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3 years ago

Why do cats have tails?

1 answer:

3 0

Cats have tails to help their balance. Similar to the stick a trapeze/high wire walker uses.

The tail helps to serve as a counterbalance when cats walk on narrow spaces such as fences or shelves. The tail also aids in balance when a cat is running after or jumping on prey.

Just so you know....Cats can live without tails.

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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

Record two activities where you would expect to have a steep learning curve, two where you would have an “average” learning curv

lora16 [44]

Answer:

Explanation:

3 0

2 years ago

Is the Earth bigger than the moon

sesenic [268]

Yes the Earth is bigger than the Moon.
The moon is one-quarter the size of Earth.

4 0

1 year ago

Read 2 more answers

Which of Newton’s Laws involves mass and acceleration? Question 1 options: 3rd 1st 2nd All of them

Feliz [49]

The answer would be Newton’s Second Law

6 0

3 years ago

Read 2 more answers

A stretched string has a mass per unit length of 4.87 g/cm and a tension of 16.7 N. A sinusoidal wave on this string has an ampl

Burka [1]

Answer:

Explanation:

Given that,

Mass per unit length is

μ = 4.87g/cm

μ=4.87g/cm × 1kg/1000g × 100cm/m

μ = 0.487kg/m

Tension

τ = 16.7N

Amplitude

A = 0.101mm

Frequency

f = 71 Hz

The wave is traveling in the negative direction

Given the wave form

y(x,t) = ym• Sin(kx + ωt)

A. Find ym?

ym is the amplitude of the waveform and it is given as

ym = A = 0.101mm

ym = 0.101mm

B. Find k?

k is the wavenumber and it can be determined using

k = 2π / λ

Then, we need to calculate the wavelength λ using

V = fλ

Then, λ = V/f

We have the frequency but we don't have the velocity, then we need to calculate the velocity using

v = √(τ/μ)

v = √(16.7/0.487)

v = 34.29

v = 5.86 m/s

Then, we can know the wavelength

λ = V/f = 5.86 / 71

λ = 0.0825 m

So, we can know the wavenumber

k = 2π/λ

k = 2π / 0.0825

k = 76.18 rad/m

C. Find ω?

This is the angular frequency and it can be determined using

ω = 2πf

ω = 2π × 71

ω = +446.11 rad/s

D. The angular frequency is positive (+) because the direction of propagation of wave is in the negative direction of x

5 0

2 years ago