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Louise Katreen V. Nimer Advanced Topics in Physics A. Define the following terms: 1. Reference point - It is a particular point from which measurements are made. Measurements only make sense if they are an interval of some sort. The philosophy of physics posits that the interval to be measured exists. Thus, it is logical to posit a reference point from which other measurements are taken. It is fixed position which helps in describing motion as a rest object should be chosen while describing motion. 2. Frame of reference - a framework that is used for the observation and mathematical description of physical phenomena and the formulation of physical laws, usually consisting of an observer, a coordinate system, and a clock or clocks assigning times at positions with respect to the coordinate system. 3. Distance - Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion in position. 4. Displacement - Displacement is a vector whose length is the shortest distance from the initial to the final position of a point P.[1] It quantifies both the distance and direction of an imaginary motion along a straight line from the initial position to the final position of the point. A displacement may be also described as a 'relative position': the final position of a point (Sf) relative to its initial position (Si), and a displacement vector can be mathematically defined as the difference between the final and initial position vectors. 5. Speed - Speed is distance traveled per unit of time. It is how fast an object is moving. Speed is the scalar quantity that is the magnitude of the velocity vector. It doesn't have a direction. A higher speed means an object is moving faster. A lower speed means it is moving slower. If it isn't moving at all, it has zero speed.

6. Velocity - Velocity is a physical vector quantity; both magnitude and direction are needed to define it. The scalar absolute value (magnitude) of velocity is called "speed", being a coherent derived unit whose quantity is measured in the SI (metric system) as metres per second (m/s) or as the SI base unit of (m⋅s−1). 7. Acceleration - Acceleration is a vector quantity that is defined as the rate at which an object changes its velocity. An object is accelerating if it is changing its velocity. B. Solve the following problems

1. A car passes a green traffic light while moving at a velocity of 5m/s. It accelerates at 0.300m/s2. What is the car’s velocity at 15.0s? Iv= 5m/s A= 0.300 m/s2 T= 15.0s Vf= vi + at = 5m/s + (0.300m/s2) (15.0s) = 5m/s + 4.5m/s = 9.5 m/s

2. A car initially traveling at 30k/s accelerates at a constant rate of 1.5m/s2. How far will the car travel in 10.0s? Vi= 30km/s A= 1.5m/s2 t= 10.0 s d= vit + 1/2at2 = (30,000m/s)(10.0s)+1/2(1.5m/s2)(10.0s)2 = 300,000m/s2 + ½(1.5m/s2) (100s) = 75m/s + 300,000 m/s = 300,075m

1. A ball is ascending at the rate of 10.0m/s at a height of 60.0m above the ground when a package is dropped. a. How long does it take the package to reach the ground? = 0+49m/s 9.8m/s2 = 5 secs. b. What is the speed of the package when it hits the ground? V= vi+vt 2 = 0+(-98m/s) 2 V= -49m/s

2. A ball was thrown vertically upward from the ground and reaches a height of 30m. a. With what speed was it thrown? D= vf2-vi2 -2g 30m= (0)2-vi2 -2g 2 Vi = -30m -2(9.8m/s2) Vi2= -30m -19.6m/s2 2 Vi = 1.53m2/s2 Vi= 1.24m/s

b. How long will it remain in air? T= vf-vi -g = 0- 1.24m/s -9.8m/s2 = 0.13s

3. A balloon is ascending at the rate of 10.0m/s at a height of 60.0m above the ground when a package is dropped. a. How long does it take the package to reach the ground? T= vf-vi -g = 0- 10.0 m/s -9.8 m/s2 = 1.02 s

b. What is the speed of the package when it hits the ground? Y= vf2-vi2 -2g = 0- 10.0 m/s -2(-9.8m/s2) = -10.0 m/s 19.6m/s2 Y=-0.51m/s 4. A bullet is fired horizontally with an initial velocity of 250m/s from a tower that is 20m high. a. Calculate the time the bullet is in the air. 2𝑦 T= √ 𝑔 2(20)

= √9.8𝑚/𝑠2 = √4.08 =2.02s b. If air resistance is neglected, find the horizontal distance the bullet travels before striking the ground. D= 250m/s(2.02s) D= 500m 5. A ball is thrown at 20.0m/s at an angle of 60⁰ above the horizontal. Neglecting friction a. How far away did it reach? R= vi2sin2(60⁰) 9.8 m/s2 = (20.0m/s)2 sin 2 (60⁰) 9.8m/s2 R= 85.47m

b. How high will it go? Y= 20.0m/s(1.77s)- ½(9.8m/s2)(1.77s)2 = 20.05m c. How long will it stay in air? 2t=2(visin60) 9.8m/s2 = 30.6 seconds

