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OLEGan [10]
3 years ago
7

Magma is classified as basaltic and andesitic or rhyolitic based on my factor

Physics
1 answer:
vladimir2022 [97]3 years ago
5 0
Magma can be classified in two ways. when it is inside the volcano it is called magma, but when it is out of the volcano it is called lava or  such as :
basalt, ashes, obsidian. magma is also called the names that have been listed because magma and lava are pretty much the same thing just magma is inside the volcano and lava is outside... hope this help you


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A heavy meterstick has a mass of 1 kg. When the meterstick is thrown like a spear past you, you measure its momentum to be 2mv.
CaHeK987 [17]

Answer:

Its length is measured to be 0.5 m

Explanation:

From theory of relativity (mass variation), we know that:

m = mo/√(1-v²/c²)

Where, m = relative mass

and, mo = rest mass

The momentum of stick while moving, will be:

P = mv

but, it is given in the form of rest mass as:

P = 2(mo)v

thus, by comparison;

2(mo)v = mv

using value of m from theory of relativity;

2(mo)v = (mo)v/√(1-v²/c²)

√(1-v²/c²) = 1/2     ______ eqn(1)

Now, for relativistic length (L), we have the formula from same theory of relativity;

L = (Lo)√(1-v²/c²)

The rest length (Lo) of meter stick is 1 m, and the remaining term on right side √(1-v²/c²), known as Lorentz Factor, can be given by eqn (1), as equal to 1/2.

Thus,

L = (1 m)(1/2)

<u>L = 0.5 m</u>

4 0
3 years ago
Please help I'll mark brainliest!!!!
madam [21]

This question is based on the fundamental assumption of  vector direction.

A vector is  a physical quantity which has  magnitude as well direction  for its complete specification.

The magnitude of a physical quantity is simply a  numerical number .Hence it can not be negative.

A negative vector is a vector which comes into existence when it is opposite to our assumed direction with respect to any other vector.  For instance, the vector is taken positive if it is along + X axis and negative if it is along - X axis.

As per the first option it is given that a vector is negative if its magnitude is greater than 1. It is not correct as magnitude play no role in it.

The second option tells that the magnitude of the vector is less than 1. Magnitude can not be negative. So this is also wrong.

Third one tells that a vector is negative if its displacement is along north. It does not give any detail information about the negativity of a vector.

In a general sense we assume that vertically downward motion  is negative and vertically upward is positive. In case of a falling object the motion is  vertically downward. So the velocity of that object is negative .

So last   option is  partially  correct  as  the vector can be negative depending on our choice of co-ordinate system.





7 0
3 years ago
Read 2 more answers
To make sure you understand how to use the equation, suppose that there are 1000 habitable planets in our galaxy, that 1 in 10 h
krok68 [10]

Answer:

5

Explanation:

Number of habitable planets = 1000

Fraction of planet with life = 1/10

Fraction of planet with life and civilization (before) = 1/4

Fraction of planet with life and civilization (now) =1/5

Therefore multiplying we have:

1000×1/10×1/4×1/5 = 5

5 0
2 years ago
Find the time t2 that it would take the charge of the capacitor to reach 99.99% of its maximum value given that r=12.0ω and c=50
defon

Answer:

Explanation:

Given that, .

R = 12 ohms

C = 500μf.

Time t =? When the charge reaches 99.99% of maximum

The charge on a RC circuit is given as

A discharging circuit

Q = Qo•exp(-t/RC)

Where RC is the time constant

τ = RC = 12 × 500 ×10^-6

τ = 0.006 sec

The maximum charge is Qo,

Therefore Q = 99.99% of Qo

Then, Q = 99.99/100 × Qo

Q = 0.9999Qo

So, substituting this into the equation above

Q = Qo•exp(-t/RC)

0.9999Qo = Qo•exp(-t / 0.006)

Divide both side by Qo

0.9999 = exp(-t / 0.006)

Take In of both sodes

In(0.9999) = In(exp(-t / 0.006))

-1 × 10^-4 = -t / 0.006

t = -1 × 10^-4 × - 0.006

t = 6 × 10^-7 second

So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge

8 0
3 years ago
Determine the critical crack length for a through crack contained within a thick plate of 7150-T651 aluminum alloy that is in un
Mila [183]

Explanation:

Formula to determine the critical crack is as follows.

          K_{IC} = \gamma \sigma_{f} \sqrt{\pi \times a}

  \gamma = 1,     K_{IC} = 24.1

  [/tex]\sigma_{y}[/tex] = 570

and,   \sigma_{f} = 570 \times \frac{3}{4}

                       = 427.5

Hence, we will calculate the critical crack length as follows.

      a = \frac{1}{\pi} \times (\frac{K_{IC}}{\sigma_{f}})^{2}

        = \frac{1}{3.14} \times (\frac{24.1}{427.5})^{2}

       = 10.13 \times 10^{-4}

Therefore, largest size is as follows.

            Largest size = 2a

                                 = 2 \times 10.13 \times 10^{-4}

                                 = 20.26 \times 10^{-4}

Thus, we can conclude that the critical crack length for a through crack contained within the given plate is 20.26 \times 10^{-4}.

4 0
3 years ago
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