Physics help,Find the speed of the block?
The force of friction, which is opposite to the pulling force, is F_f = u*m*g Where u is the coefficient of kinetic friction. F_f = 0.15*(4.69 kg)*(9.8 m/s^2) F_f = 6.8943 N So, the net force on the block is F_net = F_pull - F_friction F_net = 18 N - 6.8943 N = 11.1057 N Since F = m*a, we can solve for a, the acceleration a = F/m = 11.1057 N/4.69 kg a = 2.367953 m/s^2 We eventually want to use v = v_0 + a*t to solve for the velocity, but first we need to solve for t x = v_0*t + (1/2)*a*t^2 x = displacement (2.59 m) v_0 = 0 m/s (starting from rest) a = 2.367953 m/s^2 (from above) 2.59 m = 0 + (1/2)*(2.367953 m/s^2)*t^2 t = sqrt(2.59/((1/2)*(2.367953))) t = 1.479035 seconds Plug that into v = v_0 + a*t v = 0 + 2.367953 m/s^2*(1.479035 s) v = 3.503385 m/s v = 3.5 m/s
PHYSICS help (finding speed/height)?
kk first of all.. we need to find time so.. y=Vy*t+.5g*t^2 (notice Vy=0 as it leaves the track horizontally.) so.. 1.02=.5*9.81*t^2 t=.456 now we can find Vx=x/t (b)Vx=2.30m/s (a)Mgh(starting height)(Mgh=gravitation potential energy)=.5MVx^2(which is kinetic energy as it leaves the track) Masses cancel and we can find height to be =.2696(but dont forget to add the h2)+1.02=1.29m (c)umm for this part. u need vectors .. Vx is constant so.. Vfinal=sqrt(Vx^2+Vy^2)(pythagorean theorem) Vy^2=Vy(initial)^2+2*g*h(notice again that Vy initial is 0) so Vy(final)=20m/s so V(final)=sqrt(2.30^2+20^2) so V(final)=20.13m/s Hope this helped!!
Physics help, finding speed of masses?
the quantity which you have to calculate in this question is the acceleration of the masses . when you know the acceleration the rest is simple as a single body problem with constant acceleration draw the free body diagrams of both masses for the inclined mass : m*g*sine of angle down the plane and parallel to plane T (tension) up the plane and parallel to plane friction up the plane and parallel 0.3 * N N (normal force by the plane) perpendicular and up the plane m*g*cos of angle perpendicular and down the plane for hanging mass T in upward direction T is same as pulley is frictionless m*g in downward direction As the length of the rope is constant the accelertaion of masses are same m*g*sine - (0.31*N+T)=m*a (1) and T-m*g=m*a (2) ( i have assumed mass is moving down incline) T=m*g+m*a m*g*sine - (0.31*N+m*g+m+a)=m*a we only need value of N from FBD of inclined mass no acceleration in perpendicular to plane N=m*g*cos final eqn m*g*sine - (0.3*m*g*cos+m*g+m*a)=m*a when you know acc (a) just apply v**2=u**2 + 2*a*s s=1.6 u=0
How do I find speed in physics?
Distance travelled in a given direction divided by the time taken gives you the speed in that given direction. For example: a particle travels 200 metres, from A to B, in 10 seconds. The particle's speed in the direction AB is 200/10 = 20 metres per second.Or, as has been mentioned in the other answers, you can calculate the magnitude of the velocity. If the velocity is given as a vector, ai +bj , then the speed is sqrt(a^2 + b^2). For example: a particle's velocity vector is (3i +4j )ms^-1. Its speed is sqrt(3^2 + 4^2) = sqrt(25) = 5ms^-1.
Need Physics Help !!!?
Just equate 20 degree= 20+373=393 kinetic energy = boltmann constant* T(temperature) v= sqrt(2kT/m)=sqrt(2*1.35*373/28)=6. We need to adjust exponents k has exponent --23, proton has exponent -21 To the above value you need to divide by 10. You get 0.6 m/s I hope this matches with your value. Number density does nt matter here.
Physics HW. I need help finding speed and flight time of a bullet.?
since angle(#) = 0 and g = 9.81m/s^2 hori dist. = (v cos #)t = 50 Therefore, cos # = 1 and vt = 50 vert. dist. = (v sin #)t - 0.5gt^2 = -0.02 since the bullet goes 0.02m below the target. Therefore, sin # = 0 and (v sin #)t = 0 Thus, -0.5gt^2 = -0.02 Following up, t^2 = (-0.02/-0.5)/9.81 and t = 0.063855 (this is the bullet flight time) Now, combine this value of t with the first equation of the horizontal distance where vt = 50. Thus, bullet's speed, v = 50/0.063855 = 783.024 m/s There you go my friend. both the answers.
Velocity physics finding n?
This is an example of an inelastic collision. In all collisions, it is important to note that momentum is conserved, thus total momentum before the collision is equal to total momentum after the collision. Before the collision, total momentum is the sum of the momenta of both m1 and m2. p0 = m1v1 + m2v2 = 4m(+5v) + 4m(-2v) = 20 mv - 8 mv = 12 mv (1) We know that momentum is conserved so, p0 = pf = 12 mv = mf vf (2) We need to find what the final mass is. In an inelastic collision, the final mass is the sum of the two initial masses. mf = m1 + m2 = 4m + 4m = 8m (3) We can now substitute this in for mf in equation (2): 12 mv = 8m vf (4) Now solve for vf by dividing both side by 8m: 1.5 v = vf (5) n = 1.5 (6) Equation (6) is the answer.
Can I get help with Physics homework?
If you have a physics problem, try our online physics help.Speed your learning by having skills and knowledge you need at your fingertips. Get physics tutorials, definitions, explanations, formulas. See how to solve problems just like the ones instructors and profs often assign. Use the physics index to find help quickly for specific topics from Acceleration to Zeroeth Law of Thermodynamics. Online resources may help you develop knowledge and skills needed to do physics.
Physics help! finding speed of electron in electric field?
In an x-ray tube, electrons are accelerated in a uniform electric field and then strike a metal target. Suppose an electron starting from rest is accelerated in a uniform electric field directed horizontally and having a magnitude of 2200N/C . The electric field covers a region of space 12.0cm wide. What is the speed of the electron when it strikes the target? How far does it fall under the influence of gravity during its flight? Thanks!
General physics 1 need help .. find the amplitude, find the maximum speed of the mass?
Let the oscillation be of the form. y= a sin wt dy/dt = aw cos wt. Maximum is aw. Now w=sqrt(k/m)= sqrt(21.2/0.457)=6.8 a= 4.02/6.8=0.59 m For the second problem. use the perion w = sqrt(g/l) g=9.8 l is the length0.928 Maximum speed aw. a is the length a=2.94(radians)*0.928=0.05m w=3.25s maximum speed=0.162 m/s