Number of results: 112,111
A 30g bullet with a horizontal velocity of 500m/s comes to a stop 12 cm within a solid wall. a) what is the change in its mechanical energy? b) What is the magnitude of the average force from the wall stopping it? a) what would I use of the equation? Can you please explain ...
Wednesday, January 10, 2007 at 2:38pm by Jamie
How do i go about this question? Suppose a car is traveling at 20.0m/s, and the driver sees a traffic light turn red, After 0.530s has elapsed (the reaction time), the driver applies the brakes, and the car accelerates at 7.00m/s2. What is the stopping distance of the car, as ...
Thursday, August 16, 2007 at 6:27pm by manny
subtract from the 35 m the distance during reactiontime (v*.9) so the stopping distance is 35-.9v. That is equal to d in ... vf^2=v^2+2ad vf is zero, solve for v.
Tuesday, September 4, 2007 at 7:57pm by bobpursley
What is the maximum speed at which a car could be moving and not hit a barrier 35.0 m ahead if the average acceleration during braking is -10.0 m/s2 and it takes the driver 0.90 s before he applies the brakes? subtract from the 35 m the distance during reactiontime (v*.9) so ...
Tuesday, September 4, 2007 at 10:02pm by Sarah
A car with good tires on a dry road can decelerate at about 5 m/s^2 when breaking. Suppose a car is initially traveling at 65 mph. What is the stopping distance?(give answer in meters and in feet) I changed 65mph into 29.05m/s. so I would use d-d_0 = (v^2 -v_0^2)/2a and when I...
Wednesday, September 12, 2007 at 4:33pm by Tammy
Max friction force = M g u The equals M a when it is on the verge of slipping. Therefore a = u g is the maximum deceleration rate. u is the static coefficient of friction, 0.27 a = 2.646 m/s^2 To get the stopping distance X, use V = sqrt (2 a X)= sqrt (2 u g X) Convert the V...
Wednesday, December 5, 2007 at 10:47am by drwls
A 850.0-kg car travelling on a level road at 27.0 m/s (60.5 mi/hr) can stop, locking its wheels, in a distance of 61.0 m (200.1 ft). (a) Find the size of the horizontal force which the car applies on the road while stopping. (b) Find the stopping distance of that same car when...
Wednesday, December 5, 2007 at 4:16pm by Lindsay
(a) (Force) x (stopping distance) = kinetic energy converted to heat = (1/2) M V^2 Solve for F (b) In this case, the initial kinetic energy is converted to both gravitational potential energy and heat. The horizontal force will be whatever you get in Part (a), multiplied by ...
Wednesday, December 5, 2007 at 4:16pm by drwls
A truck driver slams on the brakes and skids to a stop through a displacement of delta x. (a) IF the truck has twice the mass, by what factor does the stopping distance change? 1) 0.25 or 2) 1 or 3) 4 (b) If the initial velocity of the truck were halved, by what factor would ...
Monday, February 25, 2008 at 6:47pm by John
(a) The answer is (2)(a factor of 1). The stopping distance does not change. That is because both the kinetic energy and the friction force double. (b) 4) 0.25
Monday, February 25, 2008 at 6:47pm by drwls
1. (avg. force)*(stopping distance) = (initial kinetic energy) = (1/2) M V^2 Solve for the initial force. The statement says that the work done frictionally heating the wood equals the in initial kinetic energy. 2. Impulse = -(intial momentum) = -M V 3. Change in momentum = -(...
Tuesday, March 25, 2008 at 1:44am by drwls
Assume that the bed sheet is of light material and deflects a lot when the egg hits it in the middle. Force = momentum change / time change = m delta v / delta t ---> m a The force on the egg is the mass of the egg * speed of egg when it hits / time to stop the time to stop...
Sunday, June 1, 2008 at 2:24pm by Damon
an arresting device on a carrier deck stops an airplane in 1.5s. the avg. acceleration was 49 m/s2. what was the stopping distance? what was the initial speed? so i think v2=0, t=1.5, a= 49, so i need d and v1. i got v1=73.5 m/sec and d= 55 m. i took 0=v1-49(1.5) to get that ...
Thursday, June 12, 2008 at 10:36am by kayla
Traumatic brain injury such as concussionn results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s^2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1000 m/s^2 lasting for at least 1 ms ...
Thursday, September 18, 2008 at 6:53pm by mb
Physics - Stopping distance
Consider two bodies, A and B, moving in the same direction with the same kinetic energy. A has a mass twice that of B. If the same retarding force is applied to each, how will the stopping distances of the bodies compare?
Wednesday, October 20, 2010 at 5:56am by Anonymous
Physics - Stopping distance
(Force)*(stopping distance) = (work done against the force) = (initial kinetic energy) Since the kinetic energies and forces are the same for the two different masses, the stopping distances are the same.
Wednesday, October 20, 2010 at 5:56am by drwls
physics hey bob
you help me solve this problem. but how am i suppose to if i dont know the mass of the car? on a level road the stopping distance for an automobile is 35.0m for an initail speed of 90 km/hr what is the stopping distance of the same automobile on a simalr road with the slope 1...
Sunday, October 31, 2010 at 8:53pm by anthony
still i cant find the normal force without mass of automabile. is normal force mass times gravity sin theta? on a level road the stopping distance for an automobile is 35.0m for an initail speed of 90 km/hr what is the stopping distance of the same automobile on a simalr road ...
Sunday, October 31, 2010 at 9:33pm by anthony
physics - stopping distance
when a driver brings a car to a stop by breaki.g as hard as possible the stopping can be regarded as the sum of reaction distance and breaking distance. tje following data are given, v1 m/s contains 10,20,30. reaction distance are 7.5, 15, 22.5. the breaking distance are 5.0, ...
Thursday, April 25, 2013 at 12:03pm by gladys
physics - stopping distance
The reaction time is t =x/v=7.5/10 = 15/20 =22.5/30 = 0.75 s. The reaction distance for v=25 m/s is x=vt=25•0.75 = 18.75 m. The acceleration is a=v²/2s=10²/2•5=20²/2•20=30²/2•45=10 m/s² The breaking distance for v=25 m/s is s= v²/2a=25²/2•10=...
Thursday, April 25, 2013 at 12:03pm by Elena