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Ashley S.
Tutor for 14 years
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SAT
TutorMe
Question:

If the sum of the consecutive odd integers from x to 15, inclusive, is -57, what is the value of x? (A) -46 (B) -31 (C) -21 (D) -15 (E) -13

Ashley S.
Answer:

There are a few different approaches to this logic problem which would work; however, you also need to remember that the SAT is timed. If you try to work out an equation for this problem, you will likely spend too many of your precious minutes. You could plug in answer choices and eliminate, but that also might take a really long time (try it and see!) The best approach is to find the trick to this type of problem. You know x must be less than 15 by the question's phrasing. What happens when you add the odd numbers less than 15? The sum keeps getting bigger and bigger (15+13+9+7+…), at least at first. However, notice the negative number in the question (-57). Working down from 15, at some point you will get into negatives and they will start cancelling out the positives. And this is the trick that the question is built around: the negatives cancel out the positives, which means you’ll get to a sum of 0 when you add all the odd numbers from -15 to 15: 15 + 13 + 11 + 9 + ... + (-9) + (-11) + (-13) + (-15) = 0 so we don't even have to think about x being in that range of values; answers (D)and (E) are obviously incorrect. Once you see that, just add the next consecutive odd integers until you get to -57: (-17) + (-19) = -36 (-17) + (-19) + (-21) = -57 so (C) x = -21 because (-21) + (-19) + (-17) + (-15) + ... + 15 = -57.

Physics
TutorMe
Question:

A stressed out graduate student creates a makeshift Beanie Baby cannon on the roof of the Physics Building just before final exam week. If she shoots the cannon horizontally with an initial velocity of 20 meters/second from a height of 30 meters, what approximate horizontal distance does the stuffed animal of doom travel?

Ashley S.
Answer:

The range the stuffed animal can travel is given by: (1) R = v0,x * t Here, v0,x is the initial velocity in the horizontal direction and t is the time spent in the air. To find the range, we must first calculate the time using the projectile motion equation for height: (2) H = (v0,y * t) - (1/2)*g*t^2 where t is the time spent in the air (what we are solving for!) and v0,y is the initial velocity in the vertical direction. However, the cannon only fires horizontally, so there is no initial vertical velocity, so the first term is zero: (3) v0,y * t = 0 This simplifies the height equation to: (4) H = - (1/2) * g * t^2 Given an initial launch height of 30 meters and the gravitational acceleration, g = -9.8 m/s^2, we use algebra to rearrange equation (4), then substitute the values in and solve for time. (5) t^2 = -2*H / g ==> t = sqrt [(-2)*H / (-9.8)] t = sqrt[ (-2) * 30 meters / (-9.8 m/s^2) ] = 2.474 seconds Now, we can find the range the Beanie Baby traveled using equation (1) and the initial horizontal velocity of 20 meters/second (6) R = 20 m/s * 2.474 s = 49.5 meters, or approximately 50 meters horizontally from the Physics building.

Physics
TutorMe
Question:

A photon strikes an electron of mass m that is initially at rest, creating an electron-positron pair. The photon is destroyed and the positron and two electrons move off at equal speeds along the initial direction of the photon. What was the energy of the photon?

Ashley S.
Answer:

There are a total of three particles to consider in the interaction. Initially, we have only a moving photon and a resting electron. After the collision, we have three particles - a positron and two electrons. We are trying to calculate the energy of the photon, which is given by its momentum times the speed of light: E = p*c Let p represent the momentum of the particle and pf is the momentum of the final electron and positron. Remember that the initial momentum of the photon is shared equally between all three final particles such that: (1) pf = p/3 The initial energy, E0, before the photon strikes the electron, is given by the energy of the photon, p*c, plus the rest energy of the electron, me*c^2 (remember, all of the electron's energy is in its rest mass because it is initially at rest) such that: (2) E0 = p*c + me*c^2 The final energy, Ef, after the collision is the sum total energy of all final particles. Because the positron is the electron's antiparticle, they have the same mass, me = mp, so I will just refer to it as me for clarity. So, (3) Ef = 3 * sqrt[ (pf*c)^2 + (me*c^2)^2 ] and we can substitute the equation for the final momentum, pf, here: (4) Ef = 3 * sqrt[ (p*c/3)^2 + (me*c^2)^2 ] Due to Conservation of Energy, E0 = Ef, so we combine equations (2) and (4) from above, giving us (5) p*c + me*c^2 = 3 * sqrt[ (p*c/3)^2 + (me*c^2)^2 ] squaring both sides to get rid of the square root gives: (6) (p*c)^2 + (me*c^2)^2 +2*p*c*me**c^2 = 9 * [(p*c/3)^2 + (me*c^2)^2] With some algebra, we can cancel the (p*c)^2 terms on both sides and rearrange the equation: (7) 2*p*c*me**c^2 = 9 *(me*c^2)^2 - (me*c^2)^2 which reduces to our solution for the energy of the photon: (8) E = p*c = 4*me*c^2

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