Tutor profile: Sam S.
Subject: SAT II Mathematics Level 1
If log base a of b = c, which of the following is true? A.) a^b = c B.) a^c = b C.) b^a = c D.) b^c = a E.) c^a = b
B is the correct answer. I find it easier to think about a common base for log, 10. If we have log base 10 of something common like 100, we know c = 2. This can be checked on your calculator quickly if you are unsure. As well, we know that 10^2 = 100. Now, substitute in 10 for a, 100 for b, and 2 for c. Now it is easier to see which variable goes into which spot in the available answers. Alternatively, you can remember the definition of what a logarithm is. The answer (c) is the power to which the base (a) must be raised to achieve the number you are taking the logarithm of (b).
Subject: Material Science
A copper wire with a 1mm diameter that is 125 meters long is measured to have a resistance of 0.7 ohms. Note the magnitude of elasticity for copper is 110 GPa and the Poisson's ratio of copper is 0.34. a.) What is the conductivity of the wire? The wire is pulled in tension with a stress of 800 MPa. b.) What is the resultant elongation? c.) What is the new diameter of the wire? d.) If the pulling of the wire did not affect it's conductivity, what is the new resistance of the wire?
a.) The equation for conductivity is conductivity = length/ (resistance * cross-sectional area) So we get: conductivity = 125 m / (0.7 ohms * pi * 0.001m ^2) conductivity = 5.68 * 10^7 (ohm-m)^-1 Answered! b.) We can use Hooke's law, which states: tensile stress = strain * modulus of elasticity, where strain is defined as the change in length divided by the original length 800 MPa = (change in length/ original length) * 110 GPa 800 Mpa * original length/ 110 GPa = change in length The original length was 125 meters, plugging that in and solving we get 800 Mpa * 125 m / 110 GPa = change in length 800 Mpa * 125 m / 110,000 Mpa = change in length change in length = 0.909 meters Answered! c.) Now we need to use Poisson's ratio to determine for a given strain along the wire (z-axis), what is the strain along the diameter of the wire (x-axis). Poisson's ratio is given as: strain_z = - strain_x / Poisson's ratio Solving for strain_x strain_x = -strain_z * Poisson's ratio Now, remember that strain = change in length / original length or dL/L dL_x / L_x = - (dL_z / L_z) *Poisson's ratio Plugging in the values we have dL_x / 0.001m = - (0.909m / 125 m) *0.34 Solving for dL_x dL_x = - 2.47 micro meters or - 0.00000247 meters Answered! d.) We now need to calculate resistance using the new values we found for length in part b (125 + 0.909 meters) and diameter in parts c (1 - 0.00247 mm) and the original conductivity we found in part a (5.68 * 10^7 (ohm-m)^-1). The conductivity equation is conductivity = length/ (resistance * cross-sectional area) Solving for resistance we get resistance = length/ (conductivity * cross-sectional area) Plugging in values and solving we get resistance = 125.909 m /(5.68 * 10^7 (ohm-m)^-1 *pi * (0.00099753 m)^2) resistance = 0.709 ohms Answered!
An astronaut (50kg and 2 meters tall) is 100 meters away from their spaceship and is floating away at 0.1 meters per second (m/s). The astronaut throws a wrench (1kg and 0.5 meters long) away from the spaceship. The wrench is spinning at one revolution per second clockwise. a.) What speed must the astronaut throw the wrench to get back to the spaceship? b.) What speed must the astronaut throw the wrench to get back to the spaceship 10 minutes after they make the throw? c.) What rotation does the astronaut now have? Note, both the astronaut and wrench can be considered rods rotating around their center (moment of inertia = Mass * Length^2 / 12). d.) If the astronaut was originally facing away from the spacecraft, after 10 minutes which way would they be facing as they come into contact with the spacecraft?
