Answer:his tension is 262 N
Explanation:1) Call T the tension of the rope2) Split T into its horizontal and vertical components.3) Horizontal component of the tension, Tx = T cos(55°)4) Vertical component of the tension, Ty = T sin (55°)5) Force equilibrium in the vertical direcction:∑Fy = 0Ty + normal force - weight of the sled = 0Call N the normal forceTy + N - 56 kg * 9.8 m/s^2Ty + N = 56 kg * 9.8 m/s^2Ty + N = 548.8NT sin(55) + N = 548.8N6) Force equilibrium in the horizontal directionconstant velocity => ∑Fx = 0Tx - Fx = 0Tcos(55) - Fx = 07) Fx is the friction force.The friction force and the normal force are related by the kinetic friction coefficient.Call μk the friction coefficientFx = μk N=> Tcos(55) - μk N = 0Tcos(55) - 0.45N = 08) Solve the system of two equations:Eq (1) T sin(55) + N = 548.8Eq (2) T cos(55) - 0.45N = 0Eq(1) 0.819T + N = 548.8Eq(2) 0.574T - 0.45N = 0The solution of the system is T = 262.01 N and N = 334.21 NThen T ≈ 262N
Answer:
velocity : velocity is a vector quantity; it is direction-aware. Velocity is the rate at which the position changes. The average velocity is the displacement or position change (a vector quantity) per time ratio.
speed : Speed is a scalar quantity that refers to "how fast an object is moving." Speed can be thought of as the rate at which an object covers distance. A fast-moving object has a high speed and covers a relatively large distance in a short amount of time. ... An object with no movement at all has a zero speed.
Explanation:
Answer:
Technician B is correct
Explanation:
An oxygen sensor will generate about 1.0 volts when the fuel mixture is rich and there is little unburned oxygen in the exhaust. When the mixture is lean, the sensor's output voltage will drop down to about 0.1 volts.
An o2 sensor cannot accurately measure how rich or how well an exhaust system is.
Therefore, Technician B is correct.
Answer:
the work done by the 30N force is 4156.92 J.
For this problem, they don´t ask you to determine the work of the total force applied in the block. They only want the work done for the force of 30N, with an angle of 30º respectively of the displacement and a traveled distance of 160m. So:
W=F·s·cos(α)=30N·160m·cos(30º)=4156.92J