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shtirl [24]
3 years ago
12

A sound wave travels in air toward the surface of a freshwater lake and enters into the water. The frequency of the sound does n

ot change when the sound enters the water. The wavelength of the sound is 2.86 m in the air, and the temperature of both the air and the water is 20 degrees C. What is the wavelength in the water
Physics
1 answer:
Shkiper50 [21]3 years ago
7 0

Answer:

Explanation:

velocity of sound in air at 20⁰C is 343 m /s

velocity of sound in water at 20⁰C is 1481 m /s

The wavelength of the sound is 2.86 m in the air so its frequency

= 343 / 2.86 = 119.93 .

This frequency of  119.93 will remain unchanged in water .

wavelength in water = velocity in water / frequency

= 1481 / 119.93

= 12. 35 m .

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ExtremeBDS [4]

Explanation:

In the given situation two forces are working. These are:

1) Electric force (acting in the downward direction) = qE

2) weight (acting in the downward direction) = mg

Therefore, work done by all the forces = change in kinetic energy

Hence, qE \times S + mg \times S = 0.5 \times mv^{2}

     1.6 \times 10^{-19} \times 1000 + 9.1 \times 10^{-31} \times 9.8 \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2}

It is known that the weight of electron is far less compared to electric force. Therefore, we can neglect the weight  and the above equation will be as follows.

   (1.6 \times 10^{-19} \times 1000) \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2
}

         v = sqrt{\frac{1.6 \times 10^{-19}}{(0.5 \times 9.1 \times 10^{-31})}

           = 592999 m/s

Since, the electron is travelling downwards it means that it looses the potential energy.

8 0
3 years ago
Determine the empirical formula for a compound that is 36.86% n and 63.14% o by mass.
olga2289 [7]

Answer:

NO_{1.499}

Explanation:

Let assume that 100 kg of the compound is tested. The quantity of kilomoles for each element are, respectively:

n_{N} = \frac{36.86\,kg}{14.006\,\frac{kg}{kmol} }

n_{N} = 2.632\,kmol

n_{O} = \frac{63.14\,kg}{15.999\,\frac{kg}{kmol} }

n_{O} = 3.946\,kmol

Ratio of kilomoles oxygen to kilomole nitrogen is:

n^{*} = \frac{3.946\,kmol}{2.632\,kmol}

n^{*}= 1.499

It means that exists 1.499 kilomole oxygen for each kilomole nitrogen.

The empirical formula for the compound is:

NO_{1.499}

8 0
3 years ago
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Frost wedging occurs when water seeps into a crack in a rock, freezes, and causes the crack in the rock to widen.
Ilia_Sergeevich [38]

Answer:

the answer is true.

3 0
3 years ago
It takes 52,000 Joules to heat a cup of coffee to boiling from room temperature. How long a piece of 20 cm wide Aluminum foil wo
vovikov84 [41]

Answer:

L = 1.11 x 10^{6} m, is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.

Explanation:

Solution:

Data Given:

Heat Energy = 52000 J

Dielectric Constant of the plastic Bag = 3.7 = K

Thickness = 2.6 x 10^{5} m =d

V = 610 volts

A = width x Length

width = 20 cm = 20 x 10^{-2} m

Length = ?

So,

we know that,

U = 1/2 C Δv^{2}

U = 52000 J

C = ?

V = 610 volts'

So,

U = 1/2 C Δv^{2}  

52000 J = (0.5) x (C) x (610^{2})

C = 0.28 F

And we also know that,

C = \frac{K*E*A}{d}

E = 8.85 x 10 ^{-12}

K = 3.7

A = 0.20 x L

d = 2.6 x 10^{5} m

Plugging in the values into the formula, we get:

0.28 = \frac{3.7 * 8.85 .10^{-12} * (0.20 . L) }{2.6 . 10^{5} }

Solving for L, we get:

L = 1.11 x 10^{6} m,

is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.  

7 0
3 years ago
What must be the length of a simple pendulum if its oscillation frequency is to be equal to that of an air-track glider of mass
Anvisha [2.4K]

Answer:

the length of the simple pendulum is 0.25 m.

Explanation:

Given;

mass of the air-track glider, m = 0.25 kg

spring constant, k = 9.75 N/m

let the length of the simple pendulum = L

let the frequency of the air-track glider which is equal to frequency of simple pendulum = F

The oscillation frequency of air-track glider is calculated as;

F = \frac{1}{2\pi } \sqrt{\frac{k}{m} } \\\\F = \frac{1}{2\pi } \sqrt{\frac{9.75}{0.25} } \\\\F = 0.994 \ Hz

The frequency of the simple pendulum is given as;

F = \frac{1}{2\pi} \sqrt{\frac{g}{l} } \\\\2\pi(F) = \sqrt{\frac{g}{l} } \\\\2\pi (0.994) = \sqrt{\frac{9.8}{l} } \\\\6.2455 = \sqrt{\frac{9.8}{l} } \\\\(6.2455)2 = \frac{9.8}{l} \\\\39.006 = \frac{9.8}{l} \\\\l = \frac{9.8}{39.006} \\\\l = 0.25 \ m

Thus, the length of the simple pendulum is 0.25 m.

8 0
2 years ago
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