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2 years ago

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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.

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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Answer: the speed at which it falls toward the Earth.

Explanation:

A bullet travelling across Earth's surface with some horizontal velocity is classical example of projectile motion.

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The observed net velocity is the vectorial sum of the vertical and horizontal velocities.

The horizontal velocity is constant, since there is not any force acting in the horizontal axis. Thi is, the object, following the first Law of Newton (inertia law) tends to continue in uniform rectilinear movement (with zero acceleration).

The vertical velocity, this is the velocity at which the bullet falls toward the Earth, is influenced (accelerated) by the action of the gravity of the Earth. So, the vertical velocity is accelerated by the pull of the Earth.

Vertical and horizontal velocities are independent of each other, which means that the speed or the magnitude of the horizontal velocity does not affect the speed at which an object (the bullet) falls toward the Earth.

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2 years ago

Briana swings a ball on the end of a rope in a circle. The rope is 1.5 m long. The ball completes a full circle every 2.2 s. Wha

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2 years ago

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Case 1: A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t₁ = 11.9 seconds to get

olga_2 [115]

Answer:

Part a)

$\omega = 8.17 rad/s$

Part b)

$N = 7.74 rev$

Part c)

$\alpha = 0.69 rad/s^2$

Part d)

$\alpha = 0.48 rad/s^2$

Part e)

$t = 9.14 s$

Explanation:

Part a)

Angular speed is given as

$\omega = 2\pi f$

$\omega = 2\pi(\frac{78}{60})$

$\omega = 8.17 rad/s$

Part b)

Since turn table is accelerating uniformly

so we will have

$\theta = \frac{\omega_f + \omega_i}{2} t$

$\theta = \frac{8.17 + 0}{2}(11.9)$

$2N\pi = 48.6$

$N = 7.74 rev$

Part c)

angular acceleration is given as

$\alpha = \frac{\omega_f - \omega_i}{t}$

$\alpha = \frac{8.17 - 0}{11.9}$

$\alpha = 0.69 rad/s^2$

Part d)

When its angular speed changes to 120 rpm

then we will have

$\omega_2 = 2\pi (\frac{120}{60})$

$\omega_2 = 12.56 rad/s$

number of turns revolved is 15 times

so we have

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

$12.56^2 - 8.17^2 = 2\alpha (2\pi\times 15)$

$\alpha = 0.48 rad/s^2$

Part e)

now for uniform acceleration we have

$\omega_f - \omega_i = \alpha t$

$12.56 - 8.17 = 0.48 t$

$t = 9.14 s$

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2 years ago

A rigid tank contains nitrogen gas at 227 °C and 100 kPa gage. The gas is heated until the gage pressure reads 250 kPa. If the a

aleksley [76]

Answer:

T₂ =602 °C

Explanation:

Given that

T₁ = 227°C =227+273 K

T₁ =500 k

Gauge pressure at condition 1 given = 100 KPa

The absolute pressure at condition 1 will be

P₁ = 100 + 100 KPa

P₁ =200 KPa

Gauge pressure at condition 2 given = 250 KPa

The absolute pressure at condition 2 will be

P₂ = 250 + 100 KPa

P₂ =350 KPa

The temperature at condition 2 = T₂

We know that

$\dfrac{T_2}{T_1}=\dfrac{P_2}{P_1}\\T_2=T_1\times \dfrac{P_2}{P_1}\\T_2=500\times \dfrac{350}{200}\ K\\$

T₂ = 875 K

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A very long uniform line of charge has charge per unit length λ1 = 4.80 μC/m and lies along the x-axis. A second long uniform li

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Answer:

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Explanation:

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Where E1 and E2 are the electric field generated at a distance of r1 and r2 respectively from the line of charges.

The net electric field at point r will be:

$E=E_1+E_2=\frac{1}{2\pi \epsilon_o}(\frac{\lambda_1}{r_1}+\frac{\lambda_2}{r_2})$

a) for y=0.200m, r1=0.200m and r2=0.200m:

$E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.200m}-\frac{2.26*10^{-6}C}{0.200m}}]=228391.8N/C$

b) for y=0.600m, r1=0.600m, r2=0.200m:

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2 years ago