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BigorU [14]
3 years ago
12

What are some of the names given to the Moon by the Algonquin?

Physics
1 answer:
frutty [35]3 years ago
6 0
The names given to the Moon by the Algonquin are as follows:

January- Full Wolf Moon
February- Full Snow Moon
March- Full Worm Moon
April- Full Pink Moon
May- Full Flower Moon
June- Full Strawberry Moon
July- Full Buck Moon
August- Full Sturgeon Moon
September- Full Corn Moon; Full Harvest Moon
October- Full Hunter's Moon
November- Full Beaver Moon
December- Full Cold Moon; Full Long Nights Moon
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Express the surface area of a cube as a function of its volume v.
vredina [299]
The surface area of a cube is:

S = 6l², making l the subject:
l = √(S/6)

While the volume is:

V = l³

If we substitute l to get:

V = (√(S/6))³

√(S/6) = ³√V

S/6 = (³√V)²

S = 6(³√V)² ⇒ 6V^(2/3)
5 0
3 years ago
One consequence of Newton's third law of motion is that __________. A. Every object that has mass has inertia B. A force acting
vovangra [49]

Answer:

B

Explanation:

a force acting upon an object increases that objects acceleration

8 0
3 years ago
If the diameter of a radar dish is doubled, what happens to its resolving power assuming that all other factors remain unchanged
kirill115 [55]

Answer:

      θ’ = θ₀ / 2

we see that the resolution angle is reduced by half

Explanation:

The resolving power of a radar is given by diffraction, for which we will use the Rayleigh criterion for the resolution of two point sources, they are considered resolved if the maximum of diffraction of one coincides with the first minimum of the other.

The first minimum occurs for m = 1, so the diffraction equation of a slit remains

        a sin θ = λ

in general, the diffraction patterns occur at very small angles, so

        sin θ = θ

          θ = λ / a

in the case of radar we have a circular aperture and the equation must be solved in polar coordinates, which introduces a numerical constant.

        θ = 1.22 λ /a

In this exercise we are told that the opening changes

         a’ = 2 a

we substitute

          θ ‘= 1.22  λ / 2a

          θ' = (1.22 λ / a) 1/2

          θ’ = θ₀ / 2

we see that the resolution angle is reduced by half

8 0
3 years ago
Give at least two reasons today’s astronomers are so interested in the discovery of additional Earth-approaching asteroids.
blondinia [14]

Answer:

1. Potential hazard

2. Mining opportunity

Explanation:

The two reason, why astronomers are so interested in the discovery of additional Earth-approaching asteroids:

1. Potential hazard: We have proof that the dinosaurs got extinct because of an asteroid/comet strike on Earth. Also we have seen the effects of the Tunguska event and Chelyabinsk tragedy. These are enough to show us that asteroids can be very dangerous and wipe out the life from Earth.

2. Mining Opportunity: We have discovered a lot of asteroids which contains a lot of metal and precious elements. There can be a possibility of mining such asteroids in the future and reducing the burden on Earth.

4 0
3 years ago
Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant
irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

a_g = \frac{GM}{R^2}

Here

M = \text{Mass inside the Orbit of the star}

R = \text{Orbital radius}

G = \text{Universal Gravitational Constant}

Mass inside the orbit in terms of Volume and Density is

M =V \rho

Where,

V = Volume

\rho =Density

Now considering the volume of the star as a Sphere we have

V = \frac{4}{3} \pi R^3

Replacing at the previous equation we have,

M = (\frac{4}{3}\pi R^3)\rho

Now replacing the mass at the gravitational acceleration formula we have that

a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho

a_g = \frac{4}{3} G\pi R\rho

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration.  So centripetal acceleration of the star is

a_c = \frac{4}{3} G\pi R\rho

At the same time the general expression for the centripetal acceleration is

a_c = \frac{\Theta^2}{R}

Where \Theta is the orbital velocity

Using this expression in the left hand side of the equation we have that

\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2

\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}

\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R

Considering the constant values we have that

\Theta = \text{Constant} \times R

\Theta \propto R

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

6 0
3 years ago
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