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galben [10]
3 years ago
14

a light beam that hits a mirror at an angle of 36°... what is the Angle of Incidence? Angle of Reflection?​

Physics
1 answer:
Kipish [7]3 years ago
6 0

Angle of incidence is 36° and so is the reflection. Both angles are equal.

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At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2(g)+O2(g)→2SO3(g) At equilibrium, the partial
elena55 [62]

Answer : The partial pressure of SO_3 is, 67.009 atm

Solution :  Given,

Partial pressure of SO_2 at equilibrium = 30.6 atm

Partial pressure of O_2 at equilibrium = 13.9 atm

Equilibrium constant = K_p=0.345

The given balanced equilibrium reaction is,

2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)

The expression of K_p will be,

K_p=\frac{(p_{SO_3})^2}{(p_{SO_2})^2\times (p_{O_2})}

Now put all the values of partial pressure, we get

0.345=\frac{(p_{SO_3})^2}{(30.6)^2\times (13.9)}

p_{SO_3}=67.009atm

Therefore, the partial pressure of SO_3 is, 67.009 atm

6 0
3 years ago
Which of the following statements is true?
Lilit [14]

A. All natural radiation is at a level low enough to be safe

8 0
4 years ago
What would happen if mass is continually added to a 1.4 solar mass neutron star?
andrey2020 [161]
Eventually wouldn't it collapse in on itself and create black hole.
8 0
3 years ago
In a football game, running back is at the 10 yard line and running up the field towards the 50 yard line, and runs for 3 second
lakkis [162]
If he keeps that pace he will be at the 34 yard line
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Using the law of conservation of angular momentum, estimate how fast a collapsed stellar core would spin if its initial spin rat
Nataly_w [17]

Answer:

\omega_{f} = 1000000\,\frac{rev}{day}

Explanation:

The law of conservation of angular momentum states that angular momentum remains constant when there is no external moment or forces applied to the system. Let assume that star can be modelled as an sphere, then:

\frac{2}{5}\cdot M\cdot R_{o}^{2} \cdot \omega_{o} = \frac{2}{5}\cdot M\cdot R_{f}^{2} \cdot \omega_{f}

The final angular speed is:

\omega_{f} = \omega_{o}\cdot (\frac{R_{o}}{R_{f}})^{2}

\omega_{f} = (1\,\frac{rev}{day} )\cdot (\frac{10000\,km}{10\,km} )^{2}

\omega_{f} = 1000000\,\frac{rev}{day}

3 0
3 years ago
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