What is the difference between average velocity and instantaneous velocity?
Q. One of the questions in my book asks the reader to consider the following motion of an object in one dimension: "A ball is thrown directly upward, rises to a highest point, and falls back into the thower's hand. The average velocity for the thrown ball is zero - the ball returns to the starting point at the end of the time interval. There is one point at which the instantaneous velocity is zero - at the top of the motion." What is the difference between average velocity and instantaneous velocity? Why does the instantaneous velocity only = 0 at the apex?
Asked by Sunday - Thu Sep 10 19:49:57 2009 - - 1 Answers - 0 Comments

A. average velocity = displacement / time Your average velocity in the course of a day is 0. You start in bed and you end in bed so your net displacement over the course of the day is 0. But that doesn't mean you haven't gotten out of bed and moved. instantaneous velocity = instantaneous speed plus direction of motion. The ball is either moving up, moving down, or not moving. When it is moving up or down, it has a non-zero instantaneous velocity. At the top of its motion, in the instant of transition from moving up to moving down, it is not moving and so has an instantaneous velocity of 0.
Answered by simplicitus - Sun Sep 13 02:43:08 2009

What is his velocity relative to the surface of the ice?
Q. A 50kg firefighter is standing on a 200kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is flat, frictionless surface. The firefighter begins to walk along the plank at a constant velocity of 2.0m/s to the right relative to the plank. (a.) What is his velocity relative to the surface of the ice? (b.) What is the velocity of the plank relative to the surface of the ice?
Asked by dezd - Sat Jan 24 23:06:52 2009 - - 3 Answers - 0 Comments

A. Let us try to solve the problem in the frame of reference of frozen lake to begin with momentum was zero. Let Vf be the velocity of fire fighter in thisframe and Vp be that of plank. Obviously they will be in opposite direction. Momentum conservation gives 200 Vp=50 Vf and Vf+Vp=2 or Vf = 2-Vp or 200 Vp = 50 (2 - Vp) or 250 Vp = 100 or Vp = 100/250 = 0.4 m/s So The velociy of firefighter with respect to lake = 1.6 m/s and that of planck with respect to lake is 0.4 m/s in teh opposite direction.
Answered by Let'slearntothink - Sun Jan 25 01:47:17 2009

What is the velocity of a roller coaster going down a slope?
Q. What is the velocity of a roller coaster going down a slope? The hill is 10m high with a downward slope of 40 degrees. There is no friction, and assume at the top of the hill the initial velocity is 0m/s. And use an equation of motion (not an energy equation)
Asked by Fred N - Mon Apr 6 16:25:59 2009 - - 4 Answers - 0 Comments

A. Doing the problem as asked... Beginning the problem we only know V(initial) = 0. First, let's solve for acceleration of the coaster going down the slope. If you sum the forces on the coaster you will find the force parallel to the slope in the direction of the coaster to be, F = mgsin(angle). Set this equation equal to (ma) and the mass of the coaster is no longer a factor. So, acceleration = gsin(angle) = 9.81m/s^2(sin40) = 6.31 m/s^2 This makes sense because it is only a fraction of acceleration due to gravity, so it must be less than 9.81. Solve for the distance down the slope using sin = opp / hyp. Slope = 15.5m Using kinematics eqn. Vf^2 = Vi^2 + 2ax Vf = sq. rt[2 (6.31)(15.5)] Vf = 14 m/s
Answered by MD2BE - Mon Apr 6 16:50:56 2009

What is the difference between velocity and displacement?
Q. Here is a question. You drop a rock into a well at the coast and hear a splash 3 seconds later. Given that acceleration is 9.8m/s^2 at sea level, calculate final velocity. Final velcoity= 9.8m*3 29.4 I have to illustrate the velocity of the rock vs. time using a scatter graph and a graph showing the fall(displacement) of the rock vs. time a scatter graph. Wont the data be the exact same? Because at every second the velocity increases and so does the displacement. What is right?
Asked by Dark angel - Wed Apr 16 03:03:10 2008 - - 4 Answers - 0 Comments

A. No, the rate of change of velocity is constant. The rate of change of displacement is not. So, your velocity vs. time graph should be a straight line. Your displacement vs. time graph should be a curved line up.
Answered by Sherwin L - Wed Apr 16 03:10:21 2008

What is the velocity of the block immediately after it is struck?
Q. A 20-g bullet moving horizontally with a velocity of 800 m/s embeds itself in a 3980-g wooden block suspended by a long cord. (a) What is the velocity of the block immediately after it is struck? (b) To what elevation does the block rise?(Show the computations pls...)Tnx!
Asked by macedon - Tue Feb 12 06:54:27 2008 - - 1 Answers - 0 Comments

