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VARVARA [1.3K]
3 years ago
14

The Pinnacles and Neenach volcanics are 23.5 million-year-old andesite to rhyolite outcrops, on either side of the San Andreas F

ault, today separated by ~320 kilometers. If they were originally products of the same eruptive center, they have been offset along the San Andreas Fault by 320 kilometers in 23.5 million years. What is the rate of offset in centimeters per year?
Physics
1 answer:
Setler79 [48]3 years ago
6 0

Answer:

v = 1.36 cm / y

Explanation:

For this exercise we must assume that the displacement of the plates is constant over time, so we will use the kinematic relationships for the uniform movement

      v = d / t

We reduce the quantities to the SI system

     d = 320 km (1000 m / 1km) (100 cm / 1 m)

     d = 3.2 107 cm

let's calculate

     v = 32.107 / 23.5 106

     v = 1.36 cm / y

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A train travels 85 kilometers in 5 hours, and then 63 kilometers in 5 hours what is its average speed?
Lesechka [4]
We know average speed =total distance/time taken
So avg speed=(85+63)/(5+5)=14.8km/hr
3 0
4 years ago
The mass of the sun is 1.9891030 kg and the mass of the Earth is 5.9721024kg. If the Earth’s acceleration toward the sun is 0.00
saul85 [17]

Answer:

this is the answer according to my calculations

Explanation:

0.001.9

5 0
3 years ago
A 3.5 kg object moving in two dimensions initially has a velocity v1 = (12.0 i^ + 22.0 j^) m/s. A net force F then acts on the o
lys-0071 [83]

Answer:

The work done by the force is 820.745 joules.

Explanation:

Let suppose that changes in potential energy can be neglected. According to the Work-Energy Theorem, an external conservative force generates a change in the state of motion of the object, that is a change in kinetic energy. This phenomenon is describe by the following mathematical model:

K_{1} + W_{F} = K_{2}

Where:

W_{F} - Work done by the external force, measured in joules.

K_{1}, K_{2} - Translational potential energy, measured in joules.

The work done by the external force is now cleared within:

W_{F} = K_{2} - K_{1}

After using the definition of translational kinetic energy, the previous expression is now expanded as a function of mass and initial and final speeds of the object:

W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})

Where:

m - Mass of the object, measured in kilograms.

v_{1}, v_{2} - Initial and final speeds of the object, measured in meters per second.

Now, each speed is the magnitude of respective velocity vector:

Initial velocity

v_{1} = \sqrt{v_{1,x}^{2}+v_{1,y}^{2}}

v_{1} = \sqrt{\left(12\,\frac{m}{s} \right)^{2}+\left(22\,\frac{m}{s} \right)^{2}}

v_{1} \approx 25.060\,\frac{m}{s}

Final velocity

v_{2} = \sqrt{v_{2,x}^{2}+v_{2,y}^{2}}

v_{2} = \sqrt{\left(16\,\frac{m}{s} \right)^{2}+\left(29\,\frac{m}{s} \right)^{2}}

v_{2} \approx 33.121\,\frac{m}{s}

Finally, if m = 3.5\,kg, v_{1} \approx 25.060\,\frac{m}{s} and v_{2} \approx 33.121\,\frac{m}{s}, then the work done by the force is:

W_{F} = \frac{1}{2}\cdot (3.5\,kg)\cdot \left[\left(33.121\,\frac{m}{s} \right)^{2}-\left(25.060\,\frac{m}{s} \right)^{2}\right]

W_{F} = 820.745\,J

The work done by the force is 820.745 joules.

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3 years ago
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Answer:

The amount of force and the angle between them.

Explanation:

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