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3 years ago

What are some ways to separate a mixture?

1 answer:

3 0

la separación mecánica de un sólido de grano grueso, insoluble, en un líquido; consiste en verter cuidadosamente el líquido, después de que se ha sedimentado el sólido. Por este proceso se separan dos líquidos miscibles, de diferente densidad, por ejemplola separación de un sólido disuelto en un líquido, por calentamiento, hasta que hierve y se transforma en vapor. Como no todas las sustancias se evaporan con la misma rapidez, el sólido disuelto se obtiene en forma pura, agua y aceite.

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On Mars a rock falls an unknown vertical distance from a resting position and lands in a crater. If it takes the rock 2.5 second

astra-53 [7]

The Answer To This Question Is B

Hope It Helped

8 0

3 years ago

What provides the force on the person on the passenger seat?

frutty [35]

When the car speeds up, slows down, or goes around a curve,
passengers need a force applied to them to make them do the
same thing, otherwise they won't keep up with the car.

The force on the passenger is applied by means of friction between
the upholstery and the seat of his pants, and also by the seat-back
or his seat-belt.

4 0

3 years ago

Read 2 more answers

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

8 0

3 years ago

A 20kg object acceleration by a force of 200N with coefficient of kineticfriction is 0.4 what is acceleration of the object?

Schach [20]

Answer:

Given - M = 20 kg

k = 0.4

F = 200 N

To find -  acceleration

Solution - 

F= kMA

200 = 0.4 * 20 * acceleration

200 = 8 * a

a = 8/200

a = 0.04 m s²

5 0

2 years ago

Suppose you have a pendulum clock that keeps correct time on Earth (acceleration due to

vekshin1

Answer:

dont know because I am a student lol

8 0

2 years ago

Read 2 more answers