User:Mathilde corazon/sandbox

Newtons Law of Motion

An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.

A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.60 seconds, what will be his final velocity and how far will he fall?

A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled.

A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s2. Determine the time for the feather to fall to the surface of the moon.

Rocket-powered sleds are used to test the human response to acceleration. If a rocket-powered sled is accelerated to a speed of 444 m/s in 1.83 seconds, then what is the acceleration and what is the distance that the sled travels?

A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4 m. Determine the acceleration of the bike.

An engineer is designing the runway for an airport. Of the planes that will use the airport, the lowest acceleration rate is likely to be 3 m/s2. The takeoff speed for this plane will be 65 m/s. Assuming this minimum acceleration, what is the minimum allowed length for the runway?

A car traveling at 22.4 m/s skids to a stop in 2.55 s. Determine the skidding distance of the car (assume uniform acceleration).

A kangaroo is capable of jumping to a height of 2.62 m. Determine the takeoff speed of the kangaroo. If Michael Jordan has a vertical leap of 1.29 m, then what is his takeoff speed and his hang time (total time to move upwards to the peak and then return to the ground)?

A bullet leaves a rifle with a muzzle velocity of 521 m/s. While accelerating through the barrel of the rifle, the bullet moves a distance of 0.840 m. Determine the acceleration of the bullet (assume a uniform acceleration).

A baseball is popped straight up into the air and has a hang-time of 6.25 s. Determine the height to which the ball rises before it reaches its peak. (Hint: the time to rise to the peak is one-half the total hang-time.) The observation deck of tall skyscraper 370 m above the street. Determine the time required for a penny to free fall from the deck to the street below.

A bullet is moving at a speed of 367 m/s when it embeds into a lump of moist clay. The bullet penetrates for a distance of 0.0621 m. Determine the acceleration of the bullet while moving into the clay. (Assume a uniform acceleration.)

A stone is dropped into a deep well and is heard to hit the water 3.41 s after being dropped. Determine the depth of the well. It was once recorded that a Jaguar left skid marks that were 290 m in length. Assuming that the Jaguar skidded to a stop with a constant acceleration of -3.90 m/s2, determine the speed of the Jaguar before it began to skid.

A plane has a takeoff speed of 88.3 m/s and requires 1365 m to reach that speed. Determine the acceleration of the plane and the time required to reach this speed.

A dragster accelerates to a speed of 112 m/s over a distance of 398 m. Determine the acceleration (assume uniform) of the dragster. With what speed in miles/hr (1 m/s = 2.23 mi/hr) must an object be thrown to reach a height of 91.5 m (equivalent to one football field)? Assume negligible air resistance

Given: a = +3.2 m/s2	t = 32.8 s	vi = 0 m/s Find: d = ?? d = vi*t + 0.5*a*t2 d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s2)*(32.8 s)2 d = 1720 m Return to Problem 1

Given: d = 110 m	t = 5.21 s	vi = 0 m/s Find: a = ?? d = vi*t + 0.5*a*t2 110 m = (0 m/s)*(5.21 s)+ 0.5*(a)*(5.21 s)2 110 m = (13.57 s2)*a a = (110 m)/(13.57 s2) a = 8.10 m/ s2 Return to Problem 2

Given: a = -9.8 m	t = 2.6 s	vi = 0 m/s Find: d = ?? vf = ?? d = vi*t + 0.5*a*t2 d = (0 m/s)*(2.60 s)+ 0.5*(-9.8 m/s2)*(2.60 s)2 d = -33.1 m (- indicates direction) vf = vi + a*t vf = 0 + (-9.8 m/s2)*(2.60 s) vf = -25.5 m/s (- indicates direction) Return to Problem 3

Given: vi = 18.5 m/s	vf = 46.1 m/s	t = 2.47 s	Find: d = ?? a = ?? a = (Delta v)/t a = (46.1 m/s - 18.5 m/s)/(2.47 s) a = 11.2 m/s2 d = vi*t + 0.5*a*t2 d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s2)*(2.47 s)2 d = 45.7 m + 34.1 m d = 79.8 m (Note: the d can also be calculated using the equation vf2 = vi2 + 2*a*d) Return to Problem 4

