All categories

3 years ago

The importance of latitude and longitude

1 answer:

7 0

The most important line of latitude is the equator, which runs horizontally around the fattest part of the earth. The most important longitude line is the Prime Meridian which runs vertically and goes through Greenwich, England

You might be interested in

Calcular la distancia recorrida por un móvil que tiene una rapidez de 35 m/s inicia un moviendo uniformemente acelerado con una

kumpel [21]

Answer:

What are the areas of the SAARC cooperation? Make their list.

5 0

3 years ago

Points A, B, and C are at the corners of an equilateral triangle of side 8 m. Equal positive charges of 4 mu or micro CC are at

aliina [53]

Answer:

a) 8.99*10³ V b) 4.5*10⁻² J c) 0 d) 0

Explanation:

The electrostatic potential V, is the work done per unit charge, by the electrostatic force, producing a displacement d from infinity (assumed to be the reference zero level).
For a point charge, it can be expressed as follows:

$V =\frac{k*q}{d}$

As the electrostatic force is linear with the charge (it is raised to first power), we can apply superposition principle.
This means that the total potential at a given point, is just the sum of the individual potentials due to the different charges, as if the others were not there.
In our case, due to symmetry, the potential, at any corner of the triangle, is just the double of the potential due to the charge located at any other corner, as follows:

$V = \frac{2*q*k}{d} = \frac{2*8.99e9N*m2/C2*4e-6C}{8m} =\\ \\ V= 8.99e3 V$

The potential at point C is 8.99*10³ V

The work required to bring a positive charge of 5μC from infinity to the point C, is just the product of the potential at this point times the charge, as follows:

$W = V * q = 8.99e3 V* 5e-6C = 4.5e-2 J$

The work needed is 0.045 J.

If we replace one of the charges creating the potential at the point C, by one of the same magnitude, but opposite sign, we will have the following equation:

$V = \frac{8.99e9N*m2/C2*(4e-6C)}{8m} + (\frac{8.99e9N*m2/C2*(-4e-6C)}{8m}) = 0$

This means that the potential due to both charges is 0, at point C.

If the potential at point C is 0, assuming that at infinity V=0 also, we conclude that there is no work required to bring the charge of 5μC from infinity to the point C, as no potential difference exists between both points.

5 0

3 years ago

A coil is wrapped with 300 turns of wire on the perimeter of a circular frame (radius = 8.0 cm). Each turn has the same area, eq

MAVERICK [17]

Answer:

Approximately 18 volts when the magnetic field strength increases from $\rm 20\; mT$ to $\rm 80\;mT$ at a constant rate.

Explanation:

By the Faraday's Law of Induction, the EMF $\epsilon$ that a changing magnetic flux induces in a coil is:

$\displaystyle \epsilon = N \cdot \frac{d\phi}{dt}$ ,

where

$N$ is the number of turns in the coil, and
$\displaystyle \frac{d\phi}{dt}$ is the rate of change in magnetic flux through this coil.

However, for a coil the magnetic flux $\phi$ is equal to

$\phi = B \cdot A\cdot \cos{\theta}$ ,

where

$B$ is the magnetic field strength at the coil, and
$A\cdot \cos{\theta}$ is the area of the coil perpendicular to the magnetic field.

For this coil, the magnetic field is perpendicular to coil, so $\theta = 0$ and $A\cdot \cos{\theta} = A$ . The area of this circular coil is equal to $\pi\cdot r^{2} = \pi\times 8.0\times 10^{-2}\approx \rm 0.0201062\; m^{2}$ .

$A\cdot \cos{\theta} = A$ doesn't change, so the rate of change in the magnetic flux $\phi$ through the coil depends only on the rate of change in the magnetic field strength $B$ . The size of the magnetic field at the instant that $B = \rm 50\; mT$ will not matter as long as the rate of change in $B$ is constant.

$\displaystyle \begin{aligned} \frac{d\phi}{dt} &= \frac{\Delta B}{\Delta t}\times A \\&= \rm \frac{80\times 10^{-3}\; T- 20\times 10^{-3}\; T}{20\times 10^{-3}\; s}\times 0.0201062\;m^{2}\\&= \rm 0.0603186\; T\cdot m^{2}\cdot s^{-1}\end{aligned}$ .

As a result,

$\displaystyle \epsilon = N \cdot \frac{d\phi}{dt} = \rm 300 \times 0.0603186\; T\cdot m^{2}\cdot s^{-1} \approx 18\; V$ .

9 0

3 years ago

A light plane is headed due south with a speed of 200 km/h relative to still air. After 1.00 hour, the pilot notices that they h

Lilit [14]

Answer: the total velocity of the air is 67.69km/h to the north and 35.4km/h to the east.

Explanation: The initial velocity of the plane is 200km/h south (supose that south is our positive x-axis here and east is the positive y-axis)

In one hour, the plane is located 137km away from the initial position, and the position in X is equal to 137km*cos(15°) = 132.33, this means that the velocity in the x axis is equal to 132.33 km/h, knowing that the initial velocity of the plane was 200km in the x-axis, this means that the velocity of the air must be:

132.33km/h - 200km/h = -67.69km/h

km and the position in "y" is equal to 137km*sin(15°) = 35.4km

This means that the velocity of the air in the y-axis is 35.4km/h

So the total velocity of the air is 67.69km/h to the north and 35.4km/h to the east.

8 0

3 years ago

Which of the following correctly describes the hierarchy that exists in the universe? A galaxy refers to all existing matter, en

Mkey [24]

The correct hierarchy would be

-<span>A solar system is a collection of planets, their moons, and other objects in orbit around a central star.

</span><span>-A galaxy refers to all existing matter, energy, and space that is held together by gravity</span>

6 0

3 years ago

Read 2 more answers