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2 years ago

Which one of the following lines best illustrates personification?

Physics

2 answers:

Korolek [52]2 years ago

8 0

The correct answer is C.

jekas [21]2 years ago

3 0

C. A narrow wind complains all day.

Personification is a figure of speech wherein animals, objects, and ideas are given human traits or characteristics.

The narrow wind, which is an unseen object, was given a human trait which is complains.

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An experiment is set up to test the angular resolution of an optical device when red light (wavelength ????r ) shines on an aper

Neko [114]

Explanation:

As per Rayleigh criterion, the angular resolution is given as follows:

$\theta=\frac{1.22 \lambda}{D}$

From this expression larger the size of aperture, smaller will be the value of angular resolution and hence, better will be the device i.e. precision for distinguishing two points at very high angular difference is higher.

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3 years ago

Water vapor enters the atmosphere through

Blizzard [7]

Answer:

Heat from the Sun causes water to evaporate from the surface of lakes and oceans. This turns the liquid water into water vapor in the atmosphere. Plants, too, help water get into the atmosphere through a process called transpiration! ... Water can also get into the atmosphere from snow and ice.

Most water vapor enters the atmosphere via evaporation and transpiration. Evaporation occurs when a single water molecule on a liquid water surface gains enough kinetic energy (often by solar radiation) to break the bond which holds the molecules together. Really Hopes this helps!

Explanation:

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2 years ago

A spaceship approaches the earth with a speed 0.50c. A passenger in the spaceship measures his heartbeat as 9- beats per minute.

JulijaS [17]

Answer:

D) 61 beats per minute

Explanation:

5 0

2 years ago

Believing that Nick will not be a good waiter because he is 60 years old is an example of O Stereotype O Prejudice Situational a

Helen [10]

Ya so quid fuh doil em yous

4 0

2 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

9 months ago