A. m = 0.02kg v = 800m/s M = 3.98kg Conserve momentum The momentum of the bullet is converted into the momentum of the combined block/bullet initial = final mv = (m+M)V so V = mv/(m+M) = 4.0m/s Conserve energy The kinetic energy of the block/bullet is converted to potential energy initial = final (m+M)V = (m+M)gh so h = V /(2g) = 0.82m .,.,.,
Answered by The Wolf - Tue Feb 12 07:03:53 2008

What is the final velocity after the carts collide?
Q. A 0.500 kg dynamics cart heading east at a velocity of 0.20 m/s collides in a head-on collision with another cart of mass 0.400 kg traveling west with a speed of 0.40 m/s. After the collision, the two carts link together. What is their final velocity?
Asked by Ruphert J - Wed Apr 11 13:12:01 2007 - - 3 Answers - 0 Comments

A. use conservation of momentum... Before collision = 0.5*0.2 - 0.4*0.4 = -0.06 kgm/s After collision = 0.9*velocity = -0.06 v = -0.06/0.9 = -0.07 m/s (minus indicates west)
Answered by Fizz_Assist - Wed Apr 11 13:18:55 2007

How do you measure velocity in the void of space?
Q. Speed is measured relative to a stationary object. Velocity is determined by calculating the time it takes to travel a known distance from a stationary object. A cars speedometer measures speed relative to the stationary ground it is traveling on. An airplane measures speed by the same thing, by how long it takes to travel between points of a known distance on the ground. So when it is said that a spacecraft is hurtling at 46,000 miles per hour beyond the orbit of Pluto towards the next star, what is that speed being compared against? In that part of space, there is no stationary object to calculate velocity. Is it still being compared to a stationary object on the surface of the Earth? At that distance, the Earth is traveling… [cont.]
Asked by cdb - Sat Jul 22 15:44:34 2006 - - 7 Answers - 0 Comments

A. kind of a weird question, but actually when you are measuring relative to a "stationary" object such as the earth, you arent really...right now i am stationary, but am still moving fast enough to move completely around the earth in one day, and even if i measured my speed relative to the axis of the earth, even the axis is rotating around the sun so that it completes one rotation every year...that is really pretty damn fast if you think about it...truthfully they can figure the distance between the earth and the planet (pluto in this case) and assume the velocity constant so they measure the time it takes to get there...it is pretty accurate to consider the velocity constant because most of the time the velocity in outer space is due to… [cont.]
Answered by sam - Sat Jul 22 15:55:40 2006

What is true about her average speed and her average velocity?
Q. Emily jogs 3 laps at the track to warm up. She jogs at 5 mph. Then she runs 3 more laps at 7 mph. What is true about her average speed and her average velocity? A. Her average speed is smaller than her average velocity. B. Her average speed and her average velocity are the same. C. Her average speed is greater than her average velocity. D. Her average speed is zero, but her average velocity is not.
Asked by Breakaway - Thu Oct 9 18:27:04 2008 - - 2 Answers - 1 Comments

A. Just remember that velocity shows direction with negative values symbolizing opposite direction! Imagine an overhead view of the circular/semi-circular track has an xy coordinate drawn over it: Going around a track, she jogs and runs to the right and upward (positive velocity) as much as she does to the left and downwards (negative velocity) so her average velocity is really 0! Speed cannot be negative, so the average speed is certainly a value greater than zero, as ANY speed is either positive or zero. (Because of what the question asks, you don't need to figure out the exact average speed) any positive number is greater than zero, so that answer is C.)
Answered by Max Power - Thu Oct 9 18:42:18 2008

What is velocity of the magician after the bullet is caught?
Q. A physicist attending a magical show by magician Jean Robert-Houdin is skeptical the magician can catch handgun bullets. Before firing the gun, the physicist asks the magician to stand on a platform which can slide on the floor without friction. The mass of the magician is 75 kg, the mass of the bullet is 10 g, and muzzle velocity is 300 m/s.
Asked by Alex Smirnoff - Thu May 14 18:36:51 2009 - - 2 Answers - 0 Comments

A. Zero, because the magician does not catch the bullet. It passes him on the upstage side, and he shows you one he put in his mouth backstage before the act started.
Answered by frediwhite@verizon.net - Thu May 14 18:43:48 2009

How do I calculate Initial, horizontal, and vertical velocity with only angle, time, and max height?
Q. I am doing a physics project and I need to calculate the initial velocity, initial x velocity, and initial y velocity. All the information I have is the maximum height, time, and angle of launch. Can you help me?
Asked by Ann S - Wed Jun 10 14:41:41 2009 - - 3 Answers - 0 Comments

