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Leviafan
2 months ago
10

According to a rule-of-thumb. every five seconds between a lightning flash and the following thunder gives the distance to the f

lash in miles. Assuming that the flash of light arrives in essentially no time at all, estimate the speed of sound in m/s from this rule. Express your answer to one significant figure and include the appropriate units. What would be the rule for kilometers?
Physics
1 answer:
ValentinkaMS [3.4K]2 months ago
3 0

Answer:

S_{s}=300 m/s

According to the guideline for kilometers, every three seconds between a lightning strike and the subsequent thunder indicates the distance to the flash in kilometers.

Explanation:

To calculate the speed of sound in meters per second, we need to utilize certain conversion factors. One mile corresponds to 5 seconds after witnessing the lightning. Furthermore, 1 mile comprises 5280 feet, and 1 foot is equivalent to 0.3048 meters. This information is sufficient to solve the issue. The conversion ratios can be set up like this:

\frac{1mi}{5s}*\frac{5280ft}{1mi}*\frac{0.3048m}{1ft}=321.87m/s

Observe how the ratios are organized such that the units cancel out during calculations. One ratio has miles in the numerator while the other has them in the denominator, leading to cancellation. The same applies to the feet.

The question requires us to provide the answer to one significant figure, resulting in the speed of sound rounding to 300m/s.

For the second part, we will again utilize conversions. This time we will set our ratios in reverse and realize that there are 1000 meters in 1 kilometer, leading us to:

\frac{5s}{1mi}*\frac{1mi}{5280ft}*\frac{1ft}{0.3048m}*\frac{1000m}{1km}=3.11s/km

This signifies that for every 3.11 seconds, the distance to the lightning strike is 1 kilometer. Since this is a fabric of general knowledge, we round to the nearest whole number for simplicity, establishing the guideline:

According to the rule for kilometers, every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.

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How many electrons does it take to make 80 μc (microcoulombs) of charge?
serg [3582]

The charge for a single electron is 1.602*10^ -19 C

80 µC can be expressed as 8*10^ - 5 C

This is basic arithmetic

Total Charge divided by the charge of one electron = Number of electrons

(8*10^ -5 C / 1.602*10^ -19 C) equals 4.99 * 10^14 electrons.

6 0
2 months ago
An organ pipe open at both ends has a radius of 4.0 cm and a length of 6.0 m. what is the frequency (in hz) of the third harmoni
inna [3103]

When air is forced into the open pipe,

L = \frac{nλ}{2}

where n represents any whole number such as 1,2,3,4, etc., and λ denotes the wavelength of the oscillation

This implies λ=\frac{2L} {n}

It is important to note that n=1 corresponds to the fundamental frequency, n=2 corresponds to the first harmonic, and so forth.

Thus, the third harmonic will be for n=4

With L=6m and n=4, solving for λ yields:

λ=\frac{(2)*(6)}{4} =3m

The connection between frequency (f), sound speed (c), and wavelength (λ) is given by:

c=f.λ or f= \frac{c}{λ}

Therefore, f=\frac{344}{3}

≈115 Hz

8 0
1 month ago
2. An airplane traveling north at 220. meters per second encounters a 50.0-meters-per-second crosswind
Yuliya22 [3333]

The speed resulting from the plane is (3) 226 m/s

Reasoning:

We can determine the plane's resultant speed using the Pythagorean Theorem because the two speeds form a right angle (creating a right triangle).

Thus, the computation is as follows:

ResultantSpeed=\sqrt{VerticalSpeed^{2}+HorizontalSpeed^{2}}\\\\ResultantSpeed=\sqrt{(220\frac{m}{s})^{2}+50\frac{m}{s})^{2}

ResulntantSpeed=\sqrt{48400\frac{m^{2} }{s^{2} }+2500\frac{m^{2} }{s^{2} } } \\\\ResultantSpeed=\sqrt{50900\frac{m^{2} }{s^{2} }}=226\frac{m}{s}

Consequently, the plane's resultant speed is (3) 226 m/s

Have a wonderful day!

5 0
2 months ago
Wire A has the same length and twice the radius of wire B. Both wires are made of the same material and carry the same current.
serg [3582]

Answer:

V_A= \frac{I_A}{n_A e A_A}= \frac{I}{ne 4A_B}= \frac{1}{4} \frac{I}{neA_B}

V_B= \frac{I_B}{n_B e A_B}= \frac{I}{ne A_B}

Consequently, we find that:

V_A = \frac{1}{4} V_B

Thus, the most suitable answer would be:

a. vA = vB/4

Explanation:

In this situation, we can establish the following conditions:

L_A = L_B =L both wires share the same length

both wires carry an identical currentI_A = I_B =I

Both wires are constructed of the same material, indicating that the electron density (n) remains constant across both wires

n_A = n_B =n

We also know that r_A = 2 r_B where r signifies the radius.

Given that wires are cylindrical in shape, we can determine the area for each case:

A_A= \pi r^2_A = \pi (2r_B)^2 = 4 \pi r^2_B= 4 A_B

A_B = \pi r^2_B

Thus, we conclude that

A_A = 4 A_B

Now we are aware that the drift velocity of an electron in a wire can be described by:

v_d = \frac{I}{neA}

Where I denotes the current, n is the electron density, e represents the electron charge, and A signifies the area.

By substituting, we arrive at:

V_A= \frac{I_A}{n_A e A_A}= \frac{I}{ne 4A_B}= \frac{1}{4} \frac{I}{neA_B}

V_B= \frac{I_B}{n_B e A_B}= \frac{I}{ne A_B}

So we observe that:

V_A = \frac{1}{4} V_B

Thus, the most fitting answer is:

a. vA = vB/4

6 0
1 month ago
A particular brand of gasoline has a density of 0.737 g/mLg/mL at 25 ∘C∘C. How many grams of this gasoline would fill a 14.9 gal
Yuliya22 [3333]

Answer:

The answer to your inquiry is Mass = 41230.7 g or 41.23 kg.

Explanation:

Data

Density = 0.737 g/ml

Mass = ?

Volume = 14.9 gal

1 gal = 3.78 l

Process

1.- Convert gallons to liters

1 gal ---------------- 3.78 l

14.9 gal ------------- x

x = 56.44 l

2.- Convert liters to milliliters

1 l ------------------- 1000 ml

56.44 l --------------- x

x = (56.44 x 1000) / 1

x = 56444 ml

3.- Calculate the mass

Formula

Density = \frac{mass}{volume}

Solving for mass

Mass = density x volume

Substituting values

Mass = 0.737 x 56444

Result

Mass = 41230.7 g or 41.23 kg.

3 0
2 months ago
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