Tutor profile: Jason T.
Subject: Physics (Newtonian Mechanics)
A ball at rest is kicked from the ground at a 45 degree angle. The initial velocity of the ball is 10.0 m/s. How long was the ball in the air? How far away from where the ball was kicked did it hit the ground? What was the maximum height of the ball? Neglect wind resistance.
To find the time the ball was in the air, we must first find the time it takes for the ball to reach it's maximum height. The equation for this is Vf = Vi*sin(x) - g*t. When the ball reaches its maximum height, the final velocity of the ball in the y direction is zero. For x = 45 degrees and g =9.81 m/s^2, we solve for t. t = (0 - 10.0*sin(45))/9.81 = 0.721 seconds. Since no other forces act on the ball and the ground is assumed to be level, the ball will reach the ground in the same amount of time as it took to reach its maximum height. Therefore, the ball remains in the air for 2*0.721 seconds, or 1.44 seconds. Knowing the time the ball was traveling, we can calculate the distance the ball traveled with the following equation: Xf = Xi + Vi*cos(x)*t. Since the ball was initially at rest, Xi is zero. Solving the equation, we find Xf = 0 + 10.0*cos(45)*1.442 = 10.2m. The ball travelled 10.2m from which the ball was kicked. Using the time it takes for reaching the ball's maximum height (or half of the total time in the air) and the equation Xf = Xi +Vi*sin(x)*t - 0.5*g*t^2, we can find the maximum height of the ball. Xf = 0 + 10*sin(45)*0.721 - 0.5*9.81*0.721^2 = 2.55m high
Given a triangle with a base length of 5 inches and a height of 30.48 cm, find the length of the hypotenuse, the area of the triangle, and the acute angle between the base and the hypotenuse.
By first converting the height to inches (30.48cm / 2.54 (cm/in)) the height is found to be 12 cm in length. Using the Pythagorean Theorem (a^2+b^2=c^2) we find the hypotenuse is 13 cm in length. This could have also been determined by simply identifying this is one of the special Pythagorean triangles, a 5-12-13 triangle. The area of the triangle is found using the formula 0.5*b*h. The area of the triangle is 0.5*5*12 = 30 cm^2 The acute angle between the base and the hypotenuse is found by using the inverse function of one of our trigonometric functions; sine, cosine, and tangent. It can be found by arctan(12/5), arcsin(12/13), or arccos(5/13) and they all come to the same answer: 67.38 degrees.
If 250mL of Sodium Hydroxide solution with a concentration of 0.400M is mixed with a 0.10M Hydrochloric Acid solution in a titration experiment, what volume of Hydrochloric Acid is used to reach the equivalence point? How many grams of Sodium Chloride are in solution at the completion of the experiment?
First, balance the chemical equation: NaOH + HCl -> NaCl + Water (H20). The number of moles of Sodium Hydroxide is equal to the molar concentration of the solution multiplied by the volume of the solution: 0.250L X 0.400M = 0.100 moles NaOH. We know by the balanced chemical equation, 1 mole NaOH reacts with 1 Mole HCl so 0.100 moles NaOH must react with 0.100 moles of HCl to completely react all of the NaOH and form a neutral aqueous solution of NaCl and water. We also know by the balanced chemical equation that 1 mole of NaCl is produced for every 1 mole of NaOH reacted. Therefore, by inspection, we know that 0.100 moles of NaCl is produced when 0.100 moles of NaOH has reacted. The amount of 0.10M HCl solution required to completely react all of the NaOH (the equivalence point) is found by dividing the number of moles of HCl by the molarity of the solution: 0.100 moles HCl / 0.10M = 1.0L. To reach the equivalence point in the titration experiment, 1.0 liter of 0.10M HCl solution is used. The number of grams of NaCl in the solution at the end of the experiment is found by multiplying the number of moles of NaCl produced by the molecular weight of NaCl. The molecular weight of NaCl is 58.443 grams/mole. Therefore, the mass of NaCl produced is 0.100 moles * 58.443 grams/mole = 5.8443 grams. By proper use of significant figures, the mass of NaCl produced is 5.84g.
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