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kicyunya [14]
3 years ago
7

A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac

e of the water. It takes a time of 2.90s for the boat to travel from its highest point to its lowest, a total distance of 0.630m . The fisherman sees that the wave crests are spaced a horizontal distance of 5.60m apart.A..How much is the wavelength?express your answers in 3 sig. fig.B. Find the period of the wave.3 sig. fig.C. How fast are the waves traveling?Express the speed v in meters per second using three significant figures.D. What is the amplitude A of each wave?Express your answer in meters using three significant figures
Physics
1 answer:
leva [86]3 years ago
5 0

Answer:

A. 5.600 m

B. 5.800 s

C. 0.966 m/s

D. 0.315 m

Explanation:

A. The wavelength is the distance between 2 crests, which is 5.600 m

B. Period of the wave is the time for the wave to complete 1 circle (highest point to next highest point). Since it takes 2.9s to travel from highest point to lowest point, it would take another 2.9 to travel from lowest point to the next highest point. So the total time is 2.9 + 2.9 = 5.8 s,

C. The wavespeed is wavelength over unit of time:

5.6 / 5.8 = 0.966 m/s

D. The amplitude would be half the length of highest point to lowest point, which is 0.63 / 2 = 0.315 m

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A tank whose bottom is a mirror is filled with water to a depth of 20.0 cm. A small fish floats motionless 7.0 cm under the surf
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Answer:

The apparent depth of (a) the fish is 5.3 cm and (b) the image of the fish is 24.8 cm.

Explanation:

According to the following equation:

\frac{n_{w} }{s} +\frac{n_{a} }{s'} = \frac{n_{a}- n_{w}}{R_{c} } \\

where <em>nw</em> and <em>na</em> is the refractive indices of water (1.33) and air (1.00) respectively; <em>s</em> is the depth of the fish below the surface of the water; s' is the apparent depth of the fish from normal incidence and Rc is the radius of curvature of the mirror at the bottom of the tank.

Note that the bottom of the tank is assumed to be a flat mirror, therefore the radius of curvature is very large (R⇒∞).

Therefore, the above equation can be expressed as:

\frac{n_{w}}{s} +\frac{n_{a}}{s'}=0

Now we can solve for the apparent depth of the fish.

(a) s'=-(\frac{n_{a}}{n_{w}})x s (Make s' subject of the formula from the above equation)

s'=(\frac{1.00}{1.33} )x7cm

∴ s'=5.3 cm.

(b) The motionless fish floats 13 cm above the mirror, therefore the image of the fish will be situated at 13 + 20 =33 cm away from the real fish.

Therefore, s = 33 cm

s'=-(\frac{n_{a}}{n_{w}})x s

s'=(\frac{1.00}{1.33} )x 33 cm

s'=24.8 cm.

NB: Here, it is assumed that the water is pure, as impurities may alter the refractive index of water.

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