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3 years ago

7

When a star goes from the main sequence to the red giant phase, it becomes larger rather than smaller. This is true because grav

ity is getting temporarily overpowered by
A) Antigravity
B) Electron Degeneracy
C) Energy of Fusion
D) Energy of Fission

1 answer:

GuDViN [60]3 years ago

4 0

I believe this is electron degeneracy, because the star is essentially having too many reactions too fast and collapses in on itself eventually.

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Unpolarizedlight of intensity I_0 is incident on three polarizingfilters. The axis of the first is vertical, that of the secondi

Marina86 [1]

To solve this problem it is necessary to apply the concepts related to the law of Malus which describe the intensity of light passing through a polarizer. Mathematically this law can be described as:

$I = I_0 cos^2\theta$

Where,

$I_0 =$ Indicates the intensity of the light before passing through the polarizer

I = Resulting intensity

$\theta$ = Indicates the angle between the axis of the analyzer and the polarization axis of the incident light

From the law of Malus when the light passes at a vertical angle through the first polarizer its intensity is reduced by half therefore

$I_1= \frac{I_0}{2}$

In the case of the second polarizer the angle is directly 60 degrees therefore

$I_2 = I_1 cos^2\theta$

$I_2 = (\frac{I_0}{2} ) cos^2(60)$

$I_2 = 0.125I_0$

In the case of the third polarizer, the angle is reflected on the perpendicular, therefore, its angle of index would be

$\theta_3 = 90-60 = 30$

Then,

$I_3 = I_2 cos^2\theta_3$

$I_3 = 0.125I_0 cos^2 (30)$

$I_3 = 0.09375I_0$

Then the intensity at the end of the polarized lenses will be equivalent to 0.09375 of the initial intensity.

5 0

2 years ago

A helicopter is hovering above the ground. Jim reaches out of the copter (with a safety harness on) at 180 m above the ground. A

vesna_86 [32]

Answer:

The maximum height of the package is 140 m above the ground. Jim Bond will not catch the package.

Explanation:

Hi there!

The equation of height and velocity of the package are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the package at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² because we consider the upward direction as positive).

v = velocity of the package at a time t.

First, let´s find the time it takes the package to reach the maximum height. For this, we will use the equation of velocity because we know that at the maximum height, the velocity of the package is zero. So, we have to find the time at which v = 0:

v = v0 + g · t

0 = 50.5 m/s - 9.8 m/s² · t

Solving for t:

-50.5 m/s / -9.81 m/s² = t

t = 5.15 s

Now, let´s find the height that the package reaches in that time using the equation of height. Let´s place the origin of the frame of reference on the ground so that the initial position of the package is 10 m above the ground:

h = h0 + v0 · t + 1/2 · g · t²

h = 10 m + 50.5 m/s · 5.15 s - 1/2 · 9.81 m/s² · (5.15 s)²

h = 140 m

The maximum height of the package is 140 m above the ground. Jim Bond will not catch the package.

6 0

3 years ago

What is meant by the term law of conservation?

GalinKa [24]

Answer:

Explanation:

any law stating that some quantity or property remains constant during and after an interaction or process, as conservation of charge or conservation of linear momentum.

4 0

2 years ago

Read 2 more answers

What idea did James Chadwick add to the atomic model in the 1930s? (1 point) Positive charge is clumped together in the nucleus.

Lelu [443]

In the 1930s James Chadwick added the neutrons to the atomic model.

8 0

2 years ago

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On a frictionless horizontal air table, puck A (with mass 0.252 kg ) is moving toward puck B (with mass 0.374 kg ), which is ini

geniusboy [140]

As we know that kinetic energy is given as

$KE = \frac{1}{2}mv^2$

Here we can find the initial speed of puck A by momentum conservation

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

$0.252\times v + 0 = 0.252\times (-0.115) + 0.374\times ( 0.650)$

$v = 0.850 m/s$

now here we will have initial kinetic energy of the mass is given as

$KE_i = \frac{1}{2}m_1v_1^2$

$KE_i = \frac{1}{2}(0.252)(0.850^2) = 0.091 J$

$KE_f = \frac{1}{2}(0.252)(-0.115)^2 + \frac{1}{2}(0.374)(0.650^2)$

$KE_f = 0.081 J$

now loss of energy is given as

$KE_i - KE_f = 0.091 - 0.081 = 0.010 J$

6 0

3 years ago