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fredd [130]
3 years ago
15

The earliest radio broadcasts on Earth were emitted about 100 years ago. Approximately where are these initial radio waves now?

(Radio waves are a type of radiation, so they travel at the speed of light)
Physics
1 answer:
Gala2k [10]3 years ago
3 0

Answer:

100 ly are d=9.454255\times10^{17}m

Explanation:

The speed of light is, by definition (we define this and derive a definition of distance from there nowadays), c=299792458m/s. We want to know, at this speed, how much distance the radio signals travel in 100 years. Since each year has 365 days (not a leap one though), each day has 24 hours, each hour has 60 minutes and each minute has 60 seconds, the number of seconds in a year will be (365)(24)(60)(60)=31536000, so the distance traveled by the waves in 100 years will be:

d=vt=(299792458m/s)(100)(31536000s)=9.454255\times10^{17}m, which, of course, are 100 light years.

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A projectile is launched horizontally from a 20-m tall edifice with a vox of 25 m/s. How long will it take for the projectile to
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a(t) = -9.8m/s^2

To get the vertical velocity we need to integrate over time to get:

v(t) = (-9.8m/s^2)*t + v0

where v0 is the initial vertical velocity because the object is thrown horizontally, we do not have any initial vertical velocity, then v0 = 0m/s

v(t) = (-9.8m/s^2)*t

To get the vertical position equation we need to integrate over time again, to get:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + p0

where p0 is the initial position, in this case is the height of the edifice, 20m

then:

p(t) = (-4.9m/s^2)*t^2+ 20m

The projectile will hit the ground when p(t) = 0m, then we need to solve:

(-4.9m/s^2)*t^2+ 20m = 0m

20m = (4.9m/s^2)*t^2

√(20m/ (4.9m/s^2)) = t = 2.02 seconds

The correct option is a.

b) The range will be the total horizontal distance traveled by the projectile, as we do not have any horizontal force, we know that the horizontal velocity is 25 m/s constant.

Now we can use the relationship:

distance = speed*time

We know that the projectile travels for 2.02 seconds, then the total distance that it travels is:

distance = 2.02s*25m/s = 50.5m

Here the correct option is a.

c) Again, the horizontal velocity never changes, is 25m/s constantly, then here the correct option is option b. 25m/s

d) Here we need to evaluate the velocity equation in t = 2.02 seconds, this is the velocity of the projectile when it hits the ground.

v(2.02s) =  (-9.8m/s^2)*2.02s = -19.796 m/s

The velocity is negative because it goes down, and it matches with option d, so I suppose that the correct option here is option d (because the sign depends on how you think the problem)

4 0
3 years ago
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