What must be present at the postsynaptic site within a chemical synapse for the signal to be transmitted?

Two children, Ferdinand and Isabella, are playing with a waterhose on a sunny summer day. Isabella is holding the hose in herhan

sweet [91]

Answer:

84.196%

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

$y_0=1\ m$

x = 10 m

t = Time taken

$v_0$ = 3.5 m/s (assumed, as it is not given)

$A_0$ = $\pi 1.5^2$

We have the equation

$y=y_0+ut+\dfrac{1}{2}gt^2\\\Rightarrow 0=y_0-\dfrac{1}{2}gt^2\\\Rightarrow t=\sqrt{\dfrac{2y_0}{g}}\\\Rightarrow t=\sqrt{\dfrac{2\times 1}{9.81}}\\\Rightarrow t=0.45152\ s$

$x=x_0+vt\\\Rightarrow 10=0+v0.45152\\\Rightarrow v=\dfrac{10}{0.45152}\\\Rightarrow v=22.14741\ m/s$

From continuity equation we have

$Av=A_0v_0\\\Rightarrow A=\dfrac{A_0v_0}{v}\\\Rightarrow A=\dfrac{\pi 1.5^2\times 3.5}{22.14741}\\\Rightarrow A=1.11706\ cm^2$

Fraction is given by

$f=\dfrac{A_0-A}{A_0}\times 100\\\Rightarrow f=\dfrac{\pi 1.5^2-1.11706}{\pi 1.5^2}\times 100\\\Rightarrow f=84.196\ \%$

The fraction is 84.196%

4 0

3 years ago

A 1.20 kg solid ball of radius 40 cm rolls down a 5.20 m long incline of 25 degrees. Ignoring any loss due to friction, how fast

sladkih [1.3K]

Answer:

1.6s

Explanation:

Given that A 1.20 kg solid ball of radius 40 cm rolls down a 5.20 m long incline of 25 degrees. Ignoring any loss due to friction,

To know how fast the ball will roll when it reaches the bottom of the incline, we need to calculate the acceleration at which it is rolling.

Since the frictional force is negligible, at the top of the incline plane, the potential energy = mgh

Where h = 5.2sin25

h = 2.2 m

P.E = 1.2 × 9.8 × 2.2

P.E = 25.84 j

At the bottom, K.E = P.E

1/2mv^2 = 25.84

Substitutes mass into the formula

1.2 × V^2 = 51.69

V^2 = 51.69/1.2

V^2 = 43.07

V = 6.56 m/s

Using the third equation of motion

V^2 = U^2 + 2as

Since the object started from rest,

U = 0

6.56^2 = 2 × a × 5.2

43.07 = 10.4a

a = 43.07/10.4

a = 4.14 m/s^2

Using the first equation of motion,

V = U + at

Where U = 0

6.56 = 4.14t

t = 6.56/4.14

t = 1.58s

Therefore, the time the ball rolls when it reaches the bottom of the incline is approximately 1.6s

6 0

3 years ago

What does acceleration always refers as

timama [110]

Answer:

Acceleration always refers to a(n) Change in velocity. A change in direction without a change in speed is called velocity.

Explanation:

6 0

4 years ago

Read 2 more answers

There is an old physics joke involving cows, and you will need to use its punchline to solve this problem

aleksandr82 [10.1K]

Answer:

The angle of deflection will be "1.07 × 10⁻⁷°".

Explanation:

The given values are:

Mass of a cow,

m = 1100 kg

Mass of bob,

mb = 1 kg

The total distance between a cow and bob will be,

d = 2 m

Let,

The tension be "t".

The angle with the verticles be " $\theta$ ".

Now,

Vertically equating forces

⇒ $T\times Cos \theta =mb\times g$ ...(equation 1)

Horizontally equating forces

⇒ $T\times Sin \theta = G\times M\times \frac{mb}{d^2}$ ...(equation 2)

From equation 1 and equation 2, we get

⇒ $tan \thata=\frac{G\times M}{g\times d^2}$

O putting the estimated values, will be

⇒ $\theta = tan(\frac{6.674\times 10^{-11}\times 1100}{9.8\times 2^2} )$

⇒ $\theta = 1.07\times 10^{-7}^{\circ}$

6 0

4 years ago

The ratio of the distance a bullet travels to the amount of time it takes to travel is known as the __________ of the bullet, de

valentinak56 [21]

The ratio of the distance a bullet travels to the amount of time it takes to travel is known as the speed or velocity of the bullet, depending on whether we need to know the direction or not. Speed is a scalar quantity. It describes to "how fast an object is moving", in our case how fast the bullet is moving.Velocity is a vector quantity. velocity describes "the rate at which an object changes its position".

5 0

3 years ago

Read 2 more answers