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# Study Guide PH2213

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This 13 page Study Guide was uploaded by Deborah Oluwadamilola Eseyin on Tuesday August 30, 2016. The Study Guide belongs to PH2213 at Mississippi State University taught by Dr. Chun Su in Fall 2016. Since its upload, it has received 18 views. For similar materials see Physics I for Scientists and Engineers in Physics at Mississippi State University.

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Date Created: 08/30/16

Study Guide for Physics 1 Test1. Worked Examples for Chapters 1, 2 and part of chapter 3. Hey guys so we made it past Chapter 1, 2 and part of chapter 3, and now we have our first test coming up. This may seem like a lot but honestly all you have to do is take a deep breath, read over your question once or twice, draw a table of what you know and what the question wants you to find. Once you go through these steps, you won’t struggle with finding the right formula to use. This Physics test maybe a little confusing for some people but I did physics for 3 years in high school, and using those steps makes it way easier to figure out what formula you need and once you’ve done that you are one step closer to finding you answer. So for this study guide I am going to solve all the practice questions he sent through the email and also the spring test 1. Examples: 1. What roughly is the percent uncertainty in the volume of a spherical beach ball whose radius is r = 0.84 + 0.04m Solution: 2 So the volume of sphere is v = 4/3(πr ) 2 And the area of a sphere is a = 4πr Next we find the derivative of the volume: 3 dv = 4/3(πr )/3 2 dv = 4πr dr Next we are going to divide dv by v Dv/v = 4πr dr ÷ 4(πr )/3 Dv/v = 4πr dr x 3/4(πr ) Dv/v = 3dr / r Dr(derivative of the radius) = 0.04 And r = 0.84 So 100 x 3(0.04/0.84) 100 x 3(0.0476) = 100 x 0.14286 The percentage therefore is equal to 14.29% 2. The diameter of the moon is 3480km. (a) what is the surface area of the moon? (b)How many times larger is the surface area of the Earth Solution: 2 (a). To find the surface area you are going to use the formula S.A. = 4πr Also r = d/2, r = 0.5(3480), r = 1740km S.A = 4(3.142)(1740) 2 S.A = 3.8 × 10 km of the moon. (b). The diameter of the earth is 12,756km. So we just repeat the steps we used in a to find the S.A of the moon then divide to find the ratio. R = d/2, r = 0.5(12,756), r = 6378km 2 S.A = 4(3.142)(6378) 8 2 S.A = 5.11 × 10 km of the earth In order, to find the ratio we are going to divide the S.A of the earth by the S.A of the moon. Ratio = 5.11 × 10 km of the earth / 3.8 × 10 km of the moon. The surface area of the earth is 13.45 times greater than the surface area of the moon. 3. Estimate how long it would take one person to mow a football field using an ordinary mower. Assume the mower moves with a 1 – km/h speed, and has a 0.5 – m width. Solution: The dimensions of a football field is 64m by 100m and the mower has a cutting width of 0.5m. So to find the time, we have to find the distance first. D = Area of field = 64 x 100 = 6400 Width of mower = 0.5m = 0.5 D = 12800m Now let’s pause for a minute our answer is given in m and we have to change it to secs. In order to do that, we have to go back through the question and then we find 1km/hr so that is going to help us convert efficiently. 12800m 1 km 1hr 1000m 1 km It’s going to take 12.8hrs to mow the field. 