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EleoNora [17]
3 years ago
5

A rock is thrown upward with a velocity of 23 meters per second from the top of a 25 meter high cliff, and it misses the cliff o

n the way back down. when will the rock be 11 meters from the water, below
Physics
1 answer:
Nana76 [90]3 years ago
8 0
Vf=vi+at
-23m/s=23m/s-9.8m/s^2(t)
t1 from throwing up back to starting position t1=4.693s
Delta(x)=vi(t)+.5a(t^2)
25-11=14
-14m=-23m/s(t)-4.8m/s^s(t^2)
t2 is from 25m to 11 m t2=.546s
Total time =5.239s
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Both atypical and traditional antipsychotics _____ levels of _____.
LiRa [457]

Answer:

Both atypical and traditional antipsychotics <u>blocks</u> levels of<u> </u><u>serotonin</u><u>.</u>

Explanation:

These drugs are capable of reducing or turning off positive psychotic symptoms such as hallucinations, delusions, and language and behavior disorder.

In addition to also acting as mood stabilizers, they have an impact on mania, depression, and produce low anxiety.

These anti-psychotics work on the 5-HT2A receptor, it belongs to the family of serotonin receptors and it is a G protein-coupled receptor. 5-HT is short for 5-hydroxy-tryptamine, the chemical name for serotonin.

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5 0
3 years ago
4
tensa zangetsu [6.8K]

Answer:

C. amount of charge on the source charge.

Explanation:

Electric field lines can be defined as a graphical representation of the vector field or electric field.

Basically, it was first introduced by Michael Faraday and it is typically a curve drawn to the tangent of a point is in the direction of the net field acting on each point.

The number, or density, of field lines on a source charge indicate the amount of charge on the source charge. Therefore, the density of field lines on a source charge is directly proportional to quantity of charge on the source.

8 0
2 years ago
An electron is moving east in a uniform electric field of 1.55 N/C directed to the west. At point A, the velocity of the electro
valkas [14]

Answer:

Final velocity of electron, v=6.45\times 10^5\ m/s    

Explanation:

It is given that,

Electric field, E = 1.55 N/C

Initial velocity at point A, u=4.52\times 10^5\ m/s

We need to find the speed of the electron when it reaches point B which is a distance of 0.395 m east of point A. It can be calculated using third equation of motion as :

v^2=u^2+2as........(1)

a is the acceleration, a=\dfrac{F}{m}

We know that electric force, F = qE

a=\dfrac{qE}{m}

Use above equation in equation (1) as:

v^2=u^2+\dfrac{2qEs}{m}

v^2=(4.52\times 10^5\ m/s)^2+2\times \dfrac{1.6\times 10^{-19}\ C\times 1.55\ N/C}{9.1\times 10^{-31}\ kg}\times 0.395\ m

v = 647302.09 m/s

or

v=6.45\times 10^5\ m/s

So, the final velocity of the electron when it reaches point B is 6.45\times 10^5\ m/s. Hence, this is the required solution.

3 0
3 years ago
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