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liq [111]
2 years ago
10

A compressed spring does not have elastic potential energy.

Physics
1 answer:
Alexus [3.1K]2 years ago
6 0

Answer:

False

Explanation:

The second you let go its gonna release kinetic energy that's why it's potential

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A bronze statue weighs 1200 N and has a base of 2m^2. What is the pressure the statue
notka56 [123]

Answer:

Explanation:

P = F/A

F = 1200 N

A = 2 m^2

P = 1200N/2 m^2

P = 600 N / m^2

7 0
2 years ago
Prefix and suffix for hydrology
Tju [1.3M]

Prefix: Hydro

Suffix: Logy

I hope I helped you! <3

8 0
3 years ago
Which of the following is not a greenhouse gas?
Harman [31]

Answer:

The answer is B

Explanation: Greenhouse gases must be able to change their vibrational state in order to absorb infrared radiation or heat. SO therefore helium is not one of them.

Hope this helps:)

6 0
2 years ago
If the Sun subtends a solid angle Ω on the sky, and the flux from the Sun just above the Earth’s atmosphere, integrated over all
Arada [10]

Answer:

A)Ω = 7.8 × 10^−5 steradians.

B) TE = 5800K

C) fλ(λ1) = (π ^2 ) /ΩBλ(T)

Explanation:

A) First of all, if we assume that the Sun emits isotropically at a luminosity (L⊙) , the flux at a given distance R from the sun would be f(d) = L⊙/ (4πd^2)

The ratio of flux at the solar photosphere to the flux at the Earth’s atmosphere would be: F⊙/{f(d⊙)} = (R⊙)^2 / (d⊙)^2

Now if we think of this relationship of the flux and the earth as a conical pattern, we'll deduce that the solid angle subtended by the sun at Earth’s surface to be;

Ω = π[(R⊙)^2 / (d⊙)^2]

Combining this with the ratio earlier gotten, well arrive at;

F⊙ = {f(d⊙ )π} /Ω

Now let's express The radius of the sun (R) in terms of its angular diameter (2α) and this gives;

R⊙ ≈ αd⊙

Now combining this with the equation for Ω earlier, we get;

Ω ≈ πα^2

So, = π((0.57/2π) /180)^2 = 7.8 × 10^−5 steradians.

B) from Stefan-Boltzmann Law,

F⊙ = σ(TE)^4

From the beginning, we know that;

F⊙ = {f(d⊙ )π} /Ω

And so replacing that in the stephan boltzmann law, we get ;

{f(d⊙ )π} /Ωσ = (TE)^4

So, (TE)^4 = {π (1.4 kWm^(−2))} / [(7.8 × 10^(−5 ) steradians x (5.66961 × 10^(−8))]

In stephan boltzmann law, σ = 5.66961 × 10^(−8)

And so, TE is approximately 5800K.

C) In order to relate fλ(λ1) with T, let's assume the sun’s surface to be an isotropically emitting blackbody, i.e its specific intensity is Iλ = Bλ(T). Hence, the flux at Sun’s surface for a given wavelength would be;

Fλ(λ1) = πBλ(T)

Now, if we combine this with the expression of F⊙ gotten earlier, well get the relation;

fλ(λ1) = (π ^2 ) /ΩBλ(T)

7 0
3 years ago
A person riding in a hot air air balloon basket ascends to 1000 ft above the ground. What happened to the atmospheric pressure a
kherson [118]
C) it decreased
because atmospheric pressure decreases as we move up.
8 0
3 years ago
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