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3 years ago

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You are hiking along a trail in a wide, dry canyon where the outdoor temperature is T = 29.5° C. To determine how far you are aw

ay from the canyon wall you yell "Hello" and hear the echo t = 1.8 s later.

1 answer:

STALIN [3.7K]3 years ago

7 0

Answer:

<em>The canyon is approximately 314 meters away</em>

Explanation:

<u>Speed of Sound</u>

If we emit sounds in an open space where a large obstacle (like a mountain) is expected to return the sounds, then it will travel forth and back at a given speed for a certain time. We can assume the speed of sound is constant, so we could know the approximate distance of the mountain (or canyon in our case) by the known formula.

$x=v_st$

Where $v_s$ is the speed of sound and t is half the time we hear our echo.

The speed of sound in m/s can be calculated from the approximate formula in terms of the temperature T in degrees Celsius

$v_s=331.3+0.606T$

We have $T=29.5^oC$ , so

$v_s=331.3+0.606(29.5)$

$v_s=349.177\ m/s$

Let's compute x, for t=1.8/2=0.9 seconds

$x=v_st=(349.177)(0.9)$

$x=314.26\ m$

The canyon is approximately 314 meters away

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Answer:

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Explanation:

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A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the

BartSMP [9]

Answer:

$r=1.14m$

Explanation:

$\theta$ is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

$F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB$

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

$F_m=F_c\\qvB=F_c(1)$

$F_c$ is the centripetal force and is defined as:

$F_c=m\frac{v^2}{r}$

Here $v$ is the proton's speed and $r$ is the radius of the circular motion. Replacing this in (1) and solving for r:

$qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}$

Recall that 1 J is equal to $6.242*10^{12}MeV$ , so:

$4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J$

We can calculate $v$ from the kinetic energy of the proton:

$K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}$

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Answer:

v ≈ 7900 m/s

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v²/R = g

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v = √(Rg)

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