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2 years ago

How does the doppler effect change wave speed?

1 answer:

4 0

It doesn't change wave speed at all.

It only changes the wavelength perceived by the observer
if he is moving toward or away from the source.

Notice that if one observer is moving toward a source, and another one
is moving away from the same source, then one observer would see an
abnormally long wavelength, and the other one would see an abnormally
short wavelength, both from the same source.

The Doppler Effect ... like beauty ... is in the eye of the observer.

You might be interested in

Suppose the coefficient of kinetic friction between mA and the plane in the figure(Figure 1) is μk = 0.15, and that mA=mB=2.7kg.

luda_lava [24]

A )
T = mB g + mB a
T + mA a - mA g sin 35° = (Mi) mA g cos 35°
------------------------------------------------------------
T = 2.7 · 9.81 + 2.7 a
T = 26.487 + 2.7 a
26.487 + 2.7 a + 2.7 a - 2.7 · 9.81 · 0.574 = 0.15 · 2.7 · 9.81 · 0.819
5.4 a + 26.487 - 15.2023 = 3.2539
5.4 a = 8.0296
a = 1.487 ≈ 1.5 m/s²
B )
T = 2,7 · 9.81 = 26.487
26.487 - 15.2035 = (Mi) · 2.7 · 9.81 · 0.819
11.2835 = (Mi) · 21.69
(Mi) = 11.2835 : 21.69 = 0.52

4 0

3 years ago

20.0 -kg cannonball is fired from a cannon with muzzle speed of 1000m/s at an angle of 37.0° with the horizontal. A second ball

Dovator [93]

The mechanical energy for the first and the second ball is

$10 ^{7} \: joules.$

Mass of the first ball = 20 kg

The initial speed at which a cannonball is fired from a cannon =1000 m/s

The angle made by the cannonball while being fired from the cannon = 37°

The maximum height reached by the first ball is,

$= \frac{ u {}^{2} _{1}sin {}^{2} θ}{2g}$

$= \frac{ {1000}^{2} sin {}^{2}37°}{2 \times 9.8}$

$= 18478.69 \: m$

The maximum height of the first cannonball is 17478.69 m.

The initial speed at which a cannonball is fired from a cannon =1000 m/s

The angle made by the cannonball while being fired from the cannon = 90 °

$= \frac{ u {}^{2} _{2}sin {}^{2} θ}{2g}$

$= \frac{ 1000{}^{2}sin^{2} 90°}{2 \times 9.8}[tex] = 51020.41 \: m$

For the first ball, total mechanical energy= Potential energy at maximum height + kinetic energy at the maximum height

So, the total mechanical energy is,

= mgh \: + \frac{1}{2}mv {}^{2} _{x}[/tex]

$= 20 \times 9.8 \times 18478.64 \times \frac{20}{2} (1000 \: cos37 °)$

$= 10 ^{7} The potential energy at the maximum height, = m _{2}gh$

$= 20 \times 9.8 \times 51020.41$

$= 10 ^{7} \:J$

Therefore, the total mechanical energy for the first and the

$\:second \: cannonball \: is \: 10 ^{7} \:joules.$

To know about energy, refer to the below link:

brainly.com/question/1932868

#SPJ4

5 0

1 year ago

Five different forces act on an object. Is it possible for the net force on the object to be zero?

bazaltina [42]

No because there must be an even # if their is an even amount one of the forces isn’t being cancelled

4 0

2 years ago

How could you increase the sleds acceleration

gladu [14]

By putting this special transportation plastic on the bottom of the sled, because the transportation plastic is slick. that is what the transport bins slide around on. (p.s) the plastic is really expensive!

6 0

2 years ago

A hollow cylinder that is rolling without slipping is given a velocity of 5.0 m/s and rolls up an incline to a vertical height o

inysia [295]

Answer:

The hollow cylinder rolled up the inclined plane by 1.91 m

Explanation:

From the principle of conservation of mechanical energy, total kinetic energy = total potential energy

$M.E_T = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 + mgh$

The total energy at the bottom of the inclined plane = total energy at the top of the inclined plane.

$\frac{1}{2}mv_i^2 + \frac{1}{2} I \omega_i^2 + mg(0) = \frac{1}{2}mv_f^2 + \frac{1}{2} I \omega_f^2 + mgh$

moment of inertia, I, of a hollow cylinder = ¹/₂mr²

substitute for I in the equation above;

$\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_i^2) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_f^2) + mgh\\\\ but \ v = r \omega\\\\\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}m v_i^2 ) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}m v_f^2) + mgh\\\\\frac{1}{2}mv_i^2 +\frac{1}{4}mv_i^2 = \frac{1}{2}mv_f^2 +\frac{1}{4}mv_f^2 +mgh\\\\\frac{3}{4}mv_i^2 = \frac{3}{4}mv_f^2 +mgh\\\\mgh = \frac{3}{4}mv_i^2 - \frac{3}{4}mv_f^2\\\\gh = \frac{3}{4}v_i^2 - \frac{3}{4}v_f^2\\\\$

$h = \frac{3}{4g}(v_1^2 -v_f^2)$

given;

v₁ = 5.0 m/s

vf = 0

g = 9.8 m/s²

$h = \frac{3}{4g}(v_1^2 -v_f^2) =\frac{3}{4*9.8}(5^2 -0) = 1.91 \ m$

Therefore, the hollow cylinder rolled up the inclined plane by 1.91 m

5 0

2 years ago