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3 years ago

8

The Sun appears at an angle of 48.4° above the horizontal as viewed by a dolphin swimming underwater. What angle does the sunlig

ht striking the water actually make with the horizon? (Assume nwater = 1.333. Enter an answer between 0° and 90°.)

1 answer:

Luden [163]3 years ago

6 0

Answer:

4.57 degree

Explanation:

As the dolphin is under water, so the angle is angle of refraction which is equal to 48.4 degree.

refractive index of water, n = 1.333

Let the angle of incidence which is the angle between the incident ray and the normal is i.

By use of Snell's law

$n = \frac{Sin i}{Sin r}$

$1.333 = \frac{Sin i}{Sin 48.4}$

$Sin i = 1.333\times Sin 48.4 = 0.9968$

i = 85.43 degree

Angle made with the horizon = 90 - 85.43 = 4.57 degree

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State three differences between solid friction and viscosity

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2 years ago

Explain, using your own words, how noise cancelling headphones work, (physics)

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Answer:

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Explanation:

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3 years ago

Determine the speed, wavelength, and frequency of light from a helium-neon laser as it travels through polystyrene. The waveleng

klemol [59]

Answer:

Speed:

$2.01x10^{8}m/s$

Wavelength:

$4.24x10^{-7}m$

Frequency:

$4.74x10^{14}Hz$

Explanation:

The speed of the laser as it travels through polystyrene can be determine by means of the equation of the refraction index:

$n = \frac{c}{v}$ (1)

Where c is the speed of light and v is the speed of the laser in the medium.

Therefore, v will be isolated from equation 1

$v = \frac{c}{n}$

$v = \frac{3x10^{8}m/s}{1.490}$

$v = 2.01x10^{8}m/s$

Hence, the speed of the laser has a value of $2.01x10^{8}m/s$

Frenquency:

Since, wavelength is the only one who depends on the media. Therefore the frequency in both medium will be the same.

To determine the frequency it can be used the following equation

$c = \nu \cdot \lambda$ (2)

Where c is the speed of light, $\nu$ is the frequency and $\lambda$ is the wavelength

Then, $\nu$ wil be isolated from equation 2.

$\nu = \frac{c}{\lambda}$ (3)

Before using equation 3 it is necessary to express $\lamba$ in units of meters.

$\lambda = 632.8nm . \frac{1m}{1x10^{9}nm}$ ⇒ $6.328x10^{-7}m$

$\nu = \frac{3x10^{8}m/s}{6.328x10^{-7}m}$

$\nu = 4.74x10^{14}s^{-1}$

$\nu = 4.74x10^{14}Hz$

Hence, the frequency of the laser has a value of $4.74x10^{14}Hz$

Wavelength:

To determine the wavelength it can be used:

$v = \nu \cdot \lambda$

$\lambda = \frac{v}{\nu}$

Where v is the speed of the laser through the polystyrene.

$\lambda = \frac{2.01x10^{8}m/s}{4.74x10^{14}s^{-1}}$

$\lambda = 4.24x10^{-7}m$

Hence, the wavelength of the laser has a value of $4.24x10^{-7}m$

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3 years ago

In the physics lab, a block of mass M slides down a frictionless incline from a height of 35cm. At the bottom of the incline it

bogdanovich [222]

Solution :

Given :

M = 0.35 kg

$m=\frac{M}{2}=0.175 \ kg$

Total mechanical energy = constant

or $K.E._{top}+P.E._{top} = K.E._{bottom}+P.E._{bottom}$

But $K.E._{top} = 0$ and $P.E._{bottom} = 0$

Therefore, potential energy at the top = kinetic energy at the bottom

$\Rightarrow mgh = \frac{1}{2}mv^2$

$\Rightarrow v = \sqrt{2gh}$

$=\sqrt{2 \times 9.8 \times 0.35}$ (h = 35 cm = 0.35 m)

= 2.62 m/s

It is the velocity of M just before collision of 'm' at the bottom.

We know that in elastic collision velocity after collision is given by :

$v_1=\frac{m_1-m_2}{m_1+m_2}v_1+ \frac{2m_2v_2}{m_1+m_2}$

here, $m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$

∴ $$v_1=\frac{0.35-0.175}{0.5250}+\frac{2 \times 0.175 \times 0}{0.525}$

$=\frac{0.175}{0.525}+0$

= 0.33 m/s

Therefore, velocity after the collision of mass M = 0.33 m/s

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3 years ago

A particular balloon can be stretched to a maximum surface area of 1257 cm2. The balloon is filled with 3.1 L of helium gas at a

chubhunter [2.5K]

Answer:

The ballon will brust at

<em>Pmax = 518 Torr ≈ 0.687 Atm </em>

<em />

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Explanation:

Hello!

To solve this problem we are going to use the ideal gass law

PV = nRT

Where n (number of moles) and R are constants (in the present case)

Therefore, we can relate to thermodynamic states with their respective pressure, volume and temperature.

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$ --- (*)

Our initial state is:

P1 = 754 torr

V1 = 3.1 L

T1 = 294 K

If we consider the final state at which the ballon will explode, then:

P2 = Pmax

V2 = Vmax

T2 = 273 K

We also know that the maximum surface area is: 1257 cm^2

If we consider a spherical ballon, we can obtain the maximum radius:

$R_{max} = \sqrt{\frac{A_{max}}{4 \pi}}$

Rmax = 10.001 cm

Therefore, the max volume will be:

$V_{max} = \frac{4}{3} \pi R_{max}^3$

Vmax = 4 190.05 cm^3 = 4.19 L

Now, from (*)

$P_{max} = P_1 \frac{V_1T_2}{V_2T_1}$

Therefore:

Pmax= P1 * (0.687)

That is:

Pmax = 518 Torr

6 0

3 years ago