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4 years ago

A free negative charge released in an electric field will

Physics

1 answer:

SashulF [63]4 years ago

5 0

Answer:

Will experience a force due to electric field.

Explanation:

When a free negative charge is released in an electric field it experiences a force due to the electric field in a direction opposite to the direction of the magnetic field.

According to Coulomb's law this force is mathematically given as:

$F=E.q$

and, electric field due to a charge is given as:

$E=\frac{1}{4\pi.\epsilon_0}.\frac{q}{r^2}$

where:

permittivity of free space $\epsilon_0=8.85\times 10^{-12}\ m^{-3}.kg^{-1}.s^4.A^2$

q = magnitude of charge

r = radial distance from the charge

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4 years ago

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4 years ago

Consider your moment of inertia about a vertical axis through the center of your body, both when you are standing straight up wi

jeka94

Answer:

I₁ / I₂ = 1.43

Explanation:

To find the relationship of the two inertial memits, let's calculate each one, let's start at the moment of inertia with the arms extended

Before starting let's reduce all units to the SI system

d₁ = 42 in (2.54 10⁻² m / 1 in) = 106.68 10⁻² m

d₂ = 38 in = 96.52 10⁻² m

The moment of inertia is a scalar quantity for which it can be added, the moment of total inertia would be the moment of inertia of the man (cylinder) plus the moment of inertia of each arm

I₁ = I_man + 2 I_ arm

Man indicates that we can approximate them to a cylinder where the average diameter is

d = (d₁ + d₂) / 2

d = (106.68 + 96.52) 10-2 = 101.6 10⁻² m

The average radius is

r = d / 2 = 50.8 10⁻² m = 0.508 m

The mass of the trunk is the mass of man minus the masses of each arm.

M = M_man - 0.2 M_man = 80 (1-0.2)

M = 64 kg

The moments of inertia are:

A cylinder with respect to a vertical axis: Ic = ½ M r²

A rod that rotates at the end: I_arm = 1/3 m L²

Let us note that the arm rotates with respect to man, but this is at a distance from the axis of rotation of the body, so we must use the parallel axes theorem for the moment of inertia of the arm with respect to e = of the body axis.

I1 = I_arm + m D²

Where D is the distance from the axis of rotation of the arm to the axis of the body

D = d / 2 = 101.6 10⁻² /2 = 0.508 m

Let's replace

I₁ = ½ M r² + 2 [(1/3 m L²) + m D²]

Let's calculate

I₁ = ½ 64 (0.508)² + 2 [1/3 8 1² + 8 0.508²]

I₁ = 8.258 + 5.33 + 4.129

I₁ = 17,717 Kg m² / s²

Now let's calculate the moment of inertia with our arms at our sides, in this case the distance L = 0,

I₂ = ½ M r² + 2 m D²

I₂ = ½ 64 0.508² + 2 8 0.508²

I₂ = 8,258 + 4,129

I₂ = 12,387 kg m² / s²

The relationship between these two magnitudes is

I₁ / I₂ = 17,717 /12,387

I₁ / I₂ = 1.43

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3 years ago