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lbvjy [14]
3 years ago
14

Between what depths does Earth's temperature increase the slowest?

Physics
1 answer:
bija089 [108]3 years ago
8 0
Temperature increases at a rate of approximately 20o C/km in the upper crust (first 10 km) but then the rate decreases to only approximately 0.3o C/km below 200km due to the homogenizing effect of mantle convection. A good approximation of the rate of pressure increase with depth is: 30 MPa/km in the crust and 35 MPa/km in the mantle.
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3 years ago
Two beetles run across flat sand, starting at the same point. Beetle 1 runs 0.58 m due east, then 0.89 m at 32o north of due eas
Ray Of Light [21]

Answer:

 d’= (0.561 i ^ - 0.634 j ^) m ,  d’= 0.847 m ,   48.5 south east

Explanation:

This is a displacement exercise, one of the easiest methods to solve it is to decompose the displacements in a coordinate system. Let's start with beetle 1

Let's use trigonometry to break down your second displacement

            d₂ = 0.89 m     θ = 32 north east

            sin  θ = d_{2y} / d₂

            d_{2y} = d2 sin 32

            d_{2y} = 0.89 sin 32

             d_{2y} = 0.472 m

             cos 32 = d₂ₓ / d₂

             d₂ₓ = d₂ cos 32

             d₂ₓ = 0.89 cos 32

             d₂ₓ = 0.755 m

We found the total displacement of the beetle 1

X axis

         d₁ = 0.58 i ^

         Dₓ = d₁ + d₂ₓ

         Dₓ = 0.58 + 0.755

         Dₓ = 1,335 m

Axis y

         D_{y} = d_{2y}

         D_{y} = 0.472 m

Now let's analyze the second beetle

        d₃ = 1.37 m     θ = 35 north east

         Sin (90-35) = d_{3y} / d₃

         d_{3y} = d₃ sin 55

         d_{3y} = 1.35 sin 55

         d_{3y} = 1,106 m

       cos 55 = d₃ₓ / d₃

         d₃ₓ = d₃ cos 55

         d₃ₓ = 1.35 cos 55

         d₃ₓ = 0.774 m

They ask us what the second displacement should be to have the same location as the beetle 1

          Dₓ = d₃ₓ + dx’

          D_{y} = d_{3y} + dy’

          dx’= Dₓ - d₃ₓ

          dx’= 1.335 - 0.774

          dx’= 0.561 m

         

         dy’= D_{y} - d_{3y}

         dy’= 0.472 - 1,106

         dy’= -0.634 m

We can give the result in two ways

          d’= (0.561 i ^ - 0.634 j ^) m

Or in the form of module and address

           d’= √ (dx’² +   dy’²)

          d’= √ (0.561² + 0.634²)

          d’= 0.847 m

          tan θ =   dy’/ dx’

          θ = tan⁻¹ dy ’/ dx’

          θ = tan⁻¹ (-0.634 / 0.561)

          θ = -48.5 º

This is 48.5 south east

5 0
2 years ago
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