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LiRa [457]
3 years ago
11

Evidence on why we don’t hear solar eruptions on Earth.

Physics
2 answers:
sukhopar [10]3 years ago
8 0
Because sound waves don't travel through the vaccume of space. Hope this helped
8_murik_8 [283]3 years ago
5 0
I can’t give five but I know on reason why if we could hear them it would be extremely loud evidence is that a scientist called Scott Mclntosh of the national center for atmospheric research (tip: if we could hear it from earth we would probs go deaf too)
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_honest person is always at peace<br>​
borishaifa [10]
Hmmm I dunno buddy sorry
5 0
3 years ago
g A boat is anchored 2000 ft from shore anddirects its searchlight towards an automobile travelingdown the straight road. At the
Darya [45]

Answer:

Explanation:

The distance of searchlight will act as radius R and velocity of car may be supposed to be tangential velocity v  . We are required to calculate  angular velocity ω .

v = 80 ft /s

R = 3000 ft

ω = v / R

= 80  / 3000 = .027 rad / s

For angular acceleration the formula is

angular acceleration α = a / R

a is linear acceleration = 15 ft / s²

α = 15 / 3000 = .005 rad / s².

3 0
3 years ago
A train travelled from city a to city b. For the first 14hours,it travelled at an average speed of 120km/h. For the 10 hours, it
Nataly [62]

Explanation:

distance covered in the first part of journey : 120×14 = 1680km

distance covered in second part of journey :

90×10 = 900 km

total distance traveled : 1680+900 = 2580km.

total time taken to complete the journey : 14+10= 24 hrs.

therefore, the average speed of the journey = 2580km/24hr = 107.5 km/h.

hope this helps you.

4 0
3 years ago
An 800-kHz radio signal is detected at a point 2.7 km distant from a transmitter tower. The electric field amplitude of the sign
dimaraw [331]

Answer:

Option D is correct: 170 µW/m²

Explanation:

Given that,

Frequency f = 800kHz

Distance d = 2.7km = 2700m

Electric field Eo = 0.36V/m

Intensity of radio signal

The intensity of radial signal is given as

I = c•εo•Eo²/2

Where c is speed of light

c = 3×10^8m/s

εo = 8.85 × 10^-12 C²/Nm²

I = 3×10^8 × 8.85×10^-12 × 0.36²/2

I = 1.72 × 10^-4W/m²

I = 172 × 10^-6 W/m²

I = 172 µW/m²

Then, the intensity of the radio wave at that point is approximately 170 µW/m²

7 0
3 years ago
The most distal part of a nail is called the ____________ . The most proximal part of the nail that is embedded in the skin is t
ankoles [38]

Answer:

1.Free edge

2.Nail Root

3.Nail matrix

4.lunula

Note: The numbering coincides respectively with the dashes in the questions.

Explanation:

Free edge is the distal white nail ending.

Nail Root is the proximal part of the nail embedded in the skin.

Nail matrix is the actively growing part of the nail were the nail root thickens

Lunula the whitish semilunar area of the proximal end of the nail body it appears whitish because stratum basala obscures. the underlying blood vessels.

7 0
3 years ago
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