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3 years ago

Which best describes the ages of the rocks on opposite sides of and the same distance from, A Seafloor spreading center?

Physics

1 answer:

Genrish500 [490]3 years ago

4 0

Answer;

-The rocks are the same age

Explanation;

Seafloor spreading is the process by which the seafloor moves apart at mid-ocean ridges. Divergent seafloor spreading occurs at this type of plate boundary.

Seafloor spreading and other tectonic activity processes are the result of mantle convection. Seafloor spreading occurs at divergent plate boundaries. As tectonic plates slowly move away from each other, heat from the mantle’s convection currents makes the crust more plastic and less dense.

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Lora (of mass 54 kg) is an expert skier. She

omeli [17]

The mechanical energy at top =Mechanical energy at bottom

Mass=m=54kg
Height=h=51m
Acceleration due to gravity=g=10m/s^2
Velocity=v=2.6m/s

$\\ \tt\longmapsto M_{initial}=M_{Final}$

Final energy at bottom=The kinetic energy

$\\ \tt\longmapsto KE=M_{initial}$

$\\ \tt\longmapsto KE=P.E_{(Top)}+K.E_{(Top)}$

$\\ \tt\longmapsto K.E=mgh+\dfrac{1}{2}mv^2$

$\\ \tt\longmapsto K.E=m\left(gh+\dfrac{v^2}{2}\right)$

$\\ \tt\longmapsto K.E=54\left((10)(51)+\dfrac{2.6^2}{2}\right)$

$\\ \tt\longmapsto K.E=54\left(510+\dfrac{6.76}{2}\right)$

$\\ \tt\longmapsto K .E=54(510+3.38)$

$\\ \tt\longmapsto K.E=54(513.38)$

$\\ \tt\longmapsto K.E=27722.52J$

$\\ \tt\longmapsto K.E=27.7KJ$

6 0

2 years ago

Problem 9: Suppose you wanted to charge an initially uncharged 85 pF capacitor through a 75 MΩ resistor to 90.0% of its final vo

Bingel [31]

Answer:

$t=14.678\times 10^{-3}s$

Explanation:

Given:

Capacitance, C = 85 pF = 85 × 10⁻¹² F

Resistance, R = 75 MΩ = 75×10⁶Ω

Charge in capacitor at any time 't' is given as:

$Q=Q_o(1-e^{-\frac{t}{RC}})$

where,

Q₀ = Maximum charge = CE

E = Initial voltage

t = time

also, Q = CV

V= Final voltage = 90% of E = 0.9E

thus, we have

$C\times 0.9E=CE(1-e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}})$

$0.9=1-e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}}$

$e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}}=1-0.9=0.1$

taking log both sides, we get

$ln(e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}})=ln(0.1)=ln(\frac{1}{10})$

$-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}=-ln(10)$

$t=75\times 10^6\ \times\ 85\times 10^{-12}\times ln{10}$

$t=14.678\times 10^{-3}s$

3 0

3 years ago

How can we determine the strength of a sonic boom?

tino4ka555 [31]

<span>There are several factors that can influence sonic booms - weight, size, and shape of the aircraft or vehicle, plus its altitude, attitude and flight path, and weather or atmospheric conditions.</span>

8 0

3 years ago

A solenoid with 3,000.0 turns is 70.0 cm long. If its self-inductance is 25.0 mH, what is its radius? (The value of μ0 is 4π x 1

nevsk [136]

Answer:

A. 2.2*10^-2m

Explanation:

Using

Area = length x L/ uo xN²

So A = 0.7m * 25 x 10^-3H /( 4π x10^-7*

3000²)

A = 17.5*10^-3/ 1.13*10^-5

= 15.5*10^-2m²

Area= π r ²

15.5E-2/3.142 = r²

2.2*10^2m

Explanation:

5 0

3 years ago

Why don’t metals break when hit with hammer

Verizon [17]

Because metallic bonds involve all of the metal atoms in a piece of metal sharing all of their valence electrons with "delocalized" bonds.

7 0

3 years ago