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saveliy_v [14]
2 years ago
10

The distance from the Moon to Earth is 3.9 x 10^8 meters. What is the time required for a light ray to travel from the Moon to E

arth?
Physics
1 answer:
kondaur [170]2 years ago
6 0

Answer:

hello I don't know the answer sorry next time I will try

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a plane is flying due east in still air at 395 km/h. suddenly, the plane is hit by wind blowing at 55km/h toward the west. what
Sphinxa [80]
Let's be clear:  The plane's "395 km/hr" is speed relative to the
air, and the wind's "55 km/hr" is speed relative to the ground.

Before the wind hits, the plane moves east at 395 km/hr relative
to both the air AND the ground.

After the wind hits, the plane still maintains the same air-speed.
That is, its velocity relative to the air is still 395 km/hr east.
But the wind vector is added to the air-speed vector, and the
plane's velocity <span>relative to the ground drops to 340 km/hr east</span>.

6 0
3 years ago
Question 30
stellarik [79]

Answer: 0.69\°

Explanation:

The angular diameter \delta of a spherical object is given by the following formula:

\delta=2 sin^{-1}(\frac{d}{2D})

Where:

d=16 m is the actual diameter

D=1338 m is the distance to the spherical object

Hence:

\delta=2 sin^{-1}(\frac{16 m}{2(1338 m)})

\delta=0.685\° \approx 0.69\° This is the angular diameter

3 0
3 years ago
The direction of centripetal force is away from the axis of rotation.
Arada [10]
I’m pretty sure it’s true x
4 0
3 years ago
Read 2 more answers
A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7
aleksandrvk [35]

(a) 1440.5 Hz

The general formula for the Doppler effect is

f'=(\frac{v+v_r}{v+v_s})f

where

f is the original frequency

f is the apparent frequency

v is the velocity of the wave

v_r is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

v_s is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

v_s = v_A = -28.7 m/s (surface A is the source, which is moving towards the receiver)

v_r = +62.2 m/s (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz

(b) 0.232 m

The wavelength of a wave is given by

\lambda=\frac{v}{f}

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m

(c) 1481.2 Hz

Again, we can use the same formula

f'=(\frac{v+v_r}{v+v_s})f

In the source frame (= on surface A), we have

v_s = v_B = -62.2 m/s (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

v_r = +28.7 m/s (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz

(d) 0.225 m

The wavelength of the wave is given by

\lambda=\frac{v}{f}

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m

8 0
3 years ago
Obtain the formula for the focal length of a lens in terms of object distance (u)
Alexxx [7]

Answer:

m=image distance÷object distance

6 0
2 years ago
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