, replace

\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ v=\sqrt[]{\frac{2(0.002kg)(9.81\text{ }\frac{m}{s^2})}{(2\cdot10^3\frac{\operatorname{kg}}{m^3})(0.0001m^2)0.8}} \\ v=\sqrt[]{\frac{0.03924\frac{\operatorname{kg}m}{s^2}}{0.16\frac{\operatorname{kg}}{m^{}}}} \\ v=\sqrt[]{0.2452\frac{m^2}{s^2}} \\ v=0.495\text{ m/s} \end{gathered}

hence, the answer is 0.495 m/s

3 0
1 year ago
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