I'm working on a group report that requires a consistent and accurate substitution of values, as well as the proper application of physics concepts in each segment. Could you please assist me by providing the appropriate sub-values and diagrams to ensure the report's clarity? Furthermore, I would greatly appreciate your help in solving some practice problems by demonstrating the step-by-step process.

\fSpiderman, the beloved superhero of comic book fans and moviegoers alike. has captured our imagination with his extraordinary web slinging abilities. from swinging through the city with ease to effortlessly lifting heavy objects. to ghting crime using his intelligence and quick thinking. In this report. we will be exploring the physics behind Spiderman's amazing web swinging abilities through the use of kinematics. newton's laws, work and energy, momentum and collisions. and rotation and moment of inertia. Kinematics Spiderman's web swing is a classic example of projectile motion. which can be analyzed using the principles of kinematics. To calculate the motion of Spiderman's web swing, we will need to know the initial velocity, the angle of projection. and the height of the starting point. Diagram or smth Let's assume that Spiderman starts his swing from rest on top of a building with a height of 100 meters. He then shoots his web at an angle of 45 degrees from the horizontal with an initial velocity of 20 mls. (Neglecting air resistance) Using the equations of kinematics, we can calculate the maximum height and the range of Spiderman's swing. The equations we will use are: - Vertical displacement (Ay) = Voyt + 0.5a't2. where Voy is the initial vertical velocity. t is time, and a is acceleration due to gravity (-9.81 misz). - Horizontal displacement (Ax) = an*t, where an is the initial horizontal velocity. - Maximum height (h) = Ay(max) = (VDyWZa - Range (R) = Ax = (Vox)2*sin(29)lg. where B is the angle of projection and g is acceleration due to gravity. First. let's find the initial vertical and horizontal velocities of Spiderman's swing. Since the angle of projection is 45 degrees, we know that the initial vertical and horizontal velocities are equal. Voy = Vosin(9) = 205in(45) = 14.14 mls Vex = Vocos(B) = 20cos(45) = 14.14 mfs Next. let's find the time it takes for Spiderman to reach the maximum height of his swing. At the maximum height. the vertical velocity is zero. so we can use the equation Ay(max) = (Voy)2!2a to find the time t. Ay(max) = (Voy)2!2a h = (14.14)2f(2*-9.81) = 10.2 m t = -Voyfa = -14.14f9.81 = 1.44 s Momentum and Collisions Spider-Man's web swing also involves the conservation of momentum and the principles of collisions as he changes direction during his swing. Let's assume Spider-Man weighs 68 kilograms and is swinging at a constant speed of 30 meters per second, and then collides with a building at a 45-degree angle with the horizontal. We can use the principle of conservation of momentum to calculate the velocity and direction of Spider-Man after the collision. The momentum of SpiderMan before the collision is given by: p1 = mv1 where m is his mass and v1 is his velocity before the collision. Substituting the values. we get: p1 : 68 kg x 30 mfs = 2,040 kg-mis After the collision, Spider-Man's momentum will still be conserved. The momentum of Spider-Man after the collision is given by: p2 = mv2 where v2 is his velocity after the collision. Since Spider-Man collides with the building at a 45degree angle, we can resolve his velocity into horizontal and vertical components. The horizontal component of his velocity will remain the same, while the vertical component will change direction. Let's assume the horizontal component of his velocity is 20 meters per second. and the vertical component is 10 meters per second (since the collision is at a 45-degree angle, the components will be equal). The momentum of Spider-Man after the collision is given by: p2 = m(20 mfs) - m(10 mfs) p2 = 680 kg-mfs Equating the momentum before and after the collision, we get: p1 = p2 2.040 kg-mls = 680 kg-m/s + m(10 mis) Solving for m, we get: m = 124 kg This means that Spider-Man's momentum after the collision is 680 kg'mfs, and his velocity can be calculated using the horizontal and vertical components. The horizontal component of his velocity after the collision is 20 meters per second. as it remains unchanged. The vertical component of his velocity after the collision can be calculated using the conservation of energy principle. Since the coilision is perfectly elastic (i.e., no energy is lost), the total energy before and after the collision will be conserved. The kinetic energy before the collision is given by: KE1 = (1.'2)mv1Z KE1 = (1(2) x 68 kg x (30 misV KE1 = 30.600 J The kinetic energy after the collision is given by: KE2 =(1.'2)mv22 KE2 = (112) x 124 kg x v22 Since the total energy is conserved. we can equate the kinetic energy before and after the collision: KE1 = KE2(1f2)mv12 =(1i2)mv22 Work and Energy Spider-Man's web swing involves the conversion of potential energy into kinetic energy, as well as the application of work to overcome the force of gravity and air resistance. Let's assume Spider-Man weighs 68 kilograms and swings from rest at a height of 50 meters above the ground. using his webbing to swing at a constant speed. We can use the principle of conservation of energy to calculate the speed at which he is swinging. The potential energy of Spider-Man at the starting point is given by: PE = mgh where m is his mass. 9 is the