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antoniya [11.8K]
3 years ago
6

A boy finds an abandoned mine shaft in the woods, and wants to know how deep the hole is. He drops in a stone, and counts 4 seco

nds before he hears the "plunk" of the stone hitting the bottom of the shaft. Approximately how deep is the shaft? (Assume a gravitational acceleration of 9.8 m/s2 and no significant air resistance.)
Physics
1 answer:
oee [108]3 years ago
4 0
S=(0x4)+(0.5x4.81x4x4)
S=0.78.48

The depth is approximately 78 meters.
(My brain hurts now) :P Good Luck!
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A sinusoidal wave is traveling along a rope. The oscillator that generates the wave completes 40.0 vibrations in 30.0s . A given
Phantasy [73]

From the solution that I have done, the wavelength in the question that we have is 31.88 cm

<h3>How to solve for the wavelength</h3>

The frequency in the question is given as 40/30 = 1.33 hz

Next we have to solve for V

= 425/10

= 42.5 cm/s

v = frequency * wavelength

we have to put in the values in the formula. This would be

42.5 = 1.33 x wavelength

we have to divide through by 1.33 to get the wavelength. This would be

42.5/1.333 = wavelength

31.88 cm = wavelength

Hence we can say that the wavelength in the question that we have here is 31.88 cm

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8 0
1 year ago
A train station bell gives off a fundamental tone of 500 Hz as the train approaches the station at a speed of 20 m/s. If the spe
Brut [27]

Answer: 529.9 Hz

Explanation:

Here we need to use the Doppler equation, so we have:

f' = f*(v + v0)/(v - vs)

Here, f is the frequency = 500Hz

v is the velocity of the wave, = 334m/s

v0 is the velocity of the observer = 20m/s

vs is the velocity of the source = 0m/s

Then we have:

f' = 500Hz*(334m/s + 20m/s)/(334m/s) = 529.9 Hz

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What is the momentum of an 18-kg object moving at 0.5 m/s?
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3 years ago
Starting from Newton’s law of universal gravitation, show how to find the speed of the moon in its orbit from the earth-moon dis
WARRIOR [948]

Answer: 1010.92 m/s

Explanation:

According to Newton's law of universal gravitation:

F=G\frac{Mm}{r^{2}} (1)

Where:

F is the gravitational force between Earth and Moon

G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} is the Gravitational Constant  

M=5.972(10)^{24} kg is the mass of the Earth

m=7.349(10)^{22} kg is the mass of the Moon

r=3.9(10)^{8} m is the distance between the Earth and Moon

Asuming the orbit of the Moon around the Earth is a circular orbit, the Earth exerts a centripetal force on the moon, which is equal to F:

F=m.a_{C} (2)

Where a_{C} is the centripetal acceleration given by:

a_{C}=\frac{V^{2}}{r} (3)  

Being V the orbital velocity of the moon

Making (1)=(2):

m.a_{C}=G\frac{Mm}{r^{2}} (4)

Simplifying:

a_{C}=G\frac{M}{r^{2}} (5)

Making (5)=(3):

\frac{V^{2}}{r}=G\frac{M}{r^{2}} (6)  

Finding V:

V=\sqrt{\frac{GM}{r}} (7)

V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24} kg)}{3.9(10)^{8} m}} (8)

Finally:

V=1010.92 m/s

5 0
2 years ago
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