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mina [271]
3 years ago
8

Pepe and Alfredo are resting on an offshore raft after a swim. They estimate that 3.0 m separates a trough and an adjacent crest

of each surface wave on the lake. They count 11 crests that pass by the raft in 21.5 s. Calculate how fast the waves are moving. (Assume the count begins and ends at the top of a crest.)

Physics
1 answer:
Agata [3.3K]3 years ago
4 0

Answer:

The velocity (v) of the wave is 3.08 ms^{-1}.

Explanation:

According to the figure, the distance (\large{L}) between a trough and its adjacent crest is \large{L = 3 m}. Also the wavelength (\large{\lambda}) of the wave is \large{\lambda = 2L}. Pepe and Alfredo count 11 crests to pass the raft in \large{t} = 21.5 s.

So, the time period (\large{T}) of oscillation of the wave is

\large{T} = \dfrac{t}{11} = \dfrac{21.5}{11} = 1.95s

So, the velocity (\large{V}) of the wave is

\large{V = \dfrac{\lambda}{T} = \dfrac{2 \times L}{T} = \dfrac{2 \times 3}{1.95}= 3.08 ms^{-1}}

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Please what is the work done by a man who is pulling a box of 45kg of mass by means of rope which makes angle of 45 degrees ?
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No work is done since no distance is given

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Since no distance is given, the force is not doing any work

No work is done by the man since we do not know the distance or displacement.

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5 0
2 years ago
2. An archer shoots an arrow at 83.0 m/s at a 62.0 degree angle. If the ground is flat, how much time is the arrow in the air?
UkoKoshka [18]

Answer:

<em>t=14.96 sec</em>

Explanation:

<u>Diagonal Launch </u>

It's a physical event that happens where an object is thrown in free air (no friction) forming an angle with the horizontal reference. The object then describes a path called a parabola.

The object will reach its maximum height and then return to the height from which it was launched. The equation for the height is :

\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}

Where vo is the initial speed, \theta is the angle, t is the time and g is the acceleration of gravity .

In this problem we'll assume the arrow was launched from the ground level (won't consider the archer's height). Thus y_o=0, and:

\displaystyle y=v_osin\theta \ t-\frac{gt^2}{2}

The value of y is zero twice: when t=0 (at launching time) and in t=t_f when it goes back to the ground. We need to find that time t_f by making y=0

\displaystyle 0=v_osin\theta\ t_f-\frac{gt_f^2}{2}

Dividing by t_f

\displaystyle v_osin\theta=\frac{gt_f}{2}

Then we find the total flight time as

\displaystyle t_f=\frac{2v_osin\theta}{g}

\displaystyle t_f=\frac{2(83)sin\ 62^o}{9.8}

\displaystyle t_f=14.96\ sec

5 0
3 years ago
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