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3 years ago

15

What type of cooling would a scientist determine formed an igneous rock found with large crystals? Slow cooling Medium rate cool

ing Quick cooling Super-fast cooling

2 answers:

Dmitriy789 [7]3 years ago

5 0

Slower cooling engenders the growth of larger crystals in igneous rocks, thus, your answer should be slow cooling!

Hope this helped!

grin007 [14]3 years ago

4 0

Your answer would be (A.) <em>slow cooking. </em>

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A 0.106-A current is charging a capacitor that has square plates 6.00 cm on each side. The plate separation is 4.00 mm. (a) Find

FrozenT [24]

Answer:

The time rate of change of flux is $1.34 \times 10^{10}$ $\frac{V}{s}$

Explanation:

Given :

Current $I = 0.106$ A

Area of plate $A = 36 \times 10^{-4}$ $m^{2}$

Plate separation $d = 4 \times 10^{-3}$ m

(A)

First find the capacitance of capacitor,

$C = \frac{\epsilon _{o} A }{d}$

Where $\epsilon _{o} = 8.85 \times 10^{-12}$

$C = \frac{8.85 \times 10^{-12 } \times 36 \times 10^{-4} }{4 \times 10^{-3} }$

$C = 7.9 \times 10^{-12}$ F

But $C = \frac{Q}{V}$

Where $Q = It$

$C = \frac{It}{V}$

$V = \frac{It}{C}$

Now differentiate above equation wrt. time,

$\frac{dV}{dt} = \frac{I}{C}$

$= \frac{0.106}{7.9 \times 10^{-12} }$

$= 1.34 \times 10^{10}$ $\frac{V}{s}$

Therefore, the time rate of change of flux is $1.34 \times 10^{10}$ $\frac{V}{s}$

8 0

3 years ago

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n200080 [17]

Answer:

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Explanation:

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3 years ago

TRUE OR FALSE! PLZ HELP

Ksju [112]

Answer:

True

Explanation:

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2 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

3 years ago

Use the mass spectrum of europium to determine the atomic mass of europium. where the peak representing eu-151 has an exact mass

slavikrds [6]

Answer: The atomic mass of a Europium atom is 151.96445 amu.

From the given information:

Percent intensity is 91.61% of Europium atom of molecular weight 150.91986 amu.

Percent intensity is 100.00% of Europium atom of molecular weight 152.92138 amu.

Abundance of Eu-151 atom:

$X_{Eu-151}=\frac{0.9161}{0.9161+1.000}=0.4781$

Abundance of Eu-153 atom:

$X_{Eu-153}=\frac{1.000}{0.9161+1.000}=0.5219$

Atomic mass of Europium atom:

$A=(X_{Eu-151}\times150.91986+X_{Eu-153}\times152.92138)amu\\A=(0.4781\times150.91986+0.5219\times152.92138)amu=151.96445 amu$

Therefore, the atomic mass of a Europium atom is 151.96445 amu.

3 0

3 years ago

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