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amm1812
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°.)
Physics
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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Answer:

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Explanation:

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2 years ago
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An unstoppable object is heading right toward an unmovable object. What's going to happen?
marshall27 [118]

Answer:

In my opinion the unstoppable object will hit the unmovable object and stop but the wheels will still be rolling and trying to move but can't.

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8 0
2 years ago
A ceiling fan with 90-cm-diameter blades is turning at 64 rpm . Suppose the fan coasts to a stop 28 s after being turned off. Wh
mixas84 [53]

Answer:

the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.

Explanation:

Given;

diameter of the ceiling fan, d = 90 cm = 0.9 m

angular speed of the fan, ω = 64 rpm

time taken for the fan to stop, t = 28 s

The distance traveled by the ceiling fan when it comes to a stop is calculated as;

d = vt = \omega r\times  t= ( \frac{64 \ rev}{\min} \times \frac{2 \pi \ rad}{rev} \times \frac{1 \min}{60 \ s} \times 0.9 \ m) \times 28 \ s\\\\d = 168.89 \ m

The speed of the tip of a blade 10 s after the fan is turned off is calculated as;

v = \frac{d}{t} \\\\v = \frac{168.89}{10} \\\\v = 16.889 \ m/s

Therefore, the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.

4 0
3 years ago
If a 80kg diver jumps off of a 5 m high dive into a regulation diving pool, how much should the temperature of the pool go up?
Agata [3.3K]

Answer:

The answer cannot be determined.

Explanation:

The energy of the diver when he hits the pool will be equal to its potential energy mgh, and for the temperature of the pool to rise up, this energy has to be converted into the heat energy of the pool.

The change in temperature {\Delta}T then will be

{\Delta}T=\frac{{\Delta}Q}{mc} .

Where m is the mass of water in the pool, c is the specific heat capacity of water, and {\Delta}Q is the added heat which in this case is the energy of the diver.

Since we do not know the mass of the water in the pool, we cannot make this calculation.

7 0
3 years ago
A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit
Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

P = \frac{1}{2} \mu \omega^2 A^2 v

Here,

\mu = Linear mass density of the string

\omega =  Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

v = \sqrt{\frac{T}{\mu}} \rightarrow Here T is the Period

For the linear mass density we have that

\mu = \frac{m}{l}

And the angular frequency can be written as

\omega = 2\pi f

Replacing this terms and the first equation we have that

P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})

P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})

P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})

PART A ) Replacing our values here we have that

P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})

P = 0.2320W

PART B) The new amplitude A' that is half ot the wavelength of the wave is

A' = \frac{1.8*10^{-3}}{2}

A' = 0.9*10^{-3}

Replacing at the equation of power we have that

P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})

P = 0.058W

8 0
3 years ago
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