PHYSICS LESSON 16 FOR
MONDAY, JUNE 28, 2010
I Introduction
II Logistics:
a. Return of papers
b. Running grades
III Review Lesson 14: Waves
IV. Lesson 16: Sound Waves, Light Waves
Sound travels through a medium
C = 300,000 km/s; or less in a medium
V. Lab Exercise #12: Thin Films
VI. Homework Set 5: Hand out, due Thursday, 7/1
VII. Essay 5: Enrico Fermi, due Thursday 7/1
Monday, June 28, 2010
Thursday, June 24, 2010
HOMEWORK SET 5
1. A wave has crests and troughs. From the top of one crest to the bottom of the next trough is 13 cm. The length from crest to trough along the x axis is 28 cm. Find the a. amplitude and b. wavelength.
2. The speed of surface waves in water decreases as the water becomes more shallow. Imagine water waves crossing the surface of an otherwise calm lake at a speed of v1 = 2.0 m/s, with a wavelength of l1 = 1.5 meters. When these same waves approach shore, the speed decreases to v2 = 1.6 m/s, while the frequency remains the same. Find the wavelength, l2, in the shallow waters.
3. Suppose that you wanted to double the wave speed, from v to 2v, on a tight string. The force that is keeping it tight is called “tension” but it is still a force and has units of Newtons. How much would you have to increase the tension so the velocity would double?
4. Waves on a particular string travel at v1 = 16 m/s. To double that speed to v2 = 32 m/s, how much would you have to increase the tension?
5. Write an expression for a harmonic wave, such that amplitude, A = 0.16 m, wavelength l = 2.1 m, and period of P = 1.8 seconds.
6. A soundwave in air has a frequency of n1 = 425 Hz.
a. Find its wavelength
b. If its frequency is increased, does the wavelength increase, decrease, or remain unchanged
c. Calculate its wavelength if the frequency is n2 = 475 Hz.
7. A man throws a rock down to the bottom of a well. From the instant that it leaves his hand until the moment he hears the rock hit bottom, a period of 1.20 seconds passes. The depth of the well is 8.85 meters. Find the initial speed, vi, of the rock (the instant that it leaves the man's hand).
8. Twenty identical violins are being played by 20 identical violinists at the same identical volume, for an aggregate sound level of 82.5 dB.
a. Find the dB level of one of the violins.
b. Find the dB level of twice as many identical violins (40).
9. Find the weight (in Newtons) of the air in your physics class. Assume that the room is a cube of 4 meters on a side, and that the density of air is rair = 1.29 kg/m3.
10. You buy a gold ring at a pawn shop, and run a “test” to determine if it's pure gold. The ring's mass is m = 0.014 grams, and has a volume of 0.022 cm3. The density of gold is rAu = 19.3 kg/m3.
11. Crutches will often have rubber tips at the bottom, so that contact with a flooring surface will decrease scratching of the floor, increase friction, and its broader tip spreads the pressure out over a larger area. Assume the radius of the tip of the cane to be 1.2 cm, and that the radius of the rubber tip is 2.5 cm. What is the ratio of the pressure of the cane alone, P1, to the pressure with the rubber tip, P2: P1/P2.
12. Two drinking glasses, G1 and G2, are filled with water to the same level of 5.0 cm. The diameter of the base of G1 is twice the diameter of the base of G2, but otherwise the two glasses are identical.
a. Is the weight of the water in G1 greater, less, or the same as weight in G2?
b. Is the water pressure at the bottom of G1 greater, less, or the same as the water pressure at the bottom of G2?
13. On Wednesday, August 15, 1934, William Beebe and Otis Barton made history by descending in the Bathysphere, a steel sphere 4.75 feet in diameter, to 3028 feet below sea level.
a. as the Bathysphere was lowered, was the buoyant force exerted on it at a depth of 10 feet greater, less, or equal to it at 50 feet?
b. Choose the best explanation from below.
I Magic
II Miracle
III Mayonnaise.
14. A fish called Wanda is carrying a pebble in its mouth swims with a small constant velocity in a small bowl. When the fish drops the pebble to the bottom of the bowl, does the water level rise, fall or stay the same?
15. Explain what a a hydrometer is.
END
2. The speed of surface waves in water decreases as the water becomes more shallow. Imagine water waves crossing the surface of an otherwise calm lake at a speed of v1 = 2.0 m/s, with a wavelength of l1 = 1.5 meters. When these same waves approach shore, the speed decreases to v2 = 1.6 m/s, while the frequency remains the same. Find the wavelength, l2, in the shallow waters.
3. Suppose that you wanted to double the wave speed, from v to 2v, on a tight string. The force that is keeping it tight is called “tension” but it is still a force and has units of Newtons. How much would you have to increase the tension so the velocity would double?
4. Waves on a particular string travel at v1 = 16 m/s. To double that speed to v2 = 32 m/s, how much would you have to increase the tension?
5. Write an expression for a harmonic wave, such that amplitude, A = 0.16 m, wavelength l = 2.1 m, and period of P = 1.8 seconds.