Given: vi = 0 m/s	d = -1.40 m	a = -1.67 m/s2 Find: t = ?? d = vi*t + 0.5*a*t2 -1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s2)*(t)2 -1.40 m = 0+ (-0.835 m/s2)*(t)2 (-1.40 m)/(-0.835 m/s2) = t2 1.68 s2 = t2 t = 1.29 s Return to Problem 5

Given: vi = 0 m/s	vf = 444 m/s	t = 1.83 s	Find: a = ?? d = ?? a = (Delta v)/t a = (444 m/s - 0 m/s)/(1.83 s) a = 243 m/s2 d = vi*t + 0.5*a*t2 d = (0 m/s)*(1.83 s)+ 0.5*(243 m/s2)*(1.83 s)2 d = 0 m + 406 m d = 406 m (Note: the d can also be calculated using the equation vf2 = vi2 + 2*a*d) Return to Problem 6

Given: vi = 0 m/s	vf = 7.10 m/s	d = 35.4 m	Find: a = ?? vf2 = vi2 + 2*a*d (7.10 m/s)2 = (0 m/s)2 + 2*(a)*(35.4 m) 50.4 m2/s2 = (0 m/s)2 + (70.8 m)*a (50.4 m2/s2)/(70.8 m) = a a = 0.712 m/s2 Return to Problem 7

Given: vi = 0 m/s	vf = 65 m/s	a = 3 m/s2 Find: d = ?? vf2 = vi2 + 2*a*d (65 m/s)2 = (0 m/s)2 + 2*(3 m/s2)*d 4225 m2/s2 = (0 m/s)2 + (6 m/s2)*d (4225 m2/s2)/(6 m/s2) = d d = 704 m Return to Problem 8

Given: vi = 22.4 m/s	vf = 0 m/s	t = 2.55 s	Find: d = ?? d = (vi + vf)/2 *t d = (22.4 m/s + 0 m/s)/2 *2.55 s d = (11.2 m/s)*2.55 s d = 28.6 m Return to Problem 9

Given: a = -9.8 m/s2	vf = 0 m/s	d = 2.62 m	Find: vi = ?? vf2 = vi2 + 2*a*d (0 m/s)2 = vi2 + 2*(-9.8 m/s2)*(2.62 m) 0 m2/s2 = vi2 - 51.35 m2/s2 51.35 m2/s2 = vi2 vi = 7.17 m/s Return to Problem 10

Given: a = -9.8 m/s2	vf = 0 m/s	d = 1.29 m	Find: vi = ?? t = ?? vf2 = vi2 + 2*a*d (0 m/s)2 = vi2 + 2*(-9.8 m/s2)*(1.29 m) 0 m2/s2 = vi2 - 25.28 m2/s2 25.28 m2/s2 = vi2 vi = 5.03 m/s To find hang time, find the time to the peak and then double it. vf = vi + a*t 0 m/s = 5.03 m/s + (-9.8 m/s2)*tup -5.03 m/s = (-9.8 m/s2)*tup (-5.03 m/s)/(-9.8 m/s2) = tup tup = 0.513 s hang time = 1.03 s Return to Problem 11

Given: vi = 0 m/s	vf = 521 m/s	d = 0.840 m	Find: a = ?? vf2 = vi2 + 2*a*d (521 m/s)2 = (0 m/s)2 + 2*(a)*(0.840 m) 271441 m2/s2 = (0 m/s)2 + (1.68 m)*a (271441 m2/s2)/(1.68 m) = a a = 1.62*105 m /s2 Return to Problem 12

Given: a = -9.8 m/s2	vf = 0 m/s	t = 3.13 s	Find: d = ?? (NOTE: the time required to move to the peak of the trajectory is one-half the total hang time - 3.125 s.) First use: vf = vi + a*t 0 m/s = vi + (-9.8 m/s2)*(3.13 s) 0 m/s = vi - 30.7 m/s vi = 30.7 m/s (30.674 m/s) Now use: vf2 = vi2 + 2*a*d (0 m/s)2 = (30.7 m/s)2 + 2*(-9.8 m/s2)*(d) 0 m2/s2 = (940 m2/s2) + (-19.6 m/s2)*d -940 m2/s2 = (-19.6 m/s2)*d (-940 m2/s2)/(-19.6 m/s2) = d d = 48.0 m Return to Problem 13

