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kvasek [131]
3 years ago
6

Donna wants to calculate the speed at which a sound wave is traveling. She knows that the frequency of the wave is 680 hertz and

that the wave has a wavelength of 0.5 meters. What is the speed of the sound wave?
Physics
2 answers:
Vera_Pavlovna [14]3 years ago
7 0

Answer:

aaaaaaaaaaaaaaaaaaaaaaa

Explanation:

Anettt [7]3 years ago
4 0

f = frequency of the sound wave = 680 hertz

λ = wavelength of the sound wave = 0.5 meters

v = speed of sound wave

we know that , speed of sound wave is given as

speed of sound wave = frequency of sound wave x wavelength of sound wave

v = f λ

inserting the above values in the formula above

v = (680 hertz) (0.5 meters)

v = 340 meter/second


hence the speed of sound wave comes out to be 340 meter/second

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stellarik [79]

Answer:

Option D: 21.8 degrees

Explanation:

In a parallel RL circuit, the current in the resistor R and that in the inductor L are separated among themselves 90 degrees as illustrated in the attached image. In the image the current in the resistor is represented in orange, that of the inductor in blue, and the total current (vector addition of the previous two) is represented  in red, forming a certain angle (theta) with respect to the current in the resistor. The output voltage is the same as the input voltage as measured over the resistor R.

Therefore, the phase angle that separated output voltage and total current can be obtained using the fact that tan(phase angle) = \frac{I_l}{I_R} = \frac{x}{y} \frac{4}{10}, therefore the angle is the arctangent of 4/10:

arctang(\frac{4}{10} )= 21.801 degrees.

8 0
3 years ago
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Which of the following is true about a planet orbiting a star in uniform circular
Mnenie [13.5K]

The velocity vector of the planet points toward the center of the  circle is the following is true about a planet orbiting a star in uniform circular  motion.

A. The velocity vector of the planet points toward the center of the  circle.

<u>Explanation:</u>

Motion of the planet around the star is mentioned to be uniform and around a circular path. Objects in uniform circular motion motion has constant angular speed but the velocity of the object will not remain constant. Since the planet is in circular motion the direction of velocity vector at a particular point is tangential to the circular path at that particular point.

Thus at every point, the direction of velocity vector changes and this means the velocity is never constant. The objects in uniform circular motion has centripetal acceleration which means that velocity vector of the planet points toward the center of the  circle.

7 0
3 years ago
I need help with this physics question
insens350 [35]
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4 0
3 years ago
A proton, starting from rest, accelerates through a potential difference of 1.0 kV and then moves into a magnetic field of 0.040
Minchanka [31]

Answer:

r = 0.11 m

Explanation:

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F_{c} = F_{B} \rightarrow \frac{m*v^{2}}{r} = qvB (1)

<u>Where:</u>

<em>m: is the proton's mass =  1.67*10⁻²⁷ kg</em>

<em>v: is the proton's velocity</em>

<em>r: is the radius of the proton's orbit</em>

<em>q: is the proton charge = 1.6*10⁻¹⁹ C</em>

<em>B: is the magnetic field = 0.040 T </em>

Solving equation (1) for r, we have:

r = \frac{mv}{qB}   (2)

By conservation of energy, we can find the velocity of the proton:

K = U \rightarrow \frac{1}{2}mv^{2} = q*\Delta V   (3)

<u>Where:</u>

<em>K: is kinetic energy</em>

<em>U: is electrostatic potential energy</em>

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Solving equation (3) for v, we have:

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Now, by introducing v into equation (2), we can find the radius of the proton's resulting orbit:

r = \frac{mv}{qB} = \frac{1.67 \cdot 10^{-27} kg*4.38 \cdot 10^{5} m/s}{1.6 \cdot 10^{-19} C*0.040 T} = 0.11 m

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I hope it helps you!  

5 0
3 years ago
An L-R-C series circuit, R = 160 Ω , L = 0.790 H , and C = 1.30×10−2 μF . The source has a voltage amplitude of 140 V and a freq
wolverine [178]

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In resonance, XL =XC, so the circuit behaves like it were purely resistive, so Z=R.

Replacing in the expression for power factor, we have:

cos φ = R/Z = R/R = 1

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b) The average power delivered by the source, in resonance, is simply the power dissipated at the resistance R, as follows:

Pavg = V² rms / R = V₀² / √2 / R = 61 W

c) If the circuit remains in resonance, the average power , which does not depends on frequency provided this condition remains, keeps the same, i.e. , 61 W.

7 0
3 years ago
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