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MArishka [77]
4 years ago
15

Humans are a part of the Earth system. List at least three examples of how you, in particular, influence one or more of Earth’s

major spheres
Physics
1 answer:
Naily [24]4 years ago
8 0

Answer:

True, The humans have modified their environment the ecological system

Explanation:

  • Human is a part of the earth system and they are included in the ecological system that in particular is concerned with the provision of the feedback mechanism and cyclic flow of the system of energy such as the atmosphere, cryosphere, and the biosphere and the hydrosphere.
  • The human has already and influences the earth's physical system alike that of the atmospheric system in terms of the climate changes and the global warming that is taking place at a massive scale.  
  • They are influencing the soil and land systems in terms of the depletion resources and the disturbing the hydrological cycles on earth that further influence the amounts of rainfall patterns.
  • <u>The three examples</u><u> </u><u>given above are those of the use of the natural resources and thereby depletion of resources which are needed to sustain the functioning of the planet and various fields of the economics and the geology and that of the socio systems are also impacted and these forms the earth's major sphere of life.</u>
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A boy throws a ball vertically up it returns the ground after 10 seconds find the maximum height reached by the ball
Akimi4 [234]

Answer:

Approximately 122.625\; {\rm m} (assuming that g = 9.81\; {\rm m\cdot s^{-2}}, the ball was launched from ground level, and that the drag on the ball is negligible.)

Explanation:

Let v_{0} denote the velocity at which the ball was thrown upward.

If the drag (air friction) on the ball is negligible, the ball would land with a velocity of exactly (-v_{0}). The velocity of the ball would be changed from v to (-v_{0})\! (such that \Delta v = (-v_{0}) - v_{0} = (-2\, v_{0})) within t = 10\; {\rm s}.

Also because the drag on the ball is negligible, the acceleration of the ball would be a = -g = -9.81\; {\rm m\cdot s^{-2}}. Thus:

\Delta v = a\, t = 10\; {\rm s} \times (-9.81\; {\rm m\cdot s^{-2}}) = -98.1\; {\rm m\cdot s^{-1}}.

Since \Delta v = (-2\, v_{0}):

-2\, v_{0} = \Delta v = -98.1\; {\rm m\cdot s^{-1}.

\begin{aligned}v_{0} &= \frac{-98.1\; {\rm m\cdot s^{-1}}}{-2}= 49.05\; {\rm m \cdot s^{-1}}\end{aligned}.

The ball reaches maximum height when its velocity is v_{1} = 0\; {\rm m\cdot s^{-1}}. Apply the SUVAT equation x = ({v_{1}}^{2} - {v_{0}}^{2}) / (2\, a) to find the displacement x between the original position (ground level, where v_{0} = 49.05\; {\rm m\cdot s^{-1}}) and the max-height position of the ball (where v_{1} = 0\; {\rm m\cdot s^{-1}}.)

\begin{aligned}x &= \frac{(0\; {\rm m\cdot s^{-1}})^{2} - (49.05\; {\rm m\cdot s^{-1}})^{2}}{2 \times (-9.81\; {\rm m\cdot s^{-2}})} \\ &\approx 122.625\; {\rm m\cdot s^{-1}}\end{aligned}.

7 0
2 years ago
5) A person holds a 1.7 kg bucket and lets it move down at
Elena L [17]

Answer:

12.5J

Explanation:

Given parameters:

Mass of bucket  = 1.7kg

Height  = 75cm  = 0.75m

Unknown;

Work done on the bucket by the person  = ?

Solution:

To solve this problem, we use the work done equation;

  Work done  = force x distance  = mgh

 m is the mass  

  g is the acceleration due to gravity

  h is the height

Now, insert parameters and solve ;

     Work done  = 1.7 x 9.8 x 0.75  = 12.5J

7 0
3 years ago
Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th
DiKsa [7]

Answer:

a.\rm -1.49\ m/s^2.

b. \rm 50.49\ m.

Explanation:

<u>Given:</u>

  • Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

<h2>(a):</h2>

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).

At time t = 3 seconds,

\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

<h2>(b):</h2>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.

For the time interval of 2 seconds,

\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.

It is the displacement of the particle in 2 seconds.

7 0
3 years ago
17. Suppose you were standing on a scale in an elevator in free fall. What would the scale read?
lianna [129]

17

What would the scale read? zero

18 In free fall you are being pulled by a gravity. "Truly" weightless presumably happens in deep space where there is nothing to pull you.

19 coasters accelerate down to simulate weight loss/zeroised. As do NASA planes,

Roller coasters are for fun seekers. NASA is for science

5 0
3 years ago
Read 2 more answers
I need to figure out A but i’m not sure how
Damm [24]

Answer:

cc

Explanation:

4 0
3 years ago
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