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2 years ago

The speed of sound in seawater is 1470 m/s. A dolphin sends out a click that reflects off of an

Physics

1 answer:

Nitella [24]2 years ago

3 0

Answer: 0.204 s

Explanation:

The speed of sound $V$ is defined as the distance traveled $d$ in a especific time $t$ :

$V=\frac{d}{t}$

Where:

$V=1470 m/s$ is the speed of sound in seawater

$t$ is the time the sound wave travels from the dolphin and then returns after the reflection

$d=2(150 m)$ is twice the distance between the dolphin and the object to which the sound waves are reflected

Finding $t$ :

$t=\frac{d}{V}$

$t=\frac{2(150 m)}{1470 m/s}$

<u>Finally:</u>

$t=0.204 s$

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A thin beam of light enters a thick plastic sheet from air at an angle of 32.0° with the normal and continues in the sheet at an

Cloud [144]

Answer:

1.36

Explanation:

$n_{air}$ = Index of refraction of air = 1

$n_{plastic}$ = Index of refraction of plastic = ?

i = angle of incidence in air = 32.0° deg

r = angle of refraction in plastic = 23.0° deg

Using Snell's law

$n_{air}$ Sini = $n_{plastic}$ Sinr

(1) SIn32.0 = $n_{plastic}$ Sin23.0

$n_{plastic}$ = 1.36

5 0

2 years ago

An object moving with a speed of 35M/s and has a kinetic energy of 1500 J what is the mass of the object

Lelu [443]

2.45 Kg

Explanation:

K.E = 0.5 mv^2

1500 = 0.5 × m × (35)^2

1500= m × 0.5 × 1225

1500 = m × 612.5

1500/612.5 = m

2.45 = m

m = 2.45kg

7 0

1 year ago

Water flowing through a cylindrical pipe suddenly comes to a section of pipe where the diameter decreases to 86% of its previous

Orlov [11]

Answer:

Explanation:

The speed of the water in the large section of the pipe is not stated

so i will assume 36m/s

(if its not the said speed, input the figure of your speed and you get it right)

Continuity equation is applicable for ideal, incompressible liquids

Q the flux of water that is Av with A the cross section area and v the velocity,

so,

$A_1V_1=A_2V_2$

$A_{1}=\frac{\pi}{4}d_{1}^{2} \\\\ A_{2}=\frac{\pi}{4}d_{2}^{2}$

the diameter decreases 86% so

$d_2 = 0.86d_1$

$v_{2}=\frac{\frac{\pi}{4}d_{1}^{2}v_{1}}{\frac{\pi}{4}d_{2}^{2}}\\\\=\frac{\cancel{\frac{\pi}{4}d_{1}^{2}}v_{1}}{\cancel{\frac{\pi}{4}}(0.86\cancel{d_{1}})^{2}}\\\\\approx1.35v_{1} \\\\v_{2}\approx(1.35)(38)\\\\\approx48.6\,\frac{m}{s}$

Thus, speed in smaller section is 48.6 m/s

3 0

1 year ago

Your friend says, “chemical changes are caused by an input in energy. In physical changes, there is no transfer of energy” is yo

nalin [4]

Answer:

Ok, let's suppose the simplest of the physical changes:

We have an object that is not moving (so it is not accelerated)

and there is change, now the object moves.

Because there was a change, means that there was an acceleration, and by the second Newton's law.

Force equals mass times acceleration:

F = m*a

There must be a force.

So suppose that you pushed the object, then some energy that you had, you transferred it to the object, that now is moving and now has kinetic energy.

Now, is kinda true that in a closed system the total energy is always constant, but it depends on what is our system.

So if we think in our system as you and the object, then in the whole system the energy does not change because the energy that you lost is now on the object, but again, there was a transfer of energy.

So no, your friend is not correct.

3 0

2 years ago

Betelgeuse is the bright red star representing the left shoulder of the constellation Orion. All the following statements about

Sav [38]

Answer:

Option A; ITS SURFACE IS COOLER THAN THE SURFACE OF THE SUN.

Explanation:

A red supergiant star is a larger and brighter type of red giant star. Red supergiants are often variable stars and are between 200 to 2,000 times bigger than the Sun. Example is Betelgeuse.

Betelgeuse is one of the largest known stars, it has a diameter of about 700 times the size of the Sun or 600 million miles, it emits almost 7,500 times as much energy as the Sun, it has a rather low surface temperature (6000F compared to the Sun's 10,000F); this means that it has a more cooler surface than the Sun's surface.

This low temperature also means that the star will appear orange-red in color, and the combination of size and temperature makes it a kind of star called a red super giant.

Although, all the statements above are correct, the only one that can be inferred from the red color of Betelgeuse is that ITS SURFACE IS COOLER THAN THE SURFACE OF THE SUN.

3 0

2 years ago