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3 years ago

How do the three types of boundary's work together to keep the Earth at equilibrium?

Physics

2 answers:

Maurinko [17]3 years ago

8 0

There are three types: divergent, convergent, and transform boundaries. I hope this helps.

icang [17]3 years ago

4 0

Answer:

Divergent , convergent and transform boundaries.

Explanation:

The three types of the boundaries that are formed on the earth are that of the divergent, convergent and the transform boundary that forms over the earth's surface due to the presence of the tectonic plate. Thus all these boundaries work in a close cyclic manner as the formation of the newer plate is balanced by the depletion of the old ones.

The rocks that are formed in the divergent boundaries of the ocean are carried away from there a source to the shoes and the plates that collide with each other undergo subduction and hence they are again regenerated n the geosphere.

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Use this formula to solve this problem:

Maslowich

Well, you gave us the formula to calculate power from work and time,
but you didn't give us the formula for work. We have to know that.

   Work = (force) x (distance)

The work to raise Sara to the top of the hill is

   Work = (300 N) x (15 meters)

   = 4,500 newton-meters = 4,500 joules .

Now we're ready to use the formula that you gave us. (Thank you.)

Power = (work) / (time)

= (4,500 joules) / (10 seconds)

450 joules/second = 450 watts.

6 0

3 years ago

Read 2 more answers

A basketball player can leap upward 0.43 m. how long does he remain in the air? use an acceleration due to gravity of 9.80 m/s2

MArishka [77]

From the equations of linear motion,
v² = u² + 2as where v is the final velocity, u is the initial velocity and a is the gravitational acceleration, and s is the displacement,
Thus, v² = u² -2gs, but v=0
hence, u² = 2gs
= 2×9.81×0.43
= 8.4366
u = √8.4366
=2.905 m/s
Hence the initial velocity is 2.905 m/s
Then using the equation v= u +gt .
Therefore, v = u -gt. (-g because the player is jumping against the gravity)
but, v = 0
Thus, u= gt
Hence, t = u/g
= 2.905/9.81
= 0.296 seconds

3 0

3 years ago

You drop a steel ball bearing, with a radius of 2.40 mm, into a beaker of honey. Note that honey has a viscosity of 6.00 Pa/s an

Stells [14]

Answer:

The “terminal speed” of the ball bearing is 5.609 m/s

Explanation:

Radius of the steel ball R = 2.40 mm

Viscosity of honey η = 6.0 Pa/s

$\text { Viscosity has Density } \sigma=1360 \mathrm{kg} / \mathrm{m}^{3}$

$\text { Steel has a density } \rho=7800 \mathrm{kg} / \mathrm{m}^{3}$

$\left.\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2} \text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^{2} \text { on Earth }\right)$

While calculating the terminal speed in liquids where density is high the stokes law is used for viscous force and buoyant force is taken into consideration for effective weight of the object. So the expression for terminal speed (Vt)

$V_{t}=\frac{2 \mathrm{R}^{2}(\rho-\sigma) \mathrm{g}}{9 \eta}$

Substitute the given values to find "terminal speed"

$\mathrm{V}_{\mathrm{t}}=\frac{2 \times 0.0024^{2}(7800-1360) 9.8}{9 \times 6}$

$\mathrm{V}_{\mathrm{t}}=\frac{0.0048 \times 6440 \times 9.8}{54}$

$\mathrm{V}_{\mathrm{t}}=\frac{302.9376}{54}$

$\mathrm{V}_{\mathrm{t}}=5.609 \mathrm{m} / \mathrm{s}$

The “terminal speed” of the ball bearing is 5.609 m/s

7 0

3 years ago

The amount of the lighted side of the moon you can see is the same during

attashe74 [19]

Answer:

The amount of the lighted side of the moon you can see is the same during "how much of the sunlit side of the moon faces Earth".

Explanation:

The Moon is in sequential rotation with Earth, and thus displays the Sun, the close side, always on the same side. Thanks to libration, Earth can display slightly greater than half (nearly 59 per cent) of the entire lunar surface.

The side of the Moon facing Earth is considered the near side, and the far side is called the reverse. The far side is often referred to as the "dark side" inaccurately but it is actually highlighted as often as the near side: once every 29.5 Earth days. During the New Moon the near side becomes blurred.

5 0

3 years ago

A centrifuge in a medical laboratory rotates at an angular speed of 3,500 rev/min, clockwise (when viewed from above). When swit

Ratling [72]

Answer:

The magnitude of angular acceleration is $232.38\ rad/s^2$ .

Explanation:

Given that,

Initial angular velocity, $\omega_i=3500\ rev/min=366.5\ rad/s$

When it switched off, it comes o rest, $\omega_f=0$

Number of revolution, $\theta=46=289.02\ rad$

We need to find the magnitude of angular acceleration. It can be calculated using third equation of rotational kinematics as :

$\omega_f^2-\omega_i^2=2\alpha \theta\\\\\alpha =\dfrac{-\omega_i^2}{2\theta}\\\\\alpha =\dfrac{-(366.51)^2}{2\times 289.02}\\\\\alpha =-232.38\ rad/s^2$

So, the magnitude of angular acceleration is $232.38\ rad/s^2$ . Hence, this is the required solution.

6 0

3 years ago