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Art [367]
3 years ago
6

In the theory of plate tectonics, various segments of Earths crust, called plates, move toward and away from each other. In one

instance, the plate that consists of the Indian subcontinent drifted from southeastern Africe to its current position in asia, traveling at a speed of 15cm/y. This plate collided with Asia, forming the Himalayan mountain range in the process. Most of this formation occurred during the last 1.00 x 10^7 years, during which time the indian subcontinents motion has slowed to about 5 cm/y. What has been the acceleration in units of cm/y^2, of the indian subcontinent during this time period.
Physics
1 answer:
GrogVix [38]3 years ago
7 0

Answer:

-1.00 × 10⁻⁶ cm/y²

Explanation:

Using a = (v - u)/t where a = acceleration of the indian subcontinent, v = final speed of the indian subcontinent = 5 cm/y, u = initial speed of the indian subcontinent = 15 cm/y, and t = time of motion of the indian subcontinent = 1.00 × 10⁷ years.

Substituting the values of the variables, we find the acceleration of the indian subcontinent, a  = (v - u)/t = (5 cm/y - 15 cm/y)/1.00 × 10⁷ years = -10 cm/y ÷ 1.00 × 10⁷ years = -1.00 × 10⁻⁶ cm/y²

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ololo11 [35]

Answer:

 v_{1f} = +3,394 103 m / s

Explanation:

We will solve this problem with the concept of the moment. Let's start by defining the system that is formed by the complete rocket before and after the explosions, bone with the two stages, for this system the moment is conserved.

The data they give is the mass of the first stage m1 = 2100 kg, the mass of the second stage m2 = 1160 kg and its final velocity v2f = +5940 m / s and the speed of the rocket before the explosion vo = +4300 m / s

The moment before the explosion

      p₀ = (m₁ + m₂) v₀

After the explosion

      pf = m₁ v_{1f} + m₂ v_{2f}

     p₀ = [texpv_{f}[/tex]

     (m₁ + m₂) v₀ = m₁ v_{1f} + m₂ v_{2f}

Let's calculate the final speed (v1f) of the first stage

     v_{1f} = ((m₁ + m₂) v₀ - m₂ v_{2f}) / m₁

     

     v_{1f} = ((2100 +1160) 4300 - 1160 5940) / 2100

     v_{1f} = (14,018 10 6 - 6,890 106) / 2100

     v_{1f} = 7,128 106/2100

     v_{1f} = +3,394 103 m / s

come the same direction of the final stage, but more slowly

4 0
3 years ago
A pulled tablecloth exerts a frictional force of 0.6 n on a plate with a mass of 0.4 kg. what is the acceleration of the plate?
Ludmilka [50]
The formula for acceleration is a = F/m; Where: F = force; m = mass
Given: F = .6n; m = .4kg; a = ?
a = F/m
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Therefore, the acceleration of the plate is 1.5 m/s^2
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jek_recluse [69]

The orbital radius is: r=\frac{GM}{v^2}

Explanation:

The problem is asking to find the radius of the orbit of a satellite around a planet, given the orbital speed of the satellite.

For a satellite in orbit around a planet, the gravitational force provides the required centripetal force to keep it in circular motion, therefore we can write:

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Re-arranging the equation, we find:

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Learn more about circular motion:

brainly.com/question/2562955

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Answer:

D) The negatively charged electrons

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:) Hoped this helped!!! Have a good day!!! <3

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