College #1 - Questions and Answers

Question 1 Three identical blocks connected by ideal strings are being pulled along a horizontal frictionless surface by a horizontal force F The magnitude of the tension in the string between blocks B and C is T=3.00N. Each block has mass m=0.400kg. What is the magnitude F of the force? What is the tension in the string between block A and block B? Answer 1 The magnitude of F of the force T=ma 3.00N = 0.400kg * a thus a = 3.00N / 0.400kg = 7.5 m/s² However, this only represents the overall acceleration; for each separate block, the a = 7.5/2 = 3.75 m/s² F can now be find by using a = F/3m and solving for F F = a * 3m F = 3.75 * 3 * 0.4 F = 4.5 N For the Tension between block A and B, all we do is T=ma, knowing that a = 3.75 m/s² for each block. T = (0.400 kg) * 3.75 m/s² = 1.5 N Answer 2 T2 = 3N M=0.4 kg T1 = 0.4a -(1) T2 - T1 = 0.4a -(2) F-T2 = 0.4a-(3) a = 3.75m/s², TAB = 1.5N Answer 3 Result: The magnitude of the force F is 3.00 N. The tension in the string between block A and block B is 3.00 N. Solution: Let's denote the magnitude of the force as F. Since the three blocks are connected by ideal strings, the tension in the string between blocks B and C is equal to the force applied on block C. Therefore, the tension T is equal to the force applied on block C, which can be expressed as: T = F Given that T = 3.00 N, we have: F = 3.00 N Now, to find the tension in the string between block A and block B, we need to consider the motion of the blocks as a system. The force applied on block A is equal to the sum of the tension in the string between A and B and the tension in the string between B and C. Therefore, the force applied on block A can be expressed as: FA = T + T Since the masses of all three blocks are identical, the acceleration of each block is the same. Applying Newton's second law to each block, we have: FA = m*a Where m is the mass of each block and a is the common acceleration of the blocks. Since the masses of the blocks are given as m = 0.400 kg, we substitute this value into the equation: FA = 0.400 kg * a Since the force applied on block A is equal to the force F, we can rewrite the equation as: F = 0.400 kg * a Now we can solve for the acceleration a. Dividing both sides of the equation by 0.400 kg, we get: a = F / 0.400 kg Since F = 3.00 N, we substitute this value into the equation: a = 3.00 N / 0.400 kg Simplifying the expression, we find: a = 7.50 m/s² Finally, to find the tension in the string between block A and block B, we use the equation: TAB = FA - T Substituting the known values, we have: TAB = 2T - T TAB = T Therefore, the tension in the string between block A and block B is also equal to 3.00 N. Answer 4 Let's denote the magnitude of the force F as F, the tension in the string between block A and B as TAB, and the mass of each block as m = 0.400 kg. Using Newton's second law, we know that the net force acting on an object is equal to its mass multiplied by its acceleration. Since the blocks are connected by ideal strings, they will all have the same acceleration. First, let's analyze the forces acting on block A: The tension TAB in the string is directed towards the right. There is no other force acting on block A horizontally. Therefore, we can write the equation of motion for block A as: ∑ Fhorizontal, A = TAB = m*a Next, let's analyze the forces acting on block B: The tension TAB in the string is directed towards the left. The tension T in the string between blocks B and C is directed towards the right. Therefore, we can write the equation of motion for block B as: ∑ Fhorizontal, B = TAB - T = m*a Finally, let's analyze the forces acting on block C: The tension T in the string between blocks B and C is directed towards the left. There is no other force acting on block C horizontally. Therefore, we can write the equation of motion for block C as: ∑ Fhorizontal, C = T = m*a Since the blocks are connected and experiencing the same acceleration, we can equate the equations of motion for each block: TAB - T = T