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Physics Exam 1 help

by: Corey Burr

Physics Exam 1 help PHYS 2350

Corey Burr
GPA 3.61

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Study problems and answers for the first exam in physics, both University and General physics.
University Physics I
Dr. E. Justiniano
Study Guide
Physics, University Physics, General Physics, Phys 1250, Phys 2350
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This 3 page Study Guide was uploaded by Corey Burr on Thursday February 18, 2016. The Study Guide belongs to PHYS 2350 at East Carolina University taught by Dr. E. Justiniano in Spring 2016. Since its upload, it has received 24 views. For similar materials see University Physics I in Physics 2 at East Carolina University.


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Date Created: 02/18/16
1. Which is a vector? Mass, temperature, volume, velocity 2. Someone falls from 25m into water below. A) How long do the fall?  h=1/2gt = 25m = ½ (10 m/s )(t)  2 2 t = 2.23 s B) How fast do they fall?  v =gt  v= 10 x 2.23 = 22m/s 3. Object thrown at 30 angle with horizontal and with speed of 20 m/s. A) How long is it in the air?  V = v sin(ᶿ) oy o o 2 2  (20)(sin 30 )= 10 m/s = upward speed. h= V t-1/2gt  0= 10oy5t = 5t(2-t)=0 t = 2s B) How high does it 2 2 get?  t=1s at top of peak. h= V t-1/2gt =oy0(1) – 5(1) = 5m = h C) How far in the horizontal direction does it travel?  V =ox cosoᶿ) = (20)(cos 30 ) = 17.3 m/s  Δx= v t = (17.3 m/s)(ox) = 35 m in horizontal direction 4. Rubber on concrete has a high coefficient of static friction (µ = 1). What ansle does the concrete need to be -1 o tilted for the rubber to slide down? µ = tan(ᶿ)s 1 = tanᶿ ᶿ= tan (1) = 45 5. What is the dimension of coefficient of static friction (µ )? no unitssb/c it is dimensionless 6. Rotating cylinder with r= 4km could be put in space and used as colony. Because of centripetal force we could walk like we do on Earth. What speed must the outer edge turn so the centripetal acceleration at the surface would equal the free fall acceleration of Earth?  (mv )/r = mg  v = (gr) = (10)(4000m) = 200 m/s 1/2 7. Motor cyclist is going through the curve with a speed that he has to lean into the curve to avoid falling. What formula describes balance that keeps him from falling?  tanᶿ= F /F gr cp 8. 2 objects, both 1kg, are 1m apart. Where’s the center of mass? -midpoint between the 2 objects 9. Wind from eat with a speed of 3 m/s. Cyclist is heading south with a speed of 4 m/s. What’s the magnitude of the wind speed he feels?  a + b = c  3 +4 = (25) 2 2 1/2= 5 m/s 10. Distance between two objects is reduced to 1/3 its original distance. What happened to the force? It increased by a factor of 9  F =G(mM)/r  r 2 2 1/3r = 9/r 2 gr 11. A car decelerates from 5 m/s to stand still in 1s. What statement is correct? – This is impossible because it 2 2 2 would take 2.5 m to come to a standstill, not 5m.  a=(v -v)/Δt = (5-0)/f=5im/s  Δx= 1/2at = ½(5)(1) = 2.5m 12. Soldier is filming how a 200 kg helicopter flies with speed of 10 m/s along a horizontal trajectory, then it explodes. The fragments fly apart. What happens to the fragments? – center of mass of fragments will keep moving forward with horizontal component of 10 m/s 13. Telecommunication satellites are often put in geostationary orbits. The revolution time of the satellite = full Earth rotation. Altitude of satellite = 35790 km above sea level. Earths radius is 6370 km. At what speed is the satellite moving around the Earth?  