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3 years ago

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Physics

1 answer:

Naddika [18.5K]3 years ago

3 0

Answer:

Explanation:

Firstly, it should noted that liquid contains particles that move free in space (but not as free as gases). When heat is applied to a liquid. The liquid begins to have an increase in thermal energy which would cause the movement of particles of the liquid to increase. As the thermal energy increases, the movement of the particles continues to increase which would eventually lead to the space of the particles in the liquid increasing. When it eventually gets to a point (of increased thermal energy, the liquid begins to boil and at this point, the liquid changes to gas.

Based on the explanation above, the flow chart in the picture above should be arranged as follows

The liquid gains thermal energy

↓

The particles of liquid move faster

↓

The space between the liquid particles increases

↓

The liquid changes to a gas

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A crate with a mass of m = 450 kg rests on the horizontal deck of a ship. The coefficient of static friction between the crate a

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Answer: $F_{v} =\mu_{k} mg$

Magnitude of the force is 2601.9 N

Explanation:

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but once crate starts moving the force of friction is reduced $F_{v} =\mu_{k} mg$ .

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Two particles, each with charge 55.3 nC, are located on the y axis at y 24.9 cm and y -24.9cm (a) Find the vector electric field

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Answer:

Ex = kq 2x / ∛ (x² + y²)² and Ex = 2008 N / C

Explanation:

a) The electric field is a vector quantity, so we must find the field for each particle and add them vectorially, as the whole process is on the X axis,

The equation for the electric field produced by a point charge is

E = k q / r²

With r the distance between the point charge and the positive test charge

We look for each electric field

Particle 1. Located at y = 24.9 m, let's use Pythagoras' theorem to find the distance

r² = x² + y²

E1 = k q / (x² + y²)

Particle 2. located at x = -24.9 m

r² = x² + y²

E2 = k q / (x² + y²)

We can see that the two fields are equal since the particles have the same charge and coordinate it and that is squared.

In the attached one we can see that the Y components of the electric fields created by each particle are always the same and it is canceled, so we only have to add the X components of the electric fields. Let's use Pythagoras' theorem to find

Let's measure the angle from axis X

cos θ = CA / H = x / (x2 + y2) ½

E1x = E1 cos θ

E2x = = E1 cos θ

The resulting field

Ey = 0

Ex = E1x + E2x 2 E1x

Ex = 2 k q / (x² + y²) cos θ) = 2 k q / (x² + y²) x / √(x² + x²)

Ex = kq 2x / ∛ (x² + y²)²

b) For this part we substitute the numerical values

Ex = 8.99 10⁹ 55.3 10⁻⁹ x / (x² + 0.249 2) ³/₂

Ex = 497.15 x / (x² + 0.062) ³/₂

Point where can the value of the electric field x = 38.1 cm = 0.381 m

Ex = 497.15 0.381 / (0.381² + 0.062) ³/₂

Ex = 497.15 0.381 / (0.1452 + 0.062) 3/2 = 189.41 / 0.2072 3/2

Ex= 189.41 /0.0943

Ex = 2008 N / C

c) E = 1.00 kN / C = 1000 N / C

To solve this part we must find x in the equation

Ex = 497.15 x / (x² + 0.062) ³/₂

Let's use some arithmetic

Ex / 497.15 = x / (x² + 0.062) ³/₂

[Ex / 497.15] ²/₃ = [x / (x² + 0.062) 3/2] ²/₃

∛[Ex / 497.15]² = (∛x²) / (x² + 0.062) (1)

The roots of this equation are the solution to the problem,

For Ex = 1.00 kN / C = 1000 N / C

[Ex / 497.15] 2/3 = 1000 / 497.15) 2/3 = 1,312

1.312 = (∛x² ) / (x² + 0.062)

1.312 (x² + 0.062) = ∛x²

1.312 X² - ∛x² + 1.312 0.062 = 0

1.312 X² - ∛x² + 0.0813 = 0

We need used computer

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