4. Hold your pencil in front of your eye at a position where its blunt end just blocks out the moon. Make the appropriate measurements to estimate the diameter of the moon, given that the earth – moon distance is 3.8 x 10 km. Solution: This is the answer given to us by Dr. Su. In this solution, he assumed the distance from -2 the eye to the pencil is 0.7m , the diameter of the pencil is 0.6 x 10 m. Right now we have two triangles one small one with base 0.7 and height 0.6 x 10 m, and a second triangle with base 3.8 x 10 km, but with no height. So we can use the ratio to ratio 5 formula to find the height. Before we do that we need to convert 3.8 x 10 km to m 3.8 x 10 km 1000m 8 1km = 3.8 x 10 m B 1 D 1 0.7 = 3.8 x 10 8 B 2 D 2 0.6 x 10-2 = D 2 Next we cross multiply, -2 8 0.7 x D 2 0.6 x 10 x 3.8 x 10 0.7 D 2 2.28 x 10 6 D 2 2.28 x 10 ÷ 0.7 6 D 2 3.26 x 10 m 6 3.26 x 10 m 1km 1000m D 2s 3.26 x 10 m or 3.26 x 10 km 3 5. Digital bits on a 12.0cm diameter audio CD are encoded along an outward spiraling the path that starts at radius R1= 2.5cm and finishes at radius R =25.8cm. The distance between -6 the centers of neighboring spiral windings is 1.6μm ( 1.6 x 10 m). (a). Determine the total length of the spiraling path.[Hint: Imagine “unwinding” the spiral into a straight path of width 1.6μm, and note that the original spiral and the straight path both occupy the same area]. (b).To read information, a CD player adjusts the rotation of the CD so that the player’s readout laser moves along the spiral path at a constant speed of 1.25 m/s. Estimate the maximum playing time of such a CD. Solution: a). We are going to find the length. Since it’s a disc it’s a circle so we are going to use the area of a circle then since there’s a straight path then we are going to use the area of a rectangle since we were given the width so we can use A = LW to find L.. A = π(R – r )2 A = LW So, π(R – r ) = LW 3.142(5.8 – 2.5 ) = L x (1.6 x 10 )-6 -6 3.142(33.64 – 6.25) = L x (1.6 x 10 ) -6 3.142(27.39) = L x (1.6 x 10 ) 86.06 = L x (1.6 x 10 )-6 L = 86.06cm / (1.6 x 10 )m-6 2 -6 2 1.6 x 10 m 100cm 100cm 1m 1m L = 86.06 / 0.016 L = 5378.75 m (b). We are going to use time equals distance/ velocity T = d/v T = 5378/1.2 T = 4482 secs 5. The position of a ball rolling in a straight line is given by x = 2.0 – 3.6t + 1.1t , where x is in meters and t in seconds. (a) Determine the position of the ball at t = 1.0s, 2.0secs, 3.0sec. (b) What is the average velocity over the interval t = 1.0s to t = 3.0secs? (c) What is its instantaneous velocity at t = 2.0 and at t = 3.0s Solution: (a) Since the ball is rolling in a straight line (vertical) we are going to use the vertical kinematic third equation which is, x = V t + 112at, and also acceleration is going to be +9.8m/s Since we have 3 times are we going to have 3 different positions, and since the ball just starts rolling the initial velocity is 0 So, x 1 V t 1 1.5(at ) 1 x 1 0.5(at )1 x 1 0.5(9.8 x 1) x 1 0.5(9.8) x 1 4.9m So, x 2 V t 1 2.5(at ) 2 X 2 0.5(at ) 2 X 2 0.5(9.8 x 2) X 2 0.5(19.6) X 2 9.8m So, x = V t + 0.5(at ) 3 1 3 3 X 3 0.5(at ) 3 X 3 0.5(9.8 x 3) X 3 0.5(29.4) X 3 14.7m (b) Average velocity = average distance / average time Average velocity = (14.7 – 4.9) / (3 – 1) Average Velocity = 9.8 / 2 Average Velocity = 4.9m/s (c) V 2 V + 1t 2 V 2 0 + 9.8(2) Instantaneous Velocity is V = 1926m/s V 3 V + 1t 3 V 3 0 + 9.8(3) Instantaneous Velocity is V = 2934 m/s Solutions for TEST 1 Spring 2016. 1. Three vectors A, B, and C are given ~s sh~wn i~ the figure below. Determine the magnitude and angle of 2A 2B + 3C. Y A (35) 48 X 60 50 C (30) B (45) So first we find x which is associated with cos and A yhich is associated with sin. Then take the tangent of A / A y x Ax= 35 cos(48) = 23.42 Ay= 35 sin(48) = 26.01 Bx= 45 cos(50) = 28.93 B = -45 sin(50) = -34.47 y Cx= -30 cos(60) = -15 Cy= -30 sin(60) = -25.98 D = 2(23.42) – 2(28.93) + 3(-15) x Dx= -56.02 D y 2(26.01) – 2(34.47) + 3(-25.98) D y 43.02 Direction: Tan θ= D y D x Tanθ = 43.02 / -56.02 Tanθ = 0.7670 Θ = tan (0.7670) 0 Direction is 37.5 in quadrant 2 Magnitude: V = D x D y V = √(-56.02) + (43.02) 2 V = 3138.2 + 1805.7 V = 4943.9 V = 70.65 So, the magnitude is 70.65 2. A lake which is roughly circular has an average depth of 7 