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Brilliant_brown [7]
3 years ago
11

Distinguish among the three places that volcanoes occur with respect to plate boundaries

Physics
1 answer:
Olin [163]3 years ago
5 0
Divergent
Convergent
And
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The image produced is magnified and real.
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You drop your frozen rock from a green bridge. The frozen rock starts from rest (initial velocity = 0ms). The rock takes 4.3s to
valentinak56 [21]

Answer:

The velocity of the frozen rock at t = 1.5\,s is -14.711 meters per second.

Explanation:

The frozen rock experiments a free fall, which is a type of uniform accelerated motion due to gravity and air viscosity and earth's rotation effect are neglected. In this case, we need to find the final velocity (v), measured in meters per second, of the frozen rock at given instant and whose kinematic formula is:

v = v_{o} + g\cdot t (Eq. 1)

Where:

v_{o} - Initial velocity, measured in meters per second.

g - Gravity acceleration, measured in meters per square second.

t - Time, measured in seconds.

If we get that v_{o} = 0\,\frac{m}{s}, g = -9.807\,\frac{m}{s^{2}} and 1.5\,s, then final velocity is:

v = 0\,\frac{m}{s}+\left(-9.807\,\frac{m}{s^{2}} \right) \cdot (1.5\,s)

v = -14.711\,\frac{m}{s}

The velocity of the frozen rock at t = 1.5\,s is -14.711 meters per second.

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A car of mass m, traveling at constant speed, rides over the top of a round hill. How do the normal force of the road on the car
dybincka [34]

Answer:

The normal force will be lower than the gravitational force acting on the car. Therefore the answer is N < mg, which is <em>option B</em>.

Explanation:

Over a round hill, the centripetal force acting toward the the radius of the hill supports the gravitational force (mg) of the car. This notion can be expressed mathematically as follows:

At the top of a round hill

Normal force = Gravitational force - centripetal force

At the foot of a round hill

Normal Force = centripetal force + Gravitational force

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Whereas the others contributed much later from around 1700's - 1900's
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