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Daniel [21]
3 years ago
11

The corpus callosum is comprised of more than 200 million axons that connect the hemispheres of the brain. Please select the bes

t answer from the choices provided T F

Physics
2 answers:
mamaluj [8]3 years ago
8 0

Hello, love! The answer is True, or T, on Edge2020.

Hope this helped!

~ V.

Pani-rosa [81]3 years ago
6 0

Answer:

<h3>The correct answer is <u>True</u></h3>
You might be interested in
If you dribble a basketball with a frequency of 1.77 Hz, how long does it take for you to complete 12 dribbles?
Licemer1 [7]
<h2>It takes 6.78 seconds to complete 12 dribbles.</h2>

Explanation:

Frequency of dribble = 1.77 Hz

That is

         Number of dribbles in 1 second = 1.77

         \texttt{Time taken for 1 dribble = }\frac{1}{1.77}=0.565s

Now we need to find how long does it take for you to complete 12 dribbles.

         Time taken for 12 dribbles = 12 x Time taken for 1 dribble

         Time taken for 12 dribbles = 12 x 0.565

         Time taken for 12 dribbles = 6.78 seconds      

It takes 6.78 seconds to complete 12 dribbles.  

8 0
3 years ago
What primary element of light helps the observer see three-dimensions?
elixir [45]
<span>Shading. When light hits an opaque surface some is absorbed, the rest is reflected, The reflected light is called shading. Reflection is not simple and varies with material. The surface’s structure defines the details of reflection. Variations produce anything from bright specular reflection</span>
7 0
3 years ago
Read 2 more answers
In a playground, there is a small merry-go-round of radius 1.20 m and mass 160 kg. Its radius of gyration is 91.0 cm. (Radius of
aksik [14]

Answer:

a) 145.6kgm^2

b) 158.4kg-m^2/s

c) 0.76rads/s

Explanation:

Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation 

(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and

(c) the angular speed of the merry-go-round and child after the child has jumped on.

a) From I = MK^2

I = (160Kg)(0.91m)^2

I = 145.6kgm^2

b) The magnitude of the angular momentum is given by:

L= r × p The raduis and momentum are perpendicular.

L = r × mc

L = (1.20m)(44.0kg)(3.0m/s)

L = 158.4kg-m^2/s

c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:

L = Iw

158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]

w = 158.6/208.96

w = 0.76rad/s

7 0
3 years ago
A weight of 30.0 N is suspended from a spring that has a force constant of 220 N/m. The system is undamped and is subjected to a
Nimfa-mama [501]

Answer:

F_0 = 393 N

Explanation:

As we know that amplitude of forced oscillation is given as

A = \frac{F_0}{ m(\omega^2 - \omega_0^2)}

here we know that natural frequency of the oscillation is given as

\omega_0 = \sqrt{\frac{k}{m}}

here mass of the object is given as

m = \frac{W}{g}

\omega_0 = \sqrt{\frac{220}{\frac{30}{9.81}}}

\omega_0 = 8.48 rad/s

angular frequency of applied force is given as

\omega = 2\pi f

\omega = 2\pi(10.5) = 65.97 rad/s

now we have

0.03 = \frac{F_0}{3.06(65.97^2 - 8.48^2)}

F_0 = 393 N

6 0
3 years ago
A 3.00 kg block moving 2.09 m/s
Talja [164]

Answer:11.64kgm/s

Explanation:

4 0
3 years ago
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