Ask question

Ask question

All categories

3 years ago

11

The type of synapse that occurs between the terminal end of the presynaptic cell and the perikaryon of the postsynaptic cell is

referred to as which kind of synapse?

1 answer:

ASHA 777 [7]3 years ago

5 0

Answer:

axosomatic synapse

Explanation:

Axosomatic synapse -

It refer to the junction between the nerves cells , is referred to as axosomatic synapse .

The junction between the axon of one neuron and the body of the other neuron is where the axosomatic synapse resides .

Via this junction the message in the form of certain chemicals or impulse gets transmitted , from one point of the by to other .

Hence , from the given information of the question ,

The correct term is axosomatic synapse .

You might be interested in

Which property describes the temperature at which a liquid changes to a gas?

Rufina [12.5K]

Answer:

Boiling Point

Explanation:

When a liquid changes to a gas is called the boiling point.

5 0

3 years ago

From the edge of a cliff, a 0.41 kg projectile is launched with an initial kinetic energy of 1430 J. The projectile's maximum up

NemiM [27]

Answer:

v₀ₓ = 63.5 m/s

v₀y = 54.2 m/s

Explanation:

First we find the net launch velocity of projectile. For that purpose, we use the formula of kinetic energy:

K.E = (0.5)(mv₀²)

where,

K.E = initial kinetic energy of projectile = 1430 J

m = mass of projectile = 0.41 kg

v₀ = launch velocity of projectile = ?

Therefore,

1430 J = (0.5)(0.41)v₀²

v₀ = √(6975.6 m²/s²)

v₀ = 83.5 m/s

Now, we find the launching angle, by using formula for maximum height of projectile:

h = v₀² Sin²θ/2g

where,

h = height of projectile = 150 m

g = 9.8 m/s²

θ = launch angle

Therefore,

150 m = (83.5 m/s)²Sin²θ/(2)(9.8 m/s²)

Sin θ = √(0.4216)

θ = Sin⁻¹ (0.6493)

θ = 40.5°

Now, we find the components of launch velocity:

x- component = v₀ₓ = v₀Cosθ = (83.5 m/s) Cos(40.5°)

<u>v₀ₓ = 63.5 m/s</u>

y- component = v₀y = v₀Sinθ = (83.5 m/s) Sin(40.5°)

<u>v₀y = 54.2 m/s</u>

7 0

3 years ago

John added 150 grams of salt to 2.5 liters of water. what is the concentration of the salt solutions

torisob [31]

Shdnejeiebx ski wiw do

4 0

3 years ago

The speed of sound in air is 345 m/s. A tuning fork vibrates above the open end of a sound resonance tube. If sound waves have w

Delvig [45]

To develop this problem it is necessary to apply the concept of Frequency based on speed and wavelength.

According to the definition the frequency can be expressed as

$f = \frac{v}{\lambda}$

Where,

v = Velocity

$\lambda =$ Wavelength

Our value are given by,

v = 345m/s

$\lambda = 63cm$

Replacing

$f = \frac{345}{0.63}$

$f = 547.61Hz$

Therefore the frequency of the tuning fork is 547.61Hz

6 0

3 years ago

A car travels 40 miles in 30 minutes.

lukranit [14]

Answer:

(a)Average velocity ,v =128.74 Km/hr

(b)Kinetic Energy , K=958546.875 Joule

(c)Distance, s=268.8m

(d)Acceleration, a= - 2.38 $m/s^2$

<u>Explanation</u>:

<u>Given</u>:

Distance travelled = 40 miles

Time taken = 30 minutes.

(A) The average velocity in kilometres/hour

Converting 40 miles into km ,

we know that,

1 mile = 1.60934

40 miles = 40 x 1.60934

so 40 miles = 64.3738 Km

similarly converting 30 minutes into hours

1 minute = $\frac{1}{60}hours$

30 minute = $\frac{30}{60}hours$

30 minute = $\frac{1}{2}hours$

Now

Average velocity = $\frac{Speed}{time}$

Substituting Values,

Average velocity = $\frac{64.3738}{\frac[1}{2}}$

Average velocity = $64.3738 \times 2$

Average velocity =128.74 Km/hr

(B) If the car weighs 1.5 tons, what is its If the car weighs 1.5 tons, what is its kinetic energy in joules (Note: you will need to convert your velocity to m/s)? in joules (Note: you will need to convert your velocity to m/s)?

Converting 1.5 tons into kg we get

1 ton = 1000 kg

so 1.5 ton =1500 kg

converting velocity to m/s

$128.74 \times \frac{5}{18}$

=>35.75 m/s----------------------------------------------------------(1)

kinetic energy K= $\frac{1}{2}mv^2$

Substituting the values,

K= $\frac{1}{2}1500(35.75)^2$

K= $\frac{1}{2}1500(1278.06)$

K= $\frac{1500 \times (1278.06)}{2}$

K= $\frac{1917093.75}{2}$

K=958546.875 Joule---------------------------------------------(2)

(c)When the driver applies the brake, it takes 15 seconds to stop. How far does the car travel (in meters) while stopping

Lets use Distance formula,

$S= ut+\frac{1}{2}at^2$

Substituting the known values,

$s= ut+\frac{1}{2}at^2$

$s= (37.75)(15)+\frac{1}{2}a(15)^2$

$s=566.25+\frac{1}{2}a(225)$

$s=566.25+\frac{(225a)}{2}$ -------------------------------------(3)

(D) What is the average acceleration of the car (in m/s2) during braking?

Using the formula

v=u +at

re arranging the formula we get,

$a = \frac{v - u}{t}$

Substituting the values

$a = \frac{0 - 35.75}{15}$

$a = \frac{- 35.75}{15}$

a= - 2.38 $m/s^2$ ----------------------------------------(4)

Now substituting 4 in 3 we get

$s=566.25+\frac{(225( - 2.38)}{2}$

$s=566.25+\frac{-535.5}{2}$

$s=536.25-267.75$

s=268.8m--------------------------------------------------------------(5)

4 0

3 years ago