All categories

3 years ago

Which sphere of Earth is associated with plate tectonics?

Physics

2 answers:

SOVA2 [1]3 years ago

8 0

Answer:

D. Lithosphere

Explanation:

Hope this helps!

Marrrta [24]3 years ago

5 0

When looking at this question, we can easily start by eliminating certain answers. In the selections you've provided, you've shown atmosphere. We can easily eliminate letter A, as that makes absolutely no sense. Moving on, you also eliminate letter B, as that deals with ecosystems and whatnot. And finally, you can eliminate hydrosphere, letter C - as that's not the same. That deals with water, like oceans or rivers.

That leaves you with D) Lithosphere for your answer. The Lithosphere is the rigid part of the earth, the outermost layer, I would say. The crust / mantle. That's why it would be letter D - plate tectonics seem to have relations with the Lithosphere. The lithosphere is affected.

You might be interested in

A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region

zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

$K_{1} = K_{2} + W_{f}$

Where:

$K_{1}$ , $K_{2}$ are the initial and final translational kinetic energies of the tobbogan, measured in joules.

$W_{f}$ - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

$f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Distance travelled by the toboggan in the rough region, measured in meters.

$m$ - Mass of the toboggan, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

If $m = 375\,kg$ , $v_{1} = 4.50\,\frac{m}{s}$ , $v_{2} = 1.20\,\frac{m}{s}$ and $\Delta s = 5.40 \,m$ , then:

$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

$f = 653.125\,N$

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

$\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%$

$\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%$

$\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%$

$\%K_{loss} = 92.889\,\%$

The rough region reduced the kinetic energy of the toboggan in 92.889 %.

c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

$\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%$

$\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%$

$\%v_{loss} = 73.333\,\%$

The speed of the toboggan is reduced in 73.333 %.

5 0

2 years ago

What is the minimum runway length that will serve? hint: you can solve this problem using ratios without having any additional i

ehidna [41]

There are many factors that determine if an aircraft can operate from a given airport. Of course the availability of certain services, such as fuel, access to air stairs and maintenance are all necessary. But before considering anything else, one must determine if the plane can physically land at an airport, and equally as important, take off.

What is the minimum runway length that will serve?

Looking at aerial views of runways can lead some to the assumption that they are all uniform, big and appropriate for any plane to land. This couldn’t be further from the truth.

A given aircraft type has its own individual set of requirements in regards to these dimensions. The classic 150’ wide runway that can handle a wide-body plane for a large group charter flight isn’t a guarantee at every airport. Knowing the width of available runways is important for a variety of reasons including runway illusion and crosswind condition.

Runways also have different approach categories based on width, and have universal threshold markings that indicate the actual width.

To learn more about runway

brainly.com/question/11553726

#SPJ4

6 0

1 year ago

If, while standing on the bank of a stream, you wished to spear a fish swimming in the water out in front of you, would you aim

Serggg [28]

Answer:

a) below the observed position

b) directly at the observed position

Explanation:

If I'm standing on the bank of a stream, and I wish to spear a fish swimming in the water out in front of me, I would aim below the observed fish to make a direct hit. This is because the phenomenon of refraction of light in water causes the light coming from the fish is refract away from the normal as it passes into the air and into my eyes.

If I'm to zap the fish with a taser, I would aim directly at the observed fish because the laser (a form of concentrated light waves) will refract into the water, taking the same path the light from the fish took to get to my eyes.

3 0

2 years ago

Could anyone help with this question I do

ra1l [238]

which of those can we not get back once it has been used, the answer is oil or petroleum

8 0

3 years ago

Read 2 more answers

A small car with mass m and speed 2v and a large car with mass 2m and speed v both travel the same circular section of an unbank

Alenkasestr [34]

Answer:

(D) F/2

Explanation:

Since the circular section is unbanked, the centripedal acceleration acting on each of the cars is

$a_s = \frac{v_s^2}{r} = \frac{(2v)^2}{r} = \frac{4v^2}{r}$

$a_l = \frac{v_l^2}{r} = \frac{v^2}{r}$

Therefore the centripetal force on each car

$F_s = m_sa_s = \frac{4mv^2}{r}$

$F_l = m_la_l = \frac{2mv^2}{r}$

Since $F_s = 2F_l$ this means the friction force required to keep the large car on the road is only half of the friction force required to keep the small car on road

So (D) F/2 is the correct answer

5 0

3 years ago