Sounds of explosions taking place on other planets is not heard by a person on the earth. give reason

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

2 years ago

A boy on a bicycle approaches a brick wall as he sounds his horn at a frequency 400 hz. the sound he hears reflected back from t

Mashutka [201]

As the question is about changing in frequency of a wave for an observer who is moving relative to the wave source, the concept that should come to our minds is "Doppler's effect."

Now the general formula of the Doppler's effect is:
$f = (\frac{g + v_{r}}{g + v_{s}})f_{o}$ -- (A)

Note: We do not need to worry about the signs, as everything is moving towards each other. If something/somebody were moving away, we would have the negative sign. However, in this problem it is not the issue.

Where,
g = Speed of sound = 340m/s.
$v_{r}$ = Velocity of the receiver/observer relative to the medium = ?.
$v_{s}$ = Velocity of the source with respect to medium = 0 m/s.
$f_{o}$ = Frequency emitted from source = 400 Hz.
$f$ = Observed frequency = 408Hz.

Plug-in the above values in the equation (A), you would get:

$408 = ( \frac{340 + v_{r}}{340 + 0})*400$

$\frac{408}{400} = \frac{340 + v_{r}}{340}$

Solving above would give you,
$v_{r}$ = 6.8 m/s

The correct answer = 6.8m/s

7 0

2 years ago

A 3kW oven supplied with 9mJ of energy.How many minutes can it run for?

andreev551 [17]

Hello!

We have the following data:

Time (T) = ? (in minutes)
Power (P) = 3 kW → 3000 W
Energy (E) = 9 MJ → 9000000 J or (W/s)

Formula of the consumption of electric energy:

$P = \frac{E}{T}$

Solving:

$P = \frac{E}{T}$

$P = \frac{E}{T} \to T = \frac{E}{P}$

$T = \frac{9000\diagup\!\!\!\!0\diagup\!\!\!\!0\diagup\!\!\!\!0\:\diagup\!\!\!\!W/s}{3\diagup\!\!\!\!0\diagup\!\!\!\!0\diagup\!\!\!\!0\:\diagup\!\!\!\!W}$

$\boxed{T = 3000\:seconds}$

How many minutes can it run for? (Let's convert in minutes)

1 minute --------- 60 seconds
y minute --------- 3000 seconds

$\frac{1}{y} = \frac{60}{3000}$

Product of extremes equals product of means

$60*y = 1*3000$

$60y = 3000$

$y = \frac{3000}{60}$

$\boxed{\boxed{y = 50\:minutes}}\end{array}}\qquad\quad\checkmark$

I hope this helps! =)

7 0

2 years ago

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What can happen to an electron when sunlight hits it? select all that apply. select all that apply. it can drop down to a lower

vlabodo [156]

There are two possible answers:
- it can move out to a higher electron shell
- it can stay in its original shell

In fact, sunlight consists of photons. When sunlight hits an electron, the electron can absorbs a photon, so it gains energy: as a result, the electron can move to a higher electron shell, which corresponds to a high energy level in the atom, if the energy given by the photon is at least equal to the energy difference between the two levels. However, if the photon energy is not large enough, the electron will stay in the same shell.

4 0

2 years ago

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How many times could Haley fly bewteen the two flowers in 1 minute (60 seconds)

dmitriy555 [2]

30 seconds is the answer

7 0

2 years ago

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