A truck moves from rest with constant acceleration 5m/sec(squared)?
828770
Did you forget the equations for velocity as a function of time and distance as a function of time, given a acceleration?
927051
The given final speed doesn't seem to compute. I get 14.8 m/s. \nMa*Va + Mb*Vb = (Ma+Mb)*Vf \nis the formula that applies to show conservation of momentum in a completely inelastic collision, which is what you have. The only explanation I could suspect is some cargo was spilled. \n\nTo continue the assignment - kinetic energy is given by\n(1/2)*m*v^2\nThere is no requirement that all the original kinetic energy be measurable as kinetic energy after a completely inelastic collision. But you should be able to point to where the missing energy went. Some of the energy probably went into cart damage. Perhaps the cargo slid around.
1163276
The angle the vehicle was traveling
602899
I always find it easier if you draw it first, which this stupid bos wont let me do.\n\nYou know one angle (theta) and one side of the triangle on parts a and b.  Use Tan Theta = x / 30 to find how far the truck has traveled in one second, and convert to MPH.\n\nFor C, you have the distance traveled in MPH and one side (30 ft).  Convert 55 MPH to ft / sec and find the angle Theta.  Tan Theta = 55 MPH / 30 ft\n\nYou are going to have to figure the answers out for yourself.  Good luck
25616
the force needed is according to newton's second law F=m.a where m=900Kg \n& the acceleration a=v/t\nthe velocity should decrease from 10 to 0 then v=10m/s & t=5 seconds\na=v/t=2m/s^2\nF=900*2=1800newtons
927736
Everyone above assumed a constant speed.  the more general formula is \n\nD = ∫v dt
92986
Slightly over 90 mph (150km/h).
329820
You can't just ask for an acceleration.  You need more factors.  All you know is that it went East 3000m and North 6000m.
1050004
Deflate the tires a little bit so that the truck has a small amount of clearance and can drive underneath it
412737
force = m*a + I*alpha\n\nthe first part is for the cart and the second is for the pulley\n\nThen relate the angular acceleration to the linear acceleration as follows:\nalpha=angular acceleration= w^2*r   in rad/s\nso a(accleration of cart)= w^2*r^2   m/s\n\nso force=m*a + I*(a/r)\n\nfrom which you can find a (acceleration)
441442
you have your units wrong, do you mean 5m/s or 5m/s^2 and you mean 20m from the ground?\nIf acc=5 when height=20 speed= 14.14 in upwards direction (using 2ax=v^2-u^2)\nIf initial speed is therefore -14.14m/s then at t=1 speed=-4.34m/s so it is still going upwards\nt=2 speed= 5.45m/s it is now going down. Both last results worked out using v=u + at\n(Sorry I assumed you meant 5m/s^2 straight up, you need the angle at which it accelerates aswell)
716944
If there is no net force (i.e. sum of forces is zero) then there will be no change in velocity. So if the object was going 15 m/s and the net force is zero, then it will still be going 15 m/, (for 2 sec = 30 meters traveled). Same concept for your jet car - constant velocity means no net force ( there may be forces acting on it, but if the velocity is constant, they must sum to zero - net force is zero).\n\nAs Newton's first law says, "An object in motion tends to stay in motion unless acted on by an outside force." So constant velocity (i.e. stays in motion at the same velocity) means no net force acting on the object.
507693
I'll get you started...\n\nFirst, determine how long the ball was in the air.  From that, determine the velocity at which the ball was thrown (the rate necessary, factoring in gravity's acceleration, to keep it airborne for that period of time).  Construct a vector using the velocity of the truck vs. the velocity of the ball to determine the angle.\n\nGood luck!
513521
the empty one,\nthe loaded truck takes more force to start moving and takes more force to stop than the empty one.\n\n-newton's first and second laws of motion
973163
Relative to a woman in the car:\n\n(a) the translational velocity is ZERO since the woman is moving at the same translational velocity as the car relative to the ground.\n\n(b) the magnitude of the translational acceleration of the center of the wheel is also ZERO because the center of the wheel is NOT moving and NOT rotating relative to the woman.\n\n(c) v, velocity of the point at the top of the tire is = 80 km/h because that is how fast the car is moving, thus the wheel must also be rotating at the same speed.\n\n(d) a, acceleration of the point at the top of the tire is given by:\n\na = v^2/r = (22.22)^2/(0.33) = 1,496.45 m/s^2\n\n(e) v, velocity of the point at the bottom of the tire is = -80 km/h because at the bottom of the tire, the tire is moving in the opposite direction\n\n(f) a, acceleration of the point at the bottom of the tire is the same as (d) a = v^2/r = 1,496.45 m/s^2\n\nNow relative to a hitch hiker on the ground:\n\n(g) v at the center of the wheel is = 80 km/h or 22.22 m/s, since the car and the center of the wheel (which is not moving relative to the car) are moving at the same speed relative to the ground.\n\n(h) a at the center of the wheel is = v^2/r = 1,496.45 m/s^2, since the center of the wheel is rotating about the bottom point of the tire as the tire rolls on the ground.\n\n(i) v at the top of the tire is v(top) = ωd, where ω is the angular velocity of the rotating tire and d is the diameter of the wheel. The reason we use diameter of the wheel is because the top of the tire is rotating abou the bottom of the tire, not the center.\n\nω = v/r = (22.22)/(0.33) = 67.33 rad/sec\nv(top) = ωd = (67.33)(0.66) = 44.44 m/s\n\n(j) a at the top of the tire is = v(top)^2/d\n    = (44.44)^2/(0.66) = 2,992.29 m/s^2  \n\n(k) v at the bottom of the tire is equal to ZERO because the bottom of the tire is not moving relativ to the hitch hiker or the ground.\n\n(l) a at the bottom of the tire is also equal to ZERO since the velocity is zero.
396138
a=F/m, lim x->inf x/x = 1. The object will accelerate.
351440
well ok. for part a, the acceleration is in the horizontal direction, because its moving horizontally. so the force is just mass x acceleration which is 70*6=420N. but thatz only in the horizontal direction.\n\nfor part b, you have to account for gravity. so gravity is pulling down at 9.45m/s^2, but youre accelerating upwards at 6m/s^2. so that means that youre overcoming the gravity and accelerating at 6m/s^2 MORE than the gravitational acceleration.\nso 9.45+6 = 15.45. this is your actual acceleration.\nso now you just do F=ma again and you get 70*15.45=1081.5N. \n\nhope this helps. let me know if you need more help. and this isnt thermodynamics.