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Sanai Y.
Math and Physics Tutor for 3 years
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Pre-Calculus
TutorMe
Question:

Factor the following 16a^2 - 16b^2

Sanai Y.
Answer:

(4a -4b)^2

Calculus
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Question:

The derivative of 4x^4 + 3x^2 + 36

Sanai Y.
Answer:

16x^3 + 6x

Physics
TutorMe
Question:

Disregarding friction, if a cube shaped object that weighed 10 kg were to be placed on an inclined plane of 30 degrees from the ground, how long would it take the box to travel 2 meters down slope?

Sanai Y.
Answer:

First, I always recommend drawing the scenario to get a visual prespective of our problem. Second, we list all our known elements of this problem We know that our object is 10 kg, so m=10kg We know that the angle of incline is 30° We also know the distance needed to be traveled, s=2m Third, we identify our variable, in this case time t=? So by using Newton's second law, F= m×a ..... Force = mass × acceleration In this case gravity substitutes acceleration since our object is placed on a surface (Note: we are calculating the force applied vertically downward, from there we can calculate the force exerted parallel to the surface) F= m×g Let us assume g=8m/s^2 F= 10kg × 8m/s^2 Kg × m/s^2 = N (Newton) F= 80 N So, 80N is the force exerted vertically downward. For us to equate the force parallel to the sloped surface we must first find out the angle between both the vertical force of the object influenced by gravity and the slope in which the object will travel. To do this imagine we draw a line from the center of our object vertically down to the ground to make a right angle. We know our angle of the slope is 30 because it was given above. All angles of a triangle must equal 180° meaning the angle we are looking for is 60°. To find the force parallel to the slope we then multiply our vertical force by cos60° because we are looking for our adjacent component of the vertical force. (Note: Fp will be the force parallel to the slope) Fp= F × cos60° Fp= 80N × 1/2 Fp= 40N Now that we have our force parallel to the slope, we find out the new acceleration the object will move at. Fp= m×a 40N = 10kg × a 4 m/s^2 = a Now that we have acceleration and distance, we use our distance time acceleration formula S is distance S= 1/2at^2 ..... we use this because the initial velocity is 0 2s/a= t^2 (2 × 2m)/ 4m/s^2 = t^2 1 s^2 = t^2 ....... t=1sec Now we know for that object to travel 2 meters on a frictionless slope it would take exactly 1 second.

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