Study the four transverse waves shown. Compare the properties of waves B, C, & D to that of wave A.

Explain the value or importance of using the international system of units when describing motion

adoni [48]

<span>It is important to use the SystĂ¨me Internationale (SI) units to describe motion, and other scientific concepts, firstly because the units are the most widely used. Unit choice is largely arbitrary, however, because many scientific units are derived from the base SI units, for example, the Newton is kg m s-2. Thus, secondly, more complex units are based on the bedrock of the SI units.</span>

5 0

3 years ago

What are some resources used to generate electricity in energy stations?

Setler79 [48]

-- Coal
-- Oil
-- Natural gas
-- Falling water
-- Sunlight
-- Nuclear fission of Uranium

3 0

3 years ago

The speed of light is the fastest in which medium

wlad13 [49]

In vacuum, going at 2.99×10^8 m/s.

3 0

3 years ago

Read 2 more answers

Which two options are examples of waves refracting?

LUCKY_DIMON [66]

A and b!
hope this helps!

3 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

TiliK225 [7]

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,

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

6 0

3 years ago