An action potential arriving at the presynaptic terminal causes what to occur?

A wheel of diameter 40.0 cm starts from rest and rotates with a constant angular acceleration of 3.00 rad/s^2. At the instant th

xxMikexx [17]

Answer:

ac= 15.07 m/s²

Explanation:

The Wheel rotates with a constant angular acceleration.:

Centripetal acceleration is calculated as follows:

ac =ω² *R Formula (1)

ac =v² / R Formula (2)

Kinematics of the wheel

We apply the equations of circular motion uniformly accelerated :

ωf²= ω₀² + 2αθ Formula (3)

v = ω* R Formula (4)

Where:

θ : angle that the wheel has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

ω₀ : initial angular speed ( rad/s)

ωf : final angular speed ( rad/s)

v: tangential velocity of a point on the rim ( m/s)

R : radius of wheel (m)

ac: centripetal acceleration, (m/s²)

Data:

D = 40.0 cm : diameter of the wheel

R = D/2= 40.0 cm/ 2 = 20 cm = 0.2m

α = 3.00 rad/s^2

ω₀ = 0

n = 2 revolutions : number of revolutions

θ =2πn (rad) = 2π*2 (rad) = 4π rad

Calculate of the ωf

We replace data in the formula (3)

ωf²= ω₀² + 2αθ

ωf²= 0 + 2(3)(4π)

ωf²= 24π

$w_{f} = \sqrt{24\pi }$

ωf = 8.68 rad/s

Calculate of the v

We replace data in the formula (4)

v = ω*R

v = (8.68)*(0.2)

v = 1.736 m/s

Calculate of the ac

We replace data in the formula (1)

ac = ( ω)²*(R)

ac = (8.68)²*(0.2)

ac = 15.07 m/s²

We replace data in the formula (2)

ac = v²/ R

ac = ( 1.736 )²/(0.2)

ac = 15.07 m/s²

7 0

4 years ago

If the mass of a material is 115 grams and the volume of the material is 16 cm3, what would the density of the material be?

hjlf

Well, density is mass/volume. So what's 115 g / 16 cm3? That'll get you your density. Remember that density will be in g/cm3!

6 0

3 years ago

A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 14 s until its motor stops. Disreg

telo118 [61]

Answer:

The maximum height of the rocket will be 1.0 × 10⁴ m.

Explanation:

Hi there!

The height of the rocket at time "t" can be calculated using the following equations:

y = y0 + v0 · t + 1/2 · a · t² (when the rocket has an upward acceleration)

y = y0 + v0 · t + 1/2 · g · t² (after the motor of the rocket stops)

Where:

y = height.

y0 = initial height.

v0 = initial velocity.

t = time.

a = acceleration due to the motors.

g = acceleration due to gravity.

The velocity of the rocket can be calculated as follows:

v = v0 + a · t (while the motor is running)

v = v0 + g · t (after the motor stops)

Where "v" is the velocity of the rocket at time "t".

The rocket rises with upward acceleration for 14 s. After that, the rocket starts being accelerated in the downward direction due to gravity. But it will continue going up after the motor stops because the rocket has initially an upward velocity that will be reduced until it becomes zero and the rocket starts to fall.

Let´s find the height reached by the rocket while it was accelerated in the upward direction (the origin of the frame of reference is located at the launching point and upward is the positive direction):

y = y0 + v0 · t + 1/2 · a · t²

y = 0 m + 0 m/s · 14 s + 1/2 · 28 m/s² · (14 s)²

y = 2.7 × 10³ m

Now let´s find the velocity reached in that time:

v = v0 + a · t

v = 28 m/s² ·14 s

v = 3.9 × 10² m/s

Now, let´s find the maximum height reached by the rocket using the equations of height and velocity after the motor stops:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Notice that now v0 and y0 will be the velocity and height reached while the rocket was being accelerated in the upward direction, respectively.

Let´s find at which time the rocket reaches its maximum height. With that time, we can calculate the max-height.

At the maximum height, the velocity of the rocket is zero, then:

v = v0 + g · t

0 = 3.9 × 10² m/s - 9.8 m/s² · t

-3.9 × 10² m/s/ -9.8 m/s² = t

t = 40 s

After the motor stops, it takes the rocket 40 s s to reach the maximum height.

Using the equation of height:

y = y0 + v0 · t + 1/2 · g · t²

y = 2.7 × 10³ m + 3.9 × 10² m/s · 40 s - 1/2 · 9.8 m/s² · (40 s)²

y = 1.0 × 10⁴ m

The maximum height of the rocket will be 1.0 × 10⁴ m

4 0

3 years ago

A hot air balloon is filled with 1.45 × 10 6 L of an ideal gas on a cool morning ( 11 ∘ C ) . The air is heated to 109 ∘ C . Wha

just olya [345]

Answer:

The volume of air after the balloon is heated = 1.95 × 10⁶ L

Explanation:

Charles Law: Charles' law states that the volume of a given mass of gas is directly proportional to the temperature in Kelvin, provided that the pressure remains constant.

It can be expressed mathematically as,

V₁/T₁ = V₂/T₂

Making V₂ The subject of the equation

V₂ = (V₁/T₁)T₂.................... Equation 1

Where V₁ = Initial Volume, T₁ = Initial Temperature, V₂ = Final Volume, T₂ = final Temperature

Given: V₁ = 1.45 × 10⁶ L, T₁ = 11 °C = (11 + 273) K = 284 K, T₂ = 109 °C = (109 + 273) = 382 K.

Substituting these values into equation 1 above,

V₂ = (1.45×10⁶)382/284

V₂ = 1.95 × 10⁶ L

Therefore the volume of air after the balloon is heated = 1.95 × 10⁶ L

6 0

3 years ago

Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J =

Sloan [31]

Answer:

The current is $I = 8.9 *10^{-5} \ A$

Explanation:

From the question we are told that

The radius is $r = 3.17 \ mm = 3.17 *10^{-3} \ m$

The current density is $J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2$

The distance we are considering is $r = 0.5 R = 0.001585$

Generally current density is mathematically represented as

$J = \frac{I}{A }$

Where A is the cross-sectional area represented as

$A = \pi r^2$

=> $J = \frac{I}{\pi r^2 }$

=> $I = J * (\pi r^2 )$

Now the change in current per unit length is mathematically evaluated as

$dI = 2 J * \pi r dr$

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

$I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr$

$I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr$

$I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.$

$I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ]$

substituting values

$I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ]$

$I = 8.9 *10^{-5} \ A$

5 0

4 years ago