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3 years ago

Glaciers begin with snowfall building up and __________________ the ice. (Choose the best answer)

Physics

1 answer:

Allisa [31]3 years ago

4 0

Answer:

compacting

Explanation:

i don't think there is very much explanation, the snow falls and compacts the ice to become giant lol

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A girl delivering newspapers covers her route by traveling 3.00 blocks west, 4.00 blocks north, and then 6.00 blocks east. What

Harrizon [31]

The concepts used to solve this problem are those related to the Pythagorean theorem for which we will calculate the distance and the pitch.

According to the attached diagram we have that the expression of the resulting displacement is

$R = \sqrt{(3b)^2+(4b)^2}$

Therefore the resultant displacement of the girl is

$R = \sqrt{(9b^2+16b^2)}$

$R = \sqrt{25b^2}$

$R = 5b$

Therefore the girl has displaced around of 5 blocks

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3 years ago

What question was asked by Faraday that the narrator calls a leap

masha68 [24]

If electricity and magnetism can create motion, can the reverse be true? Can motion and magnetism create electricity?

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3 years ago

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Time Intervals in Ice Ages

stepladder [879]

It started off with 68% less than it did at the peak, and later created a void and melted the remainder of the ice at about 92%

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3 years ago

The Coefficient of kinetic friction between the tires of your car and the roadway is \"μ\". (a) If your initial speed is \"v\" a

vazorg [7]

We make use of the equation: v^2=v0^2+2a Δd. We substitute v^2 equals to zero since the final state is halting the truck. Hence we get the equation -v0^2/2a = Δd. F = m a from the second law of motion. Rearranging, a = F/m
F = μ Fn where the force to stop the truck is the force perpendicular or normal force multiplied by the static coefficient of friction. We substitute, -v0^2/2μ Fn/m = Δd. This is equal to

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3 years ago

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As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, afte

Klio2033 [76]

Answer:

$r^2 = \frac{ 2 k \ Ze^2}{ 2m}$

Explanation:

For this exercise we must use the principle of conservation of energy

starting point. The proton very far from the nucleus

Em₀ = K = ½ m v²

final point. The point where the proton is stopped (v = 0)

Em_f = U = q V

where the potential is

V = k Ze / r²

Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching

Energy is conserved

Em₀ = Em_f

½ m v² = e ( $k \frac{Ze}{r^2}$ )

$r^2 = \frac{ 2 k \ Ze^2}{ 2m}$

with this expression we can find the closest approach distance (r)

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3 years ago