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2 years ago

which narrative point of view uses outside of the story who know. the thoug,feelings,futures and of every character?

Physics

1 answer:

notka56 [123]2 years ago

5 0

Omniscient - third person who knows how the characters feel and can give the reader unlimited information about them.

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An unmanned spacecraft leaves for Venus. Which statements about the spacecrafts journey are true?

Yuri [45]

<span>The weight of the spacecraft keeps changing.
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<span>The mass of the spacecraft remains the same.

These are the correct answers</span>

4 0

3 years ago

Read 2 more answers

A house brick has a volume of 1900 cm³ and a weight in air of 80N.What is its apparent weight in water?The density of water is 1

Shtirlitz [24]

Answer:

61 N

Explanation:

We'll begin by calculating the mass of the brick when placed in water. This can be obtained as follow:

Volume of brick = 1900 cm³

Density of water = 1 g/cm³

Mass of brick in water =…?

Density = mass / volume

1 = mass of brick in water / 1900

Cross multiply

Mass of brick in water = 1 × 1900

Mass of brick in water = 1900 g

Next, we shall convert 1900 g to Kg.

1000 g = 1 Kg

Therefore,

1900 g = 1900 g × 1 Kg / 1000 g

1900 g = 1.9 Kg

Next, we shall determine the weight in water. This can be obtained as follow:

Mass (m) = 1.9 Kg

Acceleration due to gravity (g) = 10 m/s²

Weight (W) =?

W = m × g

W = 1.9 × 10

W = 19 N

Thus, the weight of the brick in water is 19 N.

Finally, we shall determine the apparent weight of the brick in water. This can be obtained as follow:

Weight in air = 80 N

Weight in water = 19 N

Apparent weight =?

Apparent weight = weight in air – weight in water

Apparent weight = 80 – 19

Apparent weight = 61 N

3 0

3 years ago

A positron is an elementary particle identical to an electron except that its charge is . An electron and a positron can rotate

denis23 [38]

Explanation:

According to the energy conservation,

$F_{centripetal} = F_{electric}$

$\frac{mv^{2}}{r} = \frac{kq^{2}}{d^{2}}$

$v^{2} = \frac{kq^{2}r}{d^{2}m}$

= $\frac{9 \times 10^{9} N.m^{2}/C^{2} \times 1.6 \times 10^{-19} C \times 0.75 \times 10^{-9} m}{(1.50 \times 10^{-9}m)^{2} \times 9.11 \times 10^{-31} kg}$

= $8.430 \times 10^{10} m^{2}/s^{2}$

v = $\sqrt{8.430 \times 10^{10} m^{2}/s^{2}}$

= $2.903 \times 10^{5} m/s$

Formula for distance from the orbit is as follows.

S = $2 \pi r$

= $2 \times 3.14 \times 0.75 \times 10^{-9} m$

= $4.71 \times 10^{-9} m$

Now, relation between time and distance is as follows.

T = $\frac{S}{v}$

$\frac{1}{f} = \frac{S}{v}$

or, f = $\frac{v}{S}$

= $\frac{2.903 \times 10^{5} m/s}{4.71 \times 10^{-9} m}$

= $6.164 \times 10^{13} Hz$

Thus, we can conclude that the orbital frequency for an electron and a positron that is 1.50 apart is $6.164 \times 10^{13} Hz$ .

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3 years ago

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Svet_ta [14]

U hi why exactly are you trying to do

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2 years ago

Convert the following.

Alexandra [31]

Answer:

50°C whn converted into K - 323 K

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3 years ago

which narrative point of view uses outside of the story who know. the thoug,feelings,futures and of every character?​

which narrative point of view uses outside of the story who know. the thoug,feelings,futures and of every character?