m and and a 3 diameter about 1300 m. Estimate the volume of the lake (a) in m and (b)in gallons. Solution: r = d/2, r = 0.5(1300), r = 650m V = πr h V = 3.142(650)(650)(7) 6 3 9 (a). V = 9.29 x 10 m (b) V = 2.32 x 10 gallons 6 3 9.29 x 10 m 1 Lit3r 1 gallons 10 m 4 liters 3. A person throws a stone upward with a velocity of 18.0 m/s. Determine (a) how long this person will wait to catch the stone, (b) how high the stone can go, and (c) how fast it is travelling when it reaches a height of 8 m from the person's hand, and (d) how long the stone takes to reach this height. Solution: Known Wanted V 0 18.0 m/s T = ? This is what we have right now and with time we’ll fill our table up. (a) First of all, the total distance is 0 because it returns back to its origin (hand) and since we are going to use negative acceleration since the ball is going downward.(x = 0) 2 X = v 0 + 1/2at X = 18t + ½(-9.8)t 2 X = 18t – 4.9t 2 0 = t(18-4.9t) T = 0 , 18 – 4.9t = 0 T = 0, 18 = 4.9t T = 0 , t = 3.67secs (b) When the ball is at its highest point the final velocity is 0. So we can use v = v + 0ax so we want to find x 2 2 (0) = (18) + 2(-9.8)x 0 = 324 – 19.6x -324 = -19.6x X = -324/-19.6 X = 16.5m (c) We are going to use v = v + 0t to find v 2 2 V2= v +0a t2y v = (18.0) + (-9.8)(3.67) v = (18.0) – 35.966 v = ± 12.93m/s (d) V 2yV + a1y y 12.93 = 18 + (-9.8)(t) 12.93 = 18 – 9.8t 12.93 – 18 = -9.8t -5.07 = -9.8t T = 0.517secs - 12.93 = 18 + (-9.8)(t) - 12.93 = 18 – 9.8t - 12.93 – 18 = -9.8t -30.93 = -9.8t T = 3.156secs 4.A car can accelerate its velocity uniformly from 54 km/h to 72 km/h in 10 seconds. Calculate (a) the acceleration (in m=s ) of the car, (b) how far (in m) the car travels during this period, and (c) the distance the car can travel in the 4th second. Solution: Known Wanted v = 72km/hr a = ? V = 54km/hr X = ? 0 T = 10 secs X 4 ? (a).For this question we are going to use v = v + 0t, and since we are finding the acceleration we are going to solve for a (V – v 0 / t = a Velocity is usually m/s so we need to convert km/hr to m/s 72 km 1000m 1 hr 1 minute 1hr 1 km 60 minutes 60 seconds V = 20m/s 54 Km 1000 m hr 1 Minute 1 hr 1 km 60 Minutes 60 seconds V 0 15m/s A = (20 – 15) / 10 A = 0.5 m/s 2 (b). We are trying to find x so we are going to use x = v t + ½ (at ) 2 0 X = (15)(10) + ½ (0.5)(10) X = 150 + ½ (0.5)(100) X = 150 + ½ (50) X =150 +25 th X = 175m at the 10 second (c). We are trying to find the distance when the car starts moving until it’s th 4 second, so therefore, our t = 4 secs. We are still going to use the same formula used in (b) to find the distance 2 X 4 v t0+ ½ (at ) X 4 15(4) + ½ (0.5)(4)(4) X 4 60 + ½ (8) X 4 60 + 4 X 4 64m X 3 v t0+ ½ (at ) X 3 15(3) + ½ (0.5)(3)(3) X 3 45 + ½ (2.25) X 3 45 + 2.25 X 3 47.25m X Δ = X – X 4 3 X Δ = 64 – 47.25 X Δ = 16.75m 5. A helicopter is ascending vertically at a speed of 5.4 m/s. At a height of 110 m above the Earth, a package is dropped from a window. Calculate (a) the time it takes the package to reach the ground and (b) how fast the package can be when it almost hits the ground. Solution: Known Wanted V 0 5.4m/s T = ? X plane 110m, X book= -110m V f ? A plane +9.8m/s , A book= -9.8m/s2 X book V 0 + ½ (at ) -110 = 5.4(t) + ½ (-9.8)t 2 -110 = 5.4t – 4.9t 4.9t – 5.4t – 110 2 T= −????± ???? −4???????? 2???? T = -(-5.4) √ (−5.4) − 4(4.9)(−110) 2(4.9) T = (5.4) √ 29.16 + 2156 9.8 T = (5.4) ±√2185.16 9.8 T = 5.4 ± 46.75 9.8 T = 5.4 + 46.75 T = 5.4 – 46.75 1 2 9.8 9.8 T1= 5.32 secs T 2 -4.22secs Time can never be negative so the time it takes for the package to reach the ground is 5.32 secs (b). since we have initial velocity, time, acceleration. We can find the final velocity. V = v 0 at V = 5.4 + (-9.8)(5.32) V = 5.4 – 52.136 V = - 46.736 m/s (Its negative because the package is dropping down vertically)

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