6. A soundwave in air has a frequency of n1 = 425 Hz.
a. Find its wavelength
b. If its frequency is increased, does the wavelength increase, decrease, or remain unchanged
c. Calculate its wavelength if the frequency is n2 = 475 Hz.
7. A man throws a rock down to the bottom of a well. From the instant that it leaves his hand until the moment he hears the rock hit bottom, a period of 1.20 seconds passes. The depth of the well is 8.85 meters. Find the initial speed, vi, of the rock (the instant that it leaves the man's hand).
8. Twenty identical violins are being played by 20 identical violinists at the same identical volume, for an aggregate sound level of 82.5 dB.
a. Find the dB level of one of the violins.
b. Find the dB level of twice as many identical violins (40).
9. Find the weight (in Newtons) of the air in your physics class. Assume that the room is a cube of 4 meters on a side, and that the density of air is rair = 1.29 kg/m3.
10. You buy a gold ring at a pawn shop, and run a “test” to determine if it's pure gold. The ring's mass is m = 0.014 grams, and has a volume of 0.022 cm3. The density of gold is rAu = 19.3 kg/m3.
11. Crutches will often have rubber tips at the bottom, so that contact with a flooring surface will decrease scratching of the floor, increase friction, and its broader tip spreads the pressure out over a larger area. Assume the radius of the tip of the cane to be 1.2 cm, and that the radius of the rubber tip is 2.5 cm. What is the ratio of the pressure of the cane alone, P1, to the pressure with the rubber tip, P2: P1/P2.
12. Two drinking glasses, G1 and G2, are filled with water to the same level of 5.0 cm. The diameter of the base of G1 is twice the diameter of the base of G2, but otherwise the two glasses are identical.
a. Is the weight of the water in G1 greater, less, or the same as weight in G2?
b. Is the water pressure at the bottom of G1 greater, less, or the same as the water pressure at the bottom of G2?
13. On Wednesday, August 15, 1934, William Beebe and Otis Barton made history by descending in the Bathysphere, a steel sphere 4.75 feet in diameter, to 3028 feet below sea level.
a. as the Bathysphere was lowered, was the buoyant force exerted on it at a depth of 10 feet greater, less, or equal to it at 50 feet?
b. Choose the best explanation from below.
I Magic
II Miracle
III Mayonnaise.
14. A fish called Wanda is carrying a pebble in its mouth swims with a small constant velocity in a small bowl. When the fish drops the pebble to the bottom of the bowl, does the water level rise, fall or stay the same?
15. Explain what a a hydrometer is.
END
Lesson 15
PHYSICS LESSON 15 FOR
THURSDAY, JUNE 24, 2010
I Introduction
II Logistics: Return of Papers, Running Grades, Turn in Assignments Due, etc. and
III Test 4
IV Mind Game 4
V Homework Set 5 Handout
VII Essay 4: Enrico Fermi, due 7/01
THURSDAY, JUNE 24, 2010
I Introduction
II Logistics: Return of Papers, Running Grades, Turn in Assignments Due, etc. and
III Test 4
IV Mind Game 4
V Homework Set 5 Handout
VII Essay 4: Enrico Fermi, due 7/01
Wednesday, June 23, 2010
Solution Set 4
1. To tighten a spark plug, it's recommended that a torque of 15 N m be applied. If a mechanic tightens the spark plug with a wrench that is 25 cm long, what is the force needed?
Solution:
We know that torque, t, is equal to the force applied, F, multiplied by the length of the lever, or, length of the moment arm, or, length of the handle, or whatever you want to call it, r. the relationship looks like this:
t = r F
Here, we are asked to find the force, F. If we re-write the relationship above, we have:
F = t/r. Do we know the torque, t? Yes, it is given as 15 N m. Do we know the length of the moment arm, r? Yes, it is given as 25 cm = 0.25 m. So all we have to do is “plug and chug” to get the answer, and we don't have to search for clues to find anything else:
F = t/r = (15)/(0.25) = 60 N.
2. A bowling trophy of mass 1.61 kg is held at arm's length, a distance of 0.65 m from the shoulder joint. What torque does the trophy exert about the shoulder if the arm is
a. horizontal
b. 22.5° below the horizontal?
Solution:
In reality, torque is the length of the moment arm, r, multiplied by the force, F, then multiplied by the sine of the angle between them, Ɵ. Or, in other ways of putting it,
t = r F sin Ɵ. If it is a right angle, Ɵ = 90°, then the sin 90° = 1.0, and we don't worry about it.
a. t = r F = (0.65 m)(1.61)(9.8) = 10.2557 N-m, or, 10.3 N-m.
b. Since the angle is 22.5°, or (90° – 22.5°) from the axis, then the torque is t = r F sin Ɵ = 10.3 sin 67.5° = 9.52 N.