Given that T = 3.00 N, we can solve for F and TAB: TAB = T + T = 3.00 N + 3.00 N = 6.00 N F = TAB = 6.00 N Hence, the magnitude of the force F is 3.00 N, and the tension in the string between block A and B is 6.00 N. Question 2 Assuming that the random variable x has a cumulative distribution function, F(x) = { 0, x < 0, 0.25x, 0≤x<5, 1, 5 ≤ x } Determine the following: a) p (x < 2.8) b) p (x > 1.5) c) p(x < - z) d) p(x > b) Answer 1 Step 1 So we have, The cumulative distribution function of the random variable x is. F(x) = { 0, x < 0, 0.25x, 0≤x<5, 1, x ≥ 5} Step 2 Determine P (X < 2.8) That is, P(X < 2.8) = F (2.8) = 0.25 (2.8) = 0.7 Step 3 Determine P (X > 1.5) That is, P(X> 1.5) = 1-P (X ≤ 1.5) = 1-F (1.5) = 1 - [0.25 (1.5)] = 1-0.375 =0.625 Step 4 Determine P (X < -2) That is, P(X < -2) = F (-2) = 0 Step 5 Determine P (X > 6) That is, P(X > 6) = 1-P(X≤6) = 1-F(6) = 1-1 = 0 Answer 2 Step 1: a) To find P(x < 2.8), we need to evaluate the cumulative distribution function (CDF) at x = 2.8. The CDF, denoted as F(x), gives the probability that the random variable x takes on a value less than or equal to x. In this case, we have: F(x) = { 0, x < 0, 0.25x, 0≤x<5, 1, 5 ≤ x} Since 2.8 falls within the interval 0 ≤ x < 5, we can use the second case of the CDF. Thus, P(x < 2.8) = F(2.8) = 0.25 * 2.8 = 0.7. Therefore, the probability that x is less than 2.8 is 0.7. Step 2: b) To determine P(x > 1.5), we need to find the probability that x is greater than 1.5. We can approach this by calculating the complement of P(x ≤ 1.5). Since 1.5 falls within the interval 0 ≤ x < 5, we can use the second case of the CDF. Thus, P(x ≤ 1.5) = F(1.5) = 0.25 * 1.5 = 0.375. The complement of this probability is: P(x > 1.5) = 1 - P(x ≤ 1.5) = 1 - 0.375 = 0.625. Therefore, the probability that x is greater than 1.5 is 0.625. Step 3: c) The probability P(x < - z) represents the probability that x is less than a negative value -z. However, the given cumulative distribution function has a probability of 0 for all negative values of x, as stated in the first case of the CDF. Therefore, P(x < - z) = 0 for any negative value -z. Step 4: d) Similarly, the probability P(x > b) represents the probability that x is greater than a given value b. In this case, since the given cumulative distribution function has a probability of 1 for all values greater than or equal to 5, we have: P(x > b) = 1 for any value b. Answer 3 a) P(X < 2.8): Since the cumulative distribution function (CDF) is defined piecewise, we can evaluate the probability by subtracting the CDF values at the endpoints: P(X < 2.8) = F(2.8) - F(0) = 0.25(2.8) - 0 = 0.7 b) P(X > 1.5): To find the probability that X is greater than 1.5, we subtract the CDF value at 1.5 from 1 (since the CDF gives the probability of being less than or equal to a certain value): P(X > 1.5) = 1 - F(1.5) = 1 - 0.25(1.5) = 1 - 0.375 = 0.625 c) P(X < -z): Since the CDF is 0 for negative values, the probability of x being less than -z is always 0: P(X < -z) = 0 d) P(X > b): Given a value b greater than 5, the probability that x is greater than b is 0 since the CDF is 1 for x ≥ 5: P(X > b) = 0 Answer 4 a) P(x < 2.8) = 0.25(2.8) b) P(x > 1.5) = 1 - 0.25(1.5) c) P(x < -z) = 0 (for any value of z) d) If b ≥ 5, P(x > b) = 0. Otherwise, P(x > b) = 1 - 0.25b. Explanation: F(x) = { 0, x < 0, 0.25x, 0≤x<5, 1, 5 ≤ x} a) To find P(x < 2.8), we can evaluate the CDF at x = 2.8: P(x < 2.8) = F(2.8) Since 0 ≤ x < 5 for 2.8, we use the second case of the CDF: P(x < 2.8) = F(2.8) = 0.25(2.8) b) To find P(x > 1.5), we can use the complement rule. The probability of an event not occurring is equal to one minus the probability of it occurring. Therefore: P(x > 1.5) = 1 - P(x ≤ 1.5) To find P(x ≤ 1.5), we evaluate the CDF at x = 1.5: P(x ≤ 1.5) = F(1.5) Since 0 ≤ x < 5 for 1.5, we use the second case of the CDF: P(x ≤ 1.5) = F(1.5) = 0.25(1.5) Therefore, P(x > 1.5) = 1- 0.25(1.5) c) To find P(x < -z), we need to consider the definition of the CDF for x < 0. Since the CDF is zero for x < 0, it implies