PHYSICS HELP!!!!!!!!!!A 19-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force....?
a = [(2.9²)/(2*8.8)] = 0.478 m/s² Coefficient of kinetic friction = [{31 -(19*0.478)}/(19*9.81)] = 0.118 or 0.12
A 15-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 21 N.?
Net force = mass * acceleration The net force on the sled is equal to the horizontal force minus the friction force. Net force = 21 – Ff = 15 * a To determine the acceleration, use the following equation. vf^2 = vi^2 + 2 * a * d, vi = 0 1.9^2 = 2 * a * 8.1 a = 3.61÷ 16.2 This is approximately 0.22 m/s^2. 21 – Ff = 15 * 3.61÷ 16.2 Ff = 21 – 15 * 3.61÷ 16.2 = 17.65740741 N Ff = μ * 15 * 9.8 = μ *147 μ * 147 = 17.65740741 μ = 17.65740741 ÷ 147 = 0.1201417428 This is approximately 0.12.
Physics Question: A 16-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force?
Using Newton's 2nd Law Of Motion, Fn = ma where Fn = net force acting on the sled = 24 - µmg µ = coefficient of friction g = acceleration due to gravity = 9.8 m/sec^2 (constant) m = mass of the sled = 16 kg. a =acceleration Substituting values, 24 - µ(16)(9.8) = 16(a) --- call this Equation 1 The next working formula is Vf^2 - Vo^2 = 2as where Vf = final velocity = 2 m/sec. (given) Vo = initial velocity = 0 a = acceleration s = distance travelled = 8 m (given) Substituting appropriate values, 2^2 - 0 = 2(a)(8) Solving for "a", a = 4/(2*8) a = 0.25 m/sec^2 Substituting a = 0.25 m/sec^2 into Equation 1, 24 - µ(16)(9.8) = 16(0.25) Solving for µ, µ = (24 - 4)/(16*9.8) µ = 0.128 Hope this helps.
A 19-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 21 N. Starting?
Acceleration = (v^2/2d) = 0.215m/sec^2. Force = (ma) = 19 x 0.215 = 4.085N. Friction force = (21 - 4.085) = 16.915N. 16.915/(19 x g) = µ of 0.091. g = 9.8 used.
Find the coefficient of kinetic friction between sled and snow...?
The acceleration of the sled is: a = v^2/(2x) = (2.0m/s)^2/(2*12m) = 0.167m/s^2. So the net force is: Ft = ma = (20kg)*(0.167m/s^2) = 3.33N The force of friction is: Ff = Fa - Ft = 30N - 3.33N = 26.7N The weight and of the sled is: Fw = mg = (20kg)(9.81m/s^2) = 196N Since the only other vertical force is the normal force: Fn = Fw = 196N So the coefficient of friction is: mu = Ff/Fn = (26.7N)/(196N) = 0.136
Physics help please!! (:?
A plane flies along a straight line path after taking off, and it ends up 180 km farther east and 92.0 km farther north, relative to where it started. In what direction did it fly on the straight line path? 24° north of east 27° north of east 45° north of east 63° north of east A 4.2 kg sled is being pulled along a snow-covered road with a rope that exerts a horizontal force of 6.0 N and, at that moment, is accelerating at 1.1 m/s2 on level ground with friction having no significant effect. What is the normal force on the sled? 6.0 N 4.6 N 1.4 N 9.8 N 41 N How much force does it take to bring a 1,650 kg car moving at 2.20 m/s to a complete stop in 1.1 s? 830 N 1,500 N 3,300 N 4.0 × 103 N
Physics practice questions dont understand please help explain (:?
A 4.2 kg sled is being pulled along a snow-covered road with a rope that exerts a horizontal force of 6.0 N and, at that moment, is accelerating at 1.1 m/s2 on level ground with friction having no significant effect. What is the normal force on the sled? 6.0 N 4.6 N 1.4 N 9.8 N 41 N A 1.3 kg cart rests on an inclined plane. The cart is released so that it is free to roll down the frictionless surface. The inclined plane is tilted by = 14°. Find the acceleration of the cart. (Points : 3) 9.8 m/s2 9.5 m/s2 2.3 m/s2 3.1 m/s2 I dont need the answers I just need help getting there
Sled being pulled, Physics Help?
(a) The sled weighs: w = mg =55.0kg x 9.8m/s² =539N (b) To just get the sled moving apply Newton's 2nd law: ΣF = 0 = F(a) - F(s)------->Where F(a) = force applied and F(s) = force of static friction. F(a) = F(s) = μ(s)mg = 0.300(539N) =162N. (c) Same method to keep the sled moving at constant speed, except this time use coefficient of kinetic friction: F(a) = μ(k)mg = 0.100(539N) 53.9N. (d) This time, the net forces are equal to ma, so: ΣF = ma = F(a) - F(k) F(a) = ma - μ(k)mg =(55.0kg)(3.4m/s²) - 0.100(539N) = 133N.
It is *not* always easier to pull a heavy object than to push it . It depends on the direction of the application of the force with respect to the friction or opposing force . If you are pushing down , like in case of hand cart , the vertical component of the applied force is wasted in increasing friction . If you pull the vertical component effectively reduces the weight of the object and thus the friction . So in this case pulling is easier than pushing because the point of application of force is above that of deployment of the effect .Do observe the ants . They are very small creatures , so everything they transport is generally much larger than their size . So they have to apply the force from down to up . You will see ants always push large objects and do not pull them . In *their* case the vertical component of push effectively reduces the friction . Beetles also push on objects uphill .See the two examples below . The vegetable cart is being pushed , while the loaded hand cart is being pulled . Notice the difference in the height at which the force is being applied and the height of the wheels of the respective carts .
Spherical shape of the water droplets is due to the surface tension phenomenaThe cohesive forces between liquid molecules is responsible for the surface tension. The molecules of water on the surface don't have like molecules around it.So the molecules lying in the surface are attracted or pulled downward by the adjacent water molecules. So the water droplets tend to be pulled into shape of spherical due to cohesive forces.If other forces including gravity is missing,then droplets of all the fluids have spheical shape.So this spherical forms are easier to form, required less surface tension,less wall tension.More the surface tension more will be the spherical form.