Name three possible end stages of a star

A man walks along a straight path at a speed of 4 ft/s. A searchlight is located on the ground 6 ft from the path and is kept fo

BARSIC [14]

We are given that,

$\frac{dx}{dt} = 4ft/s$

We need to find $\frac{d\theta}{dt}$ when $x=8ft$

The equation that relates x and $\theta$ can be written as,

$\frac{x}{6} tan\theta$

$x = 6tan\theta$

Differentiating each side with respect to t, we get,

$\frac{dx}{dt} = \frac{dx}{d\theta} \cdot \frac{d\theta}{dt}$

$\frac{dx}{dt} = (6sec^2\theta)\cdot \frac{d\theta}{dt}$

$\frac{d\theta}{dt} = \frac{1}{6sec^2\theta} \cdot \frac{dx}{dt}$

Replacing the value of the velocity

$\frac{d\theta}{dt} = \frac{1}{6} cos^2\theta (4)^2$

$\frac{d\theta}{dt} = \frac{8}{3} cos^2\theta$

The value of $cos \theta$ could be found if we know the length of the beam. With this value the equation can be approximated to the relationship between the sides of the triangle that is being formed in order to obtain the numerical value. If this relation is known for the value of x = 6ft, the mathematical relation is obtained. I will add a numerical example (although the answer would end in the previous point) If the length of the beam was 10, then we would have to

$cos\theta = \frac{6}{10}$

$\frac{d\theta}{dt} = \frac{8}{3} (\frac{6}{10})^2$

$\frac{d\theta}{dt} = \frac{24}{25}$

Search light is rotating at a rate of 0.96rad/s

4 0

3 years ago

Explain how polarization of a cell increases the cell's internal resistance.<br>(2<br>2.

Mandarinka [93]

Answer:

Explanation: The chemical action that occurs in the cell while the current is flowing causes hydrogen bubbles to form on the surface of the anode. This action is called POLARIZATION. Some hydrogen bubbles rise to the surface of the electrolyte and escape into the air, some remain on the surface of the anode. If enough bubbles remain around the anode, the bubbles form a barrier that increases internal resistance. When the internal resistance of the cell increases, the output current is decreased and the voltage of the cell also decreases.

A cell that is heavily polarized has no useful output. There are several methods to prevent polarization or to depolarize the cell.

One method uses a vent on the cell to permit the hydrogen to escape into the air. A disadvantage of this method is that hydrogen is not available to reform into the electrolyte during recharging. This problem is solved by adding water to the electrolyte, such as in an automobile battery. A second method is to use material that is rich in oxygen, such as manganese dioxide, which supplies free oxygen to combine with the hydrogen and form water.

A third method is to use a material that will absorb the hydrogen, such as calcium. The calcium releases hydrogen during the charging process. All three methods remove enough hydrogen so that the cell is practically free from polarization.

LOCAL ACTION

When the external circuit is removed, the current ceases to flow, and, theoretically, all chemical action within the cell stops. However, commercial zinc contains many impurities, such as iron, carbon, lead, and arsenic. These impurities form many small electrical cells within the zinc electrode in which current flows between the zinc and its impurities. Thus, the chemical action continues even though the cell itself is not connected to a load.

Local action may be prevented by using pure zinc (which is not practical), by coating the zinc with mercury, or by adding a small percentage of mercury to the zinc during the manufacturing process. The treatment of the zinc with mercury is called amalgamating (mixing) the zinc. Since mercury is many times heavier than an equal volume of water, small particles of impurities weighing less than mercury will float to the surface of the mercury. The removal of these impurities from the zinc prevents local action. The mercury is not readily acted upon by the acid. When the cell is delivering current to a load, the mercury continues to act on the impurities in the zinc. This causes the impurities to leave the surface of the zinc electrode and float to the surface of the mercury. This process greatly increases the storage life of the cell.

6 0

3 years ago

Two forces, one of 100 ponds and the other 150 pounds act on the same object, at angles of 20°and 60°, respectively, withthe pos

soldi70 [24.7K]

<h2>Resultant is 235.54 pounds at an angle 44.16° to X axis.</h2>

Explanation:

Forces are 100 pound and 150 pound and angles with x axis are 20°and 60°.

That is force 1 is 100 pound with x axis at 20°

F₁ = 100 cos 20 i + 100 sin 20 j

F₁ = 93.97 i + 34.20 j

That is force 2 is 150 pound with x axis at 60°

F₂ = 150 cos 60 i + 150 sin 60 j

F₂ = 75 i + 129.90 j

F₁ + F₂ = 93.97 i + 34.20 j + 75 i + 129.90 j

F₁ + F₂ = 168.97 i + 164.10 j

$\texttt{Magnitude = }\sqrt{168.97^2+164.10^2}\\\\\texttt{Magnitude = }235.54pounds\\\\\texttt{Angle = }tan^{-1}\left ( \frac{164.10}{168.97}\right )\\\\\texttt{Angle = }44.16^0$

Resultant is 235.54 pounds at an angle 44.16° to X axis.

6 0

3 years ago

A circular loop with radius r is rotating with constant angular velocity ω in a uniform electric field with magnitude E. The axi

inn [45]

Answer:

$\Phi_{E} = E\pi r^2 \omega t$

Explanation:

The electric flux is defined as the multiple of electric field and the area that the electric field passes through, such that

$\Phi_{E} = \vec{E}\vec{A}$

When calculating the electric flux, the angle between the directions of electric field and the area becomes important, especially if the angle is changing with time.

The above formula can be rewritten as follows

$\Phi_{E} = EA\cos(\theta)$

where θ is the angle between the electric field and the area of the loop. Note that, the direction of the area of the loop is perpendicular to the plane of the loop.

If the loop is rotating with constant angular velocity ω, then the angle can be written as follows

$\theta = \omega t$

At t = 0, cos(0) = 1 and the electric flux through the loop is at its maximum value.

Therefore the electric flux can be written as a function of time

$\Phi_{E} = E\pi r^2 \omega t$

3 0

4 years ago

The experimentation step of scientific inquiry involves _______. A. Analyzing data b. Drawing a conclusion c. Choosing variables

Anna11 [10]

Answer:

C- Choosing variables and controls

Explanation:

Correct on edge.

3 0

2 years ago

Name three possible end stages of a star​

Name three possible end stages of a star