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3 years ago

A space traveller leaves Earth for 10 years at .85c. According to an observer on Earth, how much time has passed?

1 answer:

eduard3 years ago

8 0

First of all, you didn't tell us WHO measured the "10 years".

If it was the people on Earth, then 10 years passed according to them.

If it was 10 years on the space traveler's clock, then the clock in the
OTHER place, like on Earth, is subject to the relativistic 'time dilation'.

If the clocks are moving relative to each other, then the time interval measured
on either clock is equal to the interval measured on the other clock, divided by

√(1 - v²/c²) .

You said that v/c = 0.85 .

v²/c² = (0.85)² = 0.7225

1 - v²/c² = 1 - 0.7225 = 0.2775

√(1 - v²/c²) = √0.2775 = 0.5268

If one clock counts up 10 years, then the other one counts up

(10years) / 0.5268 = 18.983 years 

I believe that's the way to do this, and I'll gladly take your points,
but let me recommend that you get a second opinion before you
actually take off on your 10-year interstellar mission.

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The wavelength of a wave on a string is 1.2 meters. If the speed of the wave is 60 meters/second, what is its frequency? A. 0.20

anastassius [24]

Answer:

50 Hz

Explanation:

The frequency of a wave is given by

$f=\frac{v}{\lambda}$

where

v is the speed of the wave

$\lambda$ is the wavelength

In this problem,

v = 60 m/s

$\lambda=1.2 m$

So the frequency is

$f=\frac{60 m/s}{1.2 m}=50 Hz$

8 0

2 years ago

Lighting from the sun travels through space to earths atmosphere.

Lady bird [3.3K]

Yes it does do all that

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2 years ago

Charge flows from low potential to high potential. O True or O False

inn [45]

Answer:

False

Explanation:

Think of the electric potential in terms of potential energy. If you imagine a place with high elevation (A) and another one at sea level (B), a ball will roll from high potential to low potential (A-->B).

Everything in our universe wants to reach a lower state of energy if no external force is acted upon it. Every object tends to slow down (friction), a radioactive element dissipates energy (an unstable element releases energy to get to a stable state), water in the clouds comes down to the ground (rain experiencing difference in potential energy).

Electric potential is exactly the same, you just can't see it! It flows from higher voltage (which is a synonym for electric potential) to lower voltage.

8 0

2 years ago

Kinematics

leonid [27]

Answer:

a = 2 [m/s^2]

a = 1.6 [m/s^2]

xt = 2100 [m]

Explanation:

In order to solve this problem we must use kinematics equations. But first we must identify what kind of movement is being studied.

When the car moves from rest to 40 [m/s] by 20 [s], it has a uniformly accelerated movement, in this way we can calculate the acceleration by means of the following equation:

$v_{f} = v_{i}+(a*t)$

where:

Vf = final velocity = 40 [m/s]

Vi = initial velocity = 0 (starting from rest)

a = acceleration [m/s^2]

t = time = 20 [s]

40 = 0 + (a*20)

a = 2 [m/s^2]

The distance can be calculates as follows:

$v_{f} ^{2} = v_{i} ^{2}+(2*a*x)$

where:

x1 = distance [m]

40^2 = 0 + (2*2*x1)

x1 = 400 [m]

Now the car maintains its speed of 40 [m/s] for 30 seconds, we must calculate the distance x2 by means of the following equation, it is important to emphasize that this movement is at a constant speed.

v = x2/t2

where:

x2 = distance [m]

t2 = 30 [s]

x2 = 40*30

x2 = 1200 [m]

Immediately after a change of speed occurs, such that the previous final speed becomes the initial speed, the new Final speed corresponds to zero, since the car stops completely.

$v_{f} = v_{i}-a*t$

Note: the negative sign of the equation means that the car is stopping, i.e. slowing down.

0 = 40 - (a *25)

a = 40/25

a = 1.6 [m/s^2]

The distance can be calculates as follows:

$v_{f} ^{2} = v_{i} ^{2} -2*a*x3\\$

0 = (40^2) - (2*1.6*x3)

x3 = 500 [m]

Now we sum all the distances calculated:

xt = x1 + x2 + x3

xt = 400 + 1200 + 500

xt = 2100 [m]

8 0

2 years ago

Light rays from stars bend toward smaller angles as they enter Earth's atmosphere. a. Explain why this happens using Snell's law

Grace [21]

Answer:

Following are the answer to this question:

Explanation:

In option (a):

The principle of Snells informs us that as light travels from the less dense medium to a denser layer, like water to air or a thinner layer of the air to the thicker ones, it bent to usual — an abstract feature that would be on the surface of all objects. Mostly, on the contrary, glow shifts from a denser with a less dense medium. This angle between both the usual and the light conditions rays is referred to as the refractive angle.
Throughout in scenario, the light from its stars in the upper orbit, the surface area of both the Earth tends to increase because as light flows from the outer atmosphere towards the Earth, it defined above, to a lesser angle.

In option (b):

Rays of light, that go directly down wouldn't bend, whilst also sun source which joins the upper orbit was reflected light from either a thicker distance and flex to the usual, following roughly the direction of the curve of the earth.
Throughout the zenith specific position earlier in this thread, astronomical bodies appear throughout the right position while those close to a horizon seem to have been brightest than any of those close to the sky, and please find the attachment of the diagram.

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3 years ago