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suter [353]
3 years ago
10

What is the angle of incidence of a wave if its angle of reflection is 20 degrees?

Physics
1 answer:
konstantin123 [22]3 years ago
3 0

<em>Your answer is:</em>

<em>c) </em><u><em>20°</em></u>

<em />

<em>∝ </em><em>Sidhdi</em>

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How would you describe the movement of a transverse wave?
nlexa [21]

Answer:

Up AND Down

Explanation:

a transverse wave goes up and Down

4 0
3 years ago
What is the equivalent capacitance of the three capacitors in the figure (figure 1)?
vekshin1

The equivalent capacitance of the combination is  \dfrac{1}{C} = \dfrac{1}{C_1}+\dfrac{1}{C_2}  where C1 and C2 are the capacitance of both capacitors in series.

<h3>What is equivalent capacitor?</h3>

Let the capacitance of both capacitors be C1 and C2. For a series connected capacitors, same charge flows through the capacitors but different voltage flows through them.

Let Q be the amount of charge in each capacitors,

V be the voltage across each capacitors

C be the capacitance of the capacitor.

Using the formula Q = CV where V = Q/C... (1)

For the large capacitor with capacitance of the capacitor C1,

Q = C_1V_1;

V_1 = \dfrac{Q}{C_1}...(2)

where V_1 is the voltage across C_1,

For the small capacitor with capacitance of the capacitor C_2,

Q = C_2V_2;

V_2 = \dfrac{Q}{C_2} ... (3)

where V_2 is the voltage across C_2,

Total voltage V in the circuit will be;

V = V_1+V_2... (4)

Substituting equation 1, 2 and 3 in equation 4, we have;

\dfrac{Q}{C} = \dfrac{Q}{C_1} +\dfrac{ Q}{C_2}

\dfrac{Q}{C} = Q({\dfrac{1}{C_1}+\dfrac{1}{C_2})

Since change Q is the same for both capacitors since they are in series, they will cancel out to finally have;

\dfrac{1}{C} = ({\dfrac{1}{C_1}+\dfrac{1}{C_2})

This gives the equivalent capacitance of the combination.

To know more about equivalence capacitance follow

brainly.com/question/5626146

#SPJ4

7 0
2 years ago
The surface of a mirror is flat.
tia_tia [17]
True great job
Have a great day

3 0
3 years ago
Read 2 more answers
A tray filled with ice is removed from the freezer. After a short period of time, the ice begins to melt.
Sav [38]
A: decreases in specific heat capacity
3 0
3 years ago
A 0.09 g honeybee acquires a charge of +23 pc while flying. The earth's electric field near the surface is typically (100 N/C, d
hodyreva [135]

Answer:

2.6\times 10^{-5}

Explanation:

<u>Given:</u>

  • m = mass of the honeybee = 0.09 g = 9\times 10^{-5}\ kg
  • q = charge on the honeybee = 23 pC = 2.3\times 10^{-11}\ C
  • E = electric field near the surface of earth = 100 N/C

<u>Assume:</u>

  • g = acceleration due to gravity = 9.8\ m/s^2
  • W = weight of the honeybee
  • F = electric force on the honeybee
  • R = ratio of the electric force and the weight of the honeybee

We know that

F = qE\,\,\, and\,\,\, W = mg\\\therefore R = \dfrac{F}{W}\\\Rightarrow R = \dfrac{qE}{mg}\\\Rightarrow R = \dfrac{2.3\times 10^{-11}\ C\times 100 N/C}{9\times 10^{-5}\ kg\times 9.8\ m/s^2}\\\Rightarrow R = 2.6\times 10^{-6}

So, the ration of the electric force on the bee to its weight is 2.6\times 10^{-6}.

On multiplying this ration by 10, the ratio becomes 2.6\times 10^{-5}.

8 0
3 years ago
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