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3 years ago

14

How does sea floor spreading provide an explanation for how continents may move over earths surface?(I know this isn't Physics,

but there was no Earth Science option)

1 answer:

Neporo4naja [7]3 years ago

3 0

Alfred Wegner first proposed the idea that all the seven continents were once one super continent called "Pangea". Everyone considered his idea as a joke until he proved his theory with evidence of matching fossils, matching mountain ranges, and how all the continents fit together like puzzle pieces. Seafloor spreading is a process that occurs at mid-ocean ridges, where new oceanic crust is formed through volcanic activity and then gradually moves away from the ridge. Seafloor spreading helps explain continental drift in the theory of plate tectonics.

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Please Help Meee.. Will give brainlest

anygoal [31]

Answer:

Total resistance is 19.6 Ω for the circuit .

Explanation:

Given:

Six resistors of a circuit namely $R_1$ to $R_6$ .

Where

$R_1$ and $R_2$ are in series.

$R_3$ and $R_4$ are in parallel.

$R_5$ and $R_6$ are in parallel.

We know that continuous resistor which are in series are added up to find the equivalent resistance.

Similarly resistors which are arranged in parallel their equivalent resistance is $\frac{R_3R_5}{R_3+R_5}$ for the case of $R_3$ and $R_5$ .

According to the question:

Total resistance = R(equivalent) :

⇒ $R_(eq_) = R_1+R_2+(\frac{R_3R_4}{R_3+R_4} )+(\frac{R_5R_6}{R_5+R_6} )$

⇒ $R_(eq_)=7+8+(\frac{4\times 6}{4+6} )+(\frac{3\times 8}{3+8} )$

⇒ $R_(eq_)=7+8+(\frac{24}{10} )+(\frac{24}{11} )$

⇒ $R_(eq_)=15+(2.4 )+(2.18)$

⇒ $R_(eq_)=19.58$ Ω

So the total resistance of the circuit depicted is 19.58 Ω approximated to nearest tenth that is 19.6 Ω

6 0

4 years ago

A thin uniform rod of mass M and length L is bent at its center so that the two segments are perpendicular to each other. Find i

serg [7]

Answer:

$\frac{1}{12}ML^2$

Explanation:

The moments of the whole object is the sum of the moments of the 2 segments of rod at their ends of which length is L/2 and mass M/2:

$I = 2I_{end} = 2\frac{1}{3}\frac{M}{2}\left(\frac{L}{2}\right)^2$

$I = \frac{1}{3}M\frac{L^2}{4}$

$I = \frac{1}{12}ML^2$

5 0

3 years ago

Current is constant at all points in a parallel circuit.<br><br> True<br> False

Sergio [31]

This is false. Current is the speed of the charge, 1 amp of current is 1 coulomb per second. So you can imagine the current of a circuit as the current of a river. In a parallel circuit, the river breaks into two separate streams. Some of the water goes down one river, some goes down the other. However, the total amount of water/coulombs never changes. This means that some of the total current will go down one river, and one the other. However, with less coulombs now the current will decrease.

Long story short, since there are two paths, the charge will split and depending on the resistance of each parallel stream a different amount of charge will go down each branch.

5 0

3 years ago

What force does the water exert (in addition to that due to atmospheric pressure) on a submarine window of radius 44.0 cm at a d

Butoxors [25]

Calculate the pressure due to sea water as density*depth.
That is,
pressure = (1025 kg/m^3)*((9400 m)*(9.8 m/s^2) = 94423000 Pa = 94423 kPa

Atmospheric pressure is 101.3 kPa
Total pressure is 94423 + 101.3 = 94524 kPa (approx)

The area of the window is π(0.44 m)^2 = 0.6082 m^2

The force on the window is
(94524 kPa)*(0.6082 m^2) = 57489.7 kN = 57.5 MN approx

3 0

4 years ago

You blow across the open mouth of an empty test tube and producethe fundamental standing wave of the air column inside the testt

sertanlavr [38]

Answer:

(a). The frequency of this standing wave is 0.782 kHz.

(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.

Explanation:

Given that,

Length of tube = 11.0 cm

(a). We need to calculate the frequency of this standing wave

Using formula of fundamental frequency

$f_{1}=\dfrac{v}{4l}$

Put the value into the formula

$f_{1}=\dfrac{344}{4\times0.11}$

$f_{1}=781.81\ Hz$

$f_{1}=0.782\ kHz$

(b). If the test tube is half filled with water

When the tube is half filled the effective length of the tube is halved

We need to calculate the frequency

Using formula of fundamental frequency of the fundamental standing wave in the air

$f_{1}=\dfrac{v}{4(\dfrac{L}{2})}$

Put the value into the formula

$f_{1}=\dfrac{344}{4\times\dfrac{0.11}{2}}$

$f_{1}=1563.63\ Hz$

$f_{1}=1.563\ kHz$

Hence, (a). The frequency of this standing wave is 0.782 kHz.

(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.

6 0

4 years ago