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Ksenya-84 [330]
3 years ago
11

A bird is flying directly toward a stationary bird-watcher and emits a frequency of 1490 Hz. The bird-watcher, however, hears a

frequency of 1505 Hz. What is the speed of the bird, expressed as a percentage of the speed of sound
Physics
1 answer:
Eva8 [605]3 years ago
8 0

Answer:

The speed of the bird is 1.00% of the speed of sound.

Explanation:

The speed of the bird can be found by using the Doppler equation:

f = f_{0}(\frac{v - v_{r}}{v - v_{s}})

Where:

v: is the speed of sound = 343 m/s

f₀: is the frequency emitted = 1490 Hz

f: is the frequency observed = 1505 Hz

v_{r}: is the speed of the receiver = 0 (it is stationary)

v_{s}: is the speed of the source =?

The minus sign of v_{s} is because the source is moving towards the receiver.

By solving the above equation for v_{s} we have:

v_{s} = v - \frac{f_{0}*v}{f} = 343 - \frac{1490*343}{1505} = 3.42 m/s

The above speed in terms of the speed of sound is:

\% v_{s} = \frac{3.42}{343}\times 100 = 1.00 \%

Therefore, the speed of the bird is 1.00% of the speed of sound.  

I hope it helps you!

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borishaifa [10]

Answer:

This equation is based on twin paradox - a phenomena where one of the twin travels to space at a speed close to speed of light and the other remains on earth. the twin from the space on return discovers that the one on earth age faster.

Solution:

t_{o} = 10 years

v = 0.8c

c = speed of light in vacuum

The problem can be solved by time dilation equation:

t = \frac{t_{o}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}            (1)

where,

t = time observed from a different inertial frame

Now, using eqn (1), we get:

t = \frac{10}{\sqrt{1 - \frac{(0.8c)^{2}}{c^{2}}}}

t = 16.67 years

The age of the twin on spaceship according to the one on earth = 25+16.67 =41.66 years

8 0
3 years ago
The aorta is approximately 25 mm in diameter. The mean pressure there is about 100 mmHg and the blood flows through the aorta at
Ksenya-84 [330]

Answer:

Explanation:

25 mm diameter

r₁  = 12.5 x 10⁻³ m radius.

cross sectional area =  a₁

Pressure P₁  = 100 x 10⁻³ x 13.6 x 9.8 Pa

a )

velocity of blood v₁ = .6 m /s

Cross sectional area at blockade = 3/4 a₁

Velocity at blockade area = v₂

As liquid is in-compressible

a₁v₁ = a₂v₂

a₁ x .6 m /s  = 3/4 a₁ v₂

v₂ = .8m/s

b )

Applying Bernauli's theorem formula

P₁ + 1/2 ρv₁² =  P₂ +  1/2 ρv₂²

100 x 10⁻³ x 13.6 x10³x 9.8 + 1/2 X 1060 x .6² = P₂ +  1/2x 1060 x .8²

13328 +190.8 = P₂ + 339.2

P₂ = 13179.6 Pa

= 13179 / 13.6 x 10³ x 9.8 m of Hg

P₂ =  .09888 m of Hg

98.88 mm of Hg

8 0
4 years ago
Star temperature is indicated by?
astraxan [27]
A star's temperature is most likely indicated by the color of it. The hotter the star, the bluer it is. The colder the star, the redder it is. 
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4 years ago
Which is considered computer storage
Korolek [52]

Answer:

Second Option

Explanation:

The "hard drive" or the second option is one of the main components of storing information on a computer. You already have a hard drive built into your computer, or laptop when you buy it, and you can buy additional hard drives in the form of plugins that can store even more data if your original hard drive becomes full of data.

Hope this helps.

8 0
3 years ago
Read 2 more answers
Q7) A box sliding with a velocity of 5 m/s accelerates at 2 m/s^2. How
grigory [225]

Answer:

The box displacement after 6 seconds is 66 meters.

Explanation:

Let suppose that velocity given in statement represents the initial velocity of the box and, likewise, the box accelerates at constant rate. Then, the displacement of the object (\Delta s), in meters, can be determined by the following expression:

\Delta s = v_{o}\cdot t+\frac{1}{2}\cdot a\cdot t^{2} (1)

Where:

v_{o} - Initial velocity, in meters per second.

t - Time, in seconds.

a - Acceleration, in meters per square second.

If we know that v_{o} = 5\,\frac{m}{s}, t = 6\,s and a = 2\,\frac{m}{s^{2}}, then the box displacement after 6 seconds is:

\Delta s = 66\,m

The box displacement after 6 seconds is 66 meters.

5 0
3 years ago
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