1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Vanyuwa [196]
3 years ago
7

Conversations with astronauts on the lunar surface were characterized by a kind of echo in which the earthbound person’s voice w

as so loud in the astronaut’s space helmet that it was picked up by the astronaut’s microphone and transmitted back to Earth. It is reasonable to assume that the echo time equals the time necessary for the radio wave to travel from the Earth to the Moon and back (that is, neglecting any time delays in the electronic equipment). Calculate the distance from Earth to the Moon given that the echo time was 2.56 s and that radio waves travel at the speed of light (3.00×108 m/s). Give your answer in thousands of km.
Physics
2 answers:
svetlana [45]3 years ago
8 0

Answer: 768000km

Explanation:

Velocity is given by the relation between the distance d and the time it takes to travel that distance t:

V=\frac{d}{t}   (1)

In this problem we are told the time it takes for radio wave to travel from the Earth to the Moon and back is the "echo":

t=2.56s  (2)

In addition, we know radio waves are electromagnetic waves (light), and its velocity is:

V=3(10)^{8}m/s   (3)

Substituting (2) and (3) in (1):

3(10)^{8}m/s=\frac{d}{2.56s}   (4)

And finding d:

d=(3(10)^{8}m/s)(2.56s)   (5)

Finally we can obtain the distance:

d=768000000m=768000km  

xxTIMURxx [149]3 years ago
6 0

Answer:

384,000 km

Explanation:

Given

Velocity of radio wave v = 3.00 \times 10^{8}m/s

Duration of echo T = 2.56 s

Solution

Time taken for the radio wave to travel to moon and to travel back to earth as it was picked up by the astronaut's microphone is 2.56 s

Since any time delays in the electronic equipment can be ignored

time taken for the radio wave to reach moon

t = \frac{T}{2}\\\\t = \frac{2.56}{2} \\\\t = 1.28 s

v = \frac{d}{t}\\\\d = vt\\\\d = 3 \times 10^8 \times 1.28\\\\d = 3.84 \times 10^8 m\\\\d = 384,000 km

You might be interested in
why it is possible to estimate, to a good approximation, the energy derived from a sugar cube in the body,by doing laboratory ex
yarga [219]
The sugar cube experiment in the laboratory  gives us a good approximation of the amount of energy that can be derived from the sugar cube because the amount of energy is neither created or destroyed, it is just converted to another form. If the energy from the sugar cube is converted to other form in the lab then, it is possible that same amount of energy will be derived. 
6 0
3 years ago
Technician A says that a MAF sensor is a high-authority sensor and is responsible for determining the fuel needs of the engine b
xz_007 [3.2K]

Answer:

a. Technician A

Explanation:

Technician A says that a MAF sensor is a high-authority sensor and is responsible for determining the fuel needs of the engine based on the measured amount of air entering the engine. Technician B says that a cold wire MAF sensor uses the electronics in the sensor itself to heat a wire 20°C below the temperature of the air entering the engine. Who is right

MAF wich stands for  mass airflow sensor determines the mass of air flowing into the engine's air intake system. ... , the wire cools When air flows past the wire, decreasing its resistance, thereby more current flows through the circuit. When the MAf goes bad, it can not sense the amount of air intake into the engine.

8 0
4 years ago
A 750g aluminum pan is removed from the stove and plunged into a sink filled with 10.0kg of water at 20.0c. the water temperatur
spin [16.1K]
The concept that should be used here is that heat loss is equal to heat gain. In this item, the heat lost by the aluminum pan should be equal to the heat gained by water. Such that,
              750 g(0.215 cal/g°C)(T - 24) = (10000 g)(1 cal/g°C)(4°C)

The value of T from the equation is equal to 272°C.
3 0
3 years ago
A brick is dropped from rest from a height of 4.9 m. how long does it take the brick to reach the ground?
Igoryamba

We know the equation of motion , s =ut+\frac{1}{2} at^2, where s is the displacement, u is the initial velocity, a is the acceleration and t is the time taken.

In this case a brick is dropped from rest from a height of 4.9 m, so displacement, s = 4.9 m

Since the brick is dropped from rest, u = 0 m/s

Acceleration on brick = acceleration due to gravity = 9.81 m/s^2

Substituting

   4.9 = 0*t+\frac{1}{2} *9.81*t^2\\ \\ 4.9 = 4.9t^2\\ \\ t^2=1\\ \\ t =1 second

So the brick will reach ground after 1 second.


8 0
3 years ago
Read 2 more answers
Which of the following are in the correct order from smallest to largest?
Darina [25.2K]

Answer:

Millimeter, Centimeter, meter, kilometer

Explanation:

K H D M D C M

This is what I say to remember them

Kick Him Down Mommy Don't Commit Murder

4 0
3 years ago
Other questions:
  • The short vertical parts adjacent to it also reach into the magnetic field and should experience forces. why can we neglect them
    7·1 answer
  • A small metal bar, whose initial temperature was 30° C, is dropped into a large container of boiling water. How long will it tak
    5·2 answers
  • A soccer ball is kicked at an angle of 30° to the horizontal with an initial velocity of 16 m/s. How high does the soccer ball g
    10·1 answer
  • In a simple hydraulic press, similar to the brakes in your car, the system works on the principle that
    13·2 answers
  • Why was 6 afraid of 7​
    12·1 answer
  • Carmen is helping load furniture and boxes onto a moving truck. She picks up boxes of her things, places them on a cart, and pus
    5·1 answer
  • What is the power of ideal sunglass​
    9·2 answers
  • If a car travels 200 meters in 10 seconds, what is the speed of the car?
    6·2 answers
  • PLEASE HELP ME I REALLY NEED HELP
    5·1 answer
  • What is the electric current measured in
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!