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Lemur [1.5K]
3 years ago
10

What type of energy is involved when a river moves sediment and erodes its banks?

Physics
1 answer:
ziro4ka [17]3 years ago
8 0
<span>First question: The type of energy involved when a river moves sediment and erodes its banks is: option d. Kinetic energy. Kinetic energy is the energy associated with motion. A body (in this case the water) that moves has an energy associated with its motion that is proportional to the speed (exactly to the square of the speed). When the water collides with the banks it is the kinetic energy of the river that erodes it Second question: the answer is the option d. As gravity pulls water down a slope potential energy changes to knietic energy. This is the, water loses altitude and gains velocity. The potential energy. which is proportional to the height, decreases and the kinetic energy, which is proportional to the square of the speed, increases.</span>
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Positive feedback interactions in earth’s systems are always a result of human action. T/F
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Positive feedback interactions in earth’s systems are always a result of human action is a FALSE statement.

<u>Explanation:</u>

Earth is a unstable equilibrium which tends to move out of equilibrium, but several other factors try to bring it back in equilibrium again. Earth has a different actions going on both on its surface and also inside it.

Human can alter, or can modify a very small part of the events that occur on earth’s surface. But they don’t have any control on what’s going inside earth’s core. So positive feedback interactions are not only the results of human interaction, but also different other factors.  

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3 years ago
What is the direction of the normal contact force of the road on the wheels?
Gekata [30.6K]

Answer:

The direction of the contact forces acting on a body is not necessarily perpendicular to the contact surface. The resolution of contact forces in two components i.e. perpendicular to contact surface and along surface. Perpendicular component is normal force and parallel component is friction.

Explanation:

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3 years ago
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Read 2 more answers
A tuning fork vibrating at 508 Hz falls from rest and accelerates at 9.80 m/s^2. How far below the point of release is the tunin
JulijaS [17]

Answer:

Explanation:

given,

tuning fork vibration = 508 Hz

accelerates = 9.80 m/s²

speed of sound = 343 m/s

observed frequency = 490 Hz

f_s = f(\dfrac{v}{v-(-v_s)})

f_s = f(\dfrac{v}{v+v_s})

v_s = v[\dfrac{f_s}{f_o}-1]

      = 343[\dfrac{508}{490}-1]

      v_s=12.6 m/s

distance the tunning fork has fallen

y_1=\dfrac{v^2}{2a_y}

     =\dfrac{12.6^2}{2\times 9.8}

     =8.1 m

now, time required for the observed will be

t = \dfrac{8.1}{343} = 0.023 s

now, for the distance calculation

y_2 = u\ t + \dfrac{1}{2}at^2

  = 12.6\times 0.023 +\dfrac{1}{2}\times 9.8 \times 0.023^2

  =0.293 m

total distance

 = 8.1 + 0.293 = 8.392 m

3 0
3 years ago
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