Ask question

Ask question

All categories

3 years ago

14

What frequency must a sound wave have in air to have the same wavelength as a 750 Hz sound wave in a platinum bar? Vair = 340 m/

s, Vpt = 2800 m/s,

1 answer:

amm18123 years ago

7 0

Answer:

f = 91.1 Hz

Explanation:

As we know that it is given here that wavelength of sound in platinum is same as the wavelength of sound in air

so we can use the formula of wavelength in two mediums

$\frac{speed}{frequency} = \frac{speed}{frequency}$

now it is given that

$v_{air} = 340 m/s$

$v_{pt} = 2800 m/s$

frequency of sound in platinum is 750 Hz

now frequency of sound in air = f

now from above formula

$\frac{340}{f} = \frac{2800}{750}$

$f = 91.1 Hz$

You might be interested in

calculate the distance a plane flies on a 7.95 hour flight from Chicago to London. Assume a constant speed of 800.0 km/h

soldier1979 [14.2K]

Answer:

6360 km

Explanation:

Use the kinematics equation $x=v_ot+\frac{1}{2}at^2$ . We are given t = 7.95 hours and a = 0 m/s^2 (constant speed means there is no acceleration). Solve for x.

$x=(800)(7.95)+\frac{1}{2}(0)(7.95^2)\\x=6360 \ km$

4 0

3 years ago

A long, thin straight wire with linear charge density λ runs down the center of a thin, hollow metal cylinder of radius R. The c

Delvig [45]

Answer:

$E=\dfrac{\lambda }{2\pi \varepsilon _or}$

Explanation:

Given that

For straight wire

Charge density= λ

For hollow metal cylinder

Charge density=2 λ

We know that electric filed for wire given as

$E_w=\dfrac{\lambda_{wire} }{2\pi \varepsilon _or}$

$E_w=\dfrac{\lambda }{2\pi \varepsilon _or}$

Now the electric filed due to hollow metal cylinder

$E_c=\dfrac{\lambda_{cylinder} }{2\pi \varepsilon _or}$

$E_c=\dfrac{2\lambda }{2\pi \varepsilon _or}$

Now by considering the Gaussian surface r<R then only electric fild due to wire will present.So

At r<R

$E=\dfrac{\lambda }{2\pi \varepsilon _or}$

5 0

3 years ago

"Which of the following statements about electrons is not true?

gogolik [260]

Answer:

B) Within an atom, an electron can have only particular energies.

Explanation:

As we know that electrons have energy but apart from electrons we know that protons and neutrons inside the nucleus of atom will also have energy in them.

rest all the statements are true as we have

A) Electrons orbit the nucleus rather like planets orbiting the Sun.

TRUE, because electrons can move in stationary orbit around the nucleus

C) Electrons can jump between energy levels in an atom only if they receive or give up an amount of energy equal to the difference in energy between the energy levels.

Difference amount of energy is lost or absorbed by the electron in form of photons

D) An electron has a negative electrical charge.

Charge of an electron is given as $-1.6 \times 10^{-19} C$

E) Electrons have very little mass compared to protons or neutrons

Mass of an electron is given as

$m_e = 9.11 \times 10^{-31} kg$

mass of proton or neutron

$m_p = 1.67 \times 10^{-27} kg$

7 0

3 years ago

A car moves in a straight line at 22.0 m/s for 10.0miles, then at 30.0 m/s for another 10.0miles. Calculate the car’s average sp

maw [93]

Answer: 25.38 m/s

Explanation:

We have a straight line where the car travels a total distance $D$ , which is divided into two segments $d=10 miles$ :

$D=d+d=2d$ (1)

Where $d=10mi \frac{1609.34 m}{1 mi}=16093.4 m$

On the other hand, we know speed is defined as:

$S=\frac{d}{t}$ (2)

Where $t$ is the time, which can be isolated from (2):

$t=\frac{d}{S}$ (3)

Now, for the first segment $d=16093.4 m$ the car has a speed $S_{1}=22m/s$ , using equation (3):

$t_{1}=\frac{d}{S_{1}}$ (4)

$t_{1}=\frac{16093.4 m}{22m/s}$ (5)

$t_{1}=731.518 s$ (6) This is the time it takes to travel the first segment

For the second segment $d=16093.4 m$ the car has a speed $S_{1}=30m/s$ , hence:

$t_{2}=\frac{d}{S_{2}}$ (7)

$t_{2}=\frac{16093.4 m}{30m/s}$ (8)

$t_{2}=536.44 s$ (9) This is the time it takes to travel the secons segment

Having these values we can calculate the car's average speed $S_{ave}$ :

$S_{ave}=\frac{d + d}{t_{1} + t_{2}}=\frac{2d}{t_{1} + t_{2}}$ (10)

$S_{ave}=\frac{2(16093.4 m)}{731.518 s +536.44 s}$ (11)

Finally:

$S_{ave}=25.38 m/s$

3 0

3 years ago

When atoms _____, they break down into atoms of different elements and release energy.

SCORPION-xisa [38]

Answer:

It may be combine?

Do you have multiple choice i can see?

Explanation:

5 0

2 years ago