acceleration due to gravity (9.81 misz), and h is the height above the ground. Substituting the values. we get: PE = (68 kg) x (9.81 misz) x (50 m) = 33.366 J At the bottom of the swing. all of this potential energy has been converted into kinetic energy. The kinetic energy of Spider-Man is given by: KE = (1/2)mv2 where v is his velocity. Equating the potential energy at the starting point to the kinetic energy at the bottom of the swing, we get: PE = KE mgh =(1i'2)mv2 Solving for v, we get: v = v'(Zgh) Substituting the values. we get: v = \\l(2 x 9.81 mis2 x 50 m) = 31.3 m/s This is the speed at which Spider-Man is swinging at the bottom of his swing. To calculate the work done by Spider-Man to overcome the force of gravity and air resistance, we can use the work-energy principle. The work done by Spider-Man is given by: W=KE-PE Substituting the values. we get: W = (1!2)mv2 mgh W = (1f2) x 68 kg x (31.3 m/s)2 68 kg at 9.81 mils2 x 50 m W = 33,366 J - 16.834 J W = 16,532 J This is the work done by Spider-Man to swing from rest at a height of 50 meters above the ground. \fSo. as Spiderman pulls his legs in, his angular velocity increases, which allows him to swing faster. \fNow, we can use the range equation to find the horizontal distance Spiderman travels during his swing. R = Ax = (Vox)'sin(26)lg R = (14.14)'sin(90)i9.81 = 204.1 m Therefore, Spiderman travels a maximum height of 10.2 meters and a range of 204.1 meters during his swing. Newton's Law Spiderman's web swing can be analyzed using Newton's laws of motion. Specically, we can use Newton's second law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration: F_net = ma In this case, we can consider Spiderman as an object of mass m, swinging through the air with a certain acceleration a. The net force acting on Spiderman is the tension in the web. T, which is pulling him towards the center of the swing, minus the force of air resistance, F_air, which is pushing him backwards. We can express this as: F_net = T - F_air = ma Solving for T, we get: T = ma + F_air Now we need to consider the specific values of m. a. and F_air. Let's assume that Spiderman weighs 75 kg and is swinging in New York City, where the air resistance can be estimated as F_air = 0.5pAv2, where p is the density of air, A is the crosssectional area of Spiderman's body, and v is his velocity through the air. Assuming Spiderman is swinging at a speed of 20 mis and has a cross-sectional area of 1.5 m"2, we can calculate F_air as: F_air = 0.5pAv'2 = 0.5 x 1.2 kglmhs x 1.5 m"2 x (20 mis)"2 = 360 N Now we can substitute in our values for m. a. and F_air into our equation for T: T = ma + F_air = (75 kg) x (a) + 360 N To nd Spiderman's acceleration. we can use the formula for circular motion, which is a = v"2!r, where r is the radius of the swing. Let's assume Spiderman is swinging in a circle with a radius of 15 meters. Then we can calculate his acceleration as: a = v"2ir = (20 m!s)"2 i 15 m = 26.7 mis"2 Substituting this value into our equation for T, we get: T = (75 kg) x (26.7 mis"2) + 360 N = 2,007 N 80 the tension in Spiderman's web as he swings through the air is approximately 2,007 N. Substituting the values, we get: (172) x 68 kg x (30 mi's)2 = (172) x 124 kg x v22 v2 = v'uss kg! 124 kg) x (30 m/s)'] v2 = 20.7 mls Therefore, after the collision, Spider-Man's velocity is 20 meters per second horizontally, and 10 meters per second upwards vertically. Rotation and Moment of Inertia Let's consider some substitute values to explain the physics behind Spiderman's web-swinging. Assuming that Spiderman has a mass of 75 kg, the length of his webbing is 10 meters, and the thickness of his webbing is 1 cm. Let's also assume that his pivot point is located at the center of his chest and his body shape and position remain constant during the swing. To calculate the torque generated by the tension in the webbing, we can use the formula: Torque = Force x Distance The force acting on Spiderman is equal to the tension in the webbing, which can be calculated using the formula: Tension = Mass x Acceleration Since Spiderman is swinging in a circular path, we can use the formula for centripetal acceleration: Acceleration = (VelocityA2)f Radius Let's assume that Spiderman swings at a velocity of 30 meters per second and the radius of his swing is 20 meters. Then, we can calculate the tension in the webbing as: Tension = 75 kg x (30 mfs)"2 I 20 m = 3375 N Substituting this value into the formula for torque, we get: Torque = 3375 N x 10 m = 33,750 Nm This torque causes Spiderman to rotate around his pivot point. The rate of rotation is determined by his moment of inertia, which depends on his body position and shape. Let's assume that Spiderman's moment of inertia is 3 kg.m"2 when his legs are extended and 2 kg.m"2 when his legs are pulled in. As he swings, he changes his body position from fully extended to fully contracted and back again. Using the formula for angular momentum: Angular Momentum = Moment of Inertia x Angular Velocity We can calculate Spiderman's angular momentum as he swings. Let's assume that his angular velocity remains constant at 10 radians per second. Then, his angular momentum when his legs are extended is: Angular Momentum = 3 kg.m"2 x 10 radi's = 30 kg.m"'2.ls When his legs are pulled in, his moment of inertia decreases to 2 kg.m"2. Therefore, his angular velocity increases to maintain the same angular momentum: Angular Velocity = Angular Momentum I Moment of Inertia Angular Velocity = 30 kg.m"2!s i' 2 kg.m"2 = 15 radls