Answer:
15.3 s and 332 m
Explanation:
With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon
gm = 1/6 ge
gm = 1/6 9.8 m/s² = 1.63 m/s²
We calculate the range
R = Vo² sin 2θ / g
R = 25² sin (2 30) / 1.63
R= 332 m
We will calculate the time of flight,
Y = Voy t – ½ g t2
Voy = Vo sin θ
When the ball reaches the end point has the same initial height Y=0
0 = Vo sin t – ½ g t2
0 = 25 sin (30) t – ½ 1.63 t2
0= 12.5 t – 0.815 t2
We solve the equation
0= t ( 12.5 -0.815 t)
t=0 s
t= 15.3 s
The value of zero corresponds to the departure point and the flight time is 15.3 s
Let's calculate the reach on earth
R2 = 25² sin (2 30) / 9.8
R2 = 55.2 m
R/R2 = 332/55.2
R/R2 = 6
Therefore the ball travels a distance six times greater on the moon than on Earth
Answer:
915 Hz
Explanation:
The observed frequency from a sound source is given as
f₀ = f [(v + v₀)/(v+vₛ)]
where
f₀ = observed frequency of the sound by the observer = ?
f = actual frequency of the sound wave = 983 Hz
v = actual velocity of the sound waves = 343 m/s
vₛ = velocity of the source of the sound waves = 55.9 m/s
v₀ = velocity of the observer = 28.4 m/s
f₀ = 983 [(343+28.4)/(343+55.9)]
f₀ = 915.2 Hz = 915 Hz
Answer:
9
Explanation:
2.13 rad/s * 26.9 sec
2.13 * 26.9
57.297
3282.88 deg / 360 deg = 9.12
It makes 9 complete revolutoins
Answer:
a. 0.18Hz
b. 0.56m/s
Explanation:
From the question we can deduct the following parameters
The wavelength, λ is define as the distance between two successful crest or trough and from the question we conclude that wavelength is 3.17m.
Also the period of the wave T can be computed as
T=22.6/4
T=5.65secs.
a. To compute the frequency, recall that frequency, F=1/period.
Hence,
F=1/5.65
F=0.18Hz
b. Next we compute the wave speed.
Wave speed=frequency *wavelength
Wave speed =0.18*3.17
Wave speed =0.56m/s
Answer:the first one was x
the second one is y
Explanation:
