# Physics 6B Practice Midterm #1 Solutions 1. A block of plastic alongside a

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## Physics 6B Practice Midterm #1 Solutions 1. A block of plastic alongside a

Carlow College, US has reference to this Academic Journal, Physics 6B Practice Midterm #1 Solutions 1. A block of plastic alongside a density of 920 kg/m3 floats at the interface between oil of density 850 kg/m3 in addition to water of density 1000 kg/m3, as shown. Calculate the percentage of the plastic which is submerged in the water. The block is floating motionless, so the net force is 0. The total buoyant force includes contributions from the oil in addition to the water, as shown in the free-body diagram. Fb,water + Fb,oil ? weight = 0 The key so that this problem is so that write these forces in terms of densities: Here V is the total volume of the plastic block. Putting these into our formula, in addition to solving in consideration of Vwater: 2. A U-shaped tube contains 10 cm of alcohol (?a=800 kg/m3), 12 cm of oil (?o=850 kg/m3), in addition to water (?w=1000 kg/m3) as shown. Find the distance h between the surface of the alcohol in addition to the surface of the water. 22cm-h The gauge pressure on either side must balance at the oil/water interface. On the left, we can add the partial pressures from the alcohol in addition to the oil.

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After plunging from a bridge, a 60kg bungee-jumper oscillates up in addition to down, completing one cycle every 2 seconds. If the bungee cord is assumed so that be massless in addition to has an unstretched length of 10m, find the spring constant of the bungee cord. This is just like a typical mass-on-a-spring problem. The formula in consideration of the period of the motion is We can solve this in consideration of k, then plug in the given values. Answer c) A 50kg block is placed at the top of a cylindrical steel pole alongside a radius of 0.1m in addition to a height of 1m. Determine the change in the length of the pole. The elastic (Young?s) modulus in consideration of steel is 200*109 N/m2. Formula in consideration of Young?s modulus is F will be the weight of the block = (50kg)(9.8m/s2) = 490 N 50 kg L0=1m Answer a) You are in the front row of the Boombox Orchestra concert, standing 1m away from the speakers. The sound intensity level is an earsplitting 120 db, so you decide so that move away so that a quieter position. How far away from the speaker do you need so that be so that the level is only 80 db? The formula in consideration of intensity level in decibels is There is a great shortcut we can use in this problem. We need the level so that decrease by 40db, so the Intensity should decrease by a factor of 104. In general. For every 10db decrease in level, the intensity should be divided by 10 (if the level increases, multiply instead). The other relationship we need here is This says that the Intensity is inversely related so that the square of the distance from the source of the sound. So if the distance increases by a factor of 10, the intensity goes down by a factor of 100. In this case, we need the intensity so that change by a factor of 104 so the distance should increase by a factor of 102. ANSWER: c) 100m

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9. A loudspeaker playing a constant frequency tone is dropped off a cliff. As it accelerates downward, a person standing at the bottom of the cliff will hear a sound of: a) increasing frequency in addition to decreasing amplitude b) constant frequency in addition to increasing amplitude c) increasing frequency in addition to increasing amplitude decreasing frequency in addition to constant amplitude The sound source is getting closer so that the person, so it will get louder (increasing amplitude). Also, the source is accelerating toward the listener, so the frequency will be doppler-shifted upward more in addition to more as the speaker speeds up faster in addition to faster. Answer c) 10. A ?boat? consists of a massless hollow cube of side length 50cm, floating in a freshwater lake. When a person steps onto the ?boat? it sinks down 30cm. Find the weight of the person. BEFORE AFTER 30cm Before the person steps on, the buoyant force just equals the weight of the boat. After, the buoyant force is the total weight of the boat plus the person. So the result of the person stepping on is that the buoyant force increases by the weight of the person. The buoyant force is also the weight of the displaced water. When the person steps on, the extra water displaced has a volume of (50x50x30) =75,000 cm3 Water has a density of 1g/cm3, so the extra displaced water has a mass of 75kg. This is equal so that the mass of the person. The person?s weight is thus (75kg)(9.8m/s2) = 735N Round this up so that 750N 11. Two strings on the same guitar (same length) are tuned so that string B is one octave higher frequency than string A. Given that string A has 4 times the mass of string B, what is the ratio of the tensions in the strings? (Hint: One octave higher frequency means the frequency is twice as high). a) String A has 4 times the tension of String B b) String A has 2 times the tension of String B c) The tensions in the strings are equal d) String A has half the tension of String B The frequencies of the standing waves on a string are given by Since the lengths of the strings are the same, we see that the waves on string B must travel at twice the speed of the waves on string A. Wave speed on a string is given by Solve this in consideration of tension so that get In this case, the mass/length of string A is 4 times as much as string B. The tensions come out so that be the same since the double speed (squared) gives a factor of 4 in consideration of string B. Answer c).

One of the harmonics on a string that is 1.30 m long has a frequency of 15.60 Hz. The next higher harmonic has frequency 23.40 Hz. Find the fundamental frequency in addition to the speed of the waves on the string. The difference between any 2 successive harmonic frequencies on a particular string is always equal so that the fundamental frequency in consideration of that string. In this case that is (23.4-15.6) = 7.8Hz To get the speed, use the formula 13. Two protons are placed as follows: Proton A is placed on the X-axis, +5 cm from the origin. Proton B is placed on the Y-axis, +10 cm from the origin. Find the electric field at the origin (magnitude in addition to direction). The charge on a proton is 1.6 x 10-19C. A B EA EB Etotal There are 2 charges, so we calculate 2 contributions so that the E-field. The formula will calculate magnitudes, in addition to the direction is away from each charge since they are positive. We need so that use the Pythagorean Theorem so that find the magnitude of the resultant when these 2 vectors are added: To find the angle, we can just look at the diagram (to the left in addition to down means below the negative x-axis), or we can calculate the value using right triangle rules:

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