Given: vi = 0 m/s	d = -370 m	a = -9.8 m/s2 Find: t = ?? d = vi*t + 0.5*a*t2 -370 m = (0 m/s)*(t)+ 0.5*(-9.8 m/s2)*(t)2 -370 m = 0+ (-4.9 m/s2)*(t)2 (-370 m)/(-4.9 m/s2) = t2 75.5 s2 = t2 t = 8.69 s Return to Problem 14

Given: vi = 367 m/s	vf = 0 m/s	d = 0.0621 m	Find: a = ?? vf2 = vi2 + 2*a*d (0 m/s)2 = (367 m/s)2 + 2*(a)*(0.0621 m) 0 m2/s2 = (134689 m2/s2) + (0.1242 m)*a -134689 m2/s2 = (0.1242 m)*a (-134689 m2/s2)/(0.1242 m) = a a = -1.08*106 m /s2 (The - sign indicates that the bullet slowed down.) Return to Problem 15

Given: a = -9.8 m/s2	t = 3.41 s	vi = 0 m/s Find: d = ?? d = vi*t + 0.5*a*t2 d = (0 m/s)*(3.41 s)+ 0.5*(-9.8 m/s2)*(3.41 s)2 d = 0 m+ 0.5*(-9.8 m/s2)*(11.63 s2) d = -57.0 m (NOTE: the - sign indicates direction) Return to Problem 16

Given: a = -3.90 m/s2	vf = 0 m/s	d = 290 m	Find: vi = ?? vf2 = vi2 + 2*a*d (0 m/s)2 = vi2 + 2*(-3.90 m/s2)*(290 m) 0 m2/s2 = vi2 - 2262 m2/s2 2262 m2/s2 = vi2 vi = 47.6 m /s Return to Problem 17

Given: vi = 0 m/s	vf = 88.3 m/s	d = 1365 m	Find: a = ?? t = ?? vf2 = vi2 + 2*a*d (88.3 m/s)2 = (0 m/s)2 + 2*(a)*(1365 m) 7797 m2/s2 = (0 m2/s2) + (2730 m)*a 7797 m2/s2 = (2730 m)*a (7797 m2/s2)/(2730 m) = a a = 2.86 m/s2 vf = vi + a*t 88.3 m/s = 0 m/s + (2.86 m/s2)*t (88.3 m/s)/(2.86 m/s2) = t t = 30. 8 s Return to Problem 18

Given: vi = 0 m/s	vf = 112 m/s	d = 398 m	Find: a = ?? vf2 = vi2 + 2*a*d (112 m/s)2 = (0 m/s)2 + 2*(a)*(398 m) 12544 m2/s2 = 0 m2/s2 + (796 m)*a 12544 m2/s2 = (796 m)*a (12544 m2/s2)/(796 m) = a a = 15.8 m/s2 Return to Problem 19

Given: a = -9.8 m/s2	vf = 0 m/s	d = 91.5 m	Find: vi = ?? t = ?? First, find speed in units of m/s: vf2 = vi2 + 2*a*d (0 m/s)2 = vi2 + 2*(-9.8 m/s2)*(91.5 m) 0 m2/s2 = vi2 - 1793 m2/s2 1793 m2/s2 = vi2 vi = 42.3 m/s Now convert from m/s to mi/hr: vi = 42.3 m/s * (2.23 mi/hr)/(1 m/s) vi = 94.4 mi/h

Keplers Planetary motion 2. Galileo is often credited with the early discovery of four of Jupiter's many moons. The moons orbiting Jupiter follow the same laws of motion as the planets orbiting the sun. One of the moons is called Io - its distance from Jupiter's center is 4.2 units and it orbits Jupiter in 1.8 Earth-days. Another moon is called Ganymede; it is 10.7 units from Jupiter's center. Make a prediction of the period of Ganymede using Kepler's law of harmonies.

