That depends on what the sound is traveling through, and the density
and temperature of the substance.
In dry air, at sea-level pressure and 68° F, the speed of sound is
about 340 meters per second (761 miles per hour).
Answer:
Average speed = distance/time
From 1 to 9 seconds:
Distance covered = 1 - 0.2 = 0.8 km
Time = 9 - 1 = 8 sec
Average speed = 0.8 km / 8 sec
Average speed = 0.1 km/s .
The average speed for the whole test is 1.6 km / 20 sec = 0.08 km/sec. A graph of speed vs time would average out as a horizontal line at 0.08 km/sec from 1 sec to 21 sec. The area under it would be (0.08 km/s) x (20 sec) = 1.6 km.
Surprise surprise ! The area under a speed/time graph is the distance covered during that time !
In closing, I want to express my gratitude for the gracious bounty of 3 points with which I have been showered. Moreover, the green breadcrust and tepid cloudy water have also been refreshing.
Explanation:
Answer:
1) 65 m
2) 40 m/s downward
Explanation:
Using for both questions the kinematic equation
v² = u² + 2as
and ignoring air resistance.
1) h = 60 + √(20²/(2(9.8))) = 64.51753...
2) v = √(20² + 2(9.8)(60)) = 39.69886...
Answer:
Vd = 0.022 (V)
Explanation:
What we have as voltage drop is the difference between the voltage at source level (terminals) and the voltage at the terminal of electrical load ( in this case the toaster oven) that difference occures as consequence of the normal distributed resistance of conductor. But the voltage at the terminal of the electric load have to be measure between the two wires at the end of the circuit.
Therefore the voltage drop is = I*R = 11 (A) * 0,002 Ω = 0.022 (v)
Answer:
![\mu = 1.645](https://tex.z-dn.net/?f=%5Cmu%20%3D%201.645%20)
Explanation:
By Snell's law we know at the left surface
![\theta_i = 19^o](https://tex.z-dn.net/?f=%5Ctheta_i%20%3D%2019%5Eo)
![\theta_r = ?](https://tex.z-dn.net/?f=%5Ctheta_r%20%3D%20%3F)
![\mu_1 = 1](https://tex.z-dn.net/?f=%5Cmu_1%20%3D%201)
![\mu_2 = \mu](https://tex.z-dn.net/?f=%5Cmu_2%20%3D%20%5Cmu)
now we have
![1 sin19 = \mu sin\theta_r](https://tex.z-dn.net/?f=1%20sin19%20%3D%20%5Cmu%20sin%5Ctheta_r)
![0.33 = \mu sin\theta_r](https://tex.z-dn.net/?f=0.33%20%3D%20%5Cmu%20sin%5Ctheta_r)
now on the other surface we know that
angle of incidence = ![\theta_r'](https://tex.z-dn.net/?f=%5Ctheta_r%27)
![\theta_e = 90](https://tex.z-dn.net/?f=%5Ctheta_e%20%3D%2090%20)
so again we have
![\mu sin\theta_r' = 1 sin90](https://tex.z-dn.net/?f=%5Cmu%20sin%5Ctheta_r%27%20%3D%201%20sin90)
so we have
![\theta_r = sin^{-1}\frac{0.33}{\mu}](https://tex.z-dn.net/?f=%5Ctheta_r%20%3D%20sin%5E%7B-1%7D%5Cfrac%7B0.33%7D%7B%5Cmu%7D)
![\theta_r' = sin^{-1}\frac{1}{\mu}](https://tex.z-dn.net/?f=%5Ctheta_r%27%20%3D%20sin%5E%7B-1%7D%5Cfrac%7B1%7D%7B%5Cmu%7D)
also we know that
![\theta_r + \theta_r' = 49](https://tex.z-dn.net/?f=%5Ctheta_r%20%2B%20%5Ctheta_r%27%20%3D%2049)
![sin^{-1}\frac{0.33}{\mu} + sin^{-1}\frac{1}{\mu} = 49](https://tex.z-dn.net/?f=sin%5E%7B-1%7D%5Cfrac%7B0.33%7D%7B%5Cmu%7D%20%2B%20sin%5E%7B-1%7D%5Cfrac%7B1%7D%7B%5Cmu%7D%20%3D%2049)
By solving above equation we have
![\mu = 1.645](https://tex.z-dn.net/?f=%5Cmu%20%3D%201.645%20)