1. How many radians are there in half of a circle? - It follows that the magnitude in radians of one complete revolution (360 degrees) is the length of the entire circumference divided by the radius, or 2πr / r, or 2π. Thus 2π radians is equal to 360 degrees, meaning that one radian is equal to 180/π degrees. 2. A 1450kg car moving at 90kph goes around a curve of 30m radius. A) what is the centripetal acceleration in m/s2? B) what is the centripetal force on the car in Newton? A=v2 R = 25m/s 30m = 0.83m/s2 F= mv2 R F=1450kg (25) 30m F= 1,208 N

1. Convert the following temperatures to a) Celsius, b) Kelvin 98.6⁰F a) 98.6 ⁰F ⁰C= 5/9(98.6⁰F-32⁰) = 5/9(66.6⁰) = 37⁰C b) 98.6⁰F K= ⁰C+273 = 310 K 0⁰F a) 0⁰F ⁰C= 5/9(0⁰F-32⁰) = 5/9 (-32⁰) = -17.78⁰C b) 0⁰F K= ⁰C + 273 = -17.78⁰C +273 = 255.22 K 500⁰R a) 500⁰R ⁰R= ⁰F+460 500⁰R=⁰F+460 ⁰F= 40⁰F b) 500⁰R ⁰C=5/9(40⁰F-32⁰) = 5/9(8⁰) = 4.44⁰C

2. Convert to ⁰F and Rankine  36.8⁰C a. - ⁰F= 9/5(36.8⁰C + 32⁰) = 9/5(68.8⁰) = 123.84⁰F b. ⁰R= 123.84⁰F+460 = 583.84⁰R  5⁰C a. ⁰F= 9/5(5⁰C+ 32⁰) = 66.6⁰F b. ⁰R= 66.6⁰F+460⁰ = 526.6⁰R  -25⁰C a. ⁰F= 9/5(-25⁰C+32⁰) = 9/5(7⁰) = 12.6⁰F b. ⁰ R= 12.6⁰F+ 460⁰ = 472.6⁰K

3. At what temperature are the Celsius and Fahrenheit scales equal? - To find the temperature when both are equal, we use an old algebra trick and just set ºF =ºC and solve one of the equations. So the temperature when both the Celsius and Fahrenheit scales are the same is -40 degrees. 5/9(⁰F-32⁰)= 9/5(⁰C+32⁰) 3. A temperature change of 9⁰F corresponds to what temperature in ⁰C? - ⁰C= 59(9⁰F-32⁰) = 5/9(23⁰) = 12.78⁰F

1. How much heat is needed to raise the temperature of a 20-kg block of ice from -20⁰C to -5⁰? Q= Q1+Q2+Q3 Q1= (5 cal/g⁰C)(10⁰C-(-20⁰C)) Q1= 20,000 cal Q2= 2000g(80cal/g)= 160,000 cal Q3= mcWAt = 2000g(1cal/g⁰c) (-5⁰C-0⁰C) = - 10,000 cal Q= Q1+Q2+Q3 =20,000+160,000-10,000 Q= 170,000cal

2. Suppose at 20⁰C you have a 2.0m piece of copper wire. If you increase the temperature to 75⁰C, what will be its new length? ∞ for Cu= 1.7x10-5/⁰C L= Lo + AL = 2.0m+(1.7x10-5/C⁰)(2.0m)(55⁰C) = 110 m 3. A 50.0ml container made of Pyrex glass( ᴃ= 0.096x10-4/⁰C) is filled to the brim with benzene at 10⁰C. Will some benzene spill if the temperature is raised to 40⁰C? How much? For glass = 0.096x10-4/⁰C(50.0ml)(30⁰C) = 0.0144mL For benzene = 0.1240 × 10-6/⁰C(50.0ml)(30⁰C) = 1.86x10-4mL

4. Find the amount of thermal energy that flows through a concrete wall 10.0m long, 2.8m high, and 22.0cm wide, in a period of 24th, if the inside temperature of the wall is 20.0⁰C and the outside temperature is 5.00 ⁰C? k=3.10x10-4 kcal. ms⁰C t= 24hrs= 86,400s l= 10.0m h= 2.8m w= 0.0022m T1= 20.0⁰C T2= 5.00⁰C

a. = 3.10x10-4 kcal (2.8mx0.0022m)(86,400s)(20.0⁰C -5.00⁰C) ms⁰C = 2.4748416 kcal

5. What is the efficiency of an ideal engine that operates between the temperatures of 525k and 300k? =525k-300k 300k =.75 =75%

6. A steam engine takes superheated steam from a boiler at 200⁰C and rejects it directly into air at 100⁰C. What’s the ideal efficiency? = 200⁰C-100⁰C 100⁰C =1 = 100%

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