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romanna [79]
2 years ago
15

What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from

its siren? The speed of sound on this day is 345 m/s.
Physics
1 answer:
Strike441 [17]2 years ago
8 0

To develop this problem we will apply the concepts related to the Doppler effect. The frequency of sound perceive by observer changes from source emitting the sound. The frequency received by observer f_{obs} is more than the frequency emitted by the source. The expression to find the frequency received by the person is,

f_{obs} = f_s (\frac{v_w}{v_w-v_s})

f_s= Frequency of the source

v_w= Speed of sound

v_s= Speed of source

The velocity of the ambulance is

v_s = 119km/h (\frac{1000m}{1km})(\frac{1h}{3600s})

v_s = 30.55m/s

Replacing at the expression to frequency of observer we have,

f_{obs} = 800Hz(\frac{345m/s}{345m/s-30.55m/s})

f_{obs} = 878Hz

Therefore the frequency receive by observer is 878Hz

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maria [59]
The heat released by the water when it cools down by a temperature difference \Delta T is
Q=mC_s \Delta T
where
m=432 g is the mass of the water
C_s = 4.18 J/g^{\circ}C is the specific heat capacity of water
\Delta T =71^{\circ}C-18^{\circ}C=53^{\circ} is the decrease of temperature of the water

Plugging the numbers into the equation, we find
Q=(432 g)(4.18 J/g^{\circ}C)(53^{\circ}C)=9.57 \cdot 10^4 J
and this is the amount of heat released by the water.
7 0
2 years ago
Kiting during a storm. The legend that Benjamin Franklin flew a kite as a storm approached is only a legend — he was neither stu
Dvinal [7]

Answer:

The current is   I  =  1.1434*10^{-5}}\  A

Explanation:

From the question we are told that

   The radius of the kite string is  R =  2.02 mm =  0.00202 \ m

   The  distance it extended upward is   D =  0.823 km = 823 \  m

   The thickness of the water layer is d = 0.506 mm  =  0.000506 \  m

   The resistivity is  \rho =  159\ \Omega  \cdot m

   The potential  difference is  V  =   186 MV =  186 *10^{6} \  V

Generally the cross sectional area of the water layer is mathematically represented as

      A =  \pi r^2

Here  r is mathematically represented as

      r =  [(R + d ) - R]

=>   r =  [(0.00202 +  0.000506 ) - 0.00202]

=>  r =  0.000506

=>     A = 3.142 *  [0.000506]^2  

=>     A = 8.0447*10^{-7}\ m^2  

Generally the resistance of the water is mathematically represented as

    R =  \frac{\rho  * D }{A}

=>   R =  \frac{159  *823 }{8.0447*10^{-7}}

=>   R = 1.62662 * 10^{11} \  \Omega

Generally the current is mathematically represented as

      I  =  \frac{V}{R}

=>    I  =  \frac{186 *10^{6} }{1.62662 * 10^{11}}

=>    I  =  1.1434*10^{-5}}\  A

8 0
2 years ago
When kicking a football, the kicker rotates his leg about the hip joint. (a) If the velocity of the tip of the kicker’s shoe is
Maurinko [17]

Answer:

Part a)

\omega = 33.33 rad/s

Part b)

F = 500 N

Part c)

R_{max} = 40.8 m

Explanation:

Part a)

As we know that the velocity of the tip of the kicker's shoe is given as

v = 35 m/s

also the length of the tip of the shoe from his hip joint is given as

L = 1.05 m

now the angular speed is given as

\omega = \frac{v}{L}

\omega = \frac{35}{1.05}

\omega = 33.33 rad/s

Part b)

As we know that force on the ball is given as rate of change in momentum of the ball

so it is given as

F = \frac{\Delta P}{\Delta t}

so we have

F = \frac{m(v_f - v_i)}{\Delta t}

F = \frac{0.500(20 - 0)}{20 \times 10^{-3}}

F = 500 N

Part c)

As we know that the formula of range is given as

R = \frac{v^2 sin(2\theta)}{g}

now for maximum range we know

\theta = 45 degree

R_{max} = \frac{v^2}{g}

R_{max} = \frac{20^2}{9.8}

R_{max} = 40.8 m

6 0
2 years ago
A 3.0-cm tall statue is 48 cm in front of a convex lens having a focal length of i. (2 pts) Is the image of the statue real or v
erica [24]

Explanation:

Given that,

Height of object = 3.0 cm

Distance of object u= 48 cm

Focal length = 48 cm

We need to calculate the image distance

Using formula of lens

\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}

Put the value into the formula

\dfrac{1}{20}=\dfrac{1}{v}+\dfrac{1}{-48}

\dfrac{1}{v}=\dfrac{1}{20}+\dfrac{1}{48}

\dfrac{1}{v}=\dfrac{17}{240}

v=14.11\ cm

(I). The image is real.

(II).  The distance of the image from the lens is 14.11 cm.

(II). The image is inverted.

(IV). We need to calculate the height of the image

Using formula of magnification

m = \dfrac{v}{u}=\dfrac{h'}{h}

\dfrac{v}{u}=\dfrac{h'}{h}

Put the value into the formula

\dfrac{14.11}{48}=dfrac{h'}{3.0}

h'=\dfrac{14.11}{48}\times3.0

h'=0.88\ cm

The height of the image is 0.88 cm.

Hence, This is the required solution.

6 0
2 years ago
Songbirds often eat berries. Berry seeds are activated by the acids located in a bird's stomach. Once the bird's body eliminates
LenKa [72]
I don't think its a niche or community.My best guess is probably the habitat.
Hope that helps :)
6 0
2 years ago
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