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4 years ago

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A fisherman fishing from a pier observes that the float on his line bobs up and down, taking 2.4 s to move from its highest poin

t to its lowest point. He also estimates that the distance between adjacent wave crests is 48 m. What is the speed of the waves going past the pier?Immersive Reader

1 answer:

goblinko [34]4 years ago

4 0

Answer:

v = 10 m / s

Explanation:

For this exercise we will use the relationship between the speed of a wave and its frequency and wavelength

v = λ f

the wavelength is the distance at which the wave repeats, in this case the distance between the two ridges λ = 48 m.

the frequency is the number of oscillations per unit of time, it is also the inverse of the period which is the time in a complete oscillation, in this case they give us the time of half a period, ½ T = 2.4 s

T = 4.8 s

the frequency is

f = 1 / T

f = 1 / 4.8

f = 0.2083 Hz

let's calculate

v = 0.2083 48

v = 10 m / s

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A cord is used to vertically lower an initially stationary block of mass M = 3.6 kg at a constant downward acceleration of g/7.

dalvyx [7]

Answer:

(a) $W_c=127.008 J$

(b) $W_g=148.176 J$

(c) K.E. = 21.168 J

(d) $v=3.4293m.s^{-1}$

Explanation:

Given:

mass of a block, M = 3.6 kg

initial velocity of the block, $u=0 m.s^{-1}$

constant downward acceleration, $a_d= \frac{g}{7}$

$\Rightarrow$ That a constant upward acceleration of $\frac{6g}{7}$ is applied in the presence of gravity.

∴ $a=- \frac{6g}{7}$

height through which the block falls, d = 4.2 m

(a)

Force by the cord on the block,

$F_c= M\times a$

$F_c=3.6\times (-6)\times\frac{9.8}{7}$

$F_c=-30.24 N$

∴Work by the cord on the block,

$W_c= F_c\times d$

$W_c= -30.24\times 4.2$

We take -ve sign because the direction of force and the displacement are opposite to each other.

$W_c=-127.008 J$

(b)

Force on the block due to gravity:

$F_g= M.g$

∵the gravity is naturally a constant and we cannot change it

$F_g=3.6\times 9.8$

$F_g=35.28 N$

∴Work by the gravity on the block,

$W_g=F_g\times d$

$W_g=35.28\times 4.2$

$W_g=148.176 J$

(c)

Kinetic energy of the block will be equal to the net work done i.e. sum of the two works.

mathematically:

$K.E.= W_g+W_c$

$K.E.=148.176-127.008$

K.E. = 21.168 J

(d)

From the equation of motion:

$v^2=u^2+2a_d\times d$

putting the respective values:

$v=\sqrt{0^2+2\times \frac{9.8}{7}\times 4.2 }$

$v=3.4293m.s^{-1}$ is the speed when the block has fallen 4.2 meters.

6 0

3 years ago

Can I PLEASE get some help? I REALLY need it!

soldi70 [24.7K]

The answer is C. Hope this helps.

7 0

3 years ago

A 57 kg wagon is pulled with a constant net force of 38 N. Calculate the acceleration of the wagon. F = ma

Dahasolnce [82]

Answer:

<h2>0.67 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

f is the force

m is the mass

From the question we have

$a = \frac{38}{57} = \frac{2}{3} \\ = 0.666666...$

We have the final answer as

<h3>0.67 m/s²</h3>

Hope this helps you

4 0

3 years ago

A damped harmonic oscillator consists of a mass on a spring, with a small damping force that is proportional to the speed of the

exis [7]

Answer:

2.19 N/m

Explanation:

A damped harmonic oscillator is formed by a mass in the spring, and it does a harmonic simple movement. The period of it is the time that it does one cycle, and it can be calculated by:

T = 2π√(m/K)

Where T is the period, m is the mass (in kg), and K is the damping constant. So:

2.4 = 2π√(0.320/K)

√(0.320/K) = 2.4/2π

√(0.320/K) = 0.38197

(√(0.320/K))² = (0.38197)²

0.320/K = 0.1459

K = 2.19 N/m

4 0

3 years ago

4. Calculate the total resistance for two 180 ohm resistors connected in<br> parallel

solmaris [256]

Answer:

90 ohms

Explanation:

1/r = 1/180 + 1/180

1/r= 2/180

take the reciprocal of 2/180 which is 180/2 and its 90 ohms

3 0

3 years ago