Notation I am using: m = mass v = velocity _A = of astronaut _W = of wrench ' = after x = position t = time I = moment of inertia w = angular momentum L = length a.) Momentum is conserved: m_A * v_A + m_W * v_W = m_A' * v_A' + m_W' * v_W' The objective is to create a velocity of the astronaut after throwing the wrench (v_A') that is negative (back towards the ship). Let us isolate v_A' since that is the variable we care about. m_A * v_A + m_W * v_W - m_W' * v_W' = m_A' * v_A' (m_A * v_A + m_W * v_W - m_W' * v_W') / m_A' = v_A' Since we want the astronaut to travel backward, we can now set the v_A' to 0 and change our equal sign to a less than. (m_A * v_A + m_W * v_W - m_W' * v_W') / m_A' < 0 Now that we have our equation set up the correct way, we can isolate the variable we are trying to solve for, the final velocity of the wrench (v_W'). Let's start by dividing m_A' through each part of what is inside the parenthesis. (m_A * v_A) / m_A' + (m_W * v_W) / m_A' - (m_W' * v_W') / m_A' < 0 We know that neither the mass of the astronaut nor the wrench changes before or after the throw. So we can simplify some of this equation. v_A + (m_W * v_W) / m_A - (m_W * v_W') / m_A < 0 Now let's isolate for v_W'. - (m_W * v_W') / m_A < -v_A - (m_W * v_W) / m_A - m_W * v_W' < -v_A * m_A - m_W * v_W -v_W' < (-v_A * m_A) / m_W - v_W Multiply both sides by -1. When ever you do this with a < or > sign you must flip it to the opposite. v_W' > (v_A * m_A) / m_W + v_W Now that we have isolated v_W' we can plug in our values for everything else v_A = 0.1 m/s, m_A = 50 kg, m_W = 1 kg, v_W = 0.1 m/s v_W' > (0.1 * 50) / 1 + 0.1 v_W' > 5.1 m/s Answered! b.) Now the objective is to travel back to the spaceship in 10 minutes. We have been using seconds so we should convert that to 600 seconds. The first part of this problem is to find what final velocity of the astronaut gets them back to the spaceship in 600 seconds. Velocity = change in position over change in time v = dx / dt We can plug in for dx (100m) and dt (600s) v = 100m / 600s v = 1/6 m/s Now that we know the final velocity of the astronaut needs to be 5/3 m/s back towards the spacecraft or v_A' = -5/3 m/s, we can go back to the conservation of momentum equation we were using in part a and solve for v_W'. m_A * v_A + m_W * v_W = m_A' * v_A' + m_W' * v_W' v_W' = (m_A * v_A + m_W * v_W - m_A' * v_A') / m_W' We can now plug-in and solve for v_W'. v_W' = (50 * 0.1 + 1 * 0.1 - 50 * -1/6) / 1 v_W' = 13.43 m/s Answered! c.) Now we are concerned about the conservation of angular momentum. moment of inertia * angular speed before = moment of inertia * angular speed after I_A * w_A + I_W * w_W = I_A' * w_A' + I_W' * w_W' Initially, neither was rotating so we can simplify. I_A * 0 + I_W * 0 = I_A' * w_A' + I_W' * w_W' 0 = I_A' * w_A' + I_W' * w_W' We want to solve for the final angular momentum of the astronaut w_A'. Let's isolate that. w_A' = -I_W' * w_W' / I_A' And we need to substitute in the moment of inertia for a rod (m*L^2 /12) for both the wrench and the astronaut. w_A' = -(m_W' * L_W'^2 /12) * w_W' / (m_A' * L_A' ^2 /12) Now we can plug in values and solve the equation. w_A' = -(1 * 0.5^2 /12) * 1 / (50 * 2 ^2 /12) w_A' = 0.020833 * 1 / 16.6667 w_A' = 0.00125 rotations per second counterclockwise Answered! d.)To find which way the astronaut is facing after 10 minutes, we need to find the number of revolutions they have experienced. 0.00125 rotations per second * 10 minutes * 60 seconds per minute = 0.75 rotations counterclockwise. Given they started facing away from the ship, a 1/4 turn counterclockwise would mean their head is pointing towards the ship, a 1/2 turn would mean they are facing the ship, and a 3/4 turn would mean their feet are facing the ship. The astronaut would land on the spacecraft feet first. Answered!
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