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Vlad1618 [11]
4 years ago
12

Which planet do most known extrasolar planets most resemble?

Physics
1 answer:
tatuchka [14]4 years ago
6 0
The awnser is A>Most known exoplanets resemble gas giants known as "hot Jupiters" as a result of being large objects orbiting close to they're host star.
Most known exoplanets that we have detected are likely or confirmed to be gas giants. This is because one of the most used detection techniques is the Transit technique.

This technique involves looking at a distance star's light curve (the changes in the brightness of star over a period of time). Most stars dim and brighten over time by a small bit. But if the light curve shows a periodic and large dip in brightness, this is a sign that something large is passing (or transiting) in front of it.

The reason most detected exoplanets are gas giants is simply because they're the easiest to detect. They cause a larger dip in the curve than a smaller planet would. A small planet's dip are masked by the star's normal dimming and brightening.

Imagine having a flashlight shining on a wall. If you pass a large object through the light beam, it has a large shadow. If you pass a smaller object through, it has a smaller shadow. The change in the light you see on the wall is similar to what you will see in a star's light curve. Now if you pass an object closer to flashlight, as opposed to closer to the wall, this changes the shadow as well.

This is why most exoplanets are "hot Jupiters". They're large gas giants that are hot because they pass very close to the star, resulting in a larger dip that is easier to see in the curve. They also have shorter orbits so they're more likely to be seen in a few weeks or month's time.


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A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w
cupoosta [38]

Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

\displaystyle P=\frac{W}{t}

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

W=F.x

The second Newton's law gives us the net force as

F=m.a

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

v_f=v_o+at

v_f=at

Solving for a

\displaystyle a=\frac{v_f}{t}={5.4}{1}

a=5.4\ m/s^2

The distance traveled in the interval is given by

\displaystyle x=v_o.t+\frac{a.t^2}{2}

Since vo=0

\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}

x=2.7\ m

The force is given by

F=m.a

We don't know the value of m, so the force is

F=2.7m

Computing the work done by the sprinter

W=F.x=2.7m(5.4)

W=14.58m

The power is finally computed

\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}

P=14.58m

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}

a=8.8\ m/s^2

The distance is

\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}

x=3.8\ m

The net force is

F=m(8.8)=8.8m

The work done by the sprinter is now computed as

W=8.8m(3.8)=33.44m

At last, the output power is

\displaystyle P=\frac{33.44m}{0.5}=66.88m

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

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This force is involves the attraction between charged particles
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3 years ago
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According to Newton's law of cooling, the rate at which an object's temperature changes is directly proportional to the differen
Kryger [21]

Answer:

dT(t)/dt = k[T5 - T(t)]

Explanation:

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Introducing the constant of proportionality

dT(t)/dt = k[T5 - T(t)]

which is the desired differential equation

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