r= 35790+6370 = 42160km  v= (2πr)/(24hr) = 2π(42160)/(24 hr) = 11037 km/hr 14. Escape velocity of Mars is 5km/s. What’s the kinetic energy that a 1000 kg object needs to get from Mars’ surface to point of infinity away from Mars(escape energy of object)?E =1/2mv =½(1000)(5000) =12.5x10 J 2 9 10 2 esc 15. Steel, Young’s Modulus = 20 x 10 N/m . What is the elongation of a 25m long steel cable with a cross section -4 2 10 -4 area of 4x 10 m whit it is lifting a 800 kg object?  ΔL= 1(F)(L)/Y(A)  Y=(F/A)/(ΔL/L) = 1/(20x10 )(4x 10 ) 16. Network done in accelerating a propeller from rest to an angular speed of 200 rad/s is 3000 J. What is the 2 2 2 2 moment of Inertia of the propeller?  p=mv = L=IW; E =1/2mv = L= ½ IW kin000J= ½(I)(200) = .15 kgm 17. Billions of years from now the Sun will become White Dwarf. It will then shrink to a sphere 1% of its current radius. The sun’s full rotation is every month now. How fast will it rotate in White Dwarf stage?  I= 2/5MR ; 2 10000 times faster than it does now because angular momentum is conserved. 18. Elastic collisions: Momentum conserved, kinetic energy is NOT conserved. 19. Earth escape velocity is 11.2 km/s. Mass of Earth (M) is 4 times its original mass, but the radius is the same. 2 What would happen to the escape velocity now? – it would be twice as large.  E = 1/2mv =G(mM)/R  kin v =(2GM/R)  (4M) = 2M 1/2 esc 20. A child is perpendicularly pushing against a door with force of 100N at a distance 0.8m from hinges. What is the torque that they are applying? Τ=Fr  (100N)(0.8m)=80Nm 21. Seasaw I long, narrow, and pivoted in the middle. 2 kids on the seasaw, A & B, both 2m from pivot. If kid A is 10% lighter than kid B, what should happen for static equilibrium to occur? – Kid B that would have to move 1.8m from pivot (move closer to pivot than kid A) 22. Maximum compressive stress that iron can withstand before it fractures is 5.5 x 10 N/m . Weight density of8 2 4 3 iron is 7.9 x 10 N/m . How high can a vertical iron beam be before it crumbles under its own weight?  σ=ϱgh ; σ=(F/A)=5.5x 10 N/m 8 2 ϱ= 7.9x 10 N/m (g included in density amount)  h=(5.5x10 )/(7.9x10 )= 7000m= 8 4 7km 5 5 23. A 3 x 10 kg freight train is moving 2.5 m/s and hits a stationary car weighing 1.5x 10 kg. The two cars connect and end up moving together. What is the resulting velocity? M v = (m +m )v  (3x10 ) 5 1 1 1 2 f (2.5)=(3x10 +1.5x10 )v = 1.7 m/f 5 2 5 24. What is the involved energy loss when the two freight trains connect? E =1/2(3x10 )(2.5) = 9i38x10 J E = f 1/2(3x10 + 1.5x10 )(1.7) = 6.5x10 J  ΔE=E-E = 2.9 x 10 Ji f 5 25. Earth makes full rotation around axis in 24 hours. What’s the angular speed of rotation? Angular Velocity= -5 (angle Covered)/(Δt) = (2π)/(24)(3600s) = (1/24)(1/3600)(2π) = 7.27x10 m/s 26. Coin with radius of 2.4 cm is rolling on the edge with speed of 20 cm/s. What’s the angular speed? v = 20 cm/s r=2.4 cm  v=wr  w=v/r = 20/2.4 = 8.3 rad/s= w 27. Which has a larger moment of inertia, a solid cylinder or a hollow sphere? I = 2/5MR 2 I = 1/2MR , ½ is larger sp c than 2/5 so the cylinder has the larger moment of Inertia. 28. A reel has a radius of 0.25 m and a mass of 3 kg. Reel is a solid cylinder. What’s the moment of Inertia for the reel? I=1/2MR = ½(3kg)(0.25) = 9.4x 10 kgm -2 2 29. Force 50N is pulling on light string wrapped around the reel. What is the resulting angular acceleration of the -2 reel? Τ=(F)(r)= (50)(0.25m)=12.5  τ=Iα = 12.5=(9.4x 10 )α  α=133 rad/s 30. Trailer in meadow during a hurricane has a flat horizontal roof with an area of 10m , no wind inside, and a 3 speed of wind blowing over the roof of 60 m/s. Density of air is 1 kg/m . Which is correct? – Due to Bernoulli’s effect, 18000 N is trying to lift the roof off the trailer.  