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# Tutor profile: John G.

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John G.
Physics and Mathematics Tutor
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## Questions

### Subject:Physics (Newtonian Mechanics)

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Question:

A skydiver jumps from a plane and falls with no open parachute until the skydiver reaches terminal velocity. Explain the forces on the skydiver when he reaches terminal velocity, and then explain how the forces change when the skydiver opens the parachute.

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John G.

The skydiver has two forces that are experience. The skydiver's weight is pulling him to ground and the air resistance is opposing his downward motion. When the skydiver reaches terminal velocity he is at a constant velocity. This means he must have zero net acceleration and his zero net force. Since there are only two forces acting in opposite directions these two forces must be equal. When the skydiver opens the parachute there is a change in the air resistance of the skydiver. The shape and structure of the parachute will provide a significantly high amount of air resistance. This high air resistance will cause a net force in the upwards direction opposing the skydiver's downward motion. The net force in the upward direction will result in an acceleration in the upwards direction. This acceleration will slow the skydiver's fall towards the ground.

### Subject:Algebra

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Question:

Given the equations below find the intersection of the two lines. y = 3x 6y - 19x = 1

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John G.

This is a system of two equations. The first equation already is solved for the y variable so we will use the substitution method to find the intersection of the two lines. Step 1: (Inserted the 3x into the 2nd equation for y) 6(3x) - 19x = 1 Step 2: (Solve the equation for x) x = -1 Step 3: (Plug the value of x into the first equation to get y y= 3(-1) = -3 Your intersection of the two lines is point (-1, -3)

### Subject:Physics

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Question:

A 2 gram coin is dropped off a building and falls 20 meters to the ground. At what speed does the coin hit the ground? (assume no air resistance)

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John G.

To solve this problem we must use conservation of energy. First, when the coin is 20 meters above the ground it has gravitational potential energy stored. That potential energy can be found through the equation PE = mgh, where m is the mass, g is gravity, and h is the height above some arbitrary point. In this case the point is ground level. In order to have the right units the mass must first be converted in to kilograms. A 2 gram coin is the same as a 0.002 kilogram coin because there are 1000 grams in 1 kilogram, Gravity is a constant of 9.8 meters per second per second, lastly the 20 meter height is given in the problem PE = mgh = (0.002)(9.8)(20) = 0.392 Joules As the coin is falling the work done by gravity is converting the stored potential energy into the kinetic energy. This happens until all the stored energy is transferred and the coin hits the ground at it's fastest speed. The equation of kinetic energy is as followed. KE = (1/2)m(v^2) = 0.392 Joules The velocity in this formula can be found through five steps of algebra. Step 1: (1/2)m(v^2) = 0.392 Step 2: (Multiply by 2) m(v^2) = 0.392(2) Step 3: (Divide by the mass) (v^2) = 0.392(2) / m Step 4: (Take the square root) v = SQRT (0.392(2) / m) Step 5: (Plug in known values and solve) v = 19.8 m/s

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