that the probability of x being less than any negative value is also zero. Therefore, P(x < -z) = 0 for any value of z. d) To find P(x > b), we can use the complement rule. The probability of x being greater than b is equal to one minus the probability of x being less than or equal to b. Therefore: P(x > b) = 1-P(x ≤ b) To find P(x ≤ b), we evaluate the CDF at x = b. If b ≥ 5, then P(x ≤ b) = F(b) = 1, and therefore P(x > b) = 1 - 1 = 0. If 0 ≤ b < 5, then P(x ≤ b) = F(b) = 0.25b, and therefore P(x > b) = 1 - 0.25b. Question 3 To demonstrate that the sum of two even integers is also even, use a direct proof. Answer 1 Let n and m be even integers, and add and factor them such that n=2s and m=2t for some integers s and t, respectively. n + m = 2s + 2t = 2(s+t) = 2(z), for some integer z Answer 2 Step 1: To demonstrate that the sum of two even integers is also even, we will use a direct proof. Let n and m be two even integers. By definition, an even integer can be expressed as 2k, where k is an integer. Therefore, we can write n = 2k₁ and m = 2k₂, where k₁ and k₂ are integers. Now, we want to show that n + m is also an even integer. We can express the sum n + m as: n + m = (2k₁) + (2k₂) Step 2: Using algebraic manipulation, we can simplify this expression: n + m = 2k₁ + 2k₂ = 2(k₁ + k₂) Let k = k₁ + k₂. Since the sum of two integers is also an integer, k is an integer. Therefore, we can rewrite the expression as: n + m = 2k This shows that n + m can be expressed as 2k, where k is an integer. Hence, n + m is an even integer. Therefore, we have proven that the sum of two even integers is also even using a direct proof. Answer 3 Let's suppose we have two even integers, which we can represent as 2n and 2m, where n and m are integers. By definition, an even integer can be expressed as twice some integer. The sum of these two even integers can be written as: 2n + 2m Now, we can factor out the common factor of 2: 2(n + m) Since n and m are integers, their sum (n + m) is also an integer. Let's denote it as k: 2k By substituting k for (n + m), we can rewrite the sum of the even integers as 2k, where k is an integer. According to the definition of an even integer, 2k can be expressed as twice some integer, which means it is also even. Therefore, we have shown that the sum of two even integers, 2n and 2m, is also even, as it can be expressed as 2k, where k is an integer. Answer 4 Let a and b be two even integers. By definition, an even integer can be expressed as a = 2k and b = 2m, where k and m are integers. We want to show that a + b is also an even integer. So, we need to express a + b in the form 2n, where n is an integer. We have: a + b = (2k) + (2m) = 2(k + m) Since k and m are integers, k + m is also an integer. Let's call n = k + m. Thus, we can rewrite a + b as: a + b = 2n This shows that a + b can be expressed in the form of 2n, where n is an integer. Therefore, a + b is an even integer. Hence, we have successfully demonstrated that the sum of two even integers is also even using a direct proof. Question 4 A rectangular plot of farmland are finite on one facet by a watercourse and on the opposite 3 sides by a single-strand electrical fence. With 800 m of wire at your disposal, what's the biggest space you'll be able to enclose, and what area unit its dimensions? Answer Let the sides of the rectangle-shaped fence be a and b. We only need to use wire for the three sides of the fence because the watercourse borders the farmland on one side. We possess a + 2b = 800 The area A of the farmland is given by: A = a * b Using the first equation, we can write A as a function of b: A = A(b) = (800 - 2b) * b = 800b - 2b² We want to maximize A. First we need A' = 800 - 4b which means that the points of extrema are: A' = 0 => 800 - 4b = 0 => 4b = 800 => b = 200 m The length of the other side is a = 800 - 2 * 200 = 400 m and the largest area is A = 400 * 200 = 80000 m² Question 5 Four very long, current-carrying wires in the same plane intersect to form a square with side lengths of 45.0 cm, as shown