3. Suppose a torque rotates your body about one of three different axes of rotation: case A, an axis through your spine; case B, an axis through your hips; and case C, an axis through your ankles. Rank these three axes of rotation in increasing order of the angular acceleration produced by the torque. Indicate ties where appropriate.
Solution:
The least torque will be through the spine, the greatest through the ankles.
4. A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length 3.15 meter and mass of 8.42 kg, find the torque the person must exert on the ladder to give it an angular acceleration of 0.302 radians/sec2.
Solution:
One of our relationships is t = I a, and to find torque, we need the moment of inertia, I, and the angular acceleration, a. Do we have them? Yes! The moment of inertia around a uniform rod of length 3.15 meter and mass of 8.42 kg, is I = 1/12 m L2, or, I = (0)(8.42)(3.15)2. = (0.083)(8.42)(9.9225) = 6.96 kg m2.
Now we can find the torque: t = I a = (6.96)(0.302) = 2.1 N m.
5. A fish takes the bait and pulls on the line with a force of 2.2 N. The fishing reel, which rotates without friction, is a cylinder of radius 0.055 m and mass of 0.99 kg.
a. What is the angular acceleration of the fishing reel?
b. How much line does the fish pull from the reel in ¼ sec.?
Solution:
a. We are looking for acceleration and can use: t = I a = r F, so a = r F/I. Since I = ½ MR2 for a cylinder, then = 2F/Mr = 4.4/(0.55) = 8 rad/sec2.
b. First we use θ = θ0 + w0 t + ½ a t2 = (in radians), but initial angle and initial angular velocity are both 0.0. So, we rewrite it to be: θ = ½ a t2 = (0.50)(8)(0.25)2 = 0.25 radian. Since distance traveled is s = r , then s = (0.055m)(0.25) = 0.01375 or about 0.014m.
6. A uniform crate with a mass of 16.2 kilograms rests on the floor with a coefficient of friction of m = 0.571. The crate is a uniform cube with sides 1.21 meters in length. What horizontal force applied to the top of the crate will start the crate to tip?
Solution: This is a torque problem.
7. Calculate the angular momentum of the Earth around its own axis due to its daily motion. Assume a sphere with uniform density.
Solution: Angular momentum in this case is L = I w. where I = 2/5 M r2 = (0.4)(6 x 1024 kg)(6.4 x 106 m)2 = (2.4 x 1024)(4.1 x 1013) = 9.8 x 1037 kg m2 /s2 .
8. As an ice skater begins a spin, his angular speed is 3.17 rad/s. After pulling in his arms, his angular speed increases to 5.46 rad/s. Find the ratio of the skater’s final moment of inertia to his initial moment of inertia.
Solution: 3.17/5.46 = 0.58.
9. How much work must be done to accelerate a baton from rest to an angular speed of 7.4 rad/s about its center? Consider the baton to be a uniform rod of length 0.53 m and a mass of 0.44 kg.
Solution: W = ½ I w2, where I = ML2/12 = (0.44)(0.53)2/12 = 0.124/12 = 0.01 kg-m 2 , and w = 7.4 rad/s. So, W = ½ I w2 = (0.5)(0.01)(7.4)2 = 0.28 Joule.
10. A 6.1-kg bowling ball and a 7.2-kg bowling sphere rest on a rack 0.75 m apart.
a. What is the force of gravity on each of the spheres by the other one?
b. At what distance is the force of gravity between the spheres equal to 2.0 x 10-9 N?
Solution:
Using F = G m M/r2, we plug and chug:
a. F = [(6.67 x 10-11)(6.1)(7.2) / (0.75)2] = [(292 x 10-11)/(0.5625)] = 519 x 10-11 = 5.2 x 10-9 N.
b. Re-write F = G m M/r2 to be r2 = G m M/F, we now plug and chug, so r2 = [(6.67 x 10-11)(6.1)(7.2)]/(2.0 x 10-9) = (292 x 10-11)/(2.0 x 10-9) = (2.92 x 10-9)/(2.0 x 10-9) = (2.92)/(2.0) = 1.46. So, if r2 = 1.46, then r = ?(1.46) = 1.2 m.
11. At a certain distance from the center of Earth, a 4.6-kg object has a weight of 2.2 N.
Find the distance.
Solution:
a. First, knowing Newton's second law is F = m a, and that in this case, F = m g, and knowing both F and g, we can re-write it as m = F/g to get the mass of the object: Or, m = (2.2 N)/(9.8 m/s2) = 0.224 kg. And now, Using F = G m M/r2, we re-write it as r2 = G m M/F and then plug and chug: r2 = [(6.67 x 10-11)(0.224)(6 x 1024)/(2.2)] = (8.96 x 1013) / (2.2) = 4.1 x 1013 . So, if = 4.1 x 1013, then
r = ?(4.1 x 1013) = 6.4 x 106 m.
b. Duh. a = g = - 9.8 m/s2.