3. Suppose a small planet is discovered that is 14 times as far from the sun as the Earth's distance is from the sun (1.5 x 1011 m). Use Kepler's law of harmonies to predict the orbital period of such a planet. GIVEN: T2/R3 = 2.97 x 10-19 s2/m3

4. The average orbital distance of Mars is 1.52 times the average orbital distance of the Earth. Knowing that the Earth orbits the sun in approximately 365 days, use Kepler's law of harmonies to predict the time for Mars to orbit the sun. Orbital radius and orbital period data for the four biggest moons of Jupiter are listed in the table below. The mass of the planet Jupiter is 1.9 x 1027 kg. Base your answers to the next five questions on this information. Jupiter's Moon	Period (s)	Radius (m)	T2/R3 Io	1.53 x 105	4.2 x 108	a. Europa	3.07 x 105	6.7 x 108	b. Ganymede	6.18 x 105	1.1 x 109	c. Callisto	1.44 x 106	1.9 x 109	d.

5. Determine the T2/R3 ratio (last column) for Jupiter's moons.

6. What pattern do you observe in the last column of data? Which law of Kepler's does this seem to support?

7. Use the graphing capabilities of your TI calculator to plot T2 vs. R3 (T2 should be plotted along the vertical axis) and to determine the equation of the line. Write the equation in slope-intercept form below.

8. How does the T2/R3 ratio for Jupiter (as shown in the last column of the data table) compare to the T2/R3 ratio found in #7 (i.e., the slope of the line)?

9. How does the T2/R3 ratio for Jupiter (as shown in the last column of the data table) compare to the T2/R3 ratio found using the following equation? (G=6.67x10-11 N*m2/kg2 and MJupiter = 1.9 x 1027 kg) T2 / R3 = (4 * pi2) / (G * MJupiter ) Graph for Question #6 Return to Question #6

1.Answer: T = 7.32 days Given: Io: Rio = 4.2 and Tio = 1.8 Ganymede: Rg = 10.7 Tg=??? Use Kepler's 3rd law to solve. (Tio)^2/(Rio)3 = 0.04373; so (Tg)^2 / (Rg)3 = 0.04373 Proper algebra would yield (Tg)^2 = 0.04373 • (Rg)3 (Tg)2 = 53.57 so Tg = SQRT(53.57) = 7.32 days Answer: Tplanet = 52.4 yr Use Kepler's third law: (Te)^2/(Re)^3 = (Tp)^2/(Rp)3 Rearranging to solve for Tp: (Tp)^2=[ (Te)2 / (Re)3] • (Rp)3 or (Tp)2 = (Te)2 • [(Rp) / (Re)]3 where (Rp) / (Re) = 14 so (Tp)2 = (Te)2 • [14]3 where Te=1 yr (Tp)2 =(1 yr)2 *[14]^3 = 2744 yr2 Tp = SQRT(2744 yr2)

Answer: Tplanet = 52.4 yr Use Kepler's third law: (Te)^2/(Re)^3 = (Tp)^2/(Rp)3 Rearranging to solve for Tp: (Tp)^2=[ (Te)2 / (Re)3] • (Rp)3 or (Tp)2 = (Te)2 • [(Rp) / (Re)]3 where (Rp) / (Re) = 14 so (Tp)2 = (Te)2 • [14]3 where Te=1 yr (Tp)2 =(1 yr)2 *[14]^3 = 2744 yr2 Tp = SQRT(2744 yr2)

Given: Rmars = 1.52 • Rearth and Tearth = 365 days Use Kepler's third law to relate the ratio of the period squared to the ratio of radius cubed (Tmars)2 / (Tearth)2 • (Rmars)3 / (Rearth)3 (Tmars)2 = (Tearth)2 • (Rmars)3 / (Rearth)3 (Tmars)2 = (365 days)2 * (1.52)3 (Note the Rmars / Rearth ratio is 1.52) Tmars = 684 days

a. (T2) / (R3) = 3.16 x 10-16 s2/m3 b. (T2) / (R3) = 3.13 x 10-16 s2/m3 c. (T2) / (R3) = 2.87 x 10-16 s2/m3 d. (T2) / (R3) = 3.03 x 10-16 s2/m3

The (T2) / (R3) ratios are approximately the same for each of Jupiter's moons. This is what would be predicted by Kepler's third law!

