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3 years ago

What part of the brain has the most neurons?

Physics

2 answers:

vredina [299]3 years ago

8 0

It hink the answer is nerve cell

mash [69]3 years ago

4 0

Well, the part of the brain with the most neurons is the cerebral cortex it has about 15 to 33 neurons. Hope i helped.

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How to atoms behave in non-magnetic items?

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By altering the quantum interactions of the electrons in the atoms of a metal's atoms, scientists from the University of Leeds have generated magnetism in metals that aren’t normally magnetic.

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2 years ago

Examples of energy balance: physical education

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Answer:children burn calories to being a student

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2 years ago

Knowing that the board is 4 meters long, one child weighs 300 n and the other 250 n, where should one of the children sit so tha

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2 years ago

The wave property which is independent of all other properties is

Anna71 [15]

The wave property which is independent of all other properties is THE VELOCITY OF A WAVE.
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2 years ago

The maximum tension the wire can withstand without breaking is 300 N . A 0.800 kg projectile traveling horizontally hits and emb

topjm [15]

Answer:

Let's investigate the case where the cable breaks.

Conservation of angular momentum can be used to find the speed.

$\vec{L}_1 = \vec{L}_2\\\vec{L}_1 = m\vec{v_0} \\\vec{L}_2 = I\vec{\omega}\\$

The projectile embeds itself to the ball, so they can be treated as a combined object. <u>The moment of inertia of the combined object is equal to the sum of the moment of inertia of both objects. </u>

$I = I_{projectile} + I_{ball}\\I = mr^2 + mr^2\\I = 2mr^2$

where r is the length of the cable.

<u>After the collision, the ball and the projectile makes a circular motion because of the cable.</u> So, the force (tension) in circular motion is

$F = \frac{mv^2}{r}$

The relation between linear velocity and the angular velocity is

$v = \omega r$

So,

$F = \frac{m(\omega r)^2}{r} = m\omega^2 r = 300\\mv_0r =I\omega\\\\\omega = mv_0r/I\\300 = m(\frac{mv_0r}{I})^2r = m(\frac{mv_0r}{2mr^2})^2r = m(\frac{v_0}{2r})^2r = \frac{mv_0^2r}{4r^2} = \frac{mv_0^2}{4r}\\300 = \frac{0.8v_0^2}{4r}\\1500 = v_0^2/r\\v_0 = \sqrt{1500r}$

As can be seen, the maximum velocity for the projectile without breaking the cable is $\sqrt{1500r}$ , where r is the length of the cable.

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3 years ago