12. Find the orbital speed of a satellite in a geosynchronous circular orbit 3.58 x 107 m above Earth's surface.
Solution: Geosynchronous means it will revolve about Earth in 24 hours, or 86,400 seconds. Also, if it is 3.58 x 107 m above Earth's surface, then it is 3.58 x 107 m + 6.4 x 106 m 4.22 x 107 m = from the Earth's core. Speed is distance over time, or, v = c/P, where c = 2 p r = (2)(3.14)(4.22 x 107 m) = 26.5 x 107 m = 2.65 x 108 m. And P = 86,400 sec = 8.64 x 104 s. So, v = c/P, = (26.5 x 107)/(8.64 x 104) = 3.1 x 103 m/s.
13. Phobos, a moon of Mars, orbits at a distance of 9378 km from the center of Mars. What's its orbital period?
Solution: Here we use Kepler's 3rd Law, which is: P2 = k a3, where P is the period in seconds that we are looking for, a is the distance from the center, a = 9378 km = 9.378 x 106 m. And k = (4 p2 / G M) where G = 6.67 x 10-11. and for Mars, M = 6.4 x 1023 kg. Thus, k = (4)(p2) / (6.67 x 10-11)(6.4 x 1023) = (39.5)/(4.27 x 1013) = 9.25 x 10-13. So now we can plug and chug: P2 = k a3 = (9.25 x 10-13)(9.378 x 106 )3 = (9.25 x 10-13)(824.8 x 1018) = (9.25 x 10-13)(8.25 x 1016) = 76.3 x 103 = 7.6 x 104. Thus, if P2 = 7.6 x 104 , then P = ?(7.6 x 104 ) = 2.76 x 102 sec. (This is about 4 minutes, which sounds way too low. Maybe I need to do it again.
14. A satellite orbits the Earth in a circular orbit of radius r. At some point its rocket engine his fired in such a way that its speed increases rapidly by a small amount. As a result,
a. does the apogee distance increase, decrease, or stay the same?
b. does the perigee distance increase, decrease, or stay the same?
Solution:
a. Increase
b. Increase
15. Find the speed of the binary stars Centauri A and Centauri B. They are separated by a distance of 3.45 x 1012 m and they have an orbital period of 2.52 x 109 seconds (that's about 80 years). They have the same mass. (Binary stars are near each other and orbit each other).
Solution: Speed is distance over time, or, the circumference of the orbit divided by the about 80 years: v = 2 p r / P = (2)(3.14)(1.72 x 1012 m)/(2.52 x 109 seconds) = (10.8 x 1012)/(2.52 x 109 s) = 4.3 x 103 m/s.
16. Find the Escape Velocity, vesc, for the planet Mercury.
Solution:
Escape velocity is, ve = ?(2GM/r), with G = 6.67 x 10-11, M = 3.3 x 1023 kg, and r = 2.44 x 106 m. So, now we know that ve = ?(2GM/r) = ve = ?[(2)(6.67 x 10-11)(3.3 x 1023 kg)/(2.44 x 106 m)] = ?[(4.4 x 1013)/(2.44 x 106 m)] = ?(1.8 x 107) = 4.2 x 103 m/s.
END
Solution:
We know that torque, t, is equal to the force applied, F, multiplied by the length of the lever, or, length of the moment arm, or, length of the handle, or whatever you want to call it, r. the relationship looks like this:
t = r F
Here, we are asked to find the force, F. If we re-write the relationship above, we have:
F = t/r. Do we know the torque, t? Yes, it is given as 15 N m. Do we know the length of the moment arm, r? Yes, it is given as 25 cm = 0.25 m. So all we have to do is “plug and chug” to get the answer, and we don't have to search for clues to find anything else:
F = t/r = (15)/(0.25) = 60 N.
2. A bowling trophy of mass 1.61 kg is held at arm's length, a distance of 0.65 m from the shoulder joint. What torque does the trophy exert about the shoulder if the arm is
a. horizontal
b. 22.5° below the horizontal?
Solution:
In reality, torque is the length of the moment arm, r, multiplied by the force, F, then multiplied by the sine of the angle between them, Ɵ. Or, in other ways of putting it,
t = r F sin Ɵ. If it is a right angle, Ɵ = 90°, then the sin 90° = 1.0, and we don't worry about it.
a. t = r F = (0.65 m)(1.61)(9.8) = 10.2557 N-m, or, 10.3 N-m.
b. Since the angle is 22.5°, or (90° – 22.5°) from the axis, then the torque is t = r F sin Ɵ = 10.3 sin 67.5° = 9.52 N.
3. Suppose a torque rotates your body about one of three different axes of rotation: case A, an axis through your spine; case B, an axis through your hips; and case C, an axis through your ankles. Rank these three axes of rotation in increasing order of the angular acceleration produced by the torque. Indicate ties where appropriate.
Solution:
The least torque will be through the spine, the greatest through the ankles.
4. A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length 3.15 meter and mass of 8.42 kg, find the torque the person must exert on the ladder to give it an angular acceleration of 0.302 radians/sec2.
