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adelina 88 [10]
3 years ago
13

Which of the following statements is accurate? A. Compressions and rarefactions occur throughout a transverse wave. B. The wavel

ength of both transverse and longitudinal waves is measured parallel to the direction of the travel of the wave. C. Amplitude of longitudinal waves is measured at right angles to the direction of the travel of the wave and represents the maximum distance the molecule has moved from its normal position. D. Sound waves passing through the air will do so as transverse waves, which vibrate vertically and still retain their horizontal positions.
Physics
2 answers:
suter [353]3 years ago
6 0

Answer:

B

Explanation:

Eduardwww [97]3 years ago
4 0
<h3><u>Answer;</u></h3>

<em>B. The wavelength of both transverse and longitudinal waves is measured parallel to the direction of the travel of the wave.</em>

<h3><u>Explanation;</u></h3>
  • <em><u>Longitudinal waves are waves in which the vibration of particles is parallel to the direction of the wave motion.</u></em> The vibration of particles generates regions known as compressions and rarefactions. Therefore<em> </em><u><em>the wavelength is the distance between two successive rarefactions or compressions.</em></u>
  • <u><em>Transverse waves on the other hand are those waves in which the vibration of particles is perpendicular to the direction of the wave motion. </em></u>The vibration of particles generates regions of maximum displacement called crests and regions of minimum displacement called troughs. Thus, <em><u>the wavelength is the distance between two successive crests or troughs.</u></em>
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A 100-turn, 3.0-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 60%u2218 away from vertical increase
Xelga [282]

Answer:

Induced emf in the coil, E = 0.157 volts

Explanation:

It is given that,

Number of turns, N = 100

Diameter of the coil, d = 3 cm = 0.03 m

Radius of the coil, r = 0.015 m

A uniform magnetic field increases from 0.5 T to 2.5 T in 0.9 s.

Due to this change in magnetic field, an emf is induced in the coil which is given by :

E=-NA\dfrac{\Delta B}{\Delta t}

E=-100\times \pi (0.015)^2\times \dfrac{2.5-0.5}{0.9}

E = -0.157 volts

Minus sign shows the direction of induced emf in the coil. Hence, the induced emf in the coil is 0.157 volts.

8 0
3 years ago
Using the set-up seen here, Ms. Garcia places a golf ball between the globe and the flashlight. Turning off the lights in the ro
Julli [10]
This is a solar Eclipse.  
7 0
3 years ago
Light could be thought of as a stream of tiny particles discharged by _______ objects that travel in straight paths. *
Gelneren [198K]

Answer: Light could be thought of as a stream of tiny particles discharged by luminous objects that travel in straight paths.

Explanation:

We can define "radiation" as the transmision of energy trough waves or particles.

Particularly, light is a form of electromagnetic radiation, so the "tiny particles" of light are discharged by a radiating object, particularly we can be more explicit and call it a luminous object, in this way we are being specific about the nature of the radiation of the object.

5 0
3 years ago
When Woods hits a 0.04593 kg golf ball, the golf ball is usually traveling around 281 kilometers per hour. What average force do
Semenov [28]

Answer:

128.9 N

Explanation:

The force exerted on the golf ball is equal to the rate of change of momentum of the ball, so we can write:

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

where

F is the force

\Delta p is the change in momentum

\Delta t=0.030 s is the time interval

The change in momentum can be written as

\Delta p = m(v-u)

where

m = 0.04593 kg is the mass of the ball

u = 0 is the initial velocity of the ball

v=281 km/h =78.1 m/s is the final velocity of the ball

Substituting into the original equation, we find the force exerted on the golf ball:

F=\frac{m(v-u)}{\Delta t}=\frac{(0.04593)(78.1-0)}{0.030}=128.9 N

7 0
3 years ago
An open organ pipe of length 0.47328 m and another pipe closed at one end of length 0.702821 m are sounded together. What beat f
sineoko [7]

Answer:

fb = 240.35 Hz

Explanation:

In order to calculate the beat frequency generated by the first modes of each, organ and tube, you use the following formulas for the fundamental frequencies.

Open tube:

f=\frac{v_s}{2L}         (1)

vs: speed of sound = 343m/s

L: length of the open tube = 0.47328m

You replace in the equation (1):

f=\frac{343m/s}{2(0.47228m)}=362.36Hz      

Closed tube:

f'=\frac{v_s}{4L'}

L': length of the closed tube = 0.702821m

f'=\frac{343m/s}{4(0.702821m)}=122.00Hz

Next, you use the following formula for the beat frequency:

f_b=|f-f'|=|362.36Hz-122.00Hz|=240.35Hz

The beat frequency generated by the first overtone pf the closed pipe and the fundamental of the open pipe is 240.35Hz

7 0
3 years ago
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