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VLD [36.1K]
3 years ago
11

If you shout into the grand canyon, your voice travels at the speed of sound (340 m/s) to the bottom of the canyon and back, and

you hear an echo. how deep is the grand canyon at a spot where you can hear your echo 5.2 seconds after you shout?
Physics
1 answer:
Nostrana [21]3 years ago
4 0
Given:
The speed of sound is  340 m/s
Time for the echo is 5.2 s.

Let h = the depth of the canyon.

Because the sound of your voice travels to the bottom of the canyon and back to your ear, the total distance traveled is 2h.

By definition,
distance = velocity * time.
Therefore
2h = (340 m/s)*(5.2 s) = 1768 m
h = 884 m

Answer: The depth is 884 m.

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Answer:

Applications of zeroth law of thermodynamics:

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3. Temperature measurement with a thermometer or another device.

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3 years ago
A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike
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(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g=9.50\times 10^{3} kg

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

mu+ MU=mv+ MV

Substitute the values

9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V

3.61i=2.375j+0.150V

3.61 i-2.375j=0.150V

V=\frac{1}{0.150}(3.61 i-2.375j)

V=24.07i-15.83j

Magnitude of velocity of stone

=\sqrt{(24.07)^2+(-15.83)^2}

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

\theta=tan^{-1}(\frac{y}{x})

=tan^{-1}(\frac{-15.83}{24.07})

\theta=tan^{-1}(-0.657)

=33.3 degree below the horizontal

(b)

Initial kinetic energy

K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2

K_i=685.9 J

Final kinetic energy

K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2

=\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2

K_f=359.12 J

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

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