ΔP=1/2ϱ(v -v 2 =1½(1)(60 ) = 1800N/m 2 F=(P)(A) = (10)(1800) = 18000N 31. Piece of wood floating in water is 50% volume above the water. What is true? – There is static equilibrium where upward buoyancy force is the same as the downward gravitational force on the wood. (F =F ) B gr 32. Ball thrown straight down from balcony at a velocity of 5 m/s. It hits the ground in 2 seconds. What was the 2 2 height of the building?  h= V t+1o2gt =(5)(2)+1/2(10)(2) = 30m 33. Rock thrown up vertically on edge of cliff. Max height is 20m above top of the cliff before falling to the base of 2 2 the cliff, landing 6s after being thrown. What is the height of the cliff?  h=1/2gt  20m=1/2(10)t = t= 2s; h 1h =24s = ½(10)(6-2) =20m+x  x= 60m o o 34. Person weighing 710 N (T ) li3s in a hammock supported by ropes and 45 and 30 . Tension of the ropes? (hammock at standstill; Forces add up to zero) x =x = T 1os(20)=T1cos(45)  T 2T cos(30)/co2(451; T = 3 T 1in(30)+T si2(45) T =T 3sin145)+ cos(30)/cos(45)+ sin(45)] 710= T (1/2+0.87) =1520N = T ; T = 1 2 520N(cos30/cos45)= 640N 5 35. 32 g of low density O ga2 at 300K. What’s the internal energy? µ=5/2RT = 5/2(8.3 J/K)(300K) = 3.67x10 J 36. How high would a waterfall have to be for the water at the bottom to be 1 C more than at the top?  mgh=mC H20ΔT  h= (C H2OΔT)/g = (4200)(1)/9.8 = 427m 37. 600 cm water in aluminum can at 0 C not closed tightly. It’s heated to 30 C. How much water will overflow? -5 -5 3 -5 o 3α H2O=7.04x 10 ; 3α (liAlar expansion coefficient)= 2.4x10 ΔV =V(3α)(ΔT) Al600cm (3(2.4x10 ))(30 C)=1.3 cm ; ΔV =(600cm )(3(7.04x10 ))(30 C)=3.8 cm ; 3.8-1.3 = 2.5 cm =overflow 3 H2O 38. 1L flask contains a certain quantity of ideal gas at 300K. An equal quantity of the same gas is added after which the absolute pressure is 1.5 times its original value. What’s the final temp? PV=NkT k/V=P/NT; K/V = (1.38x10 )/(1.01x 10 )= constant; P /N T =1P 1N 1  P2=1252 ; N2=2N (P1& N2cross1out), 1/T = 1.5/2T = 1 2 1/300=1.5/2T = 225K 39. Acceleration a is the rate of change of the velocity v: average acceleration. • SI Unit: m/s² ; non-SI units: cm/s², ft/s² a= Δv/Δt = (Vf-Vi)/(tf-ti) 40. FREE FALL EX.: a stone is thrown up from the top of a building, find its motion parameters Given: with an upward y-axis: a =-9.8(m/s2), v0=20.0(m/s), y0=0, yg=-50(m) Find: (a) th=? and yh=? at highest point; (b) tt=? vt=? when back to building top; (c) tg=? when reaching ground. (a) at the highest point, vh=v0+ath=0, so th = −v0/a = -20/(-9.8) = 2.04(s) yh = y0 + v0th + ath2/2 = 0 + 20x2.04 − 9.8x(2.04)2/2 = 20.4(m) (b) when back to the building top, yt = y0 + v0tt + att2/2 = 0, so 0 + v0tt+att2/2=0, which v0+att/2=0  tt = -2v0/a = -2x20/(-9.8) = 4.08(s). and we finally find vt = v0+att = 20-9.8x4.08 = -20.0(m/s) (c) when reaching the ground: yg = y0+v0tg+atg2/2 = -50, so 0+20tg- 9.8xtg2/2 = -50 4.9tg2-20tg-50 = 0 (USE QUADRATIC EQ.) 41. Force is a vector: F = (Fx, Fy). Static friction force fs acts to keep the object from moving by another force F. Friction force fs varies depends on F to keep the object Stationary. The object will not move unless |F| exceeds maximum values of |fs|: fmax = μsN When |F| > fmax the object is in motion and the friction force decreases to: ƒk = μkN 42. The square of the orbital period of any planet is proportional to cube of the average distance from the Sun to the planet T2 = Kr3. For orbit around the Sun, K = KS = 2.97x10-19 s2/m3 F12 = F21 = (G*m1*m2)/ (r12)^2. G is the constant of universal gravitational: G = 6.673x10-11 (N.m²/kg²) 43. 