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DanielleElmas [232]
4 years ago
7

Waves are observed passing under a dock. Wave crests are 8.0 meters apart. The time for a complete wave to pass by is 4.0 second

s. The markings on the post submerged in water indicate that the water level fluctuates from a trough at 6.0 meters to a crest at 9.0 meters. What is the amplitude of the wave?
Physics
1 answer:
Harlamova29_29 [7]4 years ago
3 0
The distance between a trough and a crest is double the amplitude.

The distance between trough and crest is (9m - 6m) = 3m.

So the amplitude of those water waves is (3/2) = 1.5 meters.

For this question, we really don't need to know how far apart the crests are, or how often they pass.  There may be more parts to the question, for which that information is needed.

For example: 

-- What's the wavelength of the waves ?  8 meters.

-- What's the period of the waves ?  4 seconds.

-- What's the frequency of the waves ?  1 / (4 seconds) = 0.25 Hz.

-- What's the speed of the waves ? (8 m) x (0.25 Hz) = 2 m/s
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3 years ago
Two spherical conductors are separated by a distance much larger than either of their radii. Sphere A has a radius of 11.5 cm an
bonufazy [111]

Explanation:

As the given spheres are connected by a thin wire so, the potential on the spheres are the same.

          \frac{q_{1}}{r_{1}} = \frac{q_{2}}{r_{2}} ......... (1)

Hence, total charge will be as follows.

              q_{1} + q_{2} = Q = -95.5 nC .......... (2)

Using the above two equations, the final equation will be as follows.

          q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}

and,    q_{1} = \frac{Qr_{1}}{r_{1} + r_{2}}

Hence, we will calculate the charge on sphere B after the equilibrium is reached as follows.

          q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}

                     = \frac{-95.5 \times 74.4 cm}{(11.5 + 74.4) cm}

                     = 82.714 nC

Thus, we can conclude that the charge on sphere B after equilibrium has been reached is 82.714 nC.

                       

5 0
3 years ago
aving established that a sound wave corresponds to pressure fluctuations in the medium, what can you conclude about the directio
vovangra [49]

Answer: Pressure fluctuations travel along the direction of propagation of the sound wave.

Explanation:

Sound wave is a type of longitudinal wave. It is defined as a wave which consist of vibrations of particles traveling through a medium( such as air, or water).

Sound wave is propagated by the alternating adiabatic compression and expansion of the medium. The COMPRESSIONS are regions of high air pressure while the RAREFACTIONS are regions of low air pressure. Therefore, Since a sound wave consists of a repeating pattern of high-pressure and low-pressure regions moving through a medium, it is sometimes referred to as a PRESSURE WAVE.

The direction of the vibrating particles is parallel to the direction of propagation and that's why it's a type of LONGITUDINAL WAVE. Therefore, the correct option that

concludes about the direction in which such pressure fluctuations travel is

(Pressure fluctuations travel along the direction of propagation of the sound wave.)

8 0
4 years ago
How much electrical energy is used by a 75 W laptop that is operating for 12<br>minutes?​
Liono4ka [1.6K]

"1 watt" means 1 joule of energy per second.

75 W means 75 joules/sec .

Energy = (75 Joule/sec) x (12 min) x (60 sec/min)

Energy = (75 x 12 x 60) (Joule-<em>min-sec</em> / <em>sec-min</em>)

<em>Energy = 54,000 Joules</em>

7 0
3 years ago
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