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Sphinxa [80]
3 years ago
9

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.
Physics
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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When the temperature of 2.35 m^3 of a liquid is increased by 48.5 degrees Celsius, it expands by 0.0920 m^3. What is its coeffic
alexandr1967 [171]

Coefficient of volume expansion is 8.1 ×10⁻⁴ C⁻¹.

<u>Explanation:</u>

The volume expansion of a liquid is given by ΔV,

ΔV = α V₀ ΔT

ΔT = change in temperature  = 48.5° C

α =  coefficient of volume expansion =?

V₀ = initial volume = 2.35 m³

We need to find α , by plugin the given values in the equation and by rearranging the equation as,

\alpha=\frac{\Delta \mathrm{V}}{\mathrm{V}_{0} \Delta \mathrm{T}}=\frac{0.0920}{2.35 \times 48.5}=0.00081

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5 0
3 years ago
A student lifts their 5 kg backpack 2 meters off the floor. How much work does the student do on the backpack?
kirill115 [55]

Answer:

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4 0
2 years ago
A football is thrown from the edge of a cliff from a height of 22 m at a velocity of 18 m/s [39degrees above the horizontal]. A
mihalych1998 [28]

Answer:

The ball will get to the bottom of the cliff in 1.26 s, while the player will reach the bottom of the cliff  2 s later. Thus, it is not possible for the player to catch the ball.

Explanation:

Given;

vertical height of the cliff, h = 22 m

velocity of the ball, u = 18 m/s at an angle 39⁰

vertical component of the velocity, u_y = u \ sin  \theta

The time for the ball to get to the bottom of the cliff is calculated as;

h = ut + ¹/₂gt²

h = (u sinθ)t + ¹/₂ x 9.8 x t²

22 = (18 sin 39)t + 4.9t²

22 = 11.328t + 4.9t²

4.9t² + 11.328t  - 22 = 0

Solve the above equation with formula method;

a = 4.9, b = 11.328, c = -22

t = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-11.328\ \ +/- \ \ \sqrt{(-11.328)^2 - 4(4.9\times -22)} }{2(4.9)}\\\\t = 1.26 \ s

The time for the player to get to the bottom of the cliff;

Given maximum speed, Vx = 6.0 m/s and horizontal distance, X = 12 m;

t = \frac{X}{V_x} \\\\t = \frac{12}{6} \\\\t = 2 \ s

The ball will get to the bottom of the cliff in 1.26 s, while the player will reach the bottom of the cliff 2 s later. Thus, it is not possible for the player to catch the ball.

7 0
2 years ago
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