Suppose a 32-kg child sits in a playground swing

Light enters air from water. The angle of refraction will be A. less than the angle of incidence. B. greater than or equal to th

Rina8888 [55]

Answer:

E. greater than the angle of incidence.

Explanation:

Snell's law states that:

$n_i sin \theta_i = n_r sin \theta_r$ (1)

where

$n_i, n_r$ are the refractive index of the first and second medium

$\theta_i, \theta_r$ are the angle of incidence and refraction, respectively

For light moving from water to air, we have:

$n_i = 1.33$ (index of refraction of water)

$n_r = 1.00$ (index of refraction of air)

Substituting into (1) and re-arranging the equation, we get

$\sin \theta_r = \frac{n_i}{n_r} sin \theta_i = 1.33 sin \theta_i$

which means that

$\theta_r > \theta_i$

so, the correct answer is

E. greater than the angle of incidence.

4 0

3 years ago

Read 2 more answers

The rate at which speed changes is called

pochemuha

Answer:

Acceleration

Explanation:

I think so & hope it will help yuh!

8 0

3 years ago

Answer both and i’ll venmo u rn if ur right pls pls

Masteriza [31]

Answer:

2.5

Explanation:

25-22.5n= will give you 2.5

5 0

3 years ago

Read 2 more answers

Which steps are important when designing and conducting a scientific experiment

artcher [175]

Having your space clean. have on close toed shoes. have your hair pulled back into a ponytail. keep ur work space clean. wear gloves and goggles. do not have on droopy clothes. follow the steps on the board and double check them.

8 0

3 years ago

In a 350-m race, runner A starts from rest and accelerates at 1.6 m/s^2 for the first 30 m and then runs at constant speed. Runn

kifflom [539]

Answer:

B can take 0.64 sec for the longest nap .

Explanation:

Given that,

Total distance = 350 m

Acceleration of A = 1.6 m/s²

Distance = 30 m

Acceleration of B = 2.0 m/s²

We need to calculate the time for A

Using equation of motion

$s=ut+\dfrac{1}{2}at_{A}^2$

Put the value in the equation

$30=0+\dfrac{1}{2}\times1.6\times t_{A}^2$

$t_{A}=\sqrt{\dfrac{30\times2}{1.6}}$

$t_{A}=6.12\ sec$

We need to calculate the time for B

Using equation of motion

$s=ut+\dfrac{1}{2}at_{B}^2$

Put the value in the equation

$30=0+\dfrac{1}{2}\times2.0\times t_{B}^2$

$t_{B}=\sqrt{\dfrac{30\times2}{2.0}}$

$t_{B}=5.48\ sec$

We need to calculate the time for longest nap

Using formula for difference of time

$t'=t_{A}-t_{B}$

$t'=6.12-5.48$

$t'=0.64\ s$

Hence, B can take 0.64 sec for the longest nap .

4 0

3 years ago