How much time does it take for a bird flying at a speed of 45 miles per hour to travel a distance

A solid disk of radius 1.4 cm and mass 430 g is attached by a wire to one of its circular faces. It is twisted through an angle

Neporo4naja [7]

Answer:

f= 4,186 10² Hz

Explanation:

El sistema descrito es un pendulo de torsión que oscila con con velocidad angular, que esta dada por

w = √ k/I

donde ka es constante de torsion de hilo e I es el momento de inercia del disco

El momento de inercia de indican que giran un eje que pasa por enronqueces

I= ½ M R2

reduzcamos las cantidades al sistema SI

R= 1,4 cm = 0,014 m

M= 430 g = 0,430 kg

substituimos

w= √ (2 k/M R2)

calculemos

w = RA ( 2 370 / (0,430 0,014 2)

w = 2,963 103 rad/s

la velocidad angular esta relacionada con la frecuencia por

w =2pi f

f= w/2π

f= 2,963 10³/ (2π)

f= 4,186 10² Hz

5 0

3 years ago

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me

Sav [38]

Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

(c): $\rm 2ct-3bt^2.$

(d): $\rm 2c-6bt.$

(e): $\rm t=\dfrac{2c}{3b}.$

Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.$
$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Applying both these conditions,

$\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.$

For $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c$

Since, c is a positive constant therefore, for $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0$

Thus, particle does not reach its maximum value at $\rm t = 0\ s.$

For $\rm t_o = \dfrac{2c}{3b}$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.$

Here,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Thus, the particle reach its maximum x value at time $\rm t_o = \dfrac{2c}{3b}.$

7 0

3 years ago

A 0.49-kg cord is stretched between two supports, 7.8m apart. When one support is struck by a hammer, a transverse wave travels

katovenus [111]

To solve this problem we will apply the laws of Mersenne. Mersenne's laws are laws describing the frequency of oscillation of a stretched string or monochord, useful in musical tuning and musical instrument construction. This law tells us that the velocity in a string is directly proportional to the root of the applied tension, and inversely proportional to the root of the linear density, that is,

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