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Travka [436]
4 years ago
13

How can global warming lead to changes to the Earth’s surface? a. Global warming can lead to an increased number of earthquakes,

which change the Earth’s surface. b. Global warming can lead to glaciers melting, causing flooding to areas and the decrease of glacial land masses. c. Global warming leads to a decrease in water levels of coastal wetlands. d. Global warming cannot lead to changes to the Earth’s surface.
Physics
1 answer:
Nutka1998 [239]4 years ago
5 0

Answer:

Option:  b. Global warming can lead to glaciers melting, causing flooding to areas and the decrease of glacial land masses.

Explanation:

Global warming is the reason for the changes in environment and climate on earth. Melting of glaciers leads to an increase in water level and a decrease in landmass. One of the most climactic consequences is the decrease in Arctic sea ice. Melting polar ice along with ice sheets and glaciers across Greenland, North America, Europe, Asia, and South America suspected to increase sea levels slowly. There is an increase in the glacial retreat due to global warming, which leaves rock piles that covered with ice.  

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When an object is heated it …………… and when it cools it …………………..
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When an object is heated it Expand and when it cools it Contract

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Assume light is traveling through an optical medium of n1 and is incident on a boundary with a different optical medium with n2.
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In addition to light and chlorophyll photosynthesis requires
labwork [276]

Answer:

Water (H₂O) and Carbon dioxide (CO₂)

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Read 2 more answers
A ball is hit 1 meter above the ground with an initial speed of 40 m/s. If the ball is hit at an angle of 30 above the horizonta
Vika [28.1K]

Answer:

t_t=4.131\ s

Explanation:

Given:

height above the horizontal form where the ball is hit, y=1\ m

angle of projectile above the horizontal, \theta=30^{\circ}

initial speed of the projectile, u=40\ m.s^{-1}

<u>Firstly we find the </u><u>vertical component of the initial velocity</u><u>:</u>

u_y=u.\sin\theta

u_y=40\times \sin30^{\circ}

u_y=20\ m.s^{-1}

During the course of ascend in height of the ball when it reaches the maximum height then its vertical component of the velocity becomes zero.

So final vertical velocity during the course of ascend:

  • v_y=0\ m.s^{-1}

Using eq. of motion:

v_y^2=u_y^2-2g.h (-ve sign means that the direction of velocity is opposite to the direction of acceleration)

0^2=20^2-2\times 9.8\times h

h=20.4082\ m (from the height where it is thrown)

<u>Now we find the time taken to ascend to this height:</u>

v_y=u_y-g.t

0=20-9.8t

t=2.041\ s

<u>Time taken to descent the total height:</u>

  • we've total height, h'=h+y =20.4082+1

h'=u_y'.t'+\frac{1}{2} g.t'^2

  • during the course of descend its initial vertical velocity is zero because it is at the top height, so u_y'=0\ m.s^{-1}

21.4082=0+4.9t'^2

t'=2.09\ s

<u>Now the total time taken by the ball to hit the ground:</u>

t_t=t'+t

t_t=2.09+2.041

t_t=4.131\ s

3 0
3 years ago
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