Ask question

Ask question

All categories

alukav5142 [94]

3 years ago

15

IN WHAT SEQUENCE DOES MAGMA MOVE THROUGH A VOLCANO?

1 answer:

Natalka [10]3 years ago

6 0

Well the volcanos come from the lower crust and moves up with changes the climate

You might be interested in

Which statement best describes the difference between atoms and molecules

Rama09 [41]

What are the answers ?

6 0

2 years ago

Read 2 more answers

a crane does 9,500 J of work to lift a crate straight up using a force of 125 N. how high does the crane lift the crate?

Lina20 [59]

Answer:

<em>The crane lifts the crate up to 76 m high</em>

Explanation:

<u>Work Done by a Force</u>

The work done by a force of magnitude F that displaces an object by a distance y is given by

$W=F.y$

We know a crane does a work of 9,500 Joule to lift a crate and is using a force of 125 Nw.

The above equation can be solved to know the value of y in terms of the work and the force:

$\displaystyle y=\frac{W}{F}$

Plugging in the given values

$\displaystyle y=\frac{9,500}{125}$

$y=76\ m$

The crane lifts the crate up to 76 m high

7 0

2 years ago

How have batteries gotten better over time?

babymother [125]

The size has gotten smaller, they last longer, easier to carry around, safer, etc

5 0

2 years ago

Read 2 more answers

Energy resources that exist in limited amounts and, fonts used, cannot be replaced except over the course of millions of years a

kobusy [5.1K]

Energy resources that cannot be replaced are called "non-renewable" energy resources.

5 0

3 years ago

(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup

KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

$W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx$

Where

$x_{o}$ , $x_{f}$ - Initial and final position, respectively, measured in meters.

$F(x)$ - Force as a function of position, measured in newtons.

Given that $F = k\cdot x$ and the fact that $F = 25\,N$ when $x = 0.3\,m - 0.2\,m$ , the spring constant ( $k$ ), measured in newtons per meter, is:

$k = \frac{F}{x}$

$k = \frac{25\,N}{0.3\,m-0.2\,m}$

$k = 250\,\frac{N}{m}$

Now, the work function is obtained:

$W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx$

$W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]$

$W = 0.313\,J$

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be $r(\theta) = 2\cdot \sin 5\theta$ . The area of the region enclosed by one loop of the curve is given by the following integral:

$A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta$

$A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta$

By using trigonometrical identities, the integral is further simplified:

$A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta$

$A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta$

$A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta$

$A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)$

$A = 4\pi$

The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

5 0

2 years ago