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RUDIKE [14]
3 years ago
13

Why are lunar calendars not more widely used today?

Physics
2 answers:
solong [7]3 years ago
8 0
Because the season effect differentley
Nookie1986 [14]3 years ago
6 0
The seasons affect the scheduling of far more events the moon phases do.
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What is shot-curciting​
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A path that allows most of the current in an electric circuit to flow around or away from the principal elements or devices in the circuit.

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3 years ago
_____ in the uterus has to occur for a woman to confirm her suspicions that she is pregnant.
vlabodo [156]
Conception is the correct answer
7 0
3 years ago
How long is the photons journey from the milky way to earth
fgiga [73]
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3 years ago
A projectile is launched at an angle of 36.7 degrees above the horizontal with an initial speed of 175 m/s and lands at the same
Softa [21]

Answer:

a) The maximum height reached by the projectile is 558 m.

b) The projectile was 21.3 s in the air.

Explanation:

The position and velocity of the projectile at any time "t" is given by the following vectors:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v = (v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t"

x0 = initial horizontal position

v0 = initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration due to gravity (-9.80 m/s² considering the upward direction as positive).

v = velocity vector at time t

a) Notice in the figure that at maximum height the velocity vector is horizontal. That means that the y-component of the velocity (vy) at that time is 0. Using this, we can find the time at which the projectile is at maximum height:

vy = v0 · sin α + g · t

0 = 175 m/s · sin 36.7° - 9.80 m/s² · t

-  175 m/s · sin 36.7° /  - 9.80 m/s² = t

t = 10.7 s

Now, we have to find the magnitude of the y-component of the vector position at that time to obtain the maximum height (In the figure, the vector position at t = 10.7 s is r1 and its y-component is r1y).

Notice in the figure that the frame of reference is located at the launching point, so that y0 = 0.

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 175 m/s · 10.7 s · sin 36.7° - 1/2 · 9.8 m/s² · (10.7 s)²

y = 558 m

The maximum height reached by the projectile is 558 m

b) Since the motion of the projectile is parabolic and the acceleration is the same during all the trajectory, the time of flight will be twice the time it takes the projectile to reach the maximum height. Then, the time of flight of the projectile will be (2 · 10.7 s) 21.4 s. However, let´s calculate it using the equation for the position of the projectile.

We know that at final time the y-component of the vector position (r final in the figure) is 0 (because the vector is horizontal, see figure). Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

0 = 175 m/s · t · sin 36.7° - 1/2 · 9.8 m/s² · t²

0 = t (175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t)

0 = 175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t

-  175 m/s ·  sin 36.7 / -(1/2 · 9.8 m/s²) = t

t = 21.3 s

The projectile was 21.3 s in the air.

7 0
3 years ago
The Doppler effect using ultrasonic waves of frequency 2.25 × 106 Hz is used to monitor the heartbeat of a fetus. A (maximum) be
miss Akunina [59]

Answer:

v = 2.88 \times 10^7 m/s

Explanation:

As per Doppler's effect we know that the frequency of the sound that is observed by the detector is the reflected sound

This reflected sound is given as

f' = f (\frac{v - v_h}{v + v_h})

f' = 2.25\times 10^6(\frac{v - 1540}{v + 1540})

so we know that the beat frequency is

\Delta f = 240 Hz

so we will have

f - f' = \Delta f

2.25 \times 10^6 - 2.25\times 10^6(\frac{v - 1540}{v + 1540}) = 240

1 - (\frac{v - 1540}{v + 1540}) = 1.07 \times 10^{-4}

so we have

0.99989 = (\frac{v - 1540}{v + 1540})

1.99989\times 1540 = 1.067 \times 10^{-4} v

v = 2.88 \times 10^7 m/s

4 0
4 years ago
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