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3 years ago

How do the parallel magnetic “stripes” near mid-ocean ridges form?

2 answers:

6 0

<span>When lava erupts from mid-ocean ridges, it cools rapidly into rock. As it does, it becomes magnetised permanently in the direction of the Earth's magnetic field.</span>

MakcuM [25]3 years ago

4 0

Answer:

When oceanic crust layer of earth pulls apart, magma erupts to the surface at the mid-ocean ridges and develops new layers of ocean floor.

Magma contains ferromagnetic minerals. These minerals align themselves in the direction of earth’s magnetic field reversing its polarity every now and then. When this magma cools down, it freezes in this alignment. In this way first strip is developed.

Later when the earth magnetic field changes its polarity again, the next strip is formed but has opposite polarity to the first strip.

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1. The mass defect of iron-56 is 0.52875 amu. What is the energy equivalent of this mass?

Delvig [45]

Answer:

Hello! Your answer is BELOW

Explanation:

1.About 91.754% of all iron is iron-56. Of all nuclides, iron-56 has the lowest mass per nucleon. With 8.8 MeV binding energy per nucleon, iron-56 is one of the most tightly bound nuclei.

2.The atomic weight of lead is quite variable in nature because the three heaviest isotopes are the stable end-products of the radioactive decay of uranium (238U to 206Pb and 235U to 207Pb) and thorium (232Th to 208Pb).

3.Mass defect for uranium-238 is 3.983 × 10-25 kg.

4.Energy and Mass Are Relative

The equation E = mc^2 states that the amount of energy possessed by an object is equal to its mass multiplied by the square of the speed of light.

Hope I helped! Ask me anything if you have any questions. Brainiest plz!♥ Hope you make a 100%. Have a nice morning! -Amelia♥

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3 years ago

What term would you use to describe the movement of thermal energy between a person’s body and the frigid water she jumped into

inysia [295]

It's called heat transfer

3 0

4 years ago

Read 2 more answers

See The Picture And Anwser

pantera1 [17]

Answer:

$from \: first \: equation \: of \: motion \\ v = u + at \\ a = \frac{v - u}{t} \\ a = \frac{15 - 24}{12} \\ a = - 0.75 \: {ms}^{ - 2}$

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3 years ago

Read 2 more answers

PLZ help asap :-/<br>............................

Misha Larkins [42]

Explanation:

$\underline{\boxed{\large{\bf{Option \; A!! }}}}$

Here,

$\rm { R_1}$ = 2Ω
$\rm { R_2}$ = 2Ω
$\rm { R_3}$ = 2Ω
$\rm { R_4}$ = 2Ω

We have to find the equivalent resistance of the circuit.

Here, $\rm { R_1}$ and $\rm { R_2}$ are connected in series, so their combined resistance will be given by,

$\longrightarrow \rm { R_{(1,2)} = R_1 + R_2} \\$

$\longrightarrow \rm { R_{(1,2)} = (2 + 2) \; Omega} \\$

$\longrightarrow \rm { R_{(1,2)} = 4 \; Omega} \\$

Now, the combined resistance of $\rm { R_1}$ and $\rm { R_2}$ is connected in parallel combination with $\rm { R_3}$ , so their combined resistance will be given by,

$\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \dfrac{1}{R_{(1,2)}} + \dfrac{1}{R_3} } \\$

$\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1}{4} + \dfrac{1}{2} \Bigg ) \;\Omega} \\$

$\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1 + 2}{4} \Bigg ) \;\Omega} \\$

$\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{3}{4} \Bigg ) \;\Omega} \\$

Reciprocating both sides,

$\longrightarrow \rm {R_{(1,2,3)}= \dfrac{4}{3} \;\Omega} \\$

Now, the combined resistance of $\rm { R_1}$ , $\rm { R_2}$ and $\rm { R_3}$ is connected in series combination with $\rm { R_4}$ . So, equivalent resistance will be given by,

$\longrightarrow \rm {R_{(1,2,3,4)}= R_{(1,2,3)} + R_4} \\$

$\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4}{3} + 2 \Bigg ) \; \Omega} \\$

$\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4 + 6}{3} \Bigg ) \; \Omega} \\$

$\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{10}{3} \Bigg ) \; \Omega} \\$

$\longrightarrow \bf {R_{(1,2,3,4)}= 3.33 \; \Omega} \\$

Henceforth, Option A is correct.

$\underline{\boxed{\large{\bf{Option \; B!! }}}}$

Here, we have to find the amount of flow of current in the circuit. By using ohm's law,

$\longrightarrow$ V = IR

$\longrightarrow$ 3 = I × 3.33

$\longrightarrow$ 3 ÷ 3.33 = I

$\longrightarrow$ 0.90 Ampere = I

Henceforth, Option B is correct.

$\tt \purple{Hope \; it \; helps \; you, Army! \heartsuit } \\$

4 0

3 years ago

Suppose, you are given three capacitors C1 = 2.0μF, C2 = 4.0μF, and C3 = 6.0μF.

Ymorist [56]

Answer:

1.a) Cmax = 12μF

b) Cmin = 1.09μF

2.a) E1/E2 = 2

b) E1/E2 = 0.5

Explanation:

For the maximum Capacitance, we have to connect them in parallel. In this configuration, Ct = C1 + C2 + C3 = 12μF

For the minumum Capacitance, we have to connect them in parallel. In this configuration, $Ct = (C1^{-1} + C2^{-1} + C3^{-1})^{-1}$ = 1.09μF

To calculate the energies:

For the minimum capacitance configuration, the charge is the same on all of the capacitors, so:

$E1 / E2 = \frac{1/2*Q^2/C1}{1/2*Q^2/C2} = C2 / C1 = 2$

For the maximum capacitance configuration, the voltage is the same on all of the capacitors, so:

$E1 / E2 = \frac{1/2*C1*V^2}{1/2*C2*V^2} = C1 / C2 = 0.5$

5 0

3 years ago