The correct answer is <span>nebula forms → nebula collapses → protostar forms → fusion begins</span>
Answer:
The correct option is that (She decreases her moment of inertia, thereby increasing her angular speed.)
Explanation:
When an object is in circular motion, the vector that describes it is known as angular momentum. Angular momentum is conserved or constant when an object is spinning in a closed system and no external torques are applied to it. An example of conservation of angular momentum is seen when a woman is sitting on a spinning piano stool with her arms extended. Her angular momentum is conserved because the net torque on her is negligibly small as the friction is exerted very close to the pivot point.
When she folds her arms,her rate of spin increases greatly decreasing her moment of inertia. The work she does to pull in her arms results in an increase in rotational kinetic energy( that is, increase in her angular speed). I hope this helps, thanks!
Answer:
20.6 K
Explanation:
The root mean square velocity is the square root of the average of the square of the velocity. It can be calculated using the following expression.

where,
: root mean square velocity
R: ideal gas constant
T: absolute temperature
M: molar mass
Then, we can find the temperature,

Answer:
30m/s
Explanation:
the distance(s) covered by a body in a straight line is expressed in terms of initial velocity(u),accelaration(a),and time(t)
now since the body is dropping free by the influence of gravity,(a) is a +10ms-²
therefore S=ut +½at²
Now the stone "dropped from the top".ie.from rest,so it doesn't have an initial velocity, (since it was not pushed)
hence u=0
the velocity of the body in a straight line is
V=u+at
so we have to find t in the first equation
and put it in the second one to find V, which is the velocity.
then S=½at² since u is 0
45=½•10•t²,after solving, we get t =3s
V=at since u is 0
V =10•3 =30m/s.
<em>i</em><em> </em><em>h</em><em>o</em><em>p</em><em>e</em><em> </em><em>t</em><em>h</em><em>i</em><em>s</em><em> </em><em>h</em><em>e</em><em>l</em><em>p</em><em>s</em><em> </em><em>,</em><em> </em><em>y</em><em>o</em><em>u</em><em> </em><em>a</em><em>r</em><em>e</em><em> </em><em>f</em><em>r</em><em>e</em><em>e</em><em> </em><em>t</em><em>o</em><em> </em><em>a</em><em>s</em><em>k</em><em> </em><em>a</em><em>n</em><em>y</em><em> </em><em>f</em><em>u</em><em>t</em><em>h</em><em>u</em><em>r</em><em>e</em><em> </em><em>q</em><em>u</em><em>e</em><em>s</em><em>t</em><em>i</em><em>o</em><em>n</em><em>s</em><em>.</em>
The space station completes 2 revolutions each minute, so that it traverses a distance of 2<em>π</em> (100 m) = 200<em>π</em> m each minute, giving it a linear/tangential speed of
<em>v</em> = (200<em>π</em> m) / (60 s) ≈ 10.472 m/s
(a) The astronaut would experience an acceleration of
<em>a</em> = <em>v</em> ² / (100 m) ≈ 1.09662 m/s² ≈ 0.1119<em>g</em> ≈ 0.11<em>g</em>
<em />
(b) Now you want to find the period <em>T</em> such that <em>a</em> = <em>g</em>. This would mean the astronaut has a tangential speed of
<em>v</em> = (200<em>π</em> m) / <em>T</em>
so that her centripetal/radial acceleration would match <em>g</em> :
<em>a</em> = <em>g</em> = ((200<em>π</em> m) / <em>T </em>)² / (100 m)
Solve for <em>T</em> :
(100 m) <em>g</em> = (400<em>π</em> ² m²) / <em>T</em> ²
<em>T</em> ² = (400<em>π</em> ² m²) / ((100 m) <em>g</em>) = (4<em>π</em> ² m)/<em>g</em>
<em>T</em> = √((4<em>π</em> ² m) / (9.8 m/s²)) ≈ 2<em>π</em> √(0.102 s²) ≈ 2.007 s ≈ 2.0 s