Solution:
One of our relationships is t = I a, and to find torque, we need the moment of inertia, I, and the angular acceleration, a. Do we have them? Yes! The moment of inertia around a uniform rod of length 3.15 meter and mass of 8.42 kg, is I = 1/12 m L2, or, I = (0)(8.42)(3.15)2. = (0.083)(8.42)(9.9225) = 6.96 kg m2.
Now we can find the torque: t = I a = (6.96)(0.302) = 2.1 N m.
5. A fish takes the bait and pulls on the line with a force of 2.2 N. The fishing reel, which rotates without friction, is a cylinder of radius 0.055 m and mass of 0.99 kg.
a. What is the angular acceleration of the fishing reel?
b. How much line does the fish pull from the reel in ¼ sec.?
Solution:
a. We are looking for acceleration and can use: t = I a = r F, so a = r F/I. Since I = ½ MR2 for a cylinder, then = 2F/Mr = 4.4/(0.55) = 8 rad/sec2.
b. First we use θ = θ0 + w0 t + ½ a t2 = (in radians), but initial angle and initial angular velocity are both 0.0. So, we rewrite it to be: θ = ½ a t2 = (0.50)(8)(0.25)2 = 0.25 radian. Since distance traveled is s = r , then s = (0.055m)(0.25) = 0.01375 or about 0.014m.
6. A uniform crate with a mass of 16.2 kilograms rests on the floor with a coefficient of friction of m = 0.571. The crate is a uniform cube with sides 1.21 meters in length. What horizontal force applied to the top of the crate will start the crate to tip?
Solution: This is a torque problem.
7. Calculate the angular momentum of the Earth around its own axis due to its daily motion. Assume a sphere with uniform density.
Solution: Angular momentum in this case is L = I w. where I = 2/5 M r2 = (0.4)(6 x 1024 kg)(6.4 x 106 m)2 = (2.4 x 1024)(4.1 x 1013) = 9.8 x 1037 kg m2 /s2 .
8. As an ice skater begins a spin, his angular speed is 3.17 rad/s. After pulling in his arms, his angular speed increases to 5.46 rad/s. Find the ratio of the skater’s final moment of inertia to his initial moment of inertia.
Solution: 3.17/5.46 = 0.58.
9. How much work must be done to accelerate a baton from rest to an angular speed of 7.4 rad/s about its center? Consider the baton to be a uniform rod of length 0.53 m and a mass of 0.44 kg.
Solution: W = ½ I w2, where I = ML2/12 = (0.44)(0.53)2/12 = 0.124/12 = 0.01 kg-m 2 , and w = 7.4 rad/s. So, W = ½ I w2 = (0.5)(0.01)(7.4)2 = 0.28 Joule.
10. A 6.1-kg bowling ball and a 7.2-kg bowling sphere rest on a rack 0.75 m apart.
a. What is the force of gravity on each of the spheres by the other one?
b. At what distance is the force of gravity between the spheres equal to 2.0 x 10-9 N?
Solution:
Using F = G m M/r2, we plug and chug:
a. F = [(6.67 x 10-11)(6.1)(7.2) / (0.75)2] = [(292 x 10-11)/(0.5625)] = 519 x 10-11 = 5.2 x 10-9 N.
b. Re-write F = G m M/r2 to be r2 = G m M/F, we now plug and chug, so r2 = [(6.67 x 10-11)(6.1)(7.2)]/(2.0 x 10-9) = (292 x 10-11)/(2.0 x 10-9) = (2.92 x 10-9)/(2.0 x 10-9) = (2.92)/(2.0) = 1.46. So, if r2 = 1.46, then r = ?(1.46) = 1.2 m.
11. At a certain distance from the center of Earth, a 4.6-kg object has a weight of 2.2 N.
Find the distance.
Solution:
a. First, knowing Newton's second law is F = m a, and that in this case, F = m g, and knowing both F and g, we can re-write it as m = F/g to get the mass of the object: Or, m = (2.2 N)/(9.8 m/s2) = 0.224 kg. And now, Using F = G m M/r2, we re-write it as r2 = G m M/F and then plug and chug: r2 = [(6.67 x 10-11)(0.224)(6 x 1024)/(2.2)] = (8.96 x 1013) / (2.2) = 4.1 x 1013 . So, if = 4.1 x 1013, then
r = ?(4.1 x 1013) = 6.4 x 106 m.
b. Duh. a = g = - 9.8 m/s2.
12. Find the orbital speed of a satellite in a geosynchronous circular orbit 3.58 x 107 m above Earth's surface.
Solution: Geosynchronous means it will revolve about Earth in 24 hours, or 86,400 seconds. Also, if it is 3.58 x 107 m above Earth's surface, then it is 3.58 x 107 m + 6.4 x 106 m 4.22 x 107 m = from the Earth's core. Speed is distance over time, or, v = c/P, where c = 2 p r = (2)(3.14)(4.22 x 107 m) = 26.5 x 107 m = 2.65 x 108 m. And P = 86,400 sec = 8.64 x 104 s. So, v = c/P, = (26.5 x 107)/(8.64 x 104) = 3.1 x 103 m/s.
