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Mamont248 [21]
3 years ago
9

What are the two main factors that affect how quickly a coastline erodes?

Physics
1 answer:
Sergeu [11.5K]3 years ago
6 0
Factors Affecting the Rate of Erosion<span>. The biggest factor affecting coastal </span>erosion<span> is the </span>strength<span> of the waves breaking along the coastline. A wave's </span>strength<span> is controlled by its fetch and the </span>wind speed<span>.</span>
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The curve below shows the percentage of population of aquatic species that die in response to doses of pollutant A:
Elis [28]

Based on the trend produced by the dose - response graph, it would be best to evacuate the residents in other to prevent the increasing percentage of deaths due to the rising level of pollutant A.

  • The curve shows that the pollutant level in mg/kg of pollutant A is still increasing, hence, groundwater monitoring alone won't be the best decision to make.

  • Since the pollutant level is still increasing, then the spill level still need effective monitoring.

  • Evacuation of residents seems to be the best decision that should be taken based on the information interpreted on the graph.

Therefore, Evacuating residents to prevent rising death percentage is required as the pollutant level is yet to subside.

Learn more :brainly.com/question/24844489

8 0
3 years ago
A 70.0 kg astronaut is training for accelerations that he will experience upon reentry. He is placed in a centrifuge (r = 15.0 m
Levart [38]

Answer:

1.3823 rad/s

20.7345 m/s

28.66129935 m/s²

a=2.92164g

2006.29095 N radially outward

Explanation:

r = Radius = 15 m

m = Mass of person = 70 kg

g = Acceleration due to gravity = 9.81 m/s²

Angular velocity is given by

\omega=13.2\times \dfrac{2\pi}{60}\\\Rightarrow \omega=1.3823\ rad/s

Angular velocity is 1.3823 rad/s

Linear velocity is given by

v=r\omega\\\Rightarrow v=15\times 1.3823\\\Rightarrow v=20.7345\ m/s

The linear velocity is 20.7345 m/s

Centripetal acceleration is given by

a_c=r\omega^2\\\Rightarrow a_c=15\times 1.3823^2\\\Rightarrow a_c=28.66129935\ m/s^2

The centripetal acceleration is 28.66129935 m/s²

Acceleration in terms of g

\dfrac{a}{g}=\dfrac{28.66129935}{9.81}\\\Rightarrow a=2.92164g

a=2.92164g

Centripetal force is given by

F_c=ma_c\\\Rightarrow F_c=70\times 28.66129935\\\Rightarrow F_c=2006.29095\ N

The centripetal force is 2006.29095 N radially outward

The torque will be experienced when the centrifuge is speeding up of slowing down i.e., when it is accelerating and decelerating.

3 0
3 years ago
Un tubo cilindrico hueco de cobre mide 3 m de longitud tienen un diametro exterior de 4cm y un diametro interior de 2 cm¿cuanto
Irina-Kira [14]

Answer:

 W = 9.93 10² N

Explanation:

To solve this exercise we must use the concept of density

           ρ = m / V

the tabulated density of copper is rho = 8966 kg / m³

let's find the volume of the cylindrical tube

           V = A L

           V = π (R_ext  ² - R_int ²) L

let's calculate

          V = π (4² - 2²) 10⁻⁴  3

          V = 1.13 10⁻²  m³

         m = ρ V

        m = 8966 1.13 10⁻²

        m = 1.01 10² kg

the weight of the tube

        W = mg

         W = 1.01 10² 9.8

         W = 9.93 10² N

4 0
3 years ago
If 100ml of cold water is added to 100ml of hot water what happens
denis23 [38]
The water will be cool and steam will be created by the hot and cold water reacting together
4 0
3 years ago
A 57 g ice cube can slide without friction up and down a 33 ∘ slope. The ice cube is pressed against a spring at the bottom of t
beks73 [17]
The springs stored energy is transferred to the cube as kinetic energy and then by the slop the KE is converted to height energy. 

<span>0.5 . k . x^2 = 0.5 . m . v^2 = m . g . ∆h </span>

<span>0.5 . 50 . (0.1^2) = 0.05 . 9.8 . ∆h </span>

<span>∆h = 0.51 m = 51 cm </span>

<span>This is the height gained </span>
<span>Distance along the slope = ∆h / sin 60 = 0.589 = 59 cm </span>

<span>In the second case, the stored spring energy is converted into height energy AND frictional heat energy. </span>

<span>The height energy is m . g . d sin 60 where d is the distance the cube moves along the slope. </span>

<span>The Frictional energy converted is F . d </span>

<span>F ( the frictional force ) = µ . N </span>

<span>N ( the reaction to the component of the gravity force perpendicular to the surface of the slope ) = m . g . cos60 </span>

<span>Total energy converted </span>

<span>0.5 . k . x^2 = (m . g . dsin60) + (µ . m . g . cos60 . d ) </span>

<span>Solve for d </span>

<span>d = 0.528 = 53 cm</span>
5 0
3 years ago
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