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3 years ago

No links please, links never work, this is science not physics

Physics

1 answer:

stiks02 [169]3 years ago

7 0

Answer:

b and c are the answers

A is an opinion, D is a superstitious belief because they haven’t found Jesus obviously

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Light propagating in air strikes the surface of a lake. The reflected ray makes 62° angle with the normal. What angle does the r

blondinia [14]

Answer:

(a) 42°

Explanation:

According to the law of reflection:

Angle of incidence = Angle of reflection

So, angle of reflection = 62°

Angle of incidence = 62°

For refraction,

Using Snell's law as:

$n_i\times {sin\theta_i}={n_r}\times{sin\theta_r}$

Where,

${\theta_i}$ is the angle of incidence ( 62.0° )

${\theta_r}$ is the angle of refraction ( ? )

${n_r}$ is the refractive index of the refraction medium (water, n=1.33)

${n_i}$ is the refractive index of the incidence medium (air, n=1)

Hence,

$1\times {sin62.0^0}={1.33}\times{sin\theta_r}$

Angle of refraction = $sin^{-1}0.6639$ = 42°

4 0

3 years ago

Using the superposition method, calculate the current through R5 in Figure 8-71

Natasha2012 [34]

by superposition method we can find current in R5

here first let say only 2V battery is present in the circuit

now the equivalent resistance to be found for which we can say

2.2 k ohm and 1 k ohm is connected in parallel

$r_1 = \frac{2.2 * 1}{2.2 + 1}$

$r_1 = 0.6875 k ohm$

now it is in series with 1 k ohm and then that part is in parallel with 2.2 k ohm

$r_2 = \frac{2.2* (1+0.6875)}{2.2 + (1+0.6875)}$

$r_2 = 0.95 k ohm$

now the current flowing through the battery is

$i = \frac{2}{1 + 0.95} = 1.02 mA$

now this will divide into R3 and R2 so current flowing in R3 will be

$i_1 = \frac{2.2}{2.2+1.6875}*1.02 = 0.58 mA$

now this will again divide in R4 and R5

so current in R5 will be

$i_5 = \frac{R_4}{R_4 + R_5}* i_1$

$i_5 = 0.18 mA$

now when only 3 V battery is present in the circuit

R1 and R2 is in parallel and then it is in series with R3

so parallel combination will be

$r_1 = \frac{1*2.2}{2.2 +1} = 0.6875k ohm$

also after its series with R3

$r_2 = 1 + 0.6875 = 1.6875 k ohm$

now it is in parallel with R5 on other side

$r_3 = \frac{1.6875 * 2.2}{1.6875 + 2.2} = 0.95 k ohm$

now current through the battery will be given as

$i = \frac{3}{1 + 0.95} = 1.53 mA$

now it is divide in r2 and R5

so current in R5 is given as

$i_5 = \frac{r_2}{r_2 + R_5}*i$

$i_5 = \frac{1.6875}{2.2 + 1.6875} * 1.53$

$i_5 = 0.67 mA$

now the total current in R5 will be given by super position which is

$i = 0.67 + 0.18 = 0.85 mA$

so there is 0.85 mA current through R5 resistance

8 0

3 years ago

ASAP pls answer will mark brainiest If right

Taya2010 [7]

1. false.

2. true.

3. false.

4. true.

5. true.

6. false.

7. false. (I think).

8. true.

9. false.

10. true.

6 0

4 years ago

Read 2 more answers

State the rule that describes how a pulse will behave at a free-end.

RSB [31]

Answer:

Fixed end: inverted

Free end: not inverted

Explanation:

8 0

3 years ago

15 POINTS! What are some differences and similarities between Electromagnetic and Mechanical waves?

Dmitry_Shevchenko [17]

<span> Elecromagnetic waves can travel through a vacuum, but a mechanical wave cannot. Mechanical waves can only exist in a medium, such as air or water. Electromagnetic waves behave like waves and particles, but mechanical waves only behave like waves. </span>

8 0

3 years ago