13. Phobos, a moon of Mars, orbits at a distance of 9378 km from the center of Mars. What's its orbital period?
Solution: Here we use Kepler's 3rd Law, which is: P2 = k a3, where P is the period in seconds that we are looking for, a is the distance from the center, a = 9378 km = 9.378 x 106 m. And k = (4 p2 / G M) where G = 6.67 x 10-11. and for Mars, M = 6.4 x 1023 kg. Thus, k = (4)(p2) / (6.67 x 10-11)(6.4 x 1023) = (39.5)/(4.27 x 1013) = 9.25 x 10-13. So now we can plug and chug: P2 = k a3 = (9.25 x 10-13)(9.378 x 106 )3 = (9.25 x 10-13)(824.8 x 1018) = (9.25 x 10-13)(8.25 x 1016) = 76.3 x 103 = 7.6 x 104. Thus, if P2 = 7.6 x 104 , then P = ?(7.6 x 104 ) = 2.76 x 102 sec. (This is about 4 minutes, which sounds way too low. Maybe I need to do it again.
14. A satellite orbits the Earth in a circular orbit of radius r. At some point its rocket engine his fired in such a way that its speed increases rapidly by a small amount. As a result,
a. does the apogee distance increase, decrease, or stay the same?
b. does the perigee distance increase, decrease, or stay the same?
Solution:
a. Increase
b. Increase
15. Find the speed of the binary stars Centauri A and Centauri B. They are separated by a distance of 3.45 x 1012 m and they have an orbital period of 2.52 x 109 seconds (that's about 80 years). They have the same mass. (Binary stars are near each other and orbit each other).
Solution: Speed is distance over time, or, the circumference of the orbit divided by the about 80 years: v = 2 p r / P = (2)(3.14)(1.72 x 1012 m)/(2.52 x 109 seconds) = (10.8 x 1012)/(2.52 x 109 s) = 4.3 x 103 m/s.
16. Find the Escape Velocity, vesc, for the planet Mercury.
Solution:
Escape velocity is, ve = ?(2GM/r), with G = 6.67 x 10-11, M = 3.3 x 1023 kg, and r = 2.44 x 106 m. So, now we know that ve = ?(2GM/r) = ve = ?[(2)(6.67 x 10-11)(3.3 x 1023 kg)/(2.44 x 106 m)] = ?[(4.4 x 1013)/(2.44 x 106 m)] = ?(1.8 x 107) = 4.2 x 103 m/s.
END
LESSON 14
PHYSICS LESSON 14 FOR
WEDNESDAY, JUNE 23, 2010
I Introduction
II Logistics: Return of Papers, Running Grades, Turn in Assignments Due, etc. and
III Review of Lesson 13: Simple Harmonic Motion
Repetitive Motion
Goes in cycles
IV Lesson 14: Waves
l n = v = m/s
Sound: vs = 342 m/s
c = 300,000 km/s
V Lab 11: Waves on a String
VI Homework Set 4: due 6/24
VII Essay 3: Albert Einstein, due 6/24
WEDNESDAY, JUNE 23, 2010
I Introduction
II Logistics: Return of Papers, Running Grades, Turn in Assignments Due, etc. and
III Review of Lesson 13: Simple Harmonic Motion
Repetitive Motion
Goes in cycles
IV Lesson 14: Waves
l n = v = m/s
Sound: vs = 342 m/s
c = 300,000 km/s
V Lab 11: Waves on a String
VI Homework Set 4: due 6/24
VII Essay 3: Albert Einstein, due 6/24
LAB 11
PhysicsLab11 June 23, 2010 Name __________________
Dr Dave Menke, Instructor
I. Title: String Phone – Waves in a String
II. Purpose: to observe and hear the effects of waves on a string and to make an old fashioned “tin can” phone
III. Equipment: 2 cans each team, string – about 10 meters +/-; awl or ice pick
IV. Procedure:
1. Measure out about 10 meters of string
2. punch a small hole in the end of one can
3. carefully thread string through the hole and tie a knot
4. pull the string until it reaches the hole
5. If the knot slips through the hole, make a larger knot so that it won’t slip through.
6. If you hadn’t guessed, the knot needs to be on the inside of the can.
7. Repeat with the second can
8. You and your partner need to go to a quieter place to “talk” to each other; extraneous noise will negate the results.
9. In your “quiet place” pull the cans as far apart as possible, without making the knot pop out of either can. If that happens, oops! Start over!
10. Have one partner think of a simple sentence and write it down, but don’t show the other partner.
11. Have that one partner say that simple sentence into his/her can, in a regular voice
12. The other partner needs to cup his/her ear with the can and listen, as if listening to a sea shell for the sound of the ocean.
13. The other partner needs to listen, and remember the simple sentence, then write it down.
14. Repeat this with the second partner writing his/her own sentence, then saying it into the can, and having the other partner listen, write down what she or he hears.
