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3 years ago

Is the stomach just below the waist?

1 answer:

5 0

The stomach is above the waist, below the waist is your, yunno. the stomach and bladder sit right on top of the waist, hope this helps, have an amazing day:)

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What is the us specification for ac power?

USPshnik [31]

120 volts for most home a phone charger can convert 120 volts ac to 5 volts dc

3 0

3 years ago

A thin rod of length 0.75 m and mass 0.42 kg is suspended

MrRissso [65]

Answer:

a) K = 0.63 J, b) h = 0.153 m

Explanation:

a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is

w² = $\frac{m g d}{I}$

where d is the distance from the pivot point to the center of mass and I is the moment of inertia.

The rod is a homogeneous body so its center of mass is at the geometric center of the rod.

d = L / 2

the moment of inertia of the rod is the moment of a rod supported at one end

I = ⅓ m L²

we substitute

w = $\sqrt{\frac{mgL}{2} \ \frac{1}{\frac{1}{3} mL^2} }$

w = $\sqrt{\frac{3}{2} \ \frac{g}{L} }$

w = $\sqrt{ \frac{3}{2} \ \frac{9.8}{0.75} }$

w = 4.427 rad / s

an oscillatory system is described by the expression

θ = θ₀ cos (wt + Φ)

the angular velocity is

w = dθ /dt

w = - θ₀ w sin (wt + Ф)

In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1

In the exercise it is indicated that at the lowest point the angular velocity is

w = 4.0 rad / s

the kinetic energy is

K = ½ I w²

K = ½ (⅓ m L²) w²

K = 1/6 m L² w²

K = 1/6 0.42 0.75² 4.0²

K = 0.63 J

b) for this part let's use conservation of energy

starting point. Lowest point

Em₀ = K = ½ I w²

final point. Highest point

Em_f = U = m g h

energy is conserved

Em₀ = Em_f

½ I w² = m g h

½ (⅓ m L²) w² = m g h

h = 1/6 L² w² / g

h = 1/6 0.75² 4.0² / 9.8

h = 0.153 m

5 0

2 years ago

Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,

Liula [17]

Answer:

The third charged particle must be placed at x = 0.458 m = 45.8 cm

Explanation:

To solve this problem we apply Coulomb's law:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

$F = \frac{k*q_1*q_2}{d^2}$ Formula (1)

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁, q₂: Charges in Coulombs (C)

d: distance between the charges in meters (m)

Equivalence

1μC= 10⁻⁶C

1m = 100 cm

Data

K = 8.99 * 10⁹ N*m²/C²

q₁ = +5.00 μC = +5.00 * 10⁻⁶ C

q₂= +7.00 μC = +7.00 * 10⁻⁶ C

d₁ = x (m)

d₂ = 1-x (m)

Problem development

Look at the attached graphic.

We assume a positive charge q₃ so F₁₃ and F₂₃ are repulsive forces and must be equal so that the net force is zero:

We use formula (1) to calculate the forces F₁₃ and F₂₃

$F_{13} = \frac{k*q_1*q_3}{d_1^2}$

$F_{23} = \frac{k*q_2*q_3}{d_2^2}$

F₁₃ = F₂₃

$\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2}$ We eliminate k and q₃ on both sides

$\frac{q_1}{d_1^2}= \frac{q_2}{d_2^2}$

$\frac{q_1}{x^2}=\frac{q_2}{(1-x)^2}$

$\frac{5*10^{-6}}{x^2}=\frac{7*10^{-6}}{(1-x)^2}$ We eliminate 10⁻⁶ on both sides

$(1-x)^2 = \frac{7}{5} x^2$

$1-2x+x^2=\frac{7}{5} x^2$

$5-10x+5x^2=7 x^2$

$2x^2+10x-5=0$

We solve the quadratic equation:

$x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m$

$x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m$

In the option x₂, F₁₃ and F₂₃ will go in the same direction and will not be canceled, therefore we take x₁ as the correct option since at that point the forces are in opposite way .

x = 0.458m = 45.8cm