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2 years ago

What is the result of seafloor spreading?

Physics

1 answer:

Over [174]2 years ago

6 0

Answer:

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. The less-dense material rises, often forming a mountain or elevated area of the seafloor. Eventually, the crust cracks.

Explanation:

eventually the crust cracks.

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Plz help me plz its in the picture

natka813 [3]

Il write down the answers of the blanks which are as follows:-
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Hope this helps!

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3 years ago

Electromagnetic induction means that moving a magnet through a loop of wire creates an electric current.

MAVERICK [17]

False. An Electromagnetic induction is the production of an electromotive force across an electrical conductor in a changing magnetic field.

4 0

3 years ago

Read 2 more answers

How to find acceleration?

Yanka [14]

Acceleration = Change in Velocity / time

a = (v - u) / t

Where v = final velocity in m/s
u = initial velocity in m/s
t = time in seconds.
a = acceleration in m/s²

A proper record of the changes in velocity with the corresponding time would help find the acceleration.

4 0

3 years ago

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A hydrogen atom in a galaxy moving with a speed of 6.65×106 m/???? away from the Earth emits light with a wavelength of 5.13×10−

Mumz [18]

Answer:

The observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ .

Explanation:

Given that,

The actual wavelength of the hydrogen atom, $\lambda_a=5.13\times 10^{-7}\ m$

A hydrogen atom in a galaxy moving with a speed of, $v=6.65\times 10^6\ m/s$

We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

$v=c\times \dfrac{\lambda_o-\lambda_a}{\lambda_a}$

$\lambda_o$ is the observed wavelength

$\lambda_o=\dfrac{v\lambda_a}{c}+\lambda_a\\\\\lambda_o=\dfrac{6.65\times 10^6\times 5.13\times 10^{-7}}{3\times 10^8}+5.13\times 10^{-7}\\\\\lambda_o=5.24\times 10^{-7}\ m$

So, the observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ . Hence, this is the required solution.

8 0

3 years ago

On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a 6 iron. The acceleration due to gravity on t

kozerog [31]

Answer:

a) 6 times farther. b) 6 times longer.

Explanation:

Once released, in the horizontal direction, no other forces act on the ball, so it continues moving at the same initial velocity, which is given by the projection of the velocity vector in the horizontal direction, as follows:

vₓ = v* cos (25º) = 23 m/s * 0.906 = 20.8 m/s

In the vertical direction, the initial velocity is the projection of the velocity vector along the vertical axis, as follows:

vy = v* sin (25º) = 23 m/s * 0.422 = 9.72 m/s

Assuming that the acceleration is constant, and equal to 1/6*g, we can calculate the total time of flight, with the following kinematic equation for the vertical displacement:

$y = voy*t - (\frac{1}{2}*\frac{g}{6} * t^{2} )$

If the total displacement in the vertical direction is 0 (which means that the time if the total time of flight), we can solve for t, as follows:

$t = \frac{voy*12}{g} = \frac{9.72 m/s*12}{9.8m/s2} = 11. 9 s$

On earth, this time could be calculated in the same way:

$t = \frac{voy*12}{g} = \frac{9.72 m/s*2}{9.8m/s2} = 1.98 s$

As the time is defined by the vertical movement, we can find the horizontal distance travelled on the moon, as follows:

Δx = v₀ₓ * t = 20.8 m/s * 11. 9 s = 248.1 m

On earth, the distance travelled had been as follows:

Δx = v₀ₓ * t = 20.8 m/s * 1.98 s = 41.3 m

⇒ Δx(moon) / Δx(earth) = 248.1 / 41.3 = 6.00

b) As we have just found, the time of flight on the moon and on the earth are as follows:

tmoon = 11. 9 s

tearth = 1.98 s

⇒ t(moon) / t(earth) = 11.9 / 1.98 = 6.0

8 0

3 years ago