15. Clean up, put away toys, etc.
16. Return to your seats. Share what you heard with what your partner really said and vice versa
17. Determine how successful you were
V. Data, observations, calculations:
1. Sentence you wrote down and said: “_______________________.”
2. Sentence that your partner heard you say: “__________________.”
3. Difference, if any:
4. Sentence you heard: “__________________________.”
5. Sentence your partner wrote down and said: “__________________.”
6. Difference, if any:
VI. Results: How successful?
__This experiment was very successful as we were able to hear each other very clearly and with a high level of accuracy.
__This experiment was an abysmal failure as we were NOT able to hear each other very clearly at all.
__We were also able to approximate the speed of the sound wave in the string.
__We were also unable to approximate the speed of the sound wave in the string.
VII. Error:
A. Quantitative: Explain the reason for any differences in V 3. and V. 6.
B. Qualitative:
1. Personal –
2. Systematic –
3. Random –
VIII. Questions:
1. What causes the sound to travel along the string?
2. How fast does the sound travel on the string?
3. How can one find out how fast the sound traveled along the string?
4. What is a sonic boom? Explain in detail.
Dr Dave Menke, Instructor
I. Title: String Phone – Waves in a String
II. Purpose: to observe and hear the effects of waves on a string and to make an old fashioned “tin can” phone
III. Equipment: 2 cans each team, string – about 10 meters +/-; awl or ice pick
IV. Procedure:
1. Measure out about 10 meters of string
2. punch a small hole in the end of one can
3. carefully thread string through the hole and tie a knot
4. pull the string until it reaches the hole
5. If the knot slips through the hole, make a larger knot so that it won’t slip through.
6. If you hadn’t guessed, the knot needs to be on the inside of the can.
7. Repeat with the second can
8. You and your partner need to go to a quieter place to “talk” to each other; extraneous noise will negate the results.
9. In your “quiet place” pull the cans as far apart as possible, without making the knot pop out of either can. If that happens, oops! Start over!
10. Have one partner think of a simple sentence and write it down, but don’t show the other partner.
11. Have that one partner say that simple sentence into his/her can, in a regular voice
12. The other partner needs to cup his/her ear with the can and listen, as if listening to a sea shell for the sound of the ocean.
13. The other partner needs to listen, and remember the simple sentence, then write it down.
14. Repeat this with the second partner writing his/her own sentence, then saying it into the can, and having the other partner listen, write down what she or he hears.
15. Clean up, put away toys, etc.
16. Return to your seats. Share what you heard with what your partner really said and vice versa
17. Determine how successful you were
V. Data, observations, calculations:
1. Sentence you wrote down and said: “_______________________.”
2. Sentence that your partner heard you say: “__________________.”
3. Difference, if any:
4. Sentence you heard: “__________________________.”
5. Sentence your partner wrote down and said: “__________________.”
6. Difference, if any:
VI. Results: How successful?
__This experiment was very successful as we were able to hear each other very clearly and with a high level of accuracy.
__This experiment was an abysmal failure as we were NOT able to hear each other very clearly at all.
__We were also able to approximate the speed of the sound wave in the string.
__We were also unable to approximate the speed of the sound wave in the string.
VII. Error:
A. Quantitative: Explain the reason for any differences in V 3. and V. 6.
B. Qualitative:
1. Personal –
2. Systematic –
3. Random –
VIII. Questions:
1. What causes the sound to travel along the string?
2. How fast does the sound travel on the string?
3. How can one find out how fast the sound traveled along the string?
4. What is a sonic boom? Explain in detail.
Tuesday, June 22, 2010
LAB10
PhysicsLab10 June 22, 2010 Name __________________
Dr Dave Menke, Instructor
I Title: Simple Harmonic Motion
II Purpose: Study Simple Harmonic Motion using a plumb bob pendulum.
III Equipment
1. String, yarn, or cord ≥ 1.0 meter long (or as close as possible)
2. Weight - to make a plumb bob pendulum (washers?)
3. Stop Watch, face watch, digital watch, clock, or other chronometer
4. Meter stick or metric ruler
5. Protractor
IV Procedure (Some of this is similar to the Gravity lab)
1. Obtain, or make, a length of string or cord that is very close to 1.00 meter long. Slightly longer is better than slightly shorter.
2. Attach a weight to one end of the string to create a plumb bob, that we will call Bob.
3. Attach the other end of the string to some stationary object (door hinge, ceiling, weighted ring stand, etc.). Do NOT use a primate because it is not stable or other mammal to hold Bob because it is not stable.
4. Measure the length of Bob exactly (to the closest millimeter) after you have set it up. This will be from the point of connection on top to the middle of the weight. Record this length (we will call the length “r”) as accurately as possible.
5. Put the Data in the Table below; one column is for the number of the trials; another for the time (in seconds) for each cycle. And a third column for amplitude in the x-direction.
