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maxonik [38]
3 years ago
9

Conversations with astronauts on lunar walks had an echo that was used to estimate the distance to the Moon. The sound spoken by

the person on Earth was transformed into a radio signal sent to the Moon, and transformed back into sound on a speaker inside the astronaut’s space suit. This sound was picked up by the microphone in the space suit (intended for the astronaut’s voice) and sent back to Earth as a radio echo of sorts. If the round-trip time was 2.15 s, what was the approximate distance to the Moon, neglecting any delays in the electronics?
Physics
1 answer:
dmitriy555 [2]3 years ago
4 0

Answer:

d ≈ 320000 km

Explanation:

Echoes occurs as a result of the bouncing back of sound waves as it impinges on an obstacle. Since the echo covers two distances (i.e forth and back) the distance is written as 2d.

Velocity of radio waves is taken as the speed of light = 3*10^{8}m/s^{2}

time =2.15s

DISTANCE = VELOCITY *TIME

2d=vt

d=\frac{vt}{2}

d=\frac{3*10^{8}*2.15}{2}

    d= 322500000m

Distance to the moon is approximately 320000km

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dolphi86 [110]
The watch hand covers an angular displacement of 2π radians in 60 seconds.

ω = 2π/60
ω = 0.1 rad/s

v = ωr
v = 0.1 x 0.08
v = 8 x 10⁻³ m/s
4 0
3 years ago
A bowling ball accidentally falls out of the cargo bay of an airliner as it flies along in a horizontal direction. As seen from
valina [46]

The path the bowling ball would most closely follow after leaving the airplane is horizontal direction.

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Based on the law of inertia, which is the reluctance of an object to stop moving once in motion or start moving when it is at rest.

The bowling ball will maintain the path of the airline in the first few seconds of fall, after which it will change its path to vertical direction.

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Learn more about horizontal direction here: brainly.com/question/2534565

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3 0
2 years ago
What is the 5 quantitative research problems?<br>​
iogann1982 [59]

Answer:

There are four main types of Quantitative research: Descriptive, Correlational, Causal-Comparative/Quasi-Experimental, and Experimental Research. attempts to establish cause- effect relationships among the variables. These types of design are very similar to true experiments, but with some key differences.

Explanation:

Quantitative research is defined as a systematic investigation of phenomena by gathering quantifiable data and performing statistical, mathematical, or computational techniques.

<h3>I hope it will help you</h3>

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7 0
3 years ago
A very long uniform line of charge has charge per unit length λ1 = 4.68 μC/m and lies along the x-axis. A second long uniform li
Kitty [74]

Answer:

E_{net} = 6.44 \times 10^5 N/C

Explanation:

As we know that electric field due to infinite line charge distribution at some distance from it is given as

E = \frac{2k \lambda}{r}

now we need to find the electric field at mid point of two wires

So here we need to add the field due to two wires as they are oppositely charged

Now we will have

E_{net} = \frac{2k\lambda_1}{r} + \frac{2k\lambda_2}{r}

now plug in all data

\lambda_1 = 4.68 \muC/m

\lambda_2 = 2.48 \mu C/m

r = 0.200 m

now we have

E_{net} = \frac{2k}{r}(4.68 + 2.48)

E_{net} = \frac{2(9\times 10^9)}{0.200}(7.16 \times 10^{-6})

E_{net} = 6.44 \times 10^5 N/C

8 0
3 years ago
An alpha particle (the nucleus of a helium atom) consists of two protons and two neutrons, and has a mass of 6.64 * 10-27 kg. A
melamori03 [73]

Answer:

t = 4.21x10⁻⁷ s

Explanation:

The time (t) can be found using the angular velocity (ω):

\omega = \frac{\theta}{t}

<em>Where θ: is the angular displacement = π (since it moves halfway through a complete circle)</em>

We have:

t = \frac{\theta}{\omega} = \frac{\theta}{v/r}  

<u>Where</u>:      

<em>v: is the tangential speed </em>

<em>r: is the radius</em>

The radius can be found equaling the magnetic force with the centripetal force:

qvB = \frac{mv^{2}}{r} \rightarrow r = \frac{mv}{qB}

Where:

m: is the mass of the alpha particle = 6.64x10⁻²⁷ kg

q: is the charge of the alpha particle = 2*p (proton) = 2*1.6x10⁻¹⁹C

B: is the magnetic field = 0.155 T

Hence, the time is:

t = \frac{\theta*r}{v} = \frac{\theta}{v}*\frac{mv}{qB} = \frac{\theta m}{qB} = \frac{\pi * 6.64 \cdot 10^{-27} kg}{2*1.6 \cdot 10^{-19} C*0.155 T} = 4.21 \cdot 10^{-7} s

Therefore, the time that takes for an alpha particle to move halfway through a complete circle is 4.21x10⁻⁷ s.

I hope it helps you!    

4 0
3 years ago
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