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Inga [223]
3 years ago
9

What affects the amount of solar energy the planet receives?

Physics
2 answers:
MA_775_DIABLO [31]3 years ago
6 0

sun is facing a certain way

solmaris [256]3 years ago
4 0
The sun is facing a certain side of the earth
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A photon is absorbed by an electron that is in the n = 3 state of a hydrogen atom, causing the hydrogen atom to become ionized.
Wittaler [7]

Answer:

\lambda=3.99*10^{-7}m

Explanation:

According to the law of conservation of energy, the energy of the absorbed photon must be equal to the binding energy of the electron plus the energy of the released electron:

E_p=E_b+E_r\\\frac{hc}{\lambda}=\frac{13.6eV}{n^2}+\frac{m_ev^2}{2}

1 eV is equal to 1.6*10^{-19}J, so:

13.6eV*\frac{1.6*10^{-19}J}{1eV}=2.18*10^{-18}J

Solving for \lambda and replacing the given values:

\lambda=\frac{hc}{\frac{2.18*10^{-18}J}{n^2}+\frac{m_ev^2}{2}}\\\lambda=\frac{6.63*10^{-134}J\cdot s(3*10^8\frac{m}{s})}{\frac{2.18*10^{-18}J}{3^2}+\frac{(9.11*10^{-31}kg)(7.5*10^5\frac{m}{s})^2}{2}}\\\lambda=3.99*10^{-7}m

4 0
3 years ago
What is the final concentration of DD at equilibrium if the initial concentrations are [A][A]A_i = 1.00 MM and [B][B]B_i = 2.00
pentagon [3]

Answer:

A) Concentration of A left at equilibrium of we started the reaction with [A] = 2.00 M and [B] = 2.00 M is 0.55 M.

B) Final concentration of D at equilibrium if the initial concentrations are [A] = 1.00 M and [B] = 2.00 M is 0.90 M.

[D] = 0.90 M

Explanation:

With the first assumption that the volume of reacting mixture doesn't change throughout the reaction.

This allows us to use concentration in mol/L interchangeably with number of moles in stoichiometric calculations.

- The first attached image contains the correct question.

- The solution to part A is presented in the second attached image.

- The solution to part B is presented in the third attached image.

8 0
3 years ago
Define power and discuss how to determine it.
serious [3.7K]

there are different types of power so ill show you all the examples of power.


4 0
3 years ago
What is the density of 15 cm3 and 3 g?
lilavasa [31]

Answer:

0.2

Explanation:

3/15

6 0
3 years ago
Water flows straight down from an open faucet. The cross-sectional area of the faucet is 2.4 × 10-4m2 and the speed of the water
Ksenya-84 [330]

To solve this problem it is necessary to apply the continuity equations in the fluid and the kinematic equation for the description of the displacement, velocity and acceleration.

By definition the movement of the Fluid under the terms of Speed, acceleration and displacement is,

v_2^2 = v_1^2 + 2gh

Where,

V_i = Velocity in each state

g= Gravity

h = Height

Our values are given as,

A_1 = 2.4*10^{-4} m^2

v_1 = 0.8 m/s

h = 0.11m

Replacing at the kinetic equation to find V_2 we have,

v_2 = \sqrt{v_1^2 + 2gh}

v_2 = \sqrt{(0.8 m/s)^2 + 2(9.80 m/s2)(0.11 m)}

v_2= 1.67 m/s

Applying the concepts of continuity,

A_1v_1 = A_2v_2

We need to find A_2 then,

A_2= \frac{A_1v_1 }{v_2}

So the cross sectional area of the water stream at a point 0.11 m below the faucet is

A_2= \frac{A_1v_1 }{v_2}

A_2= \frac{(2.4*10^{-4})(0.8)}{(1.67)}

A_2= 1.14*10^{-4} m2

Therefore the cross-sectional area of the water stream at a point 0.11 m below the faucet is 1.14*10^{-4} m2

8 0
3 years ago
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