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3 years ago

Explain how an alpine glacier can change the topography of a mountainous area

1 answer:

7 0

<span>An alpine glacier can change the topography of a mountainous area through Glacial Erosion and Glacial Deposition. Glaciers are agents of erosion, it can pick up and carry large rocks and sediments. In the process, a deep cavity or hole can form when the glacier plucks a big rock from where it passed. Glaciers have shaped many Mountain Ranges and have created distinct landforms by its erosion process. In Glacial Deposition, as glaciers melt, it deposits all that it carried and a landform is developed.</span>

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A 35.8 kg box initially at rest is pushed 2.38 m along a rough, horizontal floor with a constant applied horizontal force of 108

tiny-mole [99]

Answer:

The work done by the applied force is 259.22 J.

Explanation:

The work done by the applied force is given by:

$W = F*d$

Where:

F: is the applied horizontal force = 108.915 N

d: is the distance = 2.38 m

Hence, the work is:

$W = F*d = 108.915 N*2.38 m = 259.22 J$

Therefore, the work done by the applied force is 259.22 J.

I hope it helps you!

6 0

2 years ago

A 2.5m long steel piano wire has a diameter of 0.5cm how great is the tention in the wire if it stretches by 0.45cm when tighten

UkoKoshka [18]

Answer: Their u go i found it their was about 3 pages i did not no what pages u had to do.

Explanation:

Download pdf

7 0

2 years ago

A charge of 4.5 × 10-5 C is placed in an electric field with a strength of 2.0 × 104 . If the charge is 0.030 m from the source

snow_tiger [21]

Answer:

The electrical potential energy is 0.027 Joules.

Explanation:

The values from the question are

charge (q) = $4.5 \times 10^{-5} C$

Electric Field strength (E) = $2.0 \times 10^{4} N/C$

Distance from source (d) = 0.030 m

Now the formula for the electrical potential energy (U) is given by

$U = q \times E \times d$

So now insert the values to find the answer

$U = 4.5 \times 10^{-5} C \times 2.0 \times 10^{4} N/C \times 0.030 m$

On further solving

$U = 0.027 J$

8 0

2 years ago

14.A 90 kg quarterback gets tackled by being hit by a 120 kg lineman backwards

quester [9]

The acceleration of the quarterback and the lineman is 5.55m/s² and 4.16m/s² respectively in the same direction.

As, we know, the 120 Kg lineman is moving with a force of 500N.

His net acceleration will be in the same direction as his motion.

It is already known that, If M is the mass of the body and a is the acceleration of the body, then the force F on the body can be calculated by using the formula,

F = Ma.

The weight of the quarterback is 90 Kg. He is being hit by a force of 500N.

So, the acceleration can be calculated using the formula,

500N = 90kg x a

a = 5.55 m/s².

Now, the weight if the lineman is 120kg, the force applied by him is 500N.

So, from the formula, his acceleration A will be,

500N = 120Kg x A

A = 4.16 m/s².

both of them will have acceleration in the same direction,

To know more about Force, visit,

brainly.com/question/25239010

#SPJ9

8 0

8 months ago

If the mass of the sun is 1x, at least one planet will fall into the habitable zone if I place a planet in orbits___, ____, ____

nasty-shy [4]

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2 years ago