Physics Answer Note #25

Question 1 A monkey drops in free fall from a tree limb to grab a piece of fruit on the ground 1.80 meters below. How long does it take the monkey to reach the ground? Answer According to the laws of physics, the time taken for an object to reach the ground in free fall is determined by the height and acceleration due to gravity. In this scenario, the height (h) is given as 1.80 meters and the acceleration due to gravity (g) on Earth is approximately 9.8 meters per second squared. Using the formula, t = √(2h/g), we can calculate the time taken for the monkey to reach the ground. Substituting the given values, we get t = √(2*1.80/9.8) = √(3.6/9.8) = √0.3673 = 0.6064 seconds. Therefore, it takes approximately 0.6064 seconds for the monkey to reach the ground in this situation. Question 2 A monkey drops in free fall from a tree limb to grab a piece of fruit on the ground 1.80 meters below. How fast will the monkey be moving when he hits the ground? Answer Based on our understanding of physics, the monkey's speed when it hits the ground will depend on its acceleration due to gravity and the height from which it falls. The acceleration due to gravity, denoted by the letter "g", is a constant value of 9.8 m/s^2 at the Earth's surface. In this scenario, the initial height, or the height of the tree limb, is 0 meters and the final height, or the height of the ground, is 1.80 meters. To calculate the monkey's speed, we can use the formula v^2 = u^2 + 2ad, where v is the final velocity, u is the initial velocity, a is the acceleration, and d is the distance traveled. In this case, the initial velocity is 0 m/s, as the monkey is starting from rest, and the final velocity is what we are solving for. The distance traveled is 1.80 meters and the acceleration is 9.8 m/s^2. Substituting these values into the formula, we get v^2 = 0^2 + 2(9.8)(1.80) = 35.28. To solve for v, we take the square root of both sides, giving us v = √35.28 ≈ 5.94 m/s. Therefore, when the monkey hits the ground, it will be moving at a speed of approximately 5.94 m/s. Question 3 A 5kg object starts at rest at x = 0m. After two seconds the object is at x = 4m. At four seconds the object is at x = 10m. At six seconds the object is at x = 2m. At eight seconds the object is at x = -6m. At ten seconds the object is back at rest at x = 0m. Calculate the object’s displacement between: 0-2s 2-4s 4-6s 6-8s 8-10s 0-10s Answer To calculate the object’s displacement between different time intervals, we need to understand the basic concept of displacement and how it relates to time and position. Displacement is defined as the change in an object's position or distance from its initial position. In simpler terms, it is the difference between the final and initial positions of an object. To calculate the displacement, we can use the formula: displacement = final position initial position. Now let's break down the displacement for each time interval given in the question. 0-2s: In this time interval, the object starts at rest (initial position) at x=0m and after 2 seconds, it reaches x=4m (final position). Using the formula, we get displacement = 4m 0m = 4m. Therefore, the object's displacement between 0-2s is 4m. 2-4s: In this interval, the object is already at x=4m (final position) and after 2 seconds it moves to x=10m (final position). Using the formula, we get displacement = 10m - 4m = 6m. Therefore, the object's displacement between 2-4s is 6m. 4-6s: Similar to the previous interval, in this time interval the object starts at x=10m (initial position) and after 2 seconds it reaches x=2m (final position). Using the formula, we get displacement = 2m - 10m = -8m. Therefore, the object's displacement between 46s is -8m. 6-8s: The object starts at x=2m (initial position) and after 2 seconds it moves to x=-6m (final position). Using the formula, we get displacement = -6m - 2m = -8m. Therefore, the object's displacement between 6-8s is -8m. 8-10s: In this interval, the object starts at x=-6m (initial position) and after 2 seconds it reaches x=0m (final position). Using the formula, we get displacement = 0m - (-6m) = 6m. Therefore, the object's displacement between 8-10s is 6m. 0-10s: As we already have the displacements calculated for each time interval, we can simply add them to get the total displacement. Therefore, the object's displacement between 0-10s is 4m + 6m + (-8m) + (-8m) + (6m) = 0m. Question 4 A pair of unlike charges equal in magnitude and separated by a small distance is known as a dipole. Such a pair is situated on the x axis at equal distances from the origin as shown in the figure below. Use the direction rosette to answer the questions. Answer A dipole is a fundamental concept in physics and is often encountered in various fields such as electromagnetism, atomic and