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3 years ago

Please help This very important for me you will get 40 points if you help me and explain in deatil please.

2 answers:

4 0

Well, I think mostly it is oceanic crust well because <span>The scientist noticed a symmetrical pattern of positive and negative magnetic lines as they moved along the ocean floor, and the line of symmetry was at the mid-ocean ridge. that why it is seen as the youngest on our earth</span>

frosja888 [35]3 years ago

3 0

The uppermost layer of the oceanic portion of a tectonic plate. The crust overlies the solidified and uppermost layer of the mantle. The crust and the solid mantle layer together constitute oceanic lithosphere.
Oceanic crust is the result of erupted mantle material originating from below the plate, cooled and in most instances, modified chemically by seawater. This occurs mostly at mid-ocean ridges, but also at scattered hotspots, and also in rare but powerful occurrences known as flood basalt eruptions. It is primarily composed of mafic rocks, or sima, which is rich in iron and magnesium. It is thinner than continental crust, or sial, generally less than 10 kilometers thick; however it is denser, having a mean density of about 2.9 grams per cubic centimeter as opposed to continental crust which has a density of about 2.7 grams per cubic centimeter.

(Copied and pasted because my friend wrote this on FaceBook. Hopefully this helps!)

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A plane startling from rest accelerates at 3 m/s Square. Calculate the increase in velocity after 25 sec

Tatiana [17]

Answer:

75 m/s

Explanation:

We can apply motion equations here

V = U + a * t

V = velocity @ t time

U = initial velocity

a = acceleration

t = time taken

V = U + a * t

V = 0+ 3 * 25

V = 75 m/s

After 25 seconds , subjected to the given acceleration velocity is increased from 0 to 75 m/s

3 0

3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 79.6 m/s at ground level.

Dimas [21]

Answer:

The rocket is motion above the ground for 44.7 s.

Explanation:

The equations for the height of the rocket are as follows:

y = y0 + v0 · t + 1/2 · a · t²

and, after the engine fails:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the rocket at time t

y0 = initial height

v0 = initial speed

t= time

a = acceleration due to the engines

g = acceleration due to gravity

First, let´s calculate the time the rocket is being accelerated by the engines:

(The center of the framer of reference is located at the ground, y0 = 0).

y = y0 + v0 · t + 1/2 · a · t²

1190 m = 0 m + 79.6 m/s · t + 1/2 · 4.10 m/s² · t²

0 = 2.05 m/s² · t² + 79.6 m/s · t - 1190 m

Solving the quadratic equation:

t = 11.5 s (the other value of t is discarded because it is negative).

At that time, the engines fail and the rocket starts to fall. The equation of the height will be:

y = y0 + v0 · t + 1/2 · g · t²

The initial velocity (v0) will be the velocity acquired during 11.5 s of acceleration plus the initial velocity of launch:

v = v0 + a · t

v = 79.6 m/s + 4.10 m/s² · 11.5 s

v = 126.8 m/s

Now, we can calculate the time it takes the rocket to reach the ground (y = 0) from a height of 1190 m:

y = y0 + v0 · t + 1/2 · g · t²

0 m = 1190 m + 126.8 m/s · t - 1/2 · 9.80 m/s² · t²

Solving the quadratic equation:

t = 33.2 s

Then, the total time the rocket is in motion is (33.2 s + 11.5 s) 44.7 s

5 0

3 years ago

In which scenario will the two objects have the greatest gravitational force

Brilliant_brown [7]

Answer: I think the answer C

Explanation:

7 0

2 years ago

Read 2 more answers

Consider a model of a hydrogen atom in which an electron is in a circular orbit of radius r = 5.92×10−11 m around a stationary p

DaniilM [7]

Answer:

2.068 x 10^6 m / s

Explanation:

radius, r = 5.92 x 10^-11 m

mass of electron, m = 9.1 x 10^-31 kg

charge of electron, q = 1.6 x 10^-19 C

As the electron is revolving in a circular path, it experiences a centripetal force which is balanced by the electrostatic force between the electron and the nucleus.

centripetal force = $\frac{mv^{2}}{r}$

Electrostatic force = $\frac{kq^{2}}{r^{2}}$

where, k be the Coulombic constant, k = 9 x 10^9 Nm^2 / C^2

So, balancing both the forces we get

$\frac{kq^{2}}{r^{2}}=\frac{mv^{2}}{r}$

$v=\sqrt{\frac{kq^{2}}{mr}}$

$v=\sqrt{\frac{9\times 10^{9}\times1.6\times 10^{-19}\times 1.6\times 10^{-19}}{9.1\times 10^{-31}\times 5.92\times10^{-11}}}$

v = 2.068 x 10^6 m / s

Thus, the speed of the electron is give by 2.068 x 10^6 m / s.

6 0

3 years ago

A ball of moist clay falls 15.0 m to the ground. It is in contact with the ground for 20.0 ms before stopping. (a) What is the m

rosijanka [135]

Answer:

a= -0.86 m/s²

The negative sign shows that ball down the ground or moving down

Explanation:

Vf² - Vo² = 2gS

where

Vf = velocity of clay as it hits the ground

Vo = initial velocity of clay = 0

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

S = distance travelled by clay = 15 m

Substituting appropriate values,

Vf² - 0 = 2(9.8)(15)

Vf = 17.15 m/sec.

Formula to use is,

V - Vf = aT

where

V = velocity of clay when it stops = 0

Vf = 17.15 m/sec (as determined above)

a = acceleration

T = 20 ms

Put the values to find acceleration

a=(V-Vf)/T

a=(0-17.15)/20

a= -0.86 m/s²

The negative sign shows that ball down the ground

3 0

3 years ago