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alexandr1967 [171]
3 years ago
5

What can scientist learn from studying sedimentary rocks?

Physics
2 answers:
cricket20 [7]3 years ago
7 0
They are made of rocks
Andrej [43]3 years ago
4 0
Scientists can learn by analysing the layers of the rock
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A 10kg box is sliding at 50m/s. Find the momentum
MAXImum [283]

Answer:

The momentum of the ball is 500 kg·m/s

Explanation:

The momentum is given by Mass × Velocity

The given parameters are;

The mass of the box = 10 kg

The velocity by which the box is sliding = 50 m/s

Therefore, the momentum of the ball is given as follows;

The momentum of the ball = 10 kg × 50 m/s = 500 kg·m/s

The momentum of the ball = 500 kg·m/s

5 0
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True or false? Magnetic reversals are recorded in the newly formed oceanic crust on BOTH sides of a mid-ocean ridge spreading ce
andrey2020 [161]

answer: false is my answer for thi question

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As the boat in which he is riding approaches a dock at 3.0 m/s, Jasper stands up in the boat and jumps toward the dock. Jasper a
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Im pretty sure it’s a because it makes more sense you know?.
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3 years ago
what happens to the water after it rains? a. precipitation b. runoff c. condensation user: what do organisms from the ocean use
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The rain gets evaporated in to water vapor and is returned to the clouds where they go through condensation and then they poud down as rain or A.K.A,  Precipitation.
4 0
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During an experiment of momentum, trolley, X, of mass (2.34 ± 0.01) kg is moving away from another trolley, Y, of mass (2.561 ±
Alla [95]

Answer:

P = 1 (14,045 ± 0.03 )  k gm/s

Explanation:

In this exercise we are asked about the uncertainty of the momentum of the two carriages

            Δ (Pₓ / Py) =?

 Let's start by finding the momentum of each vehicle

car X

        Pₓ = m vₓ

        Pₓ = 2.34 2.5

        Pₓ = 5.85 kg m

car Y

        Py = 2,561 3.2

        Py = 8,195 kgm

How do we calculate the absolute uncertainty at the two moments?

          ΔPₓ = m Δv + v Δm

          ΔPₓ = 2.34 0.01 + 2.561 0.01

          ΔPₓ = 0.05 kg m

         ΔP_{y} = m Δv + v Δm

         ΔP_{y} = 2,561 0.01+ 3.2 0.001

         ΔP_{y} = 0.03 kg m

now we have the uncertainty of each moment

          P = Pₓ / P_{y}

          ΔP = ΔPₓ/P_{y} + Pₓ ΔP_{y} / P_{y}²

          ΔP = 8,195 0.05 + 5.85 0.03 / 8,195²

          ΔP = 0.006 + 0.0026

          ΔP = 0.009 kg m

The result is

           P = 14,045 ± 0.039 = (14,045 ± 0.03 )  k gm/s

7 0
3 years ago
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