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Mawaba Pascal D.
Mechanical Engineering student
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French
TutorMe
Question:

Select the letter that corresponds to the correct meaning for each of the following words 1 -) Bonsoir: a. How are you? b. I am hungry. c. Good evening. 2-) Je m'appelle: a. Your name is. b. Goodluck. c. My name is. 3-) Je vais acheter du pain a la boulangerie: a. I am going to buy bread at the bakery b. I am going to buy bread at the store c. I need to use the bathroom 4-) Voulez-vous sortir avec moi ce soir? a. Would you like to dance with me? b. Will you go on a date with me tonight? c. Where do you live?

Mawaba Pascal D.

1-) c 2-) c 3-) a 4-) b

Calculus
TutorMe
Question:

CALC III Find the arc-length of the curve r(t) = < t, 4cos(t), 4sin(t) >, -3 <= t <=3 (<= signifies less than or equal).

Mawaba Pascal D.

For r(t) = < f(t), g(t), h(t)> , we find the length of the arc from t=a to t=b by evaluating the integral L = integral from t= a to t =b of ( sqrt ( f'(t)^2+g'(t)^2+h'(t)^2) dt. Which is equal to the integral from t = a to t = b of the magnitude of the derivative of r(t). We therefore start by taking the derivative of r(t): r(t) = < t, 4cos(t), 4sin(t) > r'(t) = <1, -4sin(t), 4cos(t) > We then evaluate the magnitude of r'(t): abs(r'(t)) = sqrt( 1^2 + (-4sin(t))^2 + (4cos(t))^2) abs(r'(t)) = sqrt( 1 + 16*sin^2(t) + 16*cos^2(t) ) (1) It is important to remember from trigonometry that: (sin( t ) )^2 + (cos( t ) )^2 = 1. Making the substitution, we rewrite (1) as: abs(r'(t)) = sqrt( 1 + 16) abs(r'(t)) = sqrt (17) We then evaluate the integral of the absolute value of of the derivative of r(t) as: L = integral from t= -3 to t =3 of ( sqrt(17)) dt Evaluating the integral: L = sqrt(17)*t , bounds from t = -3 to t= 3 Plugging in the bounds, we get: L = 6sqrt(17) Wich is our final answer. :)

Physics
TutorMe
Question:

A Ferrari f12 Berlinetta is lapping around a track. It comes to a flat, unbanked curve with radius R . If the coefficient of static friction between tires and road is (us) what is the maximum speed at which the driver can take the curve without sliding?

Mawaba Pascal D.

Since the Ferrari is traveling in a circular trajectory, its acceleration is written as a=v^2/R, where v is the velocity of the Ferrari, and R is the radius of the curve. By Newton's first law (F=ma), the force exerted on the Ferrari as it goes around the curve is therefore: F=m(v^2/R) (1) Notice that as this force pertains to the velocity of the Ferrari, it acts horizontally (x-direction). The only other force acting horizontally, and preventing the car from sliding is that of friction (f), caused by the contact of the tires on the asphalt. In order to find the maximum velocity of the Ferrari as it takes the turn, we use Newton's 2nd law, and write the summation of the forces is the x direction (horizontal) F - friction (f) =0 friction (f) = F (2) Substituting F from (1) we rewrite: friction (f) = m(v^2/R) (3) If friction (f) = (us)(N), where (us) is the coefficient of static friction, and N is the reaction force due to the ground we rewrite (3) as: (us)(N) = m(v^2/R) (4) We do however, not know what N is. N being a force acting vertically, we must write the summation of the forces in the y-direction to determine N: N - mg = 0 N = mg (5) Having determined N, we substitute (5) into (4) and write: (us)(mg) = m(v^2/R) Solving for v, we get: v = sqrt ((us)*g*R) (6) Which is our final answer. Given the values for coefficient of static friction(us), and the radius of the curve (R), we can compute the maximum velocity the Ferrari can go around any value of R and any value of (us). Warning: This answer is an oversimplified model as it doesn't take into account factors such as drag, down force, and center of gravity. This model basically states that an 18 wheeler with the same coefficient of static friction (us), can go around the curve at the same maximum speed as our Ferrari, which is utterly ridiculous because, come on! IT IS A FERRARRRI! :)

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