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ValentinkaMS [17]
3 years ago
14

Explain in detail why you cannot hear in space but you can see light(explanation includes particle movement of liquid, solids, g

ases, and plasma)
Physics
1 answer:
expeople1 [14]3 years ago
4 0

Answer:

The reason why one cannot hear in space but light can be seen is because light and sound are different kinds of wave

Light is an electromagnetic wave that propagates by a changing magnetic field hat creates a changing electric field. Light is created by a vibrating electric charge, and is transmitted through a medium by the absorption and reemission of the light by the medium particles

Sound is a longitudinal mechanical wave that is propagated by the vibration of the particles of the medium.

Sound requires a medium to propagate and it cannot be propagated or heard through space which is a vacuum.

Sound is created by a source that is vibrating and the vibration is transmitted to particles in the medium including solids, liquids, gases and plasma that the vibrating energy is in direct contact with and the transfer of the sound energy continues between layers of adjacent particles for the medium for the sound to be propagated from place to place

Explanation:

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What is the humidity if the dry-bulb is 10℃ and the wet-bulb is 6℃?
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Answer:

Hello,~There!

What is the humidity if the dry-bulb is 10℃ and the wet-bulb is 6℃?

<h2><u>33% According to the Graph</u></h2>

Hope this helps!

6 0
3 years ago
What is the target heart rate a female age 32
Airida [17]

Answer:

The normal resting heart rate for adults over the age of 10 years, including older adults, is between 60 and 100 beats per minute (bpm). Highly trained athletes may have a resting heart rate below 60 bpm, sometimes reaching 40 bpm. The resting heart rate can vary within this normal range.

Explanation:

I hope that answers your question!

5 0
3 years ago
Read 2 more answers
4. Derive<br>the relation, P= hd g​
Sergeu [11.5K]

p=F/A

or,P=d×V×G/A (m=d×V)

or,p= d× A×h×g/A (A and A are cut)

or,P=d×H×G

4 0
3 years ago
An oil droplet is sprayed into a uniform electric field of adjustable magnitude. The 0.11 g droplet hovers
ohaa [14]

Answer:

The direction of the field is downward, and negatively charged particles will experience an upwards force due to the field.

F = N e E     where E is the value of the field and N e the charge Q

M g = N e E      and M g is the weight of the drop

N = M g / (e E)

N = 1.1E-4 * 9.8 / (1.6E-19 * 370) = 1.1 * 9.8 / (1.6 * 370) * E15 = 1.82E13

.00011 kg is a very large drop

Q = N e = M g / E = .00011 * 9.8 / 370 = 2.91E-6 Coulombs

Check:     N = Q / e = 2.91E-6 / 1.6E-19 = 1.82E13   electrons

7 0
3 years ago
An object is launched with an initial velocity of 50.0 m/s at a launch angle of 36.9∘ above the horizontal. part a determine x-v
ankoles [38]
Given:
v = 50.0 m/s, the launch velocity
θ = 36.9°, the launch angle above the horizontal

Assume g = 9.8 m/s² and ignore air resistance.
The vertical component of the launch velocity is
Vy = (50 m/s)*sin(50°) = 30.02 m/s

The time, t, to reach maximum height is given by
(30.02 m/s) - (9.8 m/s²)*(t s) = 0
t = 3.0634 s
The time fo flight is 2*t = 6.1268 s

The horizontal velocity is
u = (50 m/s)cos(36.9°) = 39.9842 m/s
The horizontal distance traveled at time t is given in the table below.

Answer:

  t, s    x, m
------  --------
     0   0
     1   39.98
     2   79.79
     3   112.68
     4   159.58
     5   199.47
     6   239.37

5 0
3 years ago
Read 2 more answers
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