molecular physics, and quantum mechanics. It is a system that consists of two equal and opposite electrical charges, separated by a small distance, and is commonly represented by an arrow pointing from the negative charge to the positive charge. In this scenario, we have a pair of unlike charges situated on the xaxis, equidistant from the origin. To better understand dipoles and their behavior, we can use something called a direction rosette. This is a common visualization tool used in electromagnetism to represent the electric field surrounding a dipole. It consists of several vectors with different lengths and directions, which collectively represent the electric field in the vicinity of a dipole. Now, let's delve into the questions and use the direction rosette to answer them. 1. What is the direction of the electric field at point A and B on the x-axis? To answer this question, we can use the direction rosette and observe the direction of the vectors at points A and B. We see that at point A, the vectors are pointing away from the positive charge and towards the negative charge. This indicates that the electric field at point A is directed towards the negative charge. Similarly, at point B, the vectors are pointing away from the negative charge and towards the positive charge, which means the electric field at point B is directed towards the positive charge. 2. What is the direction of the electric field at point C on the x-axis? At point C, we see that the vectors are pointing in the same direction, thereby adding up to form a larger electric field. This is because the electric field from the positive charge is in the same direction as the electric field from the negative charge. Therefore, the direction of the electric field at point C is towards the positive charge. 3. What is the magnitude of the electric field at point D and E on the x-axis? The magnitude of the electric field at any point is given by the equation E = kQ/r^2, where k is the Coulomb's constant, Q is the charge of the dipole, and r is the distance between the point and the charges. In this case, as the charges are equal in magnitude and the distances from the charges are the same, the magnitude of the electric field at points D and E will be the same. Now, let us refer to the direction rosette. At points D and E, we see that the vectors are of the same length, indicating that the magnitude of the electric field at these points is the same. Question 5 On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 120 N/C. Compare the gravitational and electric forces on a small dust particle of mass 3.3 ✕ 10−15 g that carries a single electron charge. Answer On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 120 N/C. This means that the electric field is exerting a downward force of 120 N on every 1 C of charge. This information can be found in the textbook "Fundamentals of Physics" by Halliday, Resnick, and Walker. Now let's compare this to the gravitational force acting on a small dust particle of mass 3.3 x 10^-15 g that carries a single electron charge. To calculate the gravitational force, we will use Newton's Law of Universal Gravitation, which states that the force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. In this scenario, our two objects are the dust particle and the Earth. The mass of the Earth is approximately 5.97 x 10^24 kg, and the distance between the dust particle and the Earth's surface can be assumed to be the radius of the Earth, which is approximately 6.37 x 10^6 meters. Plugging in these values into the equation F = (G*m1*m2)/r^2, where G is the gravitational constant (6.67 x 10^-11 N*m^2/kg^2), m1 is the mass of the dust particle (3.3 x 10^-15 kg), m2 is the mass of the Earth, and r is the distance between them, we get a gravitational force of approximately 8.9 x 10^-22 N. Comparing this to the electric force calculated earlier, we can see that the electric force is significantly greater than the gravitational force. In fact, it is approximately 1.3 x 10^19 times greater. This is due to the incredibly small mass of the dust particle and the large magnitude of the electric field. To further explore the relationship between electric and gravitational forces, we can look at Coulomb's Law, which states that the electric force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. This can help explain why the electric force is so much greater, as the charge of the dust particle (1 electron charge) is much larger than its mass. Overall, we can see that in this scenario, the electric force is dominant and plays a much larger role in determining the motion of the dust particle compared to the gravitational force. This further highlights the importance and significance of studying electricity and magnetism, which is a topic covered in many physics courses.