All categories

3 years ago

Can someone give me some info about earthquakes I really need it to make a poster!PLEASE

1 answer:

8 0

Ummmmmmmmmm they are distructive, they mainly happen in japan or by oceans, sometimes they are small and dont cause damage but then then there is the total opposite

You might be interested in

8 m N & 5 m 30° N of E<br><br> Draw and add the vectors.

Contact [7]

by vector addition.

< 5 m, 270 degrees> + < 8 cm, 90 degrees>

=< 500 cos 270, 500 sin 270 > + < 8 cos 90, 8 sin 90 >

=< 0, −500 > + < 0, −8 >

=< 0, −492 > cm

The so-called parallelogram law lays forth the guidelines for adding two or more vectors together. By aligning the two vectors A and B head to tail and drawing the vector from the free tail to the free head, the vector sum A+B is created. If A=(a 1,a 2,...,a n) and B=(b 1,b 2,...,b n), then vector addition can be done in Cartesian coordinates by adding the respective components of the vectors.

A+B=(a 1+b 1,a 2+b 2,...,a n+b n).

The Wolfram Language uses a plus symbol to denote vector addition, as in a1, a2,..., an+b1, b2,..., bn

to learn more about vector:

brainly.com/question/13322477

#SPJ4

3 0

2 years ago

Which type of wave does NOT have a crest?

Nostrana [21]

You answer is b an oscillating wave

7 0

3 years ago

Read 2 more answers

Are fossil fuels used to produce plastics?

cupoosta [38]

Yes. Petroleum, or crude oil, is a fossil fuel and is used to create items such as bubble gum, plastic, etc.

6 0

3 years ago

A rider on a bike with the combined mass of 100 kg attains a terminal speed of 15 m/s on a 12% slope. Assuming that the only for

vlabodo [156]

Answer:

0·95

Explanation:

Given the combined mass of the rider and the bike = 100 kg

Percent slope = 12%

∴ Slope = 0·12

Terminal speed = 15 m/s

Frontal area = 0·9 m²

Let the slope angle be β

tanβ = 0·12

As it attains the terminal speed, the forces acting on the combined rider and the bike must be balanced and therefore the rider must be moving download as the directions of one of the component of weight and drag force will be in opposite directions

The other component of weight will get balance by the normal reaction and you can see the figure which is in the file attached

From the diagram m × g × sinβ = drag force

Drag force = 0·5 × d × $C_{D}$ × v² × A

where d is the density of the fluid through which it flows

$C_{D}$ is the drag coefficient

v is the speed of the object relative to the fluid

A is the cross sectional area

As tanβ = 0·12

∴ sinβ = 0·119

Let the fluid in this case be air and density of air d = 1·21 kg/m³

m × g × sinβ = 0·5 × d × $C_{D}$ × v² × A

100 × 9·8 ×0·119 = 0·5 × 1·21 × $C_{D}$ × 15² × 0·9

∴ $C_{D}$ ≈ 0·95

∴ Drag coefficient is approximately 0·95

4 0

3 years ago

Read 2 more answers

Newton's Law of Gravitation says that the magnitude F of the force exerted by a body of mass m on a body of mass M is F = GmM r2

Nookie1986 [14]

<h2>Answers:</h2>

According to Newton's Law of Gravitation, the Gravity Force is:

$F=\frac{GMm}{{r}^{2}}$ (1)

This expression can also be written as:

$F=GMm{r}^{-2}$ (2)

If we derive this force $F$ respect to the distance $r$ between the two masses:

$\frac{dF}{dr}dFdr=\frac{d}{dr}(GMm{r}^{-2})dr$ (3)

Taking into account $GMm$ are constants:

$\frac{dF}{dr}dFdr=-2GMm{r}^{-3}$ (4)

$\frac{dF}{dr}dFdr=-2\frac{GMm}{{r}^{3}}$ (5)

<h2> (b) dF/dr represents the rate of change of the force with respect to the distance between the bodies. </h2><h2 />

In other words, this means how much does the Gravity Force changes with the distance between the two bodies.

More precisely this change is inversely proportional to the distance elevated to the cubic exponent.

As the distance increases, the Force decreases.

<h2>(c) The minus sign indicates that the bodies are being forced in the negative direction. </h2>

This is because Gravity is an attractive force, as well as, a central conservative force.

This means it does not depend on time, and both bodies are mutually attracted to each other.

In the first answer we already found the decrease rate of the Gravity force respect to the distance, being its unit $N/km$ :

$\frac{dF}{dr}dFdr=-2\frac{GMm}{{r}^{3}}$ (5)

We have a force that decreases with a rate 1 $\frac{dF_{1}}{dr}dFdr=4N/km$ when $r=20000km$ :

$4N/km=-2\frac{GMm}{{(20000km)}^{3}}$ (6)

Isolating $-2GMm$ :

$-2GMm=(4N/km)({(20000km)}^{3})$ (7)

In addition, we have another force that decreases with a rate 2 $\frac{dF_{2}}{dr}dFdr=X$ when $r=10000km$ :

$XN/km=-2\frac{GMm}{{(10000km)}^{3}}$ (8)

Isolating $-2GMm$ :

$-2GMm=X({(10000km)}^{3})$ (9)

Making (7)=(9):

$(4N/km)({(20000km)}^{3})=X({(10000km)}^{3}$ (10)

Then isolating $X$ :

$X=\frac{4N/km)({(20000km)}^{3}}{{(10000km)}^{3}}$

Solving and taking into account the units, we finally have:

$X=-32N/km$ >>>>This is how fast this force changes when $r=10000 km$

7 0

4 years ago

Read 2 more answers