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Drupady [299]
4 years ago
14

Describe how the mass, luminosity, surface temperature, and radius of main-sequence stars change in value going from the “bottom

” to the “top” of the main sequence.
Physics
1 answer:
Sladkaya [172]4 years ago
4 0

Answer:

1. Least massive stars are the coolest and least luminous, lower right of main sequence, on HR diagram.  

2. Most massive are the hottest and most luminous, upper left of main sequence on Hr Diagram.  

3. The radius of stars are related to their sprectral type. having the O being the hottest upper left and M being the coolest bottom right.

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A block is resting on a platform that is rotating at an angular speed of 2.4 rad/s. The coefficient of static friction between t
Sloan [31]

Answer:

r = 0m is the Minimum distance from the axis at which the block can remain in place wothout skidding.

Explanation:

From a sum of forces:

Ff = m*a   where Ff = μ * N    and a = \frac{V^2}{r}=\omega^2*r

N - m*g = 0   So, N = m*g.   Replacing everything on the original equation:

\mu*m*g = m*\omega^2*r   (eq2)

Solving for r:

r = \frac{\mu*g}{\omega^2}=1.41m

If we analyze eq2 you can conclude that as r grows, the friction has to grow (assuming that ω is constant), so the smallest distance would be 0 and the greatest 1.41m. Beyond that distance, μ has to be greater than 0.83.

4 0
3 years ago
Which event is an example of vaporization?
oksano4ka [1.4K]
B would be an example of vaporization (liquid to gas).

———————

A is an example of deposition (gas to solid); C is an example of condensation (gas to liquid); and D is an example of condensation, deposition, or freezing—depending on the type of cloud.
6 0
3 years ago
An artillery shell is fired with an initial velocity of 300 m/s at 52.0° above the horizontal. To clear an avalanche, it explode
sasho [114]

The x- and y-coordinates are 9142.57 m and -304.425 m

<u>Explanation:</u>

As the motion of the shell is in a plane (two dimensional space) and the acceleration is that due to gravity which is vertically downward, we resolve initial velocity of the shell v_{0} in horizontal and vertical directions. If the initial velocity of the shell is making angle with the horizontal, the horizontal component of initial velocity will be

                v_{x}=v_{0} \times \cos \theta

As the acceleration of the shell is vertical having no horizontal component, the shell may be considered to move horizontally with constant velocity of v_{x} and hence the horizontal distance covered (or the x coordinate of the shell with point of projection as origin) is given by

           v_{x}=v_{o} \times \cos \theta=300 \times \cos \left(52^{\circ}\right)=184.69 \mathrm{m} / \mathrm{s}

           v_{y}=v_{o} \times \sin \theta==300 \times \sin \left(52^{\circ}\right)=236.4 \mathrm{m} / \mathrm{s}

For motion with constant acceleration, we know

            s=s_{0}+v_{0} t+\left(\frac{(1)}{2}\right) a t^{2}

Along the horizontal, x-axis, we might write this as

            x=x_{0}+v_{x 0} t+\left(\frac{1}{2}\right) a_{x} t^{2}

Measuring distances relative to the firing point means

               x_{0}=0

we know that,

              a_{x}=0

or,

             v_{x}=v_{x 0}=\text { constant }

By applying the values, we get,

           x=0+(184.69 \times 49.5)+\left(\left(\frac{1}{2}\right) \times 0 \times(49.5)^{2}\right)=9142.57 \mathrm{m}

The acceleration of gravity is vertically downward and is g=-9.8 \mathrm{m} / \mathrm{s}^{2} , hence the vertical distance covered (or y coordinate of the shell) is given by the second equation of motion

           y=y_{0}+v_{y 0} t+\left(\frac{1}{2}\right) a_{y} t^{2}

we know, y_{0}=0 and a_{y}=-9.8 \mathrm{m} / \mathrm{s}^{2}, so,

          y=0+(236.4 \times 49.5)+\left(\left(\frac{1}{2}\right) \times(-9.8) \times(49.5)^{2}\right)

                 y = 11701.8 - 4.9(2450.25)= 11701.8 - 12006.225 = - 304.425 m

7 0
4 years ago
For what angle of incidence at the first mirror will this ray strike the midpoint of the second mirror (which is 28.0 cmcm long)
loris [4]

This question is incomplete, the complete question is;

Two plane mirrors intersect at right angles. A laser beam strikes the first of them at a point 11.5 cm from their point of intersection, as shown in the figure.

For what angle of incidence at the first mirror will this ray strike the midpoint of the second mirror (which is 28.0 cmcm long) after reflecting from the first mirror

Answer: angle of incidence is 39.4°

Explanation:

Given that;

two plain mirrors intersect at right angle (90°)

distance d = 11.5 cm

S = 28.0 cm

Now the angle that the reflection ray males with first the mirror equal theta  (∅)

so

tan∅ = (S/2) / d

tan∅ = (28/2) / 11.5

tan∅ = 14 / 11.5

tan∅ = 1.2173

∅ = tan⁻¹ (1.2173)

∅ = 50.6°

so angle of incidence = 90° - ∅

= 90° - 50.6°

= 39.4°

Therefore angle of incidence is 39.4°

5 0
3 years ago
How do scientists determine the number of neutrons in an isotope of an atom?
ikadub [295]
D.  The atomic mass in amu is basically the number or nuclei since the mass of the electrons is negligible.  For a given atom (element) the number of protons is fixed.  Say the element has 10 protons.  If the atomic weight is 14 atomic mass units (amu), you know that there are 4 neutrons, since both neutrons and protons are 1 amu each and there are 10 protons
4 0
4 years ago
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