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Harman [31]
2 years ago
9

A hiker shouts toward a vertical cliff 685 m away. The echo is

Physics
1 answer:
miv72 [106K]2 years ago
7 0
To find the speed of sound you take the distance and divide it by the time so 685/4 which equals 171.25 m/s
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A 0.30-kg object connected to a light spring with a force constant of 22.6 N/m oscillates on a frictionless horizontal surface.
gtnhenbr [62]

Answer:

(a)  vmax = 0.34m/s

(b)  v = 0.13m/s

(c)  v = 0.31m/s

(d)  x = 0.039m

Explanation:

Given information about the spring-mass system:

m: mass of the object = 0.30kg

k: spring constant = 22.6 N/m

A: amplitude of the motion = 4.0cm = 0.04m

(a) The maximum speed of the object is given by the following formula:

v_{max}=\omega A       (1)

w: angular frequency of the motion.

The angular frequency is calculated with the following relation:

\omega=\sqrt{\frac{k}{m}}           (2)

You replace the expression (2) into the equation (1) and replace the values of the parameters:

v_{max}=\sqrt{\frac{k}{m}}A=\sqrt{\frac{22.6N/m}{0.30kg}}(0.04m)=0.34\frac{m}{s}

The maximum speed of the object is 0.34 m/s

(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:

v_{max}=\sqrt{\frac{22.6N/m}{0.30kg}}(0.015m)=0.13\frac{m}{s}

The speed is 0.13m/s

(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:

x=Acos(\omega t)

You solve the previous equation for t:

t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\\omega=\sqrt{\frac{22.6N/m}{0.30kg}}=8.67\frac{rad}{s}\\\\t=\frac{1}{8.67}cos^{-1}(\frac{1.5cm}{4.0cm})=0.13s

With this value of t, you can calculate the speed of the object with the following formula:

v=\omega Asin(\omega t)\\\\v=(8.67rad/s)(0.04m)sin((8.67rad/s)(0.13s))=0.31\frac{m}{s}

The speed of the object for x = 1.5cm is v = 0.31 m/s

(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:

\frac{v_{max}}{2}=\omega A sin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{v_{max}}{2\omega A})\\\\t=\frac{1}{8.67rad/s}sin^{-1}(\frac{0.13m/s}{2(8.67rad/s)(0.04m)})=0.021s

The position will be:

x=Acos(\omega t)=0.04mcos((8.67rad/s)(0.021s))=0.039m

The position of the object on which its speed is one-half its maximum velocity is 0.039

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If we're interested in knowing the rate at which light energy is received by a unit of area on a particular surface, we're reall
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