T2 = (3.03 * 10-16) * R3 - 4.62 * 10+9 Given the uncertainty in the y-intercept value, it can be approximated as 0. Thus, T2 = (3.03 * 10-16) * R3

The values are almost the same - approximately 3 x 10-16.

The values in the data table are approx. 3 x 10-16. The value of 4*pi/(G*MJupiter) is approx. 3.1 x 10-16.

It takes satellite A 7 days to complete one full orbit around a planet, at a distance of 5.0 units. Satellite B orbits it in 6 days. How far is it from the planet? The formula: (T12) / (r13) = (T22) / (r23). If we substitute values as we know them, we get 72/53 = 62/x3  adius)em, we get s of Earth years? from the Sun than the Earth is, how much longer is its year? cause of . Thus, x = 4.5 units

If Saturn is, on average, 9 times farther from the Sun than the Earth is, how long is its year in terms of Earth years? We will use the formula (T12) / (r13) = (T22) / (r23). If we substitute values as we know them, we get 12/13 = x2/93  adius)em, we get s of Earth years? from the Sun than the Earth is, how much longer is its year? cause of, so we solve for x and get x = √ (93) = 27

Satellite A is 5 times farther from a planet than satellite B. If it takes satellite A 22 weeks to complete a full orbit around the planet, how long will it take satellite B to travel around the planet once? The formula: (T12) / (r13) = (T22) / (r23). If we substitute values as we know them, we get 222/53 = x2/13  adius)em, we get s of Earth years? from the Sun than the Earth is, how much longer is its year? cause of . Thus, x = 2 weeks

If planet A and B are both circling around star A and A completes an orbit in 2 days while B needs 10 days, then what is the radius of B’s orbit if that of A is 3.0 × 108m (in meters) ? We will use the formula (T12) / (r13) = (T22) / (r23). By solving this equation, we get the radius of B: r2 = ∛ ((T22 r13) / (T12)) = ∛ ((10_ 2 (3 × 108) 3) / (2_ 2)) = 8.77 × 108 m

Momentum a.	A basketball ball having 2kg mass and 6m/s velocity moves to the east b.	A car having 15m/s velocity and 1500kg mass moves to the north c.	A child having mass 25kg and velocity 2m/s moves to the west a. Momentum of basketball; b. Momentum of car; c. Momentum of boy;

Ex 1) A 5.0 kg mass has its velocity change from 8.0 m/s east to 2.0 m/s east.  Find the objects change in momentum. m = 5.0 kg Vi = 8.0 m/s East Vf = 2.0 m/s East Δp = ? J = Ft = mΔv =Δp Δp = mΔV = (5.0 kg)(2.0 m/s - 8.0 m/s) = -30. kgm/s East or +30. kg m/s West

Ex 2) A 5.0 kg mass moving with a vector of 8.0 m/s east has an impulse applied to it which causes its velocity to change to 20. m/s East. Find Impulse: m = 5.0 kg Vi = 8.0 m/s East Vf = 20. m/s East J = ? J = Ft = mΔv =Δp

J = mΔv = (5.0 kg)(12. m/s East) = 60. kg m/s east = 60. Ns East Find the force if the impulse was applied for 3.0 sec. F = ? t = 3 seconds m = 5.0 kg Vi = 8.0 m/s East Vf = 20. m/s East J = 60. kg m/s east J = Ft = mΔv =Δp J = Ft = 60. Ns East F(3.0 sec) = 60. Ns East F = 20. N East

