review

same velocity, so same slope

V (m 3 ) P (100k Pa) a)this is an engine and the work done on the gas is +60x10 5 J b)this is a fridge and the work done on the gas is +60x10 5 J c)this is an engine and the work done on the gas is -60x10 5 J d)this is a fridge and the work done on the gas is -60x10 5 J

A gas is kept in a canister with a moveable piston. The piston is raised, while the temperature is kept constant (isothermal). Answer the following: a)the internal energy ….. (increases, decreases, remains the same) b)the work done by the gas is …. (positive, negative,zero) c)heat is …. (added, extracted, remains the same) d)entropy … (increases, decreases,remains the same) a)T is constant, so U=3/2nRT is constant b)W=p  V (by, so +p  V!)  V is positive (p is positive), so W>0 c)  U=Q+W (W: work done on the gas!!), Q=  U-W  U=0, W<0, so Q is positive (heat is added) d)  S=Q/T=+/+=+ (entropy increases)

An object of mass 0.1 kg is oscillating horizontally on a spring with spring constant k=1000 N/m. At t=0, the object is +5 cm away from the equilibrium position. The amplitude is 10 cm? a) What is the kinetic energy of the object at that point? b) What is the function describing the velocity as a function of time? a)E tot =E kin +E pot,spring =0.5mv kx 2 At maximum amplitude, only pot. energy: 0.5kx 2 max. amplitude: k=1000 N/m x=0.1 so E pot =5 J at 5 cm: E pot =0.5*1000* =1.25 J conservation of energy: E kin =5-1.25=3.75 J b)x(t)=Acos(  t+  ) at t=0, x=0.05 (A=0.1), so 0.05=0.1cos(  )  =  /3  =  (k/m)=  (1000/0.1)=100 rad/s v(t)=-A  sin(  t+  )=-0.1*100*sin(100t+  /3) v(t)=-10sin(100t+  /3)

A person is standing 1 m from a speaker. The sound level is 100 dB. He walks 99 m away from the speaker. What is the sound level (in dB)? I 2 /I 1  r 1 2 /r 2 2 r 1 =1 r 2 =100 so I 2 /I 1 =1/10000= dB=10log(I 1 /I 0 ) with I 0 = W/m 2 at 1 m distance x dB = 10log(I 2 /I 0 ) 100-x = 10log (I 1 /I 0 ) - 10log (I 2 /I 0 ) = 10log(I 1 /I 2 ) x = log(10000)=100-10*4=60 dB

A pipe is 1 m long and open on side and closed on the other. What is the fundamental frequency? What is the frequency of the 5 th possible harmonics? (v sound =340 m/s) 1 =L/4, 2 =3L/4, 3 =5L/4 ….. n =(2n-1)L/4 1 =1/4 v= f so f=v/ =340/0.25=1360 Hz 5 =9L/4 v= f so f=v/ =340/(9/4)=151 Hz

A football fan upset that his team is losing tosses out his battery powered radio out of a 20m high window. What is the frequency of the sound from the radio just before it hits the ground, relative to the frequency when it just drops out the window (assume: initial velocity 0). f’=f(v+v 0 )/(v-v s ) v: velocity of sound (343 m/s) v 0 : velocity observer (0 m/s) v s : velocity source. v s (t)=v 0 +at=0-9.8t x(t)=20-0.5gt 2 0=20-0.5*9.8*t 2 t 2 =40/9.8 t=2s so v=-19.6 s radio is moving away from the observer, so negative sign. f’/f=v/(v-v s )=343/(343-(-19.6))=0.95

A 0.2 kg aluminum plate, initially at 20 0 C slides down a 15-m long surface, inclined at 30 0 with the horizontal. The force of kinetic friction exactly balances the component of gravity down the plane. If 90% of the mechanical energy of the system is absorbed by the aluminum, what is the temperature increase at the bottom of the incline? (c Al =900 J/kg 0 C). F g =mg, along the slope: F g// =mgsin  =0.2*9.8*0.5=0.98N W froction =F  x=0.98*15=14.7 J 90% given to the plate: 0.9*14.7=13.2 J Q=cm  T  T=Q/cm=13.2/(900*0.2)  T= C.

