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german
3 years ago
10

Where would a new neuron come from

Physics
1 answer:
erica [24]3 years ago
7 0
Neurogenesis does not occur everywhere in the brain but is evident in the hippocampus and olfactory bulb and perhaps in the cerebral cortex. New neurons are born not from mature nerve cells but rather develop from neural stem cells that remain in our brains throughout life.
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You are driving your car along a country road at a speed of 27.0 m/s. as you come over the crest of a hill, you notice a farm tr
Bezzdna [24]

speed of the car = 27 m/s

speed of truck ahead = 10 m/s

relative speed of car with respect to truck

v_r = 27 - 10 = 17 m/s

relative deceleration of car

a_r = -7 m/s^2

now the distance before they stop with respect to each other is given by

v_f^2 - v_i^2 = 2 a d

0 - 17^2 = 2 *(-7)*d

d = 20.6 m

so it will come at the same speed of truck after 20.6 m distance and hence it will not hit the truck as the distance of the truck is 25 m from car

Part b)

Distance traveled by car before it stops is given by

v_f^2 - v_i^2 = 2 a s

0^2  - 27^2 = 2 * (-7)* s

s = 52.1 m

so it will stop after it will cover total 52.1 m distance

Part c)

time taken by the car to stop

v_f - v_i = at

0 - 27 = (-7) * t

t = 3.86 s

now the distance covered by truck in same time

d = 3.86 * 10 = 38.6 m

now after the car will stop its distance from the truck is

D = 25 + 38.6 - 52.1 = 11.5 m

<em>so the distance between them is 11.5 m</em>

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3 years ago
In which direction should you look to see the setting moon.
Virty [35]
Every sky object sets in the west.
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3 years ago
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What causes a tsunami
tiny-mole [99]
A tsunami is caused by earthquakes in the ocean. the earthquakes make a big rumble and the big wave travels and may go on the shore and its a called a tsunami hope this helpsBrainliest??
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3 years ago
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What are the four layers of the earth
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Answer:

earth's crust

meatle

outercore

innercore

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3 years ago
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A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma
raketka [301]

  • Let, the maximum height covered by projectile be \sf{H_m}

\purple{ \longrightarrow  \bf{h_m =  \dfrac{ {v}^{2} \: {sin}^{2} \theta  }{2g} }}

  • Projectile is thrown with a velocity = v
  • Angle of projection = θ

  • Velocity of projectile at a height half of the maximum height covered be \sf{v_0}

\qquad______________________________

Then –

\qquad \pink{  \longrightarrow \bf{ \dfrac{h_m}{2}  = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }}

\qquad \longrightarrow \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{2g} \times  \dfrac{1}{2}  =  \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{4g}  =  \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{2}  =   {v_0}^{2} \: {sin}^{2} \theta }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2} }{2}  =   {v_0}^{2} }

\qquad\longrightarrow \bf{v_0 =   \sqrt{ \dfrac{ {v}^{2} }{2} } =  \dfrac{v}{ \sqrt{2} }  }

  • Now, the vertical component of velocity of projectile at the height half of \sf{h_m} will be –

\qquad \longrightarrow   \bf{v_{(y)}=v_0 \: sin \theta }

\qquad \longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} }  \: sin \theta =  \dfrac{v \: sin \: \theta}{ \sqrt{2} }  }

Therefore, the vertical component of velocity of projectile at this height will be–

☀️\qquad\pink {\bf{ \dfrac{v \: sin \:  \theta}{ \sqrt{2} }} }

6 0
2 years ago
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