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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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NEED HELP ASAP Which of the following chemical equations shows the Law of Conservation of Mass? (1 point)
natulia [17]

Answer:

OK + MgBR arrow KBR + MG

Explanation:

I know nothing about this topic, but if it has to be balanced I am pretty certain that's the only balanced equation

4 0
2 years ago
A 190 g glider on a horizontal, frictionless air track is attached to a fixed ideal spring with force constant 160 N/m. At the i
laiz [17]

(a) Let <em>x</em> be the maximum elongation of the spring. At this point, the glider would have zero velocity and thus zero kinetic energy. The total work <em>W</em> done by the spring on the glider to get it from the given point (4.00 cm from equilibrium) to <em>x</em> is

<em>W</em> = - (1/2 <em>kx</em> ² - 1/2 <em>k</em> (0.0400 m)²)

(note that <em>x</em> > 4.00 cm, and the restoring force of the spring opposes its elongation, so the total work is negative)

By the work-energy theorem, the total work is equal to the change in the glider's kinetic energy as it moves from 4.00 cm from equilibrium to <em>x</em>, so

<em>W</em> = ∆<em>K</em> = 0 - 1/2 <em>m</em> (0.835 m/s)²

Solve for <em>x</em> :

- (1/2 (160 N/m) <em>x</em> ² - 1/2 (160 N/m) (0.0400 m)²) = -1/2 (0.190 kg) (0.835 m/s)²

==>   <em>x</em> ≈ 0.0493 m ≈ 4.93 cm

(b) The glider attains its maximum speed at the equilibrium point. The work done by the spring as it is stretched away from equilibrium to the 4.00 cm position is

<em>W</em> = - 1/2 <em>k</em> (0.0400 m)²

If <em>v</em> is the glider's maximum speed, then by the work-energy theorem,

<em>W</em> = ∆<em>K</em> = 1/2 <em>m</em> (0.835 m/s)² - 1/2 <em>mv</em> ²

Solve for <em>v</em> :

- 1/2 (160 N/m) (0.0400 m)² = 1/2 (0.190 kg) (0.835 m/s)² - 1/2 (0.190 kg) <em>v</em> ²

==>   <em>v</em> ≈ 1.43 m/s

(c) The angular frequency of the glider's oscillation is

√(<em>k</em>/<em>m</em>) = √((160 N/m) / (0.190 kg)) ≈ 29.0 Hz

3 0
3 years ago
4. What type of assessments are based on repeatable, measurable data?
Harman [31]

Answer:

  • OBJECTIVE ASSESSMENTS
  • SUBJECTIVE ASSESSMENTS
  • BODY COMPOSITION ASSESSMENTS
  • COGNITIVE ASSESSMENTS

Explanation:

PLS MAKE ME BRAINLIEST

8 0
2 years ago
An empty graduated cylinder weighs 55.26 g. When filled with 48.1 mL of an unknown liquid, it weighs 92.39 g. The density of the
katrin [286]
<h2>Density of the unknown liquid is 771.93 kg/m³</h2>

Explanation:

An empty graduated cylinder weighs 55.26 g

Weight of empty cylinder = 55.26 g = 0.05526 kg

Volume of liquid filled = 48.1 mL = 48.1 x 10⁻⁶ m³

Weight of cylinder plus liquid = 92.39 g = 0.09239 kg

Weight of liquid = 0.09239 - 0.05526

Weight of liquid = 0.03713 kg

We have

        Mass = Volume x Density

        0.03713 = 48.1 x 10⁻⁶ x Density

        Density = 771.93 kg/m³

Density of the unknown liquid is 771.93 kg/m³

5 0
3 years ago
A plane has a mass of 360,000 kg takes-off at a speed of 300 km/hr. i) What should be the minimum acceleration to take off if th
melomori [17]

Answer:

i) the minimum acceleration to take off is 22500 km/h²

ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) required force that the engine must exert to attain acceleration is 625 kN

Explanation:

Given the data in the question;

mass of plane m = 360,000 kg

take of speed v = 300 km/hr = 83.33 m/s

i)

What should be the minimum acceleration to take off if the length of the runway is 2.00 km

from Newton's equation of motion;

v² = u² + 2as

we know that a plane starts from rest, so; u = 0

given that distance S = 2 km

we substitute

(300)² = 0² + ( 2 × a × 2 )

90000 = 4 × a

a = 90000 / 4

a = 22500 km/h²

Therefore,  the minimum acceleration to take off is 22500 km/h²

ii) At this acceleration, how much time would the plane need from starting to takeoff.

from Newton's equation of motion;

v = u + at

we substitute

300 = 0 + 22500 × t

t = 300 / 22500

t = 0.0133 hrs

Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) What force must the engines exert to attain this acceleration

we know that;

F = ma

acceleration a = 22500 km/hr² = 1.736 m/s²

so we substitute

F = 360,000 kg × 1.736 m/s²

F =  624960 N

F = 625 kN

Therefore, required force that the engine must exert to attain acceleration is 625 kN

5 0
3 years ago
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