Two objects collide head on. Object 1 (m=5kg) has an initial velocity of 10m/s and object 2 (m=10 kg) has an initial velocity of -8 m/s. What is the resulting velocity a)of the combined object if the collision is fully inelastic? b)of each of the objects if the collision if fully elastic? a)inelastic: only conservation of momentum. m 1 v 1i +m 2 v 2i = (m 1 +m 2 )v f 5*10+10*(-8)=15*v f v f =-2m/s b) elastic: conservation of momentum and kinetic energy momentum: m 1 v 1i +m 2 v 2i = m 1 v 1f +m 2 v 2f 5*10-10*8=5v 1f +10v 2f kinetic energy: (v 1i -v 2i )=(v 2f -v 1f ) 10-(-8)=v 2f -v 1f -30=5v 1f +10v 2f & 18=-v 1f +v 2f v 2f =18+v 1f -30=5v 1f +10(18+v 1f ) v 1f =-14 v 2f =4 m/s

An 800-N billboard worker stands on a 4.0-m scaffold supported by vertical ropes at each end. If the scaffold weighs 500N and the worker stands 1.0m from one end, what is the tension in the ropes? 1.0m Net force in vertical direction must be 0: F L +F R -w man -w scaffold =0 F L +F R =F L +F R -1300=0 Net torque must be zero. Choose 0 at ‘R’. F L *4-800*1-500*2=4F L -1800=0 F L =1800/4=450N so F R =1300-F L =850N. LR

An engine is operated between a hot and a cold reservoir with Q hot =400J and Q cold =300J. a) what is the efficiency of the engine? The engine is modified and becomes a carnot engine. As a result the efficiency is doubled. b) what is the ratio T cold /T hot. c) what is the maximum efficiency of this engine? a)efficiency=1-Q cold /Q hot =1-300/400=0.25 b)new efficiency: 0.5=1-T cold /T hot T cold /T hot =0.5 c)0.5 (Carnot engine has maximum efficiency).

r1r1 r2r2 Two loudspeaker are connected to the same sound system and produce a sound with a frequency of 340 Hz. If v sound =340 m/s, and r 1 =3 m and r 2 =4.5 m. The interference is a)destructive b)constructive constructive: r 2 -r 1 =n destructive: r 2 -r 1 =(n+1/2) =v/f=1 m, r 2 -r 1 =1.5 m if n=1: destructive condition is met.

A canon shoots a cannonball as shown in the picture. It lands on a surface higher by 5 meter than the surface that is was shot from. If the initial velocity was 40 m/s and it was shot at an angle of 30 o, how far does it travel? 10m 5m y(t)=y 0 +v 0y t-0.5gt 2 x(t)=x 0 +v 0 t 10=5+40*sin(30)t-0.5*9.8*t 2 0=-5+20t-4.9t 2 t=0.26 s or t=3.82 s x(3.82)=0+40*cos(30)*3.82 =132 m

An nuclear power plant drives a large turbine to generate electricity while heating cooling water. 1)The heat absorbed by the water is larger than the work done by the turbine. true not true cannot say 2) The atomic energy released by nuclear fission in the powerplant is less than the absorbed heat by the cooling water plus the work done true not true cannot say

A car brakes and it takes 50 m to come to a full stop. If the initial velocity was 100 m/s, what was the deceleration? x(t)=x 0 +v 0 t+0.5at 2 v(t)=v 0 +at so: 50=0+100t+0.5*a*t 2 from x(t) 0=100+at from v(t) so t=-100/a combine: 50=-100*100/a+0.5*a*(-100/a) 2 so 0= /a+5000/a 2 or 0=-50a a =a a-100 a= m/s 2 or a=0.5 m/s 2 Must be negative, so –200.5 m/s 2

A block of material A, is floating in a liquid with a density of 0.5x10 3 kg/m 3 and 50% of its volume is submerged. The liquid is replaced with water (1.0x10 3 kg/m 3 ). 1)How much of the volume is submerged? a)less than 50% b)50% c)more than 50% B=  Vg  goes up, B goes up less volume submerged. 2) How much of the volume is submerged, exactly? With less dense liquid: F g =B mg=  1 V 1 g with water mg=  w V w g=2  1 V w g so  1 V 1 g=2  1 V w g V w =V 1 /2 only 25% submerged.