Various Math #14 - Q&A

Question 1 An underwater diver sees the sun 60° above horizontal. A) How high is the sun above the horizon to a fisherman in a boat above the diver? Express your answer using two significant figures. Answer if the diver sees the sun at 60 degrees then that means that the refracted ray is 30 degrees n₁ sin 21 = n2 sin 22 where n₁ and n₂ are refractive index of water and air respectively 1.333 sin 30 = 1 sin X2 X2 = 41.8 degrees, this is the angle of the incident ray the person above the water sees the sun at 90 - 41.8 = 48.2 degrees. the 41.8 degrees is made normal with the surface, so for horizontal to the surface subtract that from 90. Question 2 If electrical energy costs 12 cents, or $0.12, per kilowatt-hour, then what would the following events cost? (a) Burn a 40 Wlight bulb for 24 h straight. (b) Operate an electric oven for 6.0 hif it carries a current of 20.0 A at 220 V. Answer For part a) We have 0.12 / k W h r @ 24 h o u r s . T h e C o s t f o r m u l a i s : C o s t = P o w e r ( P ) × T i m e u s e d ( H r ) × C o s t p e r k W / h r P = 40 W → 0.040 k W P l u g i t a l l i n t o t h e f o r m u l a : C o s t = ( 0.040 k W ) ( 24 h r ) ( 0.12/kWhr@24hours.TheCostformulais:Cost=Power(P)×Timeused(Hr)×CostperkW/hrP=40W→0.040kWPlugitallintotheformula:Cost=(0.040kW)(24hr)( 0.12/kW hr) Cost = 11.52 cents or $0.12 (when rounded) For part b) We have the following information: 0.12 / k W h r @ 6.0 h r I = 20.0 A V = 220 V P I = ∇ → P = ( V ) ( I ) P = ( 220 V ) ( 20.0 A ) P = 4400 W → 4.4 k W N o w u s e t h e C o s t f o r m u l a f r o m a b o v e i n p a r t a . C o s t = P o w e r ( P ) × T i m e u s e d ( H r ) × C o s t p e r k W / h r C o s t = ( 4.4 k W ) ( 6.0 h r ) ( 0.12/kWhr@6.0hrI=20.0AV=220VPI=∇→P=(V)(I)P=(220V)(20.0A)P=4400W→4.4kWNowusetheCostformulafromaboveinparta.Cost=Power(P)×Timeused(Hr)×CostperkW/hrCost=(4.4kW)(6.0hr)( 0.12/kW hr) Cost = 316.8 cents = $3.17 when rounded. Question 3 A racetrack has the shape of an inverted cone. On the surface the cars race in circles that are parallel to the ground. For a speed of 34 m/, at what value of the distance should a driver locate his car if he wishes to stay on a circular path without depending on friction? Answer Nsin40 N KN N Cos40 R 40 d mg 40 40 m(v) (1) R N Sin40 mg a Ncos40 = mg (2) m(v) mgR (3) tan40 = tan40 = sin50 = v gR JR_d a R = dsin50 v tan40 = -sin50 gd d=184 m Question 4 A rescue helicopter is lifting a man (weight = 822N) from acapsized boat by means of a cable and harness. (a) What is the tension in the cable when the man is given an initial upward acceleration of 1.10m/s²? (b) What is the tension during there mainder of the rescue when he is pulled upward at a constant velocity? Answer a) F = mg m= F/g 822 = 83.79kg 9.81 because the upward acceleration is in the opposite direction as g, the man's weight (tension in the cable) becomes less. F= 83.79(9.81 – 1.10) = 729.83 N b) Tension during the remainder of the trip just W (822N) Question 5 The skateboarder in the drawing starts down the left side fthe ramp with an initial speed of 5.4 m/s. If nonconservativeforces, such as kinetic friction and air resistance, arenegligible, what would be the height h of the highestpoint reached by the skateboarder on the right side of the ramp? Answer Well, first you must realize that the speed whilst leaving theramp is going to be the same as the start because there is nochange in height. Knowing that, it will be the same type ofproblem as someone jumping in the air. So... 2 = v + 2ad (where v=final velocity, v₁ = starting velocity, a=acceleration,d= displacement) m 0 = (5.4)+2(-9.8m)d-29.16 = -19.6 $2 m m 0- 2 S d) 1.487755... m = d [using significantfigures] d=1.5 m