All categories

3 years ago

Both lead and zinc are formed as precipitates. Which of these is a step in the formation of the minerals?

1 answer:

5 0

Minerals are naturally occurring inorganic (they're not formed from amino acids, peptides, or enzymes) chemical compounds.

From the given options, the following describes a step in the formation of the minerals: materials break up due to rise in temperature.

Correct answer: D

You might be interested in

A hockey puck with a mass of 0.175 kg slides over the ice. The puck initially slides with a speed of 5.25 m/s, but it comes to a

Neko [114]

Answer:

1.70 J

Explanation:

The heat dissipated is the difference in the kinetic energies.

This is given by

$E = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

$v_i$ and $v_f$ are the initial and final velocities.

With m = 0.175 kg,

$E = \frac{1}{2}\times0.175(2.85^2 - 5.25^2) = -1.701\text{ J}$

The negative sign appears because energy is lost.

7 0

3 years ago

How many seconds are there in 13 hours?

Zina [86]

Answer:

46,800 seconds. I hope this helps

7 0

3 years ago

Read 2 more answers

A droplet of ink in an industrial ink-jet printer carries a charge of 1.6 x 10–11 coulombs and is deflected onto paper by a forc

Novosadov [1.4K]

Answer:

The Strength of the electric field produced = 2 × 10⁷ N/C

Explanation:

Electric Field: This is defined as the region where an electric force is experienced.

Electric Field Strength: The intensity of an electric field at any point is defined as the force per unit charge which it exert at that point. It direction is that of the force exerted on a positive charge.

It is represented mathematically as,

E = F/Q ................................. Equation 1

Where E = Electric field strength, F = electric force, Q = test charge.

Given: F = 3.2 × 10⁻⁴ N, Q = 1.6 × 10⁻¹¹ C

Substituting these values into equation 1

E= 3.2 × 10⁻⁴/1.6 × 10⁻¹¹

E= 2 × 10⁷ N/C

Thus the Strength of the electric field produced = 2 × 10⁷ N/C

3 0

3 years ago

A tungsten wire has resistance R at 20°C. A second tungsten wire at 20°C has twice the length and half the cross-sectional area

Colt1911 [192]

Answer:

Resistance will become 4 times of first wire resistance.

Explanation:

At the temperature of both the the tungsten wire is same so we can apply ohm's law

Let the length of first wire is $l_1$ and cross sectional area is $A_1$

Resistance of first wire $R=\frac{\rho l_1}{A_1}$ ......1

Now length of second wire is twice the length of first wire

$l_2=2l_1$ and cross sectional area $A_2=\frac{A_1}{2}$ .......2

Resistance of wire 2 $R_2=\frac{\rho l_2}{A_2}$ ........2

Dividing equation 1 by equation 1

$\frac{R}{R_2}=\frac{\rho l_1}{A_1}\times \frac{0.5A_1}{\rho 2l_1}$

$R_2=4R$

Therefore resistance will become 4 times.

8 0

3 years ago

5. Dibuja en qué posición y en qué sentido se debe aplicar una fuerza sobre la barra para que permanezca horizontal y en equilib

Vesna [10]

Answer:

La ubicación de una fuerza igual a F₁ × 5/3 (que actúa hacia abajo), que debe aplicarse para el equilibrio estático, es 3 unidades a la derecha del pivote.

Encuentre el dibujo que muestra la posición y dirección de la fuerza requerida

Explanation:

5. Draw in what position and in what direction a force must be applied to the bar so that it remains horizontal and in static equilibrium, if F1 and F2 have the same magnitude

La información dada son;

La magnitud de F₁ = La magnitud de F₂

El punto de acción de F₁ = 3 unidades a la izquierda del pivote

El punto de acción de F₂ = 2 unidades a la izquierda del pivote

Para el equilibrio, el momento en sentido horario debe ser igual al momento en sentido antihorario

El momento en sentido antihorario. Mₐ sobre el pivote = F₁ × 3 + F₂ × 2 = F₁ × 3 + F₁ × 2

El momento en sentido antihorario alrededor del pivote = F₁ × (3 + 2) = F₁ × 5

El momento en el sentido de las agujas del reloj, $M_c$ , para una fuerza F que actúa en el punto 3 unidades a la derecha del pivote si se da como sigue

$M_c$ = F × 3

Para el equilibrio, tenemos;

F₁ × 5 = F × 3

F = F₁ × 5/3

Por lo tanto, la fuerza que debe aplicarse en el lado derecho en la marca de 3 unidades de la barra para el equilibrio estático = F₁ × 5/3.

7 0

3 years ago