All categories

2 years ago

Which of the following statements is an accurate description of sound transmission through various mediums?

Physics

2 answers:

Ilia_Sergeevich [38]2 years ago

6 0

The answer is D. Sound insulators prevent sound from being transmitted better than elastic solids. just took PF test

klio [65]2 years ago

5 0

The answer is D. Sound insulators prevent sound from being transmitted better than elastic solids because sound insulators basically muffle and block the sound waves.

You might be interested in

A crate rests on a flatbed truck which is initially traveling at 17.9 m/s on a level road. The driver applies the brakes and the

Mamont248 [21]

Answer:

The minimum coefficient of friction required is 0.35.

Explanation:

The minimum coefficient of friction required to keep the crate from sliding can be found as follows:

$-F_{f} + F = 0$

$-F_{f} + ma = 0$

$\mu mg = ma$

$\mu = \frac{a}{g}$

Where:

μ: is the coefficient of friction

m: is the mass of the crate

g: is the gravity

a: is the acceleration of the truck

The acceleration of the truck can be found by using the following equation:

$v_{f}^{2} = v_{0}^{2} + 2ad$

$a = \frac{v_{f}^{2} - v_{0}^{2}}{2d}$

Where:

d: is the distance traveled = 46.1 m

$v_{f}$ : is the final speed of the truck = 0 (it stops)

$v_{0}$ : is the initial speed of the truck = 17.9 m/s

$a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2}$

If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.

$\mu = \frac{a}{g}$

$\mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}}$

$\mu = 0.35$

Therefore, the minimum coefficient of friction required is 0.35.

I hope it helps you!

4 0

3 years ago

Humans have three types of cone cells in their eyes, which are responsible for color vision. Each type absorbs a certain part of

xxMikexx [17]

Answer:

Frequency of the light will be equal to $6.97\times 10^{1}Hz$

Explanation:

We have given wavelength of the light $\lambda =430nm=430\times 10^{-9}m$

Velocity of light is equal to $v=3\times 10^8m/sec$

We have to find the frequency of light

We know that velocity is equal to $v=\lambda f$ , here $\lambda$ is wavelength and f is frequency of light

So frequency of light will be equal to $f=\frac{v}{\lambda }=\frac{3\times 10^8}{430\times 10^{-9}}=6.97\times 10^{1}Hz$

So frequency of the light will be equal to $6.97\times 10^{1}Hz$

5 0

2 years ago

M84, M87, and NGC 4258 all have accretion disks around their central black holes for which the rotational velocities have been m

givi [52]

Answer:

For M84:

M = 590.7 * 10³⁶ kg

For M87:

M = 2307.46 * 10³⁶ kg

Explanation:

1 parsec, pc = 3.08 * 10¹⁶ m

The equation of the orbit speed can be used to calculate the doppler velocity:

$v = \sqrt{\frac{GM}{r} }$

making m the subject of the formula in the equation above to calculate the mass of the black hole:

$M = \frac{v^{2} r}{G}$ .............(1)

For M84:

r = 8 pc = 8 * 3.08 * 10¹⁶

r = 24.64 * 10¹⁶ m

v = 400 km/s = 4 * 10⁵ m/s

G = 6.674 * 10⁻¹¹ m³/kgs²

Substituting these values into equation (1)

$M = \frac{( 4*10^{5}) ^{2} *24.64* 10^{16} }{6.674 * 10^{-11} }$

M = 590.7 * 10³⁶ kg

For M87:

r = 20 pc = 20 * 3.08 * 10¹⁶

r = 61.6* 10¹⁶ m

v = 500 km/s = 5 * 10⁵ m/s

G = 6.674 * 10⁻¹¹ m³/kgs²

Substituting these values into equation (1)

$M = \frac{( 5*10^{5}) ^{2} *61.6* 10^{16} }{6.674 * 10^{-11} }$

M = 2307.46 * 10³⁶ kg

The mass of the black hole in the galaxies is measured using the doppler shift.

The assumption made is that the intrinsic velocity dispersion is needed to match the line widths that are observed.

3 0

3 years ago

A runner exerts a net force of 335 n to accelerate at a rate of 2.5 m/s what is the runners mass

Olegator [25]

F=ma
M=F/a
M=335/2.5
M=134 kg

8 0

3 years ago

(1- name 1 of Faouzia’s songs from her stripped album

Mnenie [13.5K]

Answer:

I thought "Born Without a Heart" was pretty good.

Explanation:

5 0

2 years ago