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3 years ago

What do you call rocxs that form under intense heat and pressure

Physics

1 answer:

kirill [66]3 years ago

5 0

Metamorphic

Metamorphic rocks are formed under the surface of the earth from the metamorphosis (change) that occurs due to intense heat and pressure (squeezing).

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What process changes a liquid to a solid?

kipiarov [429]

Answer:

D. Freezing?

Explanation:

Get water, put it in the freezer, turns into ice after a few hours.

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2 years ago

Is the angular position of the first-order spectrum small enough for sinθ≈θ to be a good approximation?.

vesna_86 [32]

Answer:

15 to 30 so anywhere in between there.

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2 years ago

Suppose you take a piece of hard wax from an unlit candle. After you roll the wax between your fingers for a while, it becomes s

romanna [79]

Answer:

An amorphous solid does not have a definite melting point; instead, it melts gradually over a range of temperatures, because the bonds do not break all at once. This means an amorphous solid will melt into a soft, malleable state (think candle wax or molten glass) before turning completely into a liquid.

Explanation:

Hope this helps

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2 years ago

Which type of friction opposes motion?

Lady bird [3.3K]

Answer:

Friction does not oppose motion, it oppose relative motion.

Explanation:

static friction opposes relative motion

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2 years ago

Read 2 more answers

There are two identical, positively charged conducting spheres fixed in space. The spheres are 40.4 cm apart (center to center)

Nadya [2.5K]

Answer:

$q_1=5.64\times 10^{-7}\ \text{C}$ and $q_2=2.32\times 10^{-6}\ \text{C}$

Explanation:

$F_1=0.072\ \text{N}$

$F_2=0.115\ \text{N}$

r = Distance between shells = 40.4 cm

$q_1$ and $q_2$ are the charges

$k$ = Coulomb constant = $8.99\times10^{9}\ \text{Nm}^2/\text{C}^2$

Force is given by

$F_1=\dfrac{kq_1q_2}{r^2}\\\Rightarrow q_1q_2=\dfrac{F_1r^2}{k}\\\Rightarrow q_1q_2=\dfrac{0.072\times 0.404^2}{8.99\times 10^{9}}\\\Rightarrow q_1q_2=1.307\times 10^{-12}\\\Rightarrow q_1=\dfrac{1.307\times 10^{-12}}{q_2}$

$F_2=\dfrac{kq^2}{r^2}\\\Rightarrow q=\sqrt{\dfrac{F_2r^2}{k}}\\\Rightarrow q=\sqrt{\dfrac{0.115\times 0.404^2}{8.99\times 10^{9}}}\\\Rightarrow q=1.44\times 10^{-6}\ \text{C}$

$q=\dfrac{q_1+q_2}{2}\\\Rightarrow q_1+q_2=2q\\\Rightarrow q_1+q_2=2\times1.44\times 10^{-6}\\\Rightarrow q_1+q_2=2.88\times 10^{-6}$

Substituting the above value of $q_1$ we get

$\dfrac{1.307\times 10^{-12}}{q_2}+q_2=2.88\times 10^{-6}\\\Rightarrow q_2^2-2.88\times 10^{-6}q_2+1.307\times 10^{-12}=0\\\Rightarrow \frac{-\left(-0.00000288\right)\pm \sqrt{\left(-0.00000288\right)^2-4\times \:1\times \:1.307\times 10^{-12}}}{2\times \:1}\\\Rightarrow q_2=2.32\times 10^{-6}, 5.64\times 10^{-7}$

$q_1=\dfrac{1.307\times 10^{-12}}{q_2}=\dfrac{1.307\times 10^{-12}}{2.32\times 10^{-6}}\\\Rightarrow q_1=5.63\times 10^{-7}$

$q_1=\dfrac{1.307\times 10^{-12}}{q_2}=\dfrac{1.307\times 10^{-12}}{5.64\times 10^{-7}}\\\Rightarrow q_1=2.32\times 10^{-6}$

Since we know $q_1$

$q_1=5.64\times 10^{-7}\ \text{C}$ and $q_2=2.32\times 10^{-6}\ \text{C}$ .

5 0

2 years ago