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Leto [7]
2 years ago
8

A solar system may form from a spinning disk of material called a(n _____

Physics
2 answers:
Svetlanka [38]2 years ago
5 0
<h3><u>Answer;</u></h3>

<em>Accretion disk </em>

A solar system may form from a spinning disk of material called an<u><em> accretion disk </em></u>

<h3><u>Explanation</u>;</h3>
  • <em><u>The solar system is made up of the sun and several planets that orbit around it.</u></em>
  • <em><u>Accretion is the process by which planets were formed</u></em> in which they began as dust particles that orbited around a central protostar.
  • The <u><em>material in the nebula that was not absorbed by the sun moved around it and formed a flat disk of dust particles and gas that was held by the gravity of the sun. </em></u>This disk is <em><u>the accretion disk</u></em>. Each planet started as a microscopic dust particle in this accretion disk.
blsea [12.9K]2 years ago
4 0
The appropriate response is accretion disk. It is a structure (regularly a circumstellar circle) shaped by diffused material in orbital movement around a monstrous focal body. The focal body is regularly a star. Gravity makes the material in the plate winding internal towards the focal body.
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<span>You are given a submerged submarine accelerating upward at 0.325 m/s</span>² and the density of sea water is 1.025x10³ kg/m³. The submarine's average density at this time is 22 kg/m³.
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The transmutation of a radioactive uranium isotope, 234/92 U, into a radon isotope, 222/86 Rn, involves a series of three nuclea
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The answer is 26/98 how i did this is i divided them mulitiplyed well i cant really explain it but im pretty dure its right

Explanation:

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Each plate of a parallel-plate capacator is a square with side length r, and the plates are separated by a distance d. The capac
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Answer:

Explanation:

Before the dialectic was inserted the capacitor is Co

When the slab is inserted,

The capacitor becomes

C=kCo

The charge Q is given as

Q=CV

Then, when C=Co

Qo=CoV

Then, when C=kCo

Q=kCoV

Then, the change in charges is given as

Q-Qo= kCoV - CoV

∆Q= kCoV - CoV

Current is given as

I=dQ/dt

I= (kCoV - CoV) / dt

I=Co(kV-V)/dt

Note Co is the value capacitor

So, Capacitance of parallel plates capacitor is given as

Co=εoA/d

Then,

I=εoA(kV-V)/d•dt

I=VεoA(k-1)/d•dt

Where A=πr²

I = V•εo•πr²•(k-1) / d•dt

This is the required expression for current is in the required term

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A ball is thrown forward at 5 m/s. How would the path of the ball differ on Earth than on the moon? The ball would follow a curv
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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

6 0
2 years ago
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