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Artist 52 [7]
2 years ago
9

Many of the male characteristics come from which reproductive hormone?

Physics
1 answer:
motikmotik2 years ago
8 0

Answer:

The answer is C. Testosterone.

Explanation:

Thats what make us a male.

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Students in physics lab times 10 oscillations of the pendulum and get 11.50 seconds how long is a pendulum in centimeters?
allochka39001 [22]

Answer: 36.5

Explanation:

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Luis wanted to know which of his 4 toy cars is the fastest. He conducted an experime and recorded his results in the table shown
Murrr4er [49]
The dependent variable was the time and the independent variable was thecars
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Will give brainliest to right answer!
viva [34]
The correct answer is D.
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Ian’s school is exactly 6 blocks north of his house. What is his average velocity on a walk from home to school that takes him 1
choli [55]

Answer:

.5 units north

<em><u>or</u></em>

55 ft/minutes(<em>squared</em>) north

Explanation:

.5 is what your info gives me but if i take that the average distance a block is it is 660ft than the answer is 55.

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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