A person standing waist-deep in a swimming pool appears to have short legs because of light

8 points)Antireflection coating can be used on the eyeglasses to reduce the reflection of light: a) A 100nm thick coating is app

NemiM [27]

Answer:

- the coating’s index of refraction is 1.25

- the required thickness is 104.1667 nm

Explanation:

Given the data in the question;

Thickness of coating t = 100 nm

wavelength λ = 500nm

we know that refractive index is;

t = λ/4n

make n, the subject of formula

t4n = λ

n = λ / 4t

we substitute

n = 500 / ( 4 × 100 )

n = 500 / 400

n = 1.25

Therefore, the coating’s index of refraction is 1.25

2)

given that;

Index of refraction of the coating; n = 1.20

λ = 500 nm

thickness of coating t = ?

t = λ / 4n

we substitute

t = 500 / ( 4 × 1.2 )

t = 500 / 4.8

t = 104.1667 nm

Therefore, the required thickness is 104.1667 nm

4 0

3 years ago

Two loudspeakers (A and B) are 3.20m apart and emitting a sound with a frequency of 400Hz. An observer is 2.10m directly in fron

TiliK225 [7]

Answer:

The observer hears a loud sound

Explanation:

In order to know if the observer hears a loud or a quiet sound, you need to know if there is a constructive or destructive interference between the sound waves of the loudspeakers.

You first calculate the distance between the observer and the loudspeakers.

The distances are given by:

d1: distance to loudspeaker A = 2.10m

d2: distance to loudspeaker B

$d_2=\sqrt{(3.20m)^2+(2.10m)^2}=3.827m$

Next, you calculate the wavelength of the sound waves by using the following formula:

$\lambda=\frac{v_s}{f}$

vs: speed of sound = 343 m/s

f: frequency of the waves = 400Hz

λ: wavelength

$\lambda=\frac{343m/s}{400Hz}=0.8575m$

Next, you calculate the path difference between the distance from the observer to the loudspeakers:

$\Delta d=3.827m-2.10m=1.727m$

You obtain a constructive interference (loud sound) if the quotient between the wavelength of the sound and the difference path is an integer:

$\frac{\Delta d}{\lambda}=\frac{1.727m}{0.857}\approx2$

Then, there will be a constructive interference, and the sound who the observer hears is loud.

5 0

3 years ago

A girl of mass m1=60 kilograms springs from a trampoline with an initial upward velocity of v1=8.0 meters per second. At height

AleksandrR [38]

a) 5.0 m/s

This first part of the problem can be solved by using the conservation of energy. In fact, the mechanical energy of the girl just after she jumps is equal to her kinetic energy:

$E_i=\frac{1}{2}m_1v_1^2$

where m1 = 60 kg is the girl's mass and v1 = 8.0 m/s is her initial velocity.

When she reaches the height of h = 2.0 m, her mechanical energy is sum of kinetic energy and potential energy:

$E_f = \frac{1}{2}m_1 v_2 ^2 + m_1 gh$

where v2 is the new speed of the girl (before grabbing the box), and h = 2.0m. Equalizing the two equations (because the mechanical energy is conserved), we find

$\frac{1}{2}m_1 v_1^2 = \frac{1}{2}m_1 v_2 ^2 + m_1 gh\\v_1^2 = v_2^2 +2gh\\v_2 = \sqrt{v_1^2 -2gh}=\sqrt{(8.0 m/s)^2-(2)(9.8 m/s^2)(2.0 m)}=5.0 m/s$

b) 4.0 m/s

After the girl grab the box, the total momentum of the system must be conserved. This means that the initial momentum of the girl must be equal to the total momentum of the girl+box after the girl catches the box:

$p_i = p_f\\m_1 v_2 = (m_1 + m_2) v_3$

where m2 = 15 kg is the mass of the box. Solving the equation for v3, the combined velocity of the girl+box, we find

$v_3 = \frac{m_1 v_2}{m_1 + m_2}=\frac{(60 kg)(5.0 m/s)}{60 kg+15 kg}=4 m/s$

c) 2.8 m

We can use again the law of conservation of energy. The total mechanical energy of the girl after she catches the box is sum of kinetic energy and potential energy:

$E_i = \frac{1}{2}(m_1+m_2) v_3^2 + (m_1+m_2)gh=\frac{1}{2}(75 kg)(4 m/s)^2+(75 kg)(9.8 m/s^2)(2.0m)=2070 J$

While at the maximum height, the speed is zero, so all the mechanical energy is just potential energy:

$E_f = (m_1 +m_2)gh_{max}$

where h_max is the maximum height. Equalizing the two expressions (because the mechanical energy must be conserved) and solving for h_max, we find

$E_i = (m_1+m_2)gh_{max}\\h_{max}=\frac{E_i}{(m_1+m_2)g}=\frac{2070 J}{(75 kg)(9.8 m/s^2)}=2.8 m$

4 0

3 years ago

to digitize sound, a process called _____ is used to measure the amplitude of a sound wave thousands of times per second.

Leona [35]

Answer: Sound recording and production

Explanation: Digital audio is also the name for the entire technology of sound recording and reproduction using audio signals that have been encoded in digital form.

7 0

3 years ago

What is the acceleration experiance by a car that takes 10s to reach 27m/s from rest

lina2011 [118]

Magnitude of acceleration = (change in speed) / (time for the change).

Change in speed = (27 - 0) = 27 m/s
Time for the change = 10 s

Magnitude of acceleration = (27 m/s) / (10 s) = 2.7 m/s² .

4 0

3 years ago