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BaLLatris [955]

3 years ago

7

An incident ray strikes a mirror with an angle of 30 degrees to the surface of the mirror. what is the angle of the reflected ra

y?

1 answer:

kkurt [141]3 years ago

7 0

With 30degree because incident angle = reflected angle

You might be interested in

How would the model change as the atom forms bonds? The third shell would have eight electrons after the atom gains seven electr

atroni [7]

Answer:

The third shell would be empty, so the eight electrons on the second level would be the outermost after the atom lost one electron

Explanation:

When an atom is bonded with other atoms, a more stable configuration must be reached, which is why the energy of the molecule is less than the energy of the individual atoms, for this to happen in general, electrons are shared or lost and gained in each atom, depending on the electronegative of the same.

If we analyze an atom within the molecule, its last shell is full, in the case of atoms with few electrons in this shell, they are lost and in the case of many electors in this shell, it gains electrons to have eight (8) in total.

When reviewing the different answers, the correct one is:

* The third shell would be empty, so the eight electrons on the second level would be the outermost after the atom lost one electron

4 0

3 years ago

Read 2 more answers

In the diagram, ∠D is b

Degger [83]

Answer:

The answer is B

Explanation:

3 0

3 years ago

Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.00×104 Pa . Assum

cestrela7 [59]

Answer:

$T_{2}=278.80 K$

Explanation:

Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.

$(\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}$ .

Now, let's use the ideal gas equation to the initial and the final state:

$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$

Let's recall that the term nR is a constant. That is why we can match these equations.

We can find a relation between the volumes of the initial and the final state.

$\frac{V_{1}}{V_{2}}=\frac{T_{1}p_{2}}{T_{2}p_{1}}$

Combining this equation with the first equation we have:

$(\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}$

$(\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}$

Now, we just need to solve this equation for T₂.

$T_{1}\cdot (\frac{p_{2}}{p_{1}})^{\frac{\gamma - 1}{\gamma}} = T_{2}$

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.

Here,

$p_{2}=8.00\cdot 10^{4} Pa \\p_{1}=1.01\cdot 10^{5} Pa\\ T_{1}=298 K\\ \gamma=1.40$

Finally, T2 will be:

$T_{2}=278.80 K$

6 0

3 years ago

A straight line is drawn on the surface of a 14-cm-radius turntable from the center to the perimeter. A bug crawls along this li

sleet_krkn [62]

Answer:

v = $\left[\begin{array}{c}0.66&0\end{array}\right]$ m/s

Explanation:

The position vector r of the bug with linear velocity v and angular velocity ω in the laboratory frame is given by:

$\overrightarrow{r}=vtcos(\omega t)\hat{x}+vtsin(\omega t)\hat{y}$

The velocity vector v is the first derivative of the position vector r with respect to time:

$\overrightarrow{v}=[vcos(\omega t)-\omega vtsin(\omega t)]\hat{x}+[vsin(\omega t)+\omega vtcos(\omega t)]\hat{y}$

The given values are:

$t=\frac{x}{v}=\frac{14}{3.8}=3.7 s$

$\omega=\frac{45\times2\pi}{60s}=4.7\frac{1}{s}$

8 0

3 years ago

Everything is in the pic.

nikitadnepr [17]

The answer is A

Good luck!

3 0

3 years ago