A. Let's work out an example. Your projectile attains a maximum height of 1,100 meters, and is launched at an angle of 34 degrees above the horizontal. The total flight time is 29.966 seconds. Sqrt(1100 / 9.8 * 2) * 2 = 29.966 The initial vertical velocity is 146.833 m/s. Sqrt(1100 / 9.8 * 2) * 9.8 = 146.833 The initial horizontal velocity is 217.690 m/s. 146.833 / tan(34) = 217.690 The initial velocity is 262.581 m/s. Sqrt(146.833^2 + 217.690^2) = 262.581
Answered by farwallronny - Wed Jun 10 15:45:59 2009

How do you find instantaneous velocity on a linear graph?
Q. To be more precise, a graph with multiple linear (but no curved) lines. Would it simply be the slope of the line, and you enter the x coordinate (time) to that slope to find instantaneous velocity? I understand that instantaneous velocity is simply the slope of the tangent line on a curve, but this is all linear.
Asked by matt g - Sat Sep 12 21:05:04 2009 - - 2 Answers - 0 Comments

A. If it's linear then the velocity is constant so it's the same no matter where you are on the line.
Answered by hayharbr - Sat Sep 12 21:09:12 2009

How does frequency or angular velocity begin to increase or come to an end?
Q. When frequency or angular velocity does not change frequency is an integer or a number (angle l/r). But no number is frequency or angular velocity when frequency increases or decreases. When rate of counting increases or frequency increases from 0 at uniform speed, within the first cycle, period per cycle decreases from infinity. Should there be a period within the first cycle (or 0 to 1). When frequency decreases at uniform speed, can we pronounce 0, the last number, to declare that period has become infinity and frequency has become zero and counting has stopped?
Asked by Vadakkan S - Mon Jul 7 07:45:08 2008 - - 1 Answers - 0 Comments

A. This makes no sense. If the question is not clearly stated, it cannot be answered.
Answered by runningman022003 - Mon Jul 7 08:19:50 2008

What is the velocity of a particle that travels at right angles through a magnetic field?
Q. A singly-ionized particle experiences a force of 7.1 10-13 N when it travels at right angles through a 0.81-T magnetic field. What is the velocity of the particle?
Asked by Andy - Sun May 10 19:14:57 2009 - - 2 Answers - 0 Comments

A. F = (Bqv)sin(theta) theta = 90 degrees, so sin(theta) = 1, hence F=Bqv B = 0.81T F = 7.1x10^-13 N singly ionised, so its either lost or gained an electron, so the net charge of the particle is q = +/- 1.6x10^-19 C v = F/Bq = 5.47x10^6 ms^-1.
Answered by LaTeRaLuS - Sun May 10 19:21:23 2009

What happens to velocity when the frequency goes up?
Q. Please please help, I have a test tomorrow and right now im so nervouse i got hic-ups :( What happens to velocity when the frequency goes up? And what happens with the wave length?
Asked by leonamuni - Wed May 27 20:34:55 2009 - - 4 Answers - 0 Comments

A. The equation is V = F*L where V is velocity, F is frequency and L is wavelength. 1) If F increases, then V should also get bigger. 2) If V is a constant, then F*L is a constant. This means if frequency goes up, wavelength must go down in order for F*L to remain at the same value. Just read over the stuff slowly, try to understand it rather than learn it off. Remember, its just a stupid test and the equation is the important part for marks! (ha ha, 15 years of college science and he's almost got a degree, how impressive!)
Answered by unknown - Wed May 27 20:49:59 2009

What happens if you shoot high velocity ammo through a suppressor?
Q. I recently bought a Walther G22 rifle with a working suppressor, but the suppressor has a warning that it has to be used only with sub sonic ammo. I want to use high velocity ammo (even if the suppressor is less effective) because the sub sonic ammo is really messy (a lot of lead fouling and gun powder residue), but my fear is that shooting high velocity ammo throught the suppressor may damage the suppressor, the rifle or both.
Asked by daniel l - Tue May 13 12:59:42 2008 - - 6 Answers - 0 Comments

A. High velocity .22 ammo will not damage the suppressor or the rifle. The problem is sound level. With standard velocity ammo, the suppressor drops the sound level low enought to be safe without hearing protection. High velocity ammo where the bullet velocity is over the speed of sound will produce a high pitch crack that the suppressor cannot dampen. This sonic crack can be loud enough to require hearing protection. The warning is to protect your ears, not the gun or the suppressor.
Answered by corey h - Tue May 13 14:10:15 2008

