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kogti [31]
2 years ago
9

What is a subduction zone?

Physics
2 answers:
djyliett [7]2 years ago
3 0

The correct answer is:

A.

Nana76 [90]2 years ago
3 0

Answer: A

Explanation: i took the test

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skateboarder, starting from rest, rolls down a 13.5 m ramp. When she arrives at the bottom of the ramp her speed is 7.37 m/s. If
scZoUnD [109]

Answer:

1.7 m/s²

Explanation:

d = length of the ramp = 13.5 m

v₀ = initial speed of the skateboarder = 0 m/s

v = final speed of the skateboarder = 7.37 m/s

a = acceleration

Using the equation

v² = v₀² + 2 a d

7.37² = 0² + 2 a (13.5)

a = 2.01 m/s²

θ = angle of the incline relative to ground = 29.9

a' = Component of acceleration parallel to the ground

Component of acceleration parallel to the ground is given as

a' = a Cosθ

a' = 2.01 Cos29.9

a' = 1.7 m/s²

7 0
3 years ago
Do someone know the answer
JulijaS [17]
Speed x time = distance
Distance divided by time = speed
500 divided by 5
Speed = 100
4 0
3 years ago
What is the equilibrium concentration of ibr?
asambeis [7]
IBR is the thermal decomposition of iodine(I) bromide to produce iodine and bromine. This reaction takes place at a temperature of over 40,5°C and is written as: 

 <span>2IBr ⇄ I2 + Br2
</span>
Equilibrium is a state of dynamic balance where the ratio of the product and reactant concentrations is constant.<span> You can calculate the equilibrium concentration if you know the equilibrium constant Kc (Kc=I^2*Br^2/IBR^2) and the initial concentration for the reaction. The initial concentration is obtained from ICE Table.</span>
3 0
3 years ago
Which type of constraint typically requires a longer time to change?
Mars2501 [29]
Structural constraint is the answer :)
3 0
3 years ago
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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