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kifflom [539]
2 years ago
15

Earth's core is the source of the energy that drives the movement of tectonic

Physics
2 answers:
Ostrovityanka [42]2 years ago
5 0

Answer:

B. Convection

D. Conduction

Explanation:

Conduction and convection are the two most prominent processes that helps transfer energy outward to the earth's crust.

  • Energy within the core is a function of the radioactive decay and frictional heating.
  • Also, heat that accreted during the formation of the earth is a significant source of internal energy.
  • The heat is conducted away by the process of convection. This is possible due to temperature differences between different parts of the earth
  • Conduction is made made possible due to the metallic bodies in the core and other part of the inner earth.
dezoksy [38]2 years ago
4 0

Answer:

B and D

Explanation:

I just did it on a p e x. In case your not sure

You might be interested in
At what common temperature will a block of wood and a block of metal both feel neither hot nor cold to the touch ?
Anastaziya [24]

When you touch an object and heat flows OUT of it, INTO your finger, you say the object feels hot.

When you touch an object and heat flows INTO it, OUT of your finger, you say the object feels cold.

If the object has the same temperature as your finger ... <em>around the mid-90s</em> ... then no heat flows in or out of your finger when you touch the object, and the object doesn't feel hot or cold.

6 0
3 years ago
What frequency fapproach is heard by a passenger on a train moving at a speed of 18.0 m/s relative to the ground in a direction
Sergio039 [100]

Answer:

The frequency is 302.05 Hz.

Explanation:

Given that,

Speed = 18.0 m/s

Suppose a train is traveling at 30.0 m/s relative to the ground in still air. The frequency of the note emitted by the train whistle is 262 Hz .

We need to calculate the frequency

Using formula of frequency

f'=f(\dfrac{v+v_{p}}{v-v_{s}})

Where, f = frequency

v = speed of sound

v_{p} = speed of passenger

v_{s} = speed of source

Put the value into the formula

f'=262\times(\dfrac{344+18}{344-30})

f'=302.05\ Hz

Hence, The frequency is 302.05 Hz.

7 0
3 years ago
A satellite is in a circular orbit 21000 km above the Earth’s surface; i.e., it moves on a circular path under the influence of
mina [271]

Answer:

(orbital speed of the satellite) V₀ = 3.818 km

Time (t) = 4.5 × 10⁴s

Explanation:

Given that:

The radius of the Earth is 6.37 × 10⁶ m;    &

the acceleration of gravity at the satellite’s altitude is 0.532655 m/s

We can calculate the orbital speed of the satellite by using the formula:

Orbital Speed (V₀) = √(r × g)

radius of the orbit (r) = 21000 km + 6.37 × 10⁶ m

                                  = (2.1 × 10⁷ + 6.37 × 10⁶) m

                                  = 27370000

                                  = 2.737 × 10⁷m

Orbital Speed (V₀) = √(r × g)

Orbital Speed (V₀) = √(2.737 × 10⁷  × 0.532655 )

                              = 3818.215

                              = 3.818 × 10³

                             = 3.818 Km

To find the time it takes to complete one orbit around the Earth; we use the formula:

Time (t) = 2 π × \frac{r}{V_o}

            = 2 × 3.14 × \frac{2.737*10^7}{3.818*10^3}

            = 45019.28

            = 4.5 × 10 ⁴ s

6 0
3 years ago
Joe first focuses his attention (and his eyes) on the tree. The focal length of the cornea-lens system in his eye must be ______
kotykmax [81]

Answer: The focal length of the cornea-lens system in his eye must be LESS THAN the distance between the front and back of his eye.

Explanation:

The human eye the front part of the eye is the CORNEA. This is the tough white transparent part of the eye that helps in the refraction of light rays. While the backside of the eye is the RETINA. This is the part of the eye when images are focused.

When a normal eye is at rest, parallel rays from a distant object are focused on the retina. The ability of the eye - lens to focus points at different distances on the retina is known as accomodation. The adjustment of the eye lens to focus objects of varying distances is brought about by the ciliary muscles. The have the ability to change the shape of the eye which leads to change in focal length.

When a person with normal vision looks at a distant object at infinity, the lens brings parallel rays to focus on the retina. Thus, the furthest point which the eye can see distinctly is called the far point of the eye and it's infinity for a normal eye. But Joe was able to focus his eye on the tree, meaning that the tree was within his near point. This is the nearest point at which an object is clearly seen. Therefore, when the effective focal length of the cornea-lens system changes, it changes the location of the image of any object in one's field of view.

5 0
3 years ago
PLSLPSLLPSLPSLPS HELP I ONLY HAVE 5 MINS PLEASE
UkoKoshka [18]

Answer:

0.19m/s²

Explanation:

Initial velocity(u) = 50×1000/60×60

=13.88 m/s

Final velocity(v) = 36.5×1000/60×60

=10.13 m/s

Acceleration(a) = v-u/t

=10.13-13.88/19.5

a= -0.19m/s²

-a = 0.19m/s²

The magnitude of retar dation is 0.19m/s²

5 0
2 years ago
Read 2 more answers
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