What is the velocity of the 90kg player just before the tackle?
Q. A 90kg halfback runs north and is tackled by a 120kg opponent running south at 4m/s. The collision is perfectly inelastic. Just after the tackle, both players move at a velocity of 2m/s north.
Asked by Becca B - Wed Jun 6 17:08:36 2007 - - 1 Answers - 0 Comments

A. Momentum is conserved. Momemtum before = momentum after. mv1 + mv2 = (m1 + m2)v 90v + (120)(-4) = (90 + 120)2 90v - 480 = (210)2 90v - 480 = 420 90v = 900 v = 10m/s
Answered by Rec - Wed Jun 6 17:14:01 2007

What is the magnitude of the velocity of the protons?
Q. A proton source injects H(+) ions at low speed. A beam of protons (each having a rest mass 1.67*10^-27 kg) is moving in a circle of radius 80.0 cm between 2 circular flat electromagnets. If these electromagnets supply a force of 8.00*10^-13 N on each proton directed toward the center of revolution, what is the magnitude of the velocity of the protons? The proton motion is circular and the speeds involved are small compared with the velocity of light. Take the value of squareroot of 3.83 to be 1.96.
Asked by willydavidjr - Sat May 27 16:00:39 2006 - - 1 Answers - 0 Comments

A. Since the proton is moving in a circular pattern, the force given is the centripetal force. I know that it it worded weirdly, but I believe that is just to throw you off. The equation for the centripetal force due to circular motion is F = m * (v^2) / r where F is the force, m is the mass of the object, v is the velocity, and r is the radius of the circular path. So we set up the following: 8.00 * 10^-13 N = (1.67 * 10 ^-27 kg) * (v^2) / (0.8 m) I used 0.8m instead of 80cm because we need to be in standard SI units. This will make sense later. Rearrange and solve for v. V = sqrt[ (8.00 * 10^-13 N) * (0.8 m) / (1.67 * 10^-27 kg) ] Since N = (kg*m)/(s^2) we can re-write as: V = sqrt[ (8.00 * 10^-13 (kg*m)/(s^2)) * (0.8 m) / (1.6 [cont.]
Answered by bdecker202 - Wed Jun 7 22:13:17 2006

What is the magnitude of the initial velocity that player B's ball must be given?
Q. Baseball player A bunts the ball by hitting it in such a way that it acquires an initial velocity of 1.70 m/s parallel to the ground. Upon contact with the bat the ball is 1.26 m above the ground. Player B wishes to duplicate this bunt, in so far as he also wants to give the ball a velocity parallel to the ground and have his ball travel the same horizontal distance as player A's ball does. However, player B hits the ball when it is 1.42 m above the ground.
Asked by Jaime V - Thu Sep 4 11:13:33 2008 - - 1 Answers - 0 Comments

A. Vb = Va* (ya/yb) = 1.7* (1.26/1.42) = 1.6 m/s
Answered by Steve - Thu Sep 4 18:19:03 2008

What is the angular velocity and displacement of a spinning propeller?
Q. A propeller spins at 1850 rev/min. (a) What is the angular velocity in rad/s? (b) What is the angular displacement of the propeller in 0.70 s?
Asked by Mara - Wed Nov 12 16:16:30 2008 - - 1 Answers - 0 Comments

A. The data you provided is the frequency of the propeller in revolutions per minute, we must first convert it to revolutions per secons (hertzs): f = 1850 rev/min = 30.8333 rev/s (a) The angular velocity is a vector quantity with its magnitude being so called the angular frequency - it is a measure of how fast the propeller is rotating. Since there're 360 in one revolutions or 2 radians, = 2 f = 2 x 30.8333 /s = 193.7 rad/s. (b) Since the angular frequency is the ratio of angular displacement to the time interval t, = / t, so = x t = 193.7 rad/s x 0.70 s = 135.59 radians or 7769 degrees.
Answered by Dorian36 - Wed Nov 12 16:46:11 2008

How do I increase velocity of my fastball?
Q. I'm 15 years old. Been playing baseball and pitching my whole life. On a professional radar my fastball ranges from 77-81 mph. Being so into baseball I'm always looking for new ways to increase my velocity and arm strength. Any ideas?
Asked by i can't tell you - Sun Jul 5 16:23:50 2009 - - 2 Answers - 0 Comments

A. Practice. You have a long toss every few days and make sure you get arch on the ball. That builds up your arm strength and will help you throw faster.
Answered by sbraves10 - Tue Jul 7 20:54:38 2009