in the figure below. The currents running through the wires are 8.0 A, 20.0 A, 10.0 A, and I. Part A Find the current I that will make the magnetic field at the center of the square equal to zero. Part B What is the direction of the current I (up or down)? Answer 1 Solution The magnetic field is given by B = μoI/(2πr); r = 0.225 m for each wire. B1 = 1.00 * 10^-5 T, B2 = 0.80 * 10^-5 T, B3 = 2.00 * 10^-5 T Let z be the positive z-direction I4 = 10.0 A, I2 = 8.0 A, I3 = 20.0 A B1 = -1.00 * 10^-5 T, B2 = -0.80 * 10^-5 T, B3 = +2.00 * 10^-5 T B1 + B2 + B3 + B4 = 0 B4 = -(B1 + B2 + B3) = -2.0 * 10^-5 T To give B in the direction the current in wire 4 must be toward the bottom of the page. B = μoI/(2πr) so I4 = rB / (μo/2π) = (0.225 m)(2.0 * 10^-5 T) / (2 * 10^-7 T*m/A) = -2.0 A Answer 2 (A) Consider the direction 'into the page' to be negative and the direction 'out of the page' to be positive. The distance of each wire from the center of the square is half the side of the square. Thus, the distance R can be written as follows: R = a/2 Here, a is the side of the square. Substitute 45.0 cm for a in the above expression. R = 45.0 cm / 2 = 22.5 cm Use right-hand thumb rule to find the direction of the magnetic field at the center of the square due to the current 8 A. Place the thumb of right-hand in the rightward direction in which the current is flowing. Curl the fingers around the wire to see that the magnetic field at the center of the square to be into the page. According to the assumed sign convention, the direction of magnetic field in this case should be negative. The magnetic field at the center due to 8.0 A wire can be written as follows: B8A = μoI/(2πR) Substitute 8.0 A for I in the above expression. B8A = μo(8A)/(2πR) The magnetic field at the center due to 20.0 A wire can be written as follows: B20A = μoI/(2πR) Substitute 20.0 A for I in the above expression. B20A = μo(20A)/(2πR) The magnetic field at the center due to 10.0 A wire can be written as follows: B10A = μoI/(2πR) Substitute 10.0 A for I in the above expression. B10A = μo(10A)/(2πR) Now, the magnetic field at the center of the square is 0. Thus, the sum of all the magnetic fields must be equal to zero. B8.0A + B20.0A + B10.0A + B = 0 Here, B is the magnetic field in the 4th wire. Substitute μo*(8A)/(2πR) for B8A, μo*(20A)/(2πR) for B20A, and μo*(10A)/(2πR) for B10A in the above expression. μo*(8A)/(2πR) + μo*(20A)/(2πR) + μo*(10A)/(2πR) + B = 0 B = - (μo/(2πR)) * (8A - 20A + 10A) B = - (μo/(2πR)) * (-2A) The magnetic field due to the 4th wire is negative which means that the direction of magnetic field is 'into the page'. On comparing with the standard equation of magnetic field due to a current carrying wire, the current in the 4th wire is as follows: B = μoI/(2πR) Substitute -(μo/(2πR)) * 2A for B in the above expression. -(μo/(2πR)) * 2A = μoI/(2πR) I = -2 A (B) Due to wire with 8.0 A current, the magnetic field points 'into the page'. This can be verified by right-hand thumb rule. Place the right hand in the direction of current 8.0 A. Curl the finger around it. The direction in which the fingers point at the center of the square gives the direction of the magnetic field due to this wire at the center. The direction comes out to be 'into the page'. Similarly, the direction of 20.0 A wire is 'out of the page'. Also, the direction of 10.0 A wire is 'into the page'. Since, the magnetic fields due to 8.0 A and 10.0 A points 'into the page', they will add up. This added quantity is still less than the magnetic field due to 20.0 A wire. Thus, there should be another 2.0 A current in the 4th wire, in a direction such that the remaining 'out of the page' magnetic field gets cancelled. For this, the magnetic field due to the 4th wire should point 'into the page'. Again, by right-hand thumb rule, curl the fingers around the 4th wire in such a way that the fingers goes into the page, the direction of the thumb will give the direction of the current in the 4th wire. The direction of current comes out to be 'downwards'.