These are the problems I need help on. I really don\'t understandthese. Problem

Question

These are the problems I need help on. I really don't understandthese.

Problem 1: A water tank standing on the floor has to smallholes vertically above one another punched in one side. The holesare 2 cm and 8 cm above the floor.
a)How high does the water stand in the tank (calculate H)
b) Calculate the distance x.
c) At what depth should a single hole be made to maximize x?

Problem 2: A horizontal block-spring system has a totalmechanical energy of 10J, the amplitude of 0.2m, and a maximumspeed of 1.5m/s. Find:
a) the spring contant, k.
b) the mass of the block, m
c) the frequency of oscillation, f.

Problem 3: 1 kg piece of metal is heated to a hightemperature T metal, and then dropped into a beaker containing 0.1kg of water at 20 degrees Celsius and 0.5 kg of ice at 0 degreesCelsius. Calculate the initial temperature of the metal T metal, ifthe specific heat of the water is CH20 = 4186 J/kg degrees Celsiusand the final equilibrium temperature of the mixture is 80 degreesCelsius. (Lf = 333,000 J/kg for water, cmetal = 400 J/kg degreesCelsius)

Problem 4: An oscillator consists of a block attached to aspring (k = 40 N/m). At some time t, the position, velocity andacceleration of the block are: x = 5m, v = 10 m/s and a =-1.48m/s^2. Calculate the frequency of oscillation, the mass of theblock and the amplitude of the motion.

Problem 5: A horizontal spring oscillates with an amplitudeA = 5cm. What is the magnitude of the displacement "x" when thekinetic energy KE is half the elastic potential energy Uspring?

Explanation / Answer

I'll try to answer some of them and give hints to others as this isactually multiple problems (and shouldn't be placed on onepost) 1)There is not enough information to calculate the answer plain andsimple: However, problem 65 of chapter 14 in Fundamentals ofPhysics is extremely similar by the sounds of it. 2) .5kx^2=10j at max x k=20/.04 .5mv^2=10 at max v m=20/(1.5)^2 v=-Asin(t+) vmax=A =2f f=vmax/(A*2pi)= 1.5/(2*.2) 3) use 0=Lm(ice)+Cm(60)(water)+Cm(79.5)water- Cm(80-Ti)(metal)solving fo Ti 4) same formulas as problem 2 but we must work backwards a=-^2Acos(t+) a=-^2x solve for /2=f v=-Asin(t+) divide velocity by - so you have -v/=Asin() x=Acos(t+) -(v/x)=tan solve for A=x/cos() .5kx^2+.5mv^2=.5kA^2 solve for m 5).5kx^2+.5mv^2=.5kA^2 .75kx^2+.5mv^2=.5kA^2 solve for x

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