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3 years ago

Why can light travel through outer space, but sound cannot?

2 answers:

6 0

because sound needs to travel through a medium , like air . much of outer space is a vacuum, with nothing there , so sound has nothing to travel through . Light does not need a medium to travel through .

I hope that's help !

WINSTONCH [101]3 years ago

5 0

Usually, sound needs a medium to travel through, like a vacuum for example.

Light does not need a medium to travel, and since air is considered a medium, light is not dependent on that.

On the other hand, sound needs a medium to travel through, and outer space doesn't contain space, therefore, no sound.

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Instantaneous speed is measured

VMariaS [17]

Answer:

C. At a particular instant

Explanation:

Speed is the defined as the ratio between the distance covered by an object and the time taken:

$v=\frac{d}{t}$

where d is the distance and t the time.

However, there are two possible measurements of speed:

- Average speed: this is the speed measured over a non-zero time interval (for example: a car moving 100 metres in 5 seconds; its average speed is

$v=\frac{100 m}{5 s}=20 m/s$

- Instantaneous speed: this is the speed of an object measured at a particular instant in time, so for a time interval that tends to zero. So, in the previous example, the average speed is 20 m/s but the instantaneous speed of the car at various instants of time can be different from that value.

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3 years ago

Having landed on a newly discovered planet, an astronaut sets up a simple pendulum of length 1.38 m and finds that it makes 441

Tasya [4]

The period of a simple pendulum is given by:
$T=2 \pi \sqrt{ \frac{L}{g} }$
where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:
$g= \frac{4 \pi^2}{T^2}L$ (1)

We already know the length of the pendulum, L=1.38 m, however we need to find its period of oscillation.

We know it makes N=441 oscillations in t=1090 s, therefore its frequency is
$f= \frac{N}{t}= \frac{441}{1090 s}=0.40 Hz$
And its period is the reciprocal of its frequency:
$T= \frac{1}{f}= \frac{1}{0.40 Hz}=2.47 s$

So now we can use eq.(1) to find the gravitational acceleration of the planet:
$g= \frac{4 \pi^2}{T^2}L = \frac{4 \pi^2}{(2.47 s)^2} (1.38 m) =8.92 m/s^2$

3 0

3 years ago

____ are solid while lines stretching across one or more lanes in the same direction, indicating the proper place to come to a s

Natalka [10]

Stop lines are solid white lines painted across the traffic lanes at intersections and pedestrian crosswalks, indicating the exact place to stop.

8 0

3 years ago

A solid non-conducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a radius r (r <

Svet_ta [14]

Answer:

Hence the answer is E inside $= KQr_{1} /R^{3}$ .

Explanation:

E inside $= KQr_{1} /R^{3}$

so if r1 will be the same then

E $\begin{bmatrix}Blank Equation\end{bmatrix}$ proportional to 1/R3

so if R become 2R

E becomes 1/8 of the initial electric field.

8 0

2 years ago

Read 2 more answers

A satellite circles the Earth in an orbit whose radius is twice the Earth’s radius. The Earth’s mass is 5.98 x 1024 kg, and its

gavmur [86]

Hello!

Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:

$d = 2\pi r$ ← The circumference of the orbit