Physics distance and velocity questions.?
a) Here, we use this equation of motion: v^2 = u^2 + 2as Substitute the corresponding values given in the question: v (final velocity) = 0 m/s u (initial velocity) = 8 m/s a (acceleration) = -2m/s/s So, we have, 0^2 = 8^2 + 2(-2)s s = 16 m b) Here we use this equation of motion: s = ut + 0.5a(t^2) Substitute the corresponding values given in the question, and the answer in a): s (displacement) = 16 m u (initial velocity) = 8 m/s a (acceleration) = -2 m/s/s So, we have, 16 = 8t + 0.5(-2)(t^2) t^2 - 8t + 16 = 0 <-- this is a quadratic equation (t - 4)(t - 4) = 0 t = 4 s (repeated roots) c) Here we use this equation of motion: v = u + at Substutite the corresponding values given in the question: t (time) = (5.0 - 4) = 1 s u (initial velocity) = 0 m/s a (acceleration) = -2 m/s/s So, we have, v = 0 + (-2)(1) v = -2 m/s The velocity here is negative because it is opposite to the initial direction of motion (i.e. when t = 0) Hope that helps!
Please answer these questions asap?
Suppose you need to remove a nail from a board by using a claw hammer. What is the input distance for a claw hammer if the output distance is 2.0 cm and the mechanical advantage is 5.5? 2. What would be the mechanical efficiency if a rope is pulled 5.0 m, and a box is lifted 3.0 m in the air? 3. An axe used to split wood is driven into a piece of wood a distance of 3.0 cm. If the mechanical advantage of the axe is 0.85, how far apart is the wood split? 4. The mechanical advantage of an automobiles wheel and axle is 8.93. If the wheels output force is 2.22 N, what is the input force that turns the axle? 5. A mover uses a ramp to load a crate of nails onto a truck. The crate, which must be lifted 1.4 m from the street to the bed of the truck, is pushed along the length of the ramp. If the ramp is 4.6 m long and friction between the ramp and crate can be ignored, what is the mechanical advantage of the ramp? 6. A complex arrangement of pulleys forms what is called the block in a block and tackle. The rope used to lift the pulleys and the load is the tackle. A block and tackle is used to lift a truck engine, which has a weight of nearly 7406 N. The force required to lift this weight using the block and tackle is 309 N. What is the mechanical advantage of the block and tackle? 7. block and tackle (pulley system) is used to lift an engine. The engine weights 6000 N and is raised 0.9 m. The operator pulls with a force of 1000 N over a distance of 6 m. What is the work input? 8. In the question above, what is the work output? 9. In problem 7, what is the efficiency? 10. A simple machine is used to lift a load weighing 3000 N. The operator lifts the load a distance of 1.25 m by pulling with a force of 1000 N over a distance of 4 m. What is the work input? 11. What is the work output in the problem above? 12. What is the efficiency in problem 10?
Physics question, using F=ma...?
This may or may not be correct, but I'm seeing this as more of trigonometry than physics. Well, yes, it's a physics problem, but I think you can just use trig. If you make it a triangle, your angle is 22 degrees and your "opposite" from that angle is 25 cm. You can just use sin22= 25/hypotenuse In this case, the hypotenuse would be the distance the box has moved up the ramp. I don't have a calculator near me or else I'd plug it in for you. If this is wrong because you're required to use F=ma to get it then I'm sorry. Physics is a drag.
How do I calculate stopping distance for problems in physics?
1) Acceleration = (f/m) = 1400/51, = -27.451m/sec^2. Initial V at water entry = sqrt.(2ah) = sqrt.(54.902 x 4.4) = 15.542m/sec. Drop height from board to water = (v^2/2g) = 12.3242m. metres. Total distance = (12.3242 + 4.4) = 16.724 metres. 2) Height from which it starts = (sin 30) x 2.92 = 1.46 metres. GPE = (mgh) = 8.8 x 9.8 x 1.46, = 125.91 Joules. This becomes KE at bottom of ramp. V at bottom = sqrt.(2KE/m) = 5.3494m/sec. Acceleration = (v^2/2d) = -2.611m/sec^2. Friction force = (ma) = 22.9768N. Normal force = (8.8 x 9.8) = 86.24N. a) Coefficient = (22.9768/86.24) = 0.266. b) Change in KE due to friction = 125.91 Joules (above).
If a block slides down a 30 degree incline at a constant speed, what is the coefficient of dynamic friction between the block and the incline?
θ = 30 degreesvelocity is constant therefore acceleration a = 0Please refer to my analysis using the diagram below.The coefficient of dynamic friction is equal to 0.5773.
A ramp is inclined at an angle of 41.3 degree w/ the ground. One end of a board, 20.6 ft inch. length is?
Draw triangle PQR with angle Q=41.3, QR=12.2, PR=20.6 and draw the perpendicular from R to QR to meet it at S. If length QR=x then RS=xsin41.3 and QS=xcos41.3 so SP=12.2-xcos41.3. From Pythagoras in triangle PRS 20.6^2=(12-xcos41.3)^2+ (xsin41.3)^3 giving 20.6^2=144-(24cos41.3)x+x^2cos^2(41.3) +x^2(sin^2(41.3) so 20.6^2=144-18.03x+x^2 since cos^2(41.3)+sin^2(41.3)=1 so x^2-18.03x-280.36=0 Solve this quadratic equation and choose the positive value
A 50-kg box is pushed along a horizontal surface. The coefficient of kinetic friction is 0.35. What horizontal force would accelerate it at 1.2 m/s^2?
The situation involves taking into account two main contributing forces. These are the Friction Forces and the Accelerating forces.1. If the box has to be moved with constant velocity then only the frictional forces will be existing. ( 50. 9.8. 0.35) this assumes that the frictional forces remain constant and that the force is horizontal. If the force was at a down or up angle pushing or at an up or down angle pulling then the problem would add other components due to the up angle or down angle of the pushing force.2. When accelerating in addition to the friction forces one also needs to add an accelerating force which is M.a or (50. 1,2)Again it would add a little more effort if one considers the pushing or the pulling forces, not only when horizontally applied but at an angle. This would give an idea of what to select in actual real life when one is faced with this dilemma.