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2 years ago

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If the charge that enters each meter of the axon gets distributed uniformly along it, how many coulombs of charge enter a 0.100

mm length of the axon?

1 answer:

SCORPION-xisa [38]2 years ago

3 0

Answer:

Charge enter a 0.100 mm length of the axon is $8.98\times 10^{-12} C$

Explanation:

Electric field E at a point due to a point charge is given by

$E=k \frac{q}{r^2}$

where $k$ is the constant = $9.0 \times 10^9 Nm^2 / C^2$

$q$ is the magnitude of point charge and $r$ is the distance from the point charge

Charges entering one meter of axon is $5.\times 10^{11} \times (+e)$

Charges entering 0.100 mm of axon is $5.\times 10^{11} \times (+e) \times (0.1 \times 10^{-3}$

substituting the value of $+e=1.6\times 10^{-19} C$ in above equation, we get charge enter a 0.100 mm length of the axon is

$q=5.\times 10^{11} \times1.6\times 10^{-19} \times (0.1 \times 10^{-3}\\q=8.98\times 10^{-12} C$

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Modern wind turbines generate electricity from wind power. The large, massive blades have a large moment of inertia and carry a

horsena [70]

Answer:

a)106.48 x 10⁵ kg.m²

b)144.97 x 10⁵ kgm² s⁻¹

Explanation:

a)Given

m = 5500 kg

l = 44 m

Moment of inertia of one blade

$I$ = 1/3 x m l²

where m is mass of the blade

l is length of each blade.

Putting all the required values, moment of inertia of one blade will be

$I$ = 1/3 x 5500 x 44²

$I$ = 35.49 x 10⁵ kg.m²

Moment of inertia of 3 blades

$I$ = 3 x 35.49 x 10⁵ kg.m²

$I$ = 106.48 x 10⁵ kg.m²

b) Angular momentum 'L' is given by

L = $I$ x ω

where,

$I$ = moment of inertia of turbine i.e 106.48 x 10⁵ kg.m²

ω=angular velocity =2π f

f is frequency of rotation of blade i.e 13 rpm

f = 13 rpm=>= 13 / 60 revolution per second

ω = 2π f => 2π x 13 / 60 rad / s

L= $I$ x ω =>106.48 x 10⁵ x 2π x 13 / 60

= 144.97 x 10⁵ kgm² s⁻¹

7 0

2 years ago

You are working as an assistant to an air-traffic controller at the local airport, from which small airplanes take off and land.

Alika [10]

Answer:

d = 2021.6 km

Explanation:

We can solve this distance exercise with vectors, the easiest method s to find the components of the position of each plane and then use the Pythagorean theorem to find distance between them

Airplane 1

Height y₁ = 800m

Angle θ = 25°

cos 25 = x / r

sin 25 = z / r

x₁ = r cos 20

z₁ = r sin 25

x₁ = 18 103 cos 25 = 16,314 103 m = 16314 m

z₁ = 18 103 sin 25 = 7,607 103 m= 7607 m

2 plane

Height y₂ = 1100 m

Angle θ = 20°

x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m

z₂ = 20 103 without 25 = 8.452 103 m = 8452 m

The distance between the planes using the Pythagorean Theorem is

d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2

Let's calculate

d² = (18126-16314)² + (1100-800)² + (8452-7607)²

d² = 3,283 106 +9 104 + 7,140 105

d² = (328.3 + 9 + 71.40) 10⁴

d = √(408.7 10⁴)

d = 20,216 10² m

d = 2021.6 km

7 0

2 years ago

A torsional pendulum consists of a disk of mass 450 g and radius 3.5 cm, hanging from a wire. If the disk is given an initial an

Montano1993 [528]

To solve this problem we will use the kinematic equations of angular motion, starting from the definition of angular velocity in terms of frequency, to verify the angular displacement and its respective derivative, let's start:

$\omega = 2\pi f$

$\omega = 2\pi (2.5)$

$\omega = 5\pi rad/s$

The angular displacement is given as the form:

$\theta (t) = \theta_0 cos(\omega t)$

In the equlibrium we have to $t=0, \theta(t) = \theta_0$ and in the given position we have to

$\theta(t) = \theta_0 cos(5\pi t)$

Derived the expression we will have the equivalent to angular velocity

$\frac{d\theta}{dt} = 2.7rad/s$

Replacing,

$\theta_0(sin(5\pi t))5\pi = 2.7$

Finally

$\theta_0 = \frac{2.7}{5\pi}rad = 9.848\°$

Therefore the maximum angular displacement is 9.848°

6 0

2 years ago

The equilibrium fraction of lattice sites that are vacant in silver (Ag) at 600°C is 1 × 10-6. Calculate the number of vacancies

algol [13]

Answer :

The number of vacancies (per meter cube) = 5.778 × 10^22/m^3.

Explanation:

Given,

Atomic mass of silver = 107.87 g/mol

Density of silver = 10.35 g/cm^3

Converting to g/m^3,

= 10.35 g/cm^3 × 10^6cm^3/m^3

= 10.35 × 10^6 g/m^3

Avogadro's number = 6.022 × 10^23 atoms/mol

Fraction of lattice sites that are vacant in silver = 1 × 10^-6

Nag = (Na * Da)/Aag

Where,

Nag = Total number of lattice sites in Ag

Na = Avogadro's number

Da = Density of silver

Aag = Atomic weight of silver

= (6.022 × 10^23 × (10.35 × 10^6)/107.87

= 5.778 × 10^28 atoms/m^3

The number of vacancies (per meter cube) = 5.778 × 10^28 × 1 × 10^-6

= 5.778 × 10^22/m^3.

6 0

2 years ago

Two long, parallel wires carry unequal currents in the same direction. The ratio of the currents is 3 to 1. The magnitude of the

astraxan [27]

Answer:

3A is the larger of the two currents.

Explanation:

Let the currents in the two wires be I₁ and I₂

given:

Magnitude of the electric field, B = 4.0μT = 4.0×10⁻⁶T

Distance, R = 10cm = 0.1m

Ratio of the current = I₁ : I₂ = 3 : 1

Now, the magnitude of a magnetic field at a distance 'R' due to the current 'I' is given as

$B = \frac{\mu_oI}{2\pi R}$

Where $\mu_o$ is the magnitude constant = 4π×10⁻⁷ H/m

Thus, the magnitude of a magnetic field due to I₁ will be

$B_1 = \frac{\mu_oI_1}{2\pi R}$

$B_2 = \frac{\mu_oI_2}{2\pi R}$

given,

B = B₁ - B₂ (since both the currents are in the same direction and parallel)

substituting the values of B, B₁ and B₂

we get

4.0×10⁻⁶T = $\frac{\mu_oI_1}{2\pi R}$ - $\frac{\mu_oI_2}{2\pi R}$

or

4.0×10⁻⁶T = $\frac{\mu_o}{2\pi R}\times (I_1-I_2 )$

also

$\frac{I_1}{I_2} = \frac{3}{1}$

⇒ $I_1 = 3\times I_2$

substituting the values in the above equation we get

4.0×10⁻⁶T = $\frac{4\pi\times 10^{-7}}{2\pi 0.1}\times (3 I_2-I_2)$

⇒ $I_2 = 1A$

also

$I_1 = 3\times I_2$

⇒ $I_1 = 3\times 1A$

⇒ $I_1 = 3A$

Hence, the larger of the two currents is 3A

3 0

2 years ago