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hammer [34]
2 years ago
8

4. Going back to the dog whistle in question 1, what is the minimum riding speed needed to be able to hear the whistle? Remember

, you can assume the following things: The whistle you use to call your hunting dog has a frequency of 21.0 kHz, but your dog is ignoring it. You suspect the whistle may not be working, but you can't hear sounds above 20.0 kHz. The speed of sound is 330 m/s at the current air temperature.
Physics
1 answer:
jeyben [28]2 years ago
8 0

Answer:

The minimum riding speed relative to the whistle (stationary) to be able to hear the sound at 21.0 kHz frequency is 15.7  m/s

Explanation:

The Doppler shift equation is given as follows;

f' = \dfrac{v - v_o}{v + v_s} \times f

Where:

f' = Required observed frequency = 20.0 kHz

f = Real frequency = 21.0 kHz

v = Sound wave velocity = 330 m/s

v_o = Observer velocity = X m/s

v_s = Source velocity = 0 m/s (Assuming the source is stationary)

Which gives;

20 = \dfrac{330- v_o}{330+0} \times 21

330 - v_o = (20/21)*330

v_o = 330 - (20/21)*330 = 15.7 m/s

The minimum riding speed relative to the whistle (stationary) to be able to hear the sound at 21.0 kHz frequency = 15.7  m/s.

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The position function x(t) of a particle moving along an x axis is x = 4.00 - 6.00t2, with x in meters and t in seconds. (a) at
elena-14-01-66 [18.8K]

The position function x(t) of a particle moving along an x axis is x=4.00 - 6.00t^2

a) The point at which particle stop, it's velocity = 0 m/s

  So dx/dt = 0

        0 = 0- 12t = -12t

  So when time t= 0, velocity = 0 m/s

    So the particle is starting from rest.

At t = 0 the particle is (momentarily) stop

b) When t = 0

 x=4.00 - 6.00*0^2 = 4m

SO at x = 4m the particle is (momentarily) stop

c) We have x=4.00 - 6.00t^2

   At origin x = 0

  Substituting

         0 = 4.00 - 6.00t^2\\ \\ t^2 = \frac{2}{3}

         t = 0.816 seconds or t = - 0.816 seconds

So when  t = 0.816 seconds and t = - 0.816 seconds, particle pass through the origin.

5 0
2 years ago
A block rests on a flat plate that executes vertical simple harmonic motion with a period of 0.74 s. What is the maximum amplitu
Mumz [18]

Answer:

maximum amplitude  = 0.13 m

Explanation:

Given that

Time period T= 0.74 s

acceleration of gravity g= 10 m/s²

We know that time period of simple harmonic motion given as

T=\dfrac{2\pi}{\omega}

0.74=\dfrac{2\pi}{\omega}

ω = 8.48 rad/s

ω=angular frequency

Lets take amplitude = A

The maximum acceleration given as

a= ω² A

The maximum acceleration should be equal to g ,then block does not separate

a= ω² A

10= 8.48² A

A=0.13 m

maximum amplitude  = 0.13 m

8 0
2 years ago
A tennis player serves a tennis ball such that it is moving horizontally when it leaves the racquet. When the ball travels a hor
nalin [4]

Answer:

u_x=38.13\ m/s

Explanation:

Given that initially ball moves in the horizontal direction ,it means that the velocity in the vertical direction is zero.

Horizontal distance = 13 m

Vertical distance = 57 cm

Lets take time to cover 57 cm distance in vertical direction is t.

We know that g is the constant acceleration in the vertical direction so we can apply the equation of motion in the vertical direction.

S=u_yt+\dfrac{1}{2}gt^2

Here u_y=0

S= 57 cm

0.57=0\times t+\dfrac{1}{2}\times 9.81\times t^2

t=0.34 s

Now in the horizontal direction

x=u_xt

Here x=13 m

t= 0.34 s

So

13=u_x\times 0.34

u_x=38.13\ m/s

So the initial speed of ball is 38.13 m/s.

7 0
1 year ago
A projectile has an initial horizontal velocity of 15 meters per second and an initial vertical velocity of 25 meters per second
Artyom0805 [142]

Answer:

75 m

Explanation:

The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.

The horizontal component of the velocity of the projectile is

v_x = 15 m/s

and it is constant during the motion;

the total time of flight is

t = 5 s

Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:

d= v_x t =(15 m/s)(5 s)=75 m

5 0
2 years ago
Temperature difference in the body. The surface temperature of the body is normally about 7.00 ∘C lower than the internal temper
egoroff_w [7]

Answer:

7 K.

12. 6 °F

Explanation:

Convert the individual temperatures to Kelvin (Surface temperature and internal temperature) before calculating the temperature difference of the body,

Let The Surface temperature Be = X °C

And the internal Temperature will be = (X + 7) °C

Converting the surface and the internal temperature to temperature in Kelvin

Surface Temperature of the body (K) = (X + 273) K

Internal Temperature of the body (K) = (X + 7) + 273 = (X + 280) K.

∴ Temperature difference of the body (K) = Internal temperature(K) - surface temperature(K) = (X + 279) - (X + 280)

   = X - X + 280 - 273 = 7 K.

∴Temperature difference of the body (K) = 7 K

Also for Fahrenheit, Convert the individual temperatures (Surface temperature and internal temperature) to Fahrenheit before calculating the temperature difference of the body.

We use , F = 1.8C + 32

Where C = temperature in Celsius.

also,

Let The Surface temperature Be = X °C

And the internal Temperature of the body will be = (X + 7) °C

Converting to Fahrenheit

Surface Temperature of the body = 1.8X + 32 °F

Internal Temperature of the body = 1.8(X+7) + 32 = 1.8X + 12.6 + 32

Internal Temperature of the body = 1.8X + 44.6 °F

∴ The temperature difference of the body (°F) = Internal temperature(°F) - surface temperature(°F) = (1.8X + 44.6) - (1.8X + 32)

      surface temperature(°F) = 1.8X - 1.8X  + 44.6 - 32

       surface temperature(°F) = 12. 6 °F.

   

3 0
2 years ago
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