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1 month ago

Compare and contrast the strength of the forces between two objects with a mass of 1 kg each, a charge of 10

1 answer:

3 0

Let's explore the commonalities between the two forces: * They are proportional to the product of either mass or charge. * Their relationship to distance is inverse, meaning the forces weaken with the square of the distance. * Both forces are long-range, as a distance of zero does not exist up to infinity. Gravitational force is always attractive, while electric force can be either attractive or repulsive. Now, let's look at the distinctions: * The electric force is significantly stronger than the gravitational force. * The gravitational force is perpetually attractive, in contrast to the electric force, which can either attract or repel.

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A thin electrical heating element provides a uniform heat flux qo" to the outer surface of a duct through which air flows. The d

Ostrovityanka [3204]

Answer:

a) q = 2800 W/m²

b) To = 59.4°C

Explanation:

Given the following parameters:

L = 10 mm

K = 20 W/m·K

T = 30°C

h = 100 W/m²K

Ti = 58°C

Calculating heat flux q:

q = h ΔT

q = 100 x (58 - 30)

q = 2800 W/m²

According to Fourier's law, heat transfer is defined as:

Q = K A ΔT/L

Assuming the outer temperature is To:

To = Ti + qL/K

Substituting the given values:

To = Ti + qL/K

$To= 58 + 2800 \times \dfrac{ 0.01}{20}$

To = 59.4°C

5 0

2 months ago

When you urinate, you increase pressure in your bladder to produce the flow. For an elephant, gravity does the work. An elephant

ValentinkaMS [3465]

Answer:

a) v = 1.19 m/s, b) P₁ = 0.922 x 10⁵ Pa

Explanation:

1) We will apply the fluid continuity equation

Q = A v

The area of a circle is

A = π r² = π d²/4

v = Q / A = Q 4 / π d²

v = 0.006 4/π 0.08²

v = 1.19 m/s

2) Apply Bernoulli's equation, considering point 1 as the bladder and point 2 as where the urine exits

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

The problem states that

P₂ = 1.0013 x 10⁵ Pa

v₁ = 0

y₁ = 1 m

y₂ = 0

Density of water (ρ) = 1000 kg/m³

P₁ + ρ y₁ = P₂ + ½ ρ v₂²

P₁ = P₂ + ½ ρ v₂² - ρ g y₁

P₁ = 1.013 x 10⁵ + ½ 1000 (1.19)² - 1000 9.8 1

P₁ = 1.013 x 10⁵ + 708.5 - 9800

P₁ = 92208.5 Pa

P₁ = 0.922 x 10⁵ Pa

8 0

2 months 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,

Sav [3153]

Answer:

The third charged particle needs to be positioned at x = 0.458 m = 45.8 cm

Explanation:

To address this inquiry, we refer to Coulomb's law:

Two point charges (q₁, q₂) spaced apart by a distance (d) impart a mutual force (F) with a magnitude defined by the equation:

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

F: Electric force in Newtons (N)

K: Coulomb's constant in N*m²/C²

q₁, q₂: Charges quantified in Coulombs (C)

d: distance in meters (m)

Equivalence

1μC= 10⁻⁶C

1m = 100 cm

Information

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 analysis

Refer to the accompanying graphic.

We assume a positive charge q₃, thus F₁₃ and F₂₃ exhibit repulsive forces that need to be equivalent so the resultant force is zero:

Utilizing equation (1), we will determine 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 from 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 discard 10⁻⁶ from 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 for the quadratic equation:

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

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

Choosing x₂ results in F₁₃ and F₂₃ acting in the same direction and unable to cancel each other, therefore we select x₁ as the accurate choice since at this stage the forces counteract each other.

x = 0.458m = 45.8cm