Solutions By Email

Physics problems, solutions to email in these areas: measurement and units; kinematics in one, two and three dimensions; vectors; force, movement, dynamics, circular movement; gravitation; work, energy and power; momentum, collisions, impulse; rotational movement; static of a rigid body; fluids, density, pressure; oscillatory movement, vibration; sound; temperature, kinetic theory; heat: first thermodynamics law, second thermodynamics law; relativity; electromagnetism: Coulomb's Law, electric field, electric potential, Gauss' Law for the electric field, Ohm's law, Ampere Law, Faraday's Law, Maxwell Equations, electromagnetic waves, photons; optics: reflection, refraction, polarization, interference, diffraction. Physics Homework: Problems - Solutions: for more details into what type of problems we can attend, visit: Physics Problems, Assignments Sent Via EmailPHYSICS PROBLEMS: Kinematics As an illustration, here is a problem with its solution for you to see how it is sent by email. Some artillery men place an old cannon over the sea level at the top of a cliff. The projectile is shooted horizontally with an initial velocity Vi. The cannon is 60 m above the sea level. The time taken since the instant the projectile is shot until its impact is heard by the artillery men, is 4.0 s. Knowing that the sound velocity in the air is 340 m/s, calculate the horizontal distance x from the impact point on the sea to the base of the cliff and the initial velocity Vi of the projectile. ANSWERS : x = 159 m; V_{i}= 45.4 m/s Placing the origin of the coordinate system in the sea along with the cliff and making x to show the horizontal distance and y for the altitude, we have: (A) x = V_{i}cost (B) y = y_{i}+ V_{i}sen t - (1/2)gt^{2}. y_{i}is the initial vertical position. In this case y_{i}= 60 m, V_{i}is the magnitude of the initial velocity, is the angle that the initial velocity vector makes with the x axis, in this case = 0º because the shot is horizontal. So sen0º = 0 y cos0º = 1. The time the projectile takes to reach the water is obtained doing y = 0. y = 0 = 60m + V_{i}sen0º t - (1/2)gt^{2}. Then, (1/2)gt^{2}= 60m, from where t = 3,5 seconds. Then, the time the sound takes to travel the distance between the impact point and the cannon is 4s - 3,5s = 0,5s. The distance between the impact point and the cannon is calculated using the known relation: distance = velocity*time = 340 m/s*0.5 s = 170 m. These 170 m form the hypotenuse of the rectangle triangle with 60 m and x as the sides, then x^{2}= (170m)^{2}- (60m)^{2}. Solving x in this triangle rectangle, x = 159 meters. The initial velocity, when the shooting ocurred, can be calculated with the equation (A) x = V_{i}cos t 159m = V_{i}*1*3,5s Then V_{i}= 45,4 m/s.

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