1. Calculate the momentum of a 12ookg car with a velocity of 25m/s. p = mv = 1200 X 25 = 30,000kg.m/s 2. What is the momentum of a child and wagon if the total mass of the child and wagon is 22kg and the velocity is 1.5m/s? p = mv = 22 X 1.5 = 33kg.m/s 3. The parking brake on a 1200kg automobile has broken, and the vehicle has reached a momentum of 7800kg.m/s. What is the velocity of the vehicle? V = p/m = 7800/1200 = 6.5m/s 4. A toy dart gun generates a dart with .140kg.m/s momentum and a velocity of 4m/s. What is the mass of the dart in grams? (hint: figure kg, then convert answer     to grams) M = p/v = .140/4 = .035kg     conversion:  .035  X 1000 = 35grams 5. A bowling ball of 35.2kg, generates 218 kg.m/s units of momentum. What is the velocity of the bowling ball? V = p/m = 218/35.2 = 6.2m/s 6. A school bus traveling at 40 km/hr. (11.1m/s) has a momentum of 152625 kg.m/s. What is the mass of the bus? M = p/v = 152625/11.1 = 13,750kg Conservation of Momentum Problems (Collision Problems) 7. A 12,000kg. railroad car is traveling at 2m/s when it strikes another 10,000kg.railroad car that is at rest. If the cars lock together, what is the final speed of the two railroad cars? p1 = p2     m1 v1 = m2 v2      (12,000) (2)  = (22,000)  v2             m2 =mass of both cars 12,000 + 10,000 24,000 = 22,000 v2      24,000/22,000 = v2      v2 = 1.1m/s 8. A 9,300 kg. railroad car traveling at a velocity of 15m/s strikes a second  boxcar at rest. If the two cars stick together and move off with a velocity of 6m/s, what is the mass of the second car? p1 = p2     m1 v1  =  m2  v2      (9,300) (15)  =  (m2)  (6)              m2 = mass of both cars = 9,300  + X       139,500  = (9,300 + X) (6)                                                                               X = mass of second boxcar 139,500 = 55,800 +6X 139,500 - 55,800 = 6X 83,700 =  6X 83,700/6 = X       X = 13,950 kg 9.  A 25 gram bullet is fired from a gun with a speed of 230m/s. If the gun has a mass of .9kg. what is the recoil speed of the gun? p1 = p2    m1 v1 = m2 v2                                          Uses Newton's Third Law (action = reaction) (.025) (230) =  (.9)  v2                         Convert 25grams to kg = 25/1000 = .025kg 5.75 = (.9)  v2       5.75/.9  = v2       6.4m/s = v2 10. A 20 gram bullet traveling at 250m/s strikes a block of wood that weighs 2kg. With what velocity will the block and bullet move after the collision? p1 = p2 m1 v1 = m2 v2 (.020) (250) = (2.02)  v2 5 = (2.02)  v2 5/2.02 = v2 2.5m/s = v2

WAVE AND FREQUENCY

Problem 1. A saxophone is playing a steady note of frequency 266 Hz. The temperature in the room is 25 C. Suppose that at some instant the varying pressure at your eardrum is at a maximum. How far away in meters is the next pressure maximum? Solution: The distance between the pressure maxima is an integer number of wavelength. Therefore the shortest distance is the wavelength. The speed of sound at 25 C is 346.33 m/s. Then we can find the wavelength: Problem 2. (Inquiry into Physics-5th ed.,Ostdiek,Bord) The quartz crystal used in an electric watch vibrates with frequency 32,768 Hz. What is the period of the crystals motion? Solution: By definition the frequency is the inverse period (see equations). Then the period is

Problem 3. A sound wave traveling at 350 m/s has a frequency of 500 Hz. What is its wavelength? Solution: The wavelength is related to the frequency and the speed by the following relation (see equations):

Problem 4. Estimate how far away a cicada can be heard if the lowest possible audible intensity of a sound it produces is and the power of the cicada's sound source is. Solution: We can estimate that the total power of the cicada's sound is distributed uniformly over the spherical surface of radius. Then at distance the intensity of the sound is

The largest radius is achieved when the intensity is. Then

Then

Problem 5. Light of wavelength 497.0 nm appers to have a wavelength of 500.2 nm when it reaches eart from a distance star. find the velocity of the star if the speed of light is 300000000 m/s. Solution: In this problem we need to use the expression for the Doppler shift of the frequency of the wave for an observer moving relative to the source of the waves.. If the source of the light (wave) is moving with a speed then the change in the frequency is

where is the speed of light. Since

Then

and

From this expression we can find the speed of the source