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2 years ago

Describe how pieces of rope move as waves pass

1 answer:

6 0

Well, as the waves move it moves the rope as if its trying to take shape of it. Since the rope it light it will move along the ocean and the ocean will keep pushing up on the rope. (even without the waves the water is pushing the rope up so it can take its shape)

Maybe that'll help

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A 1.30-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v

atroni [7]

Answer:

1. t = 0.0819s

2. W = 0.25N

3. n = 36

4. y(x , t)= Acos[172x + 2730t]

Explanation:

1) The given equation is

$y(x, t) = Acos(kx -wt)$

The relationship between velocity and propagation constant is

$v = \frac{\omega}{k}=\frac{2730rad/sec}{172rad/m}\\\\$

v = 15.87m/s

Time taken, $t = \frac{\lambda}{v}$

$= \frac{1.3}{15.87}\\\\=0.0819 sec$

t = 0.0819s

The velocity of transverse wave is given by

$v = \sqrt{\frac{T}{\mu}}$

$v = \sqrt{\frac{W}{\frac{m}{\lambda}}}$

mass of string is calculated thus

mg = 0.0125N

$m = \frac{0.0125N}{9.8N/s}$

m = 0.00128kg

$\omega = \frac{v^2m}{\lambda}$

$\omega = \frac{(15.87^2)(0.00128)}{1.30}$

$\omega =$ 0.25N

The propagation constant k is

$k=\frac{2\pi}{\lambda}$

hence

$\lambda = \frac{2\pi}{k}\\\\\lambda = \frac{2 \times 3.142}{172}$

$\lambda =$ 0.036 m

No of wavelengths, n is

$n = \frac{L}{\lambda}\\\\n = \frac{1.30m}{0.036m}\\$

n = 36

The equation of wave travelling down the string is

$y(x, t)=Acos[kx -wt]\\\\becomes\\\\y(x , t)= Acos[(172 rad.m)x + (2730 rad.s)t]$

$without, unit\\\\y(x , t)= Acos[172x + 2730t]$

7 0

2 years ago

From the top of a cliff, a person uses a slingshot to fire a pebble straight downward, which is the negative direction. The init

Talja [164]

Answer:

$\vec{a} = -(9.8~{\rm m/s^2})\^y$

Explanation:

Regardless of the initial velocity of the pebble, the acceleration of the pebble is equal to the gravitational acceleration which is equal to 9.8 m/s2 towards downwards direction.

This can be shown by Newton's Second Law. According to the law, the net force applied on an object is equal to mass times acceleration of that object.

During the downward motion, the only force acting on the pebble is the gravitational force, hence its acceleration is equal to gravitational acceleration.

8 0

3 years ago

Without steering, a car (mass 1501 kg) can travel with a certain speed around a banked frictionless corner angled at 25.0° to th

natima [27]

Answer:

Its tricky but am still working on it

6 0

2 years ago

While frying French fries, Wanda burned her hand. Burns are examples of a:

Alja [10]

Answer:

d. homeostatic imbalance

Explanation:

8 0

2 years ago

Read 2 more answers

A weightlifter presses a 400 n weight 0.5 m over his head in 2 seconds, what is the power of the weight lifter

Scrat [10]

There are two ways to solve this problem. First we write the given.

Given: Force F = 400 N; Height h = 0.5 m; Time t = 2 s

Formula: P = W/t; but Work W = Force x distance or W = f x d

Weight is also a Force, therefore: W = mg, solve for Mass m = ?

m = w/g m = 400 N/9.8 m/s² m = 40.82 Kg

P = W/t = F x d/t = mgh/t P = (40.82 Kg)(9.8 m/s²)/2 s

P = 100 J/s or 100 Watts

3 0

2 years ago