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vodka [1.7K]
4 years ago
7

12.

Physics
2 answers:
KATRIN_1 [288]4 years ago
7 0

Answer:

31.25 is the correct answer

AveGali [126]4 years ago
7 0
Let t seconds be the time taken for the first stone to reach the ground.

Using s = ut + 0.5at^2, with s = H, u = 0, a = 10:

H = 0.5 x 10 x t^2

H = 5t^2

The time taken for the second stone to reach the ground will be (t - 1) seconds. The stone will fall from a height of H - 20 metres.

Use s = ut + 0.5at^2, with s = H - 20, u = 0, a = 10, t = (t - 1):

H - 20 = 0.5 x 10 (t - 1)^2

H - 20 = 5(t - 1)^2

We know that H = 5t^2:

5t^2 - 20 = 5(t - 1)^2

5t^2 - 20 = 5(t^2 - 2t +1)

5t^2 - 20 = 5t^2 - 10t + 5

-20 = -10t +5

10t = 25

t = 2.5

Substituting for H:

H = 5 x 2.5^2 = 31.25 metres
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Answer:

3.82746e+26 watts

Explanation:

There are two ways to solve this problem. One way is to use the equation

L = 4πσR²T⁴

where

L = the sun's bolometric (all-spectrum) luminous power

σ = 5.670374419e-8 W m⁻² K⁻⁴ = the Stefan-Boltzmann constant

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You find that

L = 3.82746e+26 watts

The other way to solve the problem is to use the Planck integral for radiant flux.

L = 4π²R ∫(v₁,v₂) 2hv³/{c² exp[hv/(kT)]−1} dv

where

h = 6.62607015e-34 J sec

c = 299792458 m sec⁻¹

k = 1.380649e-23 J K⁻¹

v₁ = 0 = frequency band lower bound, in Hz

v₂ = ∞ = frequency band upper bound, in Hz

You find, once again, that

L = 3.82746e+26 watts

The advantage of using the Planck integral becomes clear when you want to calculate the sun's luminous power only in a specific band, rather than across the entire spectrum. For example, if we do the calculation again, except that we use

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L (visible) = 1.56799e+26 watts

So the fraction of the sun's luminosity that is in the visible spectrum is

L (visible) / L = 0.4096686

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4 years ago
_______ says the universe began as a point and expanded.
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3 years ago
Air is heated in a glass bottle. The heat energy added to the air is 2.0 × 104 joules. What is the change in internal energy of
Liono4ka [1.6K]
The change in internal energy of the gas is \Delta U = 2.0 \cdot 10^4 J.

In fact, the 1st law of thermodynamics states that the change in internal energy of a system is equal to the amount of heat given to the system (Q) plus the work done on the system (W):
\Delta U = Q+W
In this example, no work is done on the bottle so W=0, while the heat given to the system is Q=2.0 \cdot 10^4 J, so the change in internal energy of the gas is
\Delta U = Q = 2.0 \cdot 10^4 J
4 0
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