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BigorU [14]
2 years ago
13

The process of star and planet formation begins with a large cloud of gas and dust called a solar nebula. Rank the formation eve

nts that occur within a cloud from earliest to latest.
Rank from earliest to latest. To rank items as equivalent, overlap them.
A. The cloud is large, cool, and slowly rotating
B. The cloud collapses into a disk
C. Competing rotational and gravitational forces begin to flatten the cloud
D. The cloud becomes denser, heats up, and rotates faster
E. The cloud starts to contract under the influence of gravity
Physics
1 answer:
likoan [24]2 years ago
8 0

Answer:

c

Explanation:

You might be interested in
Assume the motions and currents mentioned are along the x axis and fields are in the y direction. (a) does an electric field exe
matrenka [14]
<span> (a) does an electric field exert a force on a stationary charged object? 
Yes. The force exerted by an electric field of intensity E on an object with charge q is
</span>F=qE
<span>As we can see, it doesn't depend on the speed of the object, so this force acts also when the object is stationary.

</span><span>(b) does a magnetic field do so?
No. In fact, the magnetic force exerted by a magnetic field of intensity B on an object with  charge q and speed v is
</span>F=qvB \sin \theta
where \theta is the angle between the direction of v and B.
As we can see, the value of the force F depends on the value of the speed v: if the object is stationary, then v=0, and so the force is zero as well.

<span>(c) does an electric field exert a force on a moving charged object? 
Yes, The intensity of the electric force is still
</span>F=qE
<span>as stated in point (a), and since it does not depend on the speed of the charge, the electric force is still present.

</span><span>(d) does a magnetic field do so?
</span>Yes. As we said in point b, the magnetic force is
F=qvB \sin \theta
And now the object is moving with a certain speed v, so the magnetic force F this time is different from zero.

<span>(e) does an electric field exert a force on a straight current-carrying wire?
Yes. A current in a wire consists of many charges traveling through the wire, and since the electric field always exerts a force on a charge, then the electric field exerts a force on the charges traveling through the wire.

</span><span>(f) does a magnetic field do so? 
Yes. The current in the wire consists of charges that are moving with a certain speed v, and we said that a magnetic field always exerts a force on a moving charge, so the magnetic field is exerting a magnetic force on the charges that are traveling through the wire.

</span><span>(g) does an electric field exert a force on a beam of moving electrons?
Yes. Electrons have an electric charge, and we said that the force exerted by an electric field is
</span>F=qE
<span>So, an electric field always exerts a force on an electric charge, therefore on an electron beam as well.

</span><span>(h) does a magnetic field do so?
Yes, because the electrons in the beam are moving with a certain speed v, so the magnetic force
</span>F=qvB \sin \theta
<span>is different from zero because v is different from zero.</span>
6 0
2 years ago
A bicycle tire rotates 25 times in 10 seconds. What is it’s average angular velocity?
ryzh [129]
<h2>Answer:</h2>

<u>Angular velocity of bicycle tire is 15.78 radians per second.</u>

<h3>Explanation:</h3>

Angular velocity is the change in angular speed of an object with respect to time take for change or it is the rate of change of circular motion.

In the given question the circular displacement is 25 rounds around a central point.

The angular displacement is measured in degrees and 1 round is equal to 360 degrees.

25 Rounds = 25 × 360 = 9000 degrees.

Angular velocity = angular displacement /time = 9000/10 = 900 degrees per second.

In SI,angular velocity is represented in radians per second.

So, 1 radian = 57.29 degrees

Angular velocity = 15.78 radians per second

3 0
2 years ago
A 0.200-kg mass attached to the end of a spring causes it to stretch 5.0 cm. If another 0.200-kg mass is added to the spring, th
ziro4ka [17]

Answer:

A: 4 times as much

B: 200 N/m

C: 5000 N

D: 84,8 J

Explanation:

A.

In the first question, we have to caculate the constant of the spring with this equation:

m*g=k*x

Getting the k:

k=\frac{m*g}{x} =\frac{0,2[kg]*9,81[\frac{m}{s^{2} } ]}{0,05[m]} =39,24[\frac{N}{m}]

Then we can calculate how much the spring stretch whith the another mass of 0,2kg:

x=\frac{m*g}{k} =\frac{0,4[kg]*9,81[\frac{m}{s^{2} } ]}{39,24[\frac{N}{m}]} =0,1[m]\\

The energy of a spring:

E=\frac{1}{2}*k*x^{2}

For the first case:

E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,05[m])^{2} =0,049 [J]

For the second case:

E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,1[m])^{2} =0,0196 [J]

If you take the relation E2/E1 = 4.

B.

We have the next facts:

x=0,005 m

E = 0,0025 J

Using the energy equation for a spring:

E=\frac{1}{2}*k*x^{2}⇒k=\frac{E*2}{x^{2} } =\frac{0,0025[J]*2}{(0,005[m])^{2} } =200[\frac{N}{m} ]

C.

The potential energy of the diver will be equal to the kinetic energy in the moment befover hitting the watter.

E=W*h=500[N]*10[m]=5000[J]

Watch out the units in this case, the 500 N reffer to the weighs of the diver almost relative to the earth, thats equal to m*g.

D.

The work is equal to the force acting in the direction of the motion. so we have to do the diference beetwen angles to obtain the effective angle where the force is acting: 47-15=32 degree.

The force acting in the direction of the ramp will be the projection of the force in the ramp, equal to F*cos(32). The work will be:

W=F*d=F*cos(32)*d=10N*cos(32)*10m=84,8J

7 0
2 years ago
A muon formed high in the Earth's atmosphere is measured by an observer on the Earth's surface to travel at speed V - 0.983c for
Alex_Xolod [135]

Answer:

The moun lives 2.198*10^-6 s as measured by its own frame of reference

The Earth moved 648 m as measured by the moun's frame of reference

Explanation:

From the point of view of the observer on Earth the muon traveled 3.53 km at 0.983c

0.983 * 3*10^8 = 2.949*10^8 m/s

Δt = d/v = 3530 / 2.949*10^8 = 1.197*10^-5 s

The muon lived 1.197*10^-5 s from the point of view of the observer.

The equation for time dilation is:

\Delta t' = \Delta t * \sqrt{1 - \frac{v^2}{c^2}}

Then:

\Delta t' = 1.197*10^-5 * \sqrt{1 - \frac{(0.983c)^2}{c^2}} = 2.198*10^-6 s

From the point of view of the moun Earth moved at 0.983c (2.949*10^8 m/s) during a time of 2.198*10^-6, so it moved

d = v*t = 2.949*10^8 * 2.198*10^-6 = 648 m

7 0
2 years ago
A rod with a rest length of 1m whizzes past an observer at a speed of 0.995c, where c is the speed of light. Is it possible that
Tresset [83]

Answer:

B. no

Explanation:

  • When any body moves at a speed comparable to the speed of light (<em>i.e. relativistic speed</em>) then the observer sees a contraction in length of the body along the axis of motion.

Assuming that the motion of the body is along the axis of the rod, then the observer will measure its length to be lesser than its length at rest.

<u>Then according to Einstein's theory of relativity:</u>

L=\frac{L_0}{\gamma}

where:

L_0=  original length of the object (along the direction of motion)

L = observed length of the rod

\gamma = Lorentz factor

\gamma\equiv \frac{1}{\sqrt{1-\frac{v^2}{c^2} } }

6 0
1 year ago
Read 2 more answers
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