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lions [1.4K]
3 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 period of the wave?
Physics
2 answers:
Debora [2.8K]3 years ago
4 0
<h3><u>Answer;</u></h3>

The period of the wave is <u><em>4 seconds</em></u>

<h3><em><u>Explanation;</u></em></h3>
  • <em><u>The period of a wave or periodic time is the time taken for one complete oscillation to occur.</u></em> In this case, one complete oscillation occurs when the wave moves from one crest to the next or a trough to the next. <em><u>This takes 4 seconds. Therefore the period is 4 seconds.</u></em>
  • <em><u>Frequency on the other hand is the number of oscillations by a wave in one second. Thus, f = 1/T, that is frequency is the reciprocal of periodic time.</u></em>
AlexFokin [52]3 years ago
4 0
Period of wave is time takes wave to pass the fixed point.
you have been said that it's 4s, that's the answer 
T = 4s

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A water balloon is thrown horizontally from a tower that is 45 m high. It strikes the shoes of an unsuspecting passerby who is 4
LekaFEV [45]

Answer:

14.85 m/s

Explanation:

From the question given above, the following data were obtained:

Height (h) of tower = 45 m

Horizontal distance (s) moved by the balloon = 45 m

Horizontal velocity (u) =?

Next, we shall determine the time taken for the balloon to hit the shoe of the passerby. This is illustrated below:

Height (h) of tower = 45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

45 = ½ × 9.8 × t²

45 = 4.8 × t²

Divide both side by 4.9

t² = 45/4.9

Take the square root of both side

t = √(45/4.9)

t = 3.03 s

Finally, we shall determine the magnitude of the horizontal velocity of the balloon as shown below:

Horizontal distance (s) moved by the balloon = 45 m

Time (t) = 3.03 s

Horizontal velocity (u) =?

s = ut

45 = u × 3.03

Divide both side by 3.03

u = 45/3.03

u = 14.85 m/s

Thus, the magnitude of the horizontal velocity of the balloon was 14.85 m/s

4 0
3 years ago
A machine is currently set to a feed rate of 5.921 inches per minute (IPM). Te machinist changes this setting to 6.088 IPM. By h
lukranit [14]

Answer:

By 16.7% or 0.167 IPM

Explanation:

Substracting the final IPM (6.088) to the initial IPM (5.921) gives us the net difference, which is how much did it increase in IPM. Multiplying this number by 100 gives us the percentual increase in the feed rate.

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3 years ago
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I can't seem to get the right angular acceleration and also not sure how to do part b. Help will be much appreciated.
Tcecarenko [31]

Answer:It’s 5 I believe

Explanation: it says to round to the nearest thousandths, so it’ll be 5.

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andreev551 [17]

Answer: Teeth is an example of a wedge.

Explanation :

the machines that make our work easier are called simple machines. Some machines can be compound because they are a combination of more than two simple machines. For example, stapler.

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The mechanical advantage of a wedge is more than 1.

So, the correct option is (b) " Wedge".

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Observer 1 rides in a car and drops a ball from rest straight downward, relative to the interior of the car. The car moves horiz
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a) 10.5 m/s

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As the ball falls down, it also gains speed along the vertical direction (due to the effect of gravity). The vertical speed is given by

v_y = u_y + gt

where

u_y =0 is the initial vertical speed

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Therefore, after t = 1.00 s, the vertical speed is

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b) \theta = -68.8^{\circ}

As we discussed in previous part, according to observer 2 the ball is travelling both horizontally and vertically.

The direction of travel of the ball, according to observer 2, is given by

\theta = tan^{-1} (\frac{v_y}{v_x})=tan^{-1} (\frac{-9.8}{3.8})=-68.8^{\circ}

We have to understand in which direction is this angle measured. In fact, the car is moving forward, so v_x has forward direction (we can say it is positive if we take forward as positive direction).

Also, the ball is moving downward, so v_y is negative (assuming upward is the positive direction). This means that the direction of the ball is forward-downward, so the angle above is measured as angle below the positive horizontal direction:

\theta = -68.8^{\circ}

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