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3 years ago

8

Hi, I need help with this! I'll give five stars and thanks to all the helpful answers! I need five examples of events causing we

athering and erosion. When I looked up the question, I didn't find any events, only examples! Thank you! :)

1 answer:

Alex17521 [72]3 years ago

4 0

Example of weathering and erosion:

- The ocean beating up against the cliff of a wall and slowly wearing away rock

- A river flowing through river banks and the bank slowly stretching apart further and further as time goes on

- Water freezes in a crack on a rock. As the water freezes, the crack expands on the rock.

- A plant is planted in the same spot every year and slowly breaks up the soil.

- Acid rain slowly breaks down the structure of a building.

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ASAP WILL GIVE BRAINLIEST Take a piece of paper, some nails and a magnet bar, place a magnet at the centre of the paper, and spr

Zanzabum

Answer: find the answer in the explanation

Explanation:

When a magnet is placed at the centre of the paper, and the nails are sprinkled on the paper, what will happen to the nails is that, the nails will form a pattern on the paper according to the magnetic field of the bar magnetic pole.

Other phenomena you can observe are:

1.) The nails will align themselves and show some lines of forces which is equivalent to the magnetic field lines

2.) The direction of the line of forces

3.) The strength of the magnetic field pole.

7 0

2 years ago

Name the fundamental units involved in pascal<br>

gtnhenbr [62]

Explanation:

Hey there!

Here,

Pascal is a unit of pressure.

$pressure = \frac{f}{a}$

Now, As per the formula the units are:

kg, m and s^2.

<em><u>Hope it helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

5 0

3 years ago

A rod of length Lo moves iwth a speed v along the horizontal direction. The rod makes an angle of (θ)0 with respect to the x' ax

Colt1911 [192]

Answer:

From the question we are told that

The length of the rod is $L_o$

The speed is v

The angle made by the rod is $\theta$

Generally the x-component of the rod's length is

$L_x = L_o cos (\theta )$

Generally the length of the rod along the x-axis as seen by the observer, is mathematically defined by the theory of relativity as

$L_xo = L_x \sqrt{1 - \frac{v^2}{c^2} }$

=> $L_xo = [L_o cos (\theta )] \sqrt{1 - \frac{v^2}{c^2} }$

Generally the y-component of the rods length is mathematically represented as

$L_y = L_o sin (\theta)$

Generally the length of the rod along the y-axis as seen by the observer, is also equivalent to the actual length of the rod along the y-axis i.e $L_y$

Generally the resultant length of the rod as seen by the observer is mathematically represented as

$L_r = \sqrt{ L_{xo} ^2 + L_y^2}$

=> $L_r = \sqrt{[ (L_o cos(\theta) [\sqrt{1 - \frac{v^2}{c^2} }\ \ ]^2+ L_o sin(\theta )^2)}$

=> $L_r= \sqrt{ (L_o cos(\theta)^2 * [ \sqrt{1 - \frac{v^2}{c^2} } ]^2 + (L_o sin(\theta))^2}$

=> $L_r = \sqrt{(L_o cos(\theta) ^2 [1 - \frac{v^2}{c^2} ] +(L_o sin(\theta))^2}$

=> $L_r = \sqrt{L_o^2 * cos^2(\theta) [1 - \frac{v^2 }{c^2} ]+ L_o^2 * sin(\theta)^2}$

=> $L_r = \sqrt{ [cos^2\theta +sin^2\theta ]- \frac{v^2 }{c^2}cos^2 \theta }$

=> $L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }$

Hence the length of the rod as measured by a stationary observer is

$L_r = L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }$

Generally the angle made is mathematically represented

$tan(\theta) = \frac{L_y}{L_x}$

=> $tan {\theta } = \frac{L_o sin(\theta )}{ (L_o cos(\theta ))\sqrt{ 1 -\frac{v^2}{c^2} } }$

=> $tan(\theta ) = \frac{tan\theta}{\sqrt{1 - \frac{v^2}{c^2} } }$

Explanation:

7 0

3 years ago

Some compounds, like carbon dioxide, are gaseous at room temperature, while others are solid at room temperature. Why is this?

Vesnalui [34]

The answer of a & b are force of cohesion and force of adhesion
Of rest two answers I don't know

8 0

3 years ago

Read 2 more answers

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