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4 years ago

9

Which of the following natural features is formed by weathering, erosion, and deposition?

2 answers:

Otrada [13]4 years ago

7 0

The answer is d. sand dunes

Black_prince [1.1K]4 years ago

7 0

Answer:

Sand dunes

Explanation:

Dunes are hills of sand which are formed by the action of winds which is an agent of weathering. Sand dunes usually formed along the beach or in a desert. Sand dunes form when the wind blows, they carry particles of sand before they drop to the ground. As they roll and bounce on the ground, sand particles create small, wave-shaped ripples of sand. These ripples can build up into larger structures.

Dunes can grow very high say up to 4000 feet and be many miles long. Dunes can be various shapes such as crescents, stars, or just repeated lines.

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Aaron walks 5 miles every day for exercise, leaving his front porch at 5:00 am. And returning to his porch at 6:00 am. What is t

taurus [48]

Answer:

Total displacement = 0

Explanation:

He lives his front porch and still returns to his front porch.

Now, displacement is a vector quantity and as such, it is the distance between the initial point of movement and the final point of movement.

In this case the initial point is the same as the final point and thus the displacement is zero.

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3 years ago

A power station burns 75 kilograms of coal per second. Each kg of coal contains 27 million joules of energy.

olasank [31]

A) $P = 75 \times (2.7\times10^7) = 2.025\times10^9 W$

b) $Efficiency = \frac{8\times10^8}{2.025\times10^9} \times 100 \approx 39.5\%$

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3 years ago

Read 2 more answers

Find velocity vx(t) and coordinate x(t) for a particle of mass m which is subject to the force given by: Fx = F0 e −kt , where F

muminat

Answer:

$v_{x} (t)=1+\frac{ F_{0}}{km}(1-e^{-kt})$

$x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}(e^{-kt}-1)$

Explanation:

Given that the force of the particle is,

$F_{x}=F_{0}e^{-kt}$

Now it can be further written as

$m\frac{dv}{dt}= F_{0}e^{-kt}\\\frac{dv}{dt}=\frac{ F_{0}}{m} e^{-kt}\\dv=\frac{ F_{0}}{m}e^{-kt}dt\\ v=\frac{ F_{0}}{-km}e^{-kt}+C$

Now the initial conditions are v=1 at t=0.

So,

$1=\frac{ F_{0}}{-km}e^{0}+C\\C=1+\frac{ F_{0}}{km}$

Now the velocity will become.

$v_{x} (t)=\frac{ F_{0}}{-km}e^{-kt}+1+\frac{ F_{0}}{km}\\v_{x} (t)=1+\frac{ F_{0}}{km}(1-e^{-kt})$

And,

$\frac{dx}{dt} =1+\frac{ F_{0}}{km}(1-e^{-kt})\\dx=(1+\frac{ F_{0}}{km}(1-e^{-kt}))dt\\x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}e^{-kt}+C\\$

And, another initial condition is x=0 at t=0

$0=0+\frac{ F_{0}0}{km}+\frac{ F_{0}t}{k^{2} m}e^{0}+C\\C=-\frac{ F_{0}t}{k^{2} m}$

Now,

$x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}e^{-kt}+-\frac{ F_{0}t}{k^{2} m}\\x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}(e^{-kt}-1)$

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3 years ago

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ehidna [41]

aisnhsjdjxbxbxhxbxjx

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3 years ago

PLEASE HELP BRAINLIEST

Leni [432]

B is the answer!!!!!!!!!!!!!

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3 years ago