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valentina_108 [34]
2 years ago
10

Why are movie scenes in outer space often unrealistic?

Physics
2 answers:
kifflom [539]2 years ago
8 0

Answer:

see explanation

Explanation:

by my calculations, i believe it may be because they have to pretend to be in space.

kari74 [83]2 years ago
6 0

one of them is Space & Interstellar Travel

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the atomic number of cesium (Cs) is 55. If an atom of cesium has 78 neutrons, what is the atomic mass of cesium?
Slav-nsk [51]
Atomic mass= number of protons + number of neutrons
55 + 78 = 133
hope this helps
7 0
2 years ago
Read 2 more answers
Imagine that the earth and the-moon have positive charges of the same magnitud. How big isäºthe charge necesary to produce an el
lions [1.4K]

Answer:

5.7 x 10^12 C

Explanation:

Let the charge on earth and moon is q.

mass of earth, Me = 5.972 x 10^24 kg

mass of moon, Mm = 7.35 x 10^22 kg

Let d be the distance between earth and moon.

the gravitational force between them is

F_{g}=G\frac{M_{e} \times M_{m}}{d^{2}}

The electrostatic force between them is

F_{e}=\frac{Kq^{2}}{d^{2}}

According to the question

1 % of Fg = Fe

0.01 \times 6.67\times10^{-11}\frac{5.97 \times 10^{24}\times7.35 \times 10^{22}}{d^{2}}=9 \times 10^{9}\frac{q^{2}}{d^{2}}

2.927 \times 10^{35}=9 \times10^{9}q^{2}

3.25 \times 10^{25}=q^{2}

q = 5.7 x 10^12 C

Thus, the charge on earth and the moon is 5.7 x 10^12 C.

6 0
3 years ago
2 QUESTIONS HELP PLZ...:/
algol [13]
The first one is: head
Second one is: 10 trillion km
5 0
3 years ago
A speaker is designed for wide dispersion for a high frequency sound. What should the diameter of the circular opening be for a
andriy [413]

To solve this problem we will apply the concepts related to wavelength, as well as Rayleigh's Criterion or Optical resolution, the optical limit due to diffraction can be calculated empirically from the following relationship,

sin\theta = 1.22\frac{\lambda}{d}

Here,

\lambda = Wavelength

d= Diameter of aperture

\theta = Angular resolution or diffraction angle

Our values are given as,

\theta = 11\°

The frequency of the sound is f = 9100 Hz

The speed of the sound is v = 343 m/s

The wavelength of the sound is

\lambda = \frac{v}{f}

Here,

v = Velocity of the wave

f = Frequency

Replacing,

\lambda = \frac{(343 m/s)}{(9100 Hz)}

\lambda = 0.0377 m

The diffraction condition is then,

sin\theta = 1.22\frac{\lambda}{d}

Replacing,

sin(11\°) = 1.22\frac{(0.0377 m)}{(d)}

d = 0.24 m

Therefore the diameter should be 0.24m

6 0
3 years ago
Beth moves a 15 N book 20 meters in 10 seconds. How much power was produced?
Mice21 [21]

Answer:

30 Watts

Explanation:

 Power = Work/Time

Work = Force*Distance

Power = Force * Distance / Time

Power = 15 N * 20 meters / 10 sec

Power = 30 Watts

5 0
3 years ago
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