6. Have one of the lab partners pull the pendulum back, to an angle of θ = 45° (try to be exact; use protractor) as seen in the diagram a
7. Measure the x-component of the Bob's motion. If the string pendulum is exactly 1.0 meter long, and if the angle is exactly 45°, then the x-component will be (1.0 m) Sin 45° = 0.707 m = 70.7 cm. Thus, when you measure the x-component, it will be very close to 70 cm. This will be your original (and maximum) amplitude.
8. Simultaneously, release Bob and depress the stop watch button to start the time “running.” It is best to have the same homosapien release Bob and operate the stopwatch (the same brain controls both hands).
9. Allow Bob to swing freely as long it can. Every time that it returns to its starting point, note and record the time, and the distance. For example, at t = 0.00 s, the x-component will be (about) 70 cm. The next time that it comes back, about 1.8 seconds later, the x-component will be less, maybe 65 cm; next time, maybe 60 cm; and so forth. If you find it very difficult to do both the time and the x-component distance, you may substitute that data that you gathered for Lab #2 about Gravity. Keep the pendulum swinging until it stops, or, nearly stops. If you need to create another table to extend the data, please do so.
10. When done with gathering the data, plot the data points on a graph, with time in seconds, t, along the horizontal (left-right) and amplitude (length of x) along the vertical. Connect the dots as smoothly as you can.
11. Determine the period of oscillation using the graph of data.
V Data & Calculations:
Trial Time (sec) Amplitude (cm)
1 0 70
2 2 65
3 4 60
VI Results:
The purpose if this laboratory (Example of Simple Harmonic Motion) was / was not achieved due to:
VII Error Analysis:
A. Quantitative Error: NA
B. Qualitative Error:
1. Personal -
2. Random -
3. Systematic –
VIII Questions:
1. What is the period of oscillation?
2. What is the maximum amplitude?
3. What is the average periodic decrease in amplitude for each cycle?
4. List 4 items in your life and your world that oscillate.
Dr Dave Menke, Instructor
I Title: Simple Harmonic Motion
II Purpose: Study Simple Harmonic Motion using a plumb bob pendulum.
III Equipment
1. String, yarn, or cord ≥ 1.0 meter long (or as close as possible)
2. Weight - to make a plumb bob pendulum (washers?)
3. Stop Watch, face watch, digital watch, clock, or other chronometer
4. Meter stick or metric ruler
5. Protractor
IV Procedure (Some of this is similar to the Gravity lab)
1. Obtain, or make, a length of string or cord that is very close to 1.00 meter long. Slightly longer is better than slightly shorter.
2. Attach a weight to one end of the string to create a plumb bob, that we will call Bob.
3. Attach the other end of the string to some stationary object (door hinge, ceiling, weighted ring stand, etc.). Do NOT use a primate because it is not stable or other mammal to hold Bob because it is not stable.
4. Measure the length of Bob exactly (to the closest millimeter) after you have set it up. This will be from the point of connection on top to the middle of the weight. Record this length (we will call the length “r”) as accurately as possible.
5. Put the Data in the Table below; one column is for the number of the trials; another for the time (in seconds) for each cycle. And a third column for amplitude in the x-direction.
6. Have one of the lab partners pull the pendulum back, to an angle of θ = 45° (try to be exact; use protractor) as seen in the diagram a
7. Measure the x-component of the Bob's motion. If the string pendulum is exactly 1.0 meter long, and if the angle is exactly 45°, then the x-component will be (1.0 m) Sin 45° = 0.707 m = 70.7 cm. Thus, when you measure the x-component, it will be very close to 70 cm. This will be your original (and maximum) amplitude.
8. Simultaneously, release Bob and depress the stop watch button to start the time “running.” It is best to have the same homosapien release Bob and operate the stopwatch (the same brain controls both hands).
9. Allow Bob to swing freely as long it can. Every time that it returns to its starting point, note and record the time, and the distance. For example, at t = 0.00 s, the x-component will be (about) 70 cm. The next time that it comes back, about 1.8 seconds later, the x-component will be less, maybe 65 cm; next time, maybe 60 cm; and so forth. If you find it very difficult to do both the time and the x-component distance, you may substitute that data that you gathered for Lab #2 about Gravity. Keep the pendulum swinging until it stops, or, nearly stops. If you need to create another table to extend the data, please do so.
10. When done with gathering the data, plot the data points on a graph, with time in seconds, t, along the horizontal (left-right) and amplitude (length of x) along the vertical. Connect the dots as smoothly as you can.
11. Determine the period of oscillation using the graph of data.
V Data & Calculations:
Trial Time (sec) Amplitude (cm)
1 0 70
2 2 65
3 4 60
VI Results:
The purpose if this laboratory (Example of Simple Harmonic Motion) was / was not achieved due to:
VII Error Analysis:
A. Quantitative Error: NA
B. Qualitative Error:
1. Personal -
2. Random -
3. Systematic –
VIII Questions:
1. What is the period of oscillation?
2. What is the maximum amplitude?
3. What is the average periodic decrease in amplitude for each cycle?
4. List 4 items in your life and your world that oscillate.
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