1. A force of 1000 N acts to stop a 1000 kg car moving at 20 m/s. How much time does it take the car to stop? Given: F = -1000 N ; Change in v = -20m/s ; m=1000 kg Want: Time (t) Eqn: F t = m (change in v)  t = m(change in v)/F = (1000kg)(-20m/s)/(-1000N) = 20 sec 44. 2. A 1000 kg car is at rest initially. A 500 Newton force is applied to the car in the forward direction for 10 seconds. How fast is the car going at the end of the 10 seconds? Given: m=1000 kg ; F=500 N ; t=10 sec ; initial v = 0 m/s Want: final v Eqn: F t = m v ; (since initial v=0)  v = (F t)/m = (500N x10s)/ (1000 kg) = 5 m/s. 45. 3. Ex: MOTION OF THE BLOCK ON A SLOPE. Given: v0=0, θ=30o, μk=0.1 Find: (a) a=? (b) v=?@t=4s (a) mg+fk+n=ma  x: mgsinθ – fk = ma y: -mgcosθ +n = 0  a = gsinθ – μkgcosθ = 0.41g=4.0(m/s2) and it points in x-axis (b) v = v0+at = 0+4.0x4 = 16(m/s) 46. 4. Ex: Circulating car on leveled ground. Given: v=13.4(m/s), r=50.0(m). Find: minimum μs=?ΣF = n + mg + fs but n + mg=0 in the vertical direction, so we have |ΣF| = |fs| and n=mg, which yields (mv^2)/r = f = μN 47. 5.Ex: MOVING A CRATE Given: v0=(0,0), w=300(N) F=(Fx,Fy)=(20.0,0)(N) Δt=2.00(s) Find: (a) a=? (b) Δx=? (a) a=F/m and m=w/g, so ax=Fx/m=Fxg/w=20x9.8/300=0.654(m/s2) ay=Fy/m=0  a=(0.654, 0) (m/s2) (b) Δx= v0xΔt+axΔt2/2 = axΔt2/2=1.31(m) 48. 6. Ex: MASS ON A SRPING: Find the instantaneous acceleration of a 1.00 kg mass suspended from a spring of force constant 5.00 (N/cm), when the spring is stretched 10.0 cm. The mass is initially at rest. Given: m=1.00(kg), k=5.00(N/cm)=500(N/m), v0=0, y=−10.0(cm) = −0.100(m) Find: a=? a = ΣF/m  ΣF = Fs+Fg --> a = (Fs+Fg)/m a = {−ky−mg}/m = {−500x(−0.1) − 1x9.8}/1= 40.2(m/s2) 49. 7. Example: MOTION OF CONNECTED OBJECTS: Given: m1=4.00(kg), m2=7.00(kg), μk=0.300 neglect friction by pulley to the rope and the mass of rope Find: (a) a=|a1|=|a2|=? (b) T=|T|=|T’|? (a) use the free body analysis, we separate m1 and m2 as Shown T–μkm1g= m1a T’+m2g = m2a2 T = m1μkg+m1a and T= m2g–m2a  m1μkg+m1a = m2g–m2a So, a = (m2g–m1μkg)/ (m1+m2)=5.17(m/s2) Use the answer for a to find T with any of these two equations  T–μkm1g = m1a and T–m2g = –m2a  T = m2g–m2a =32.4(N) 50. 8. GRAVITY ACCELERATIONS AT THE EARTH’S & MOON’S SURFACE. Given: Earth: RE=6.38x106(m), ME=5.97x1024(kg) Moon: RM=1.74x106(m), MM=7.35x1022(kg) Find: gE=? and gM=? at the surface of Earth and Moon gE= GME/RE2 = 9.80 (m/s2) gM= GMM/RM2 = 1.62 (m/s2) 51. 9. FREE FALL EX.: a stone is thrown up from the top of a building, find its motion parameters Given: with an upward y-axis: a =-9.8(m/s2), v0=20.0(m/s), y0=0, yg=-50(m) Find: (a) th=? and yh=? at highest point; (b) tt=? vt=? when back to building top; (c) tg=? when reaching ground. (a) at the highest point, vh=v0+ath=0, so th = −v0/a = -20/(-9.8) = 2.04(s) yh = y0 + v0th + ath2/2 = 0 + 20x2.04 − 9.8x(2.04)2/2 = 20.4(m) (b) when back to the building top, yt = y0 + v0tt + att2/2 = 0, so 0 + v0tt+att2/2=0, which v0+att/2=0  tt = -2v0/a = -2x20/(-9.8) = 4.08(s). and we finally find vt = v0+att = 20-9.8x4.08 = -20.0(m/s) (c) when reaching the ground: yg = y0+v0tg+atg2/2 = -50, so 0+20tg- 9.8xtg2/2 = -50 4.9tg2-20tg-50 = 0 (USE QUADRATIC EQ.)


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