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2 years ago

What are igneous intrusions and extrusions

1 answer:

3 0

Igneous intrusions form when magma cools and solidifies before it can reach the surface. An extrusion consists of extrusive rock; which forms above the surface of the crust.

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State the objects in the universe that reflect light from stars

stepan [7]

Everything in the universe that is not a star reflects light from Stars. Otherwise you can't see it.

7 0

3 years ago

Please help me with this i need your help

krek1111 [17]

Answer:
Speed of the wave is 7.87 m/s.
Explanation:
It is given that, tapping the surface of a pan
of water generates 17.5 waves per second
We know that the number of waves per
second is called the frequency of a wave.
So, f= 17.5 HZ
Wavelength of each wave,
A = 45 cm = 0.45 m
Speed of the wave is given by:
175 × 0.45
V= 7.87 m/s
So, the speed of the wave is 7.87 m/s
Hence, this is the required solution.

3 0

1 year ago

The circumference of a sphere was measured to be

professor190 [17]

To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are

$\phi = 76cm$

$Error (dr) = 0.5cm$

The radius then would be

$\phi = 2\pi r \\76cm = 2\pi r\\r = \frac{38}{\pi} cm$

And

$\frac{d\phi}{dr} = 2\pi \\d\phi = 2\pi dr \\0.5 = 2\pi dr$

PART A ) For the Surface Area we have that,

$A = 4\pi r^2 \\A = 4\pi (\frac{38}{\pi})^2\\A = \frac{5776}{\pi}$

Deriving we have that the change in the Area is equivalent to the maximum error, therefore

$\frac{dA}{dr} = 4\pi (2r) \\dA = 4r (2\pi dr)$

Maximum error:

$dA = 4(\frac{38}{\pi})(0.5)$

$dA = \frac{76}{\pi}cm^2$

The relative error is that between the value of the Area and the maximum error, therefore:

$\frac{dA}{A} = \frac{\frac{76}{\pi}}{\frac{5776}{\pi}}$

$\frac{dA}{A} = 0.01315 = 1.31\%$

PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so

$V = \frac{4}{3} \pi r^3$

$V = \frac{4}{3} \pi (\frac{38}{\pi})^3$

$V = \frac{219488}{3\pi^2}$

Therefore the Maximum Error would be,

$\frac{dV}{dr} = \frac{4}{3} 3\pi r^2$

$dV = 2r^2 (2\pi dr)$

$dV = 4r^2 (\pi dr)$

Replacing the value for the radius

$dV = 4(\frac{38}{\pi})^2(0.5)$

$dV = \frac{2888}{\pi^2} cm^3$

And the relative Error

$\frac{dV}{V} = \frac{ \frac{2888}{\pi^2}}{ \frac{219488}{3\pi^2} }$

$\frac{dV}{V} = 0.03947$

$\frac{dV}{V} = 3.947\%$

3 0

2 years ago

The half-life of the radioactive element beryllium-13 is 5 × 10-10 seconds, and half-life of the radioactive element beryllium-1

telo118 [61]

<h2>Answer: The half-life of beryllium-15 is 400 times greater than the half-life of beryllium-13.</h2>

Explanation:

The half-life $h$ of a radioactive isotope refers to its decay period, which is the average lifetime of an atom before it disintegrates.

In this case, we are given the half life of two elements:

beryllium-13: $h_{B-13}=5(10)^{-10}s=0.0000000005s$

beryllium-15: $h_{B-15}=2(10)^{-7}s=0.0000002s$

As we can see, the half-life of beryllium-15 is greater than the half-life of beryllium-13, but how great?

We can find it out by the following expression:

$h_{B-15}=X.h_{B-13}$

Where $X$ is the amount we want to find:

$X=\frac{h_{B-15}}{h_{B-13}}$

$X=\frac{2(10)^{-7}s}{5(10)^{-10}s}$

Finally:

$X=400$

Therefore:

The half-life of beryllium-15 is <u>400 times greater than</u> the half-life of beryllium-13.

8 0

2 years ago

Apparent magnitude depends mainly upon _________. our sun has a large apparent magnitude.

Ksenya-84 [330]

Apparent magnitude depends mainly on the brightness of the object as seen from an observer on Earth. This is taken into account without the effects of the atmosphere.

7 0

3 years ago