Will mark the brainliest When a car of mass 1000kg is travelling along a level road at a steady speed of 20ms −1
its engine is working at 18kW.
Resultant force = 900N Braking Force = 3100N Gravity = 9.81
By how much does this change if the car is travelling down the same hill at 20ms −1 ?

Adelivery truck leaves a warehouse and travels 2.60 km north. the truck makes a left turn and travels 1.25 km west before making another right turn to travel 1.40 km north to arrive at its destination. what is the magnitude and direction of the truck's displacement from the warehouse?

Part f - example: finding two forces (part i) two dimensional dynamics often involves solving for two unknown quantities in two separate equations describing the total force. the block in (figure 1) has a mass m=10kg and is being pulled by a force f on a table with coefficient of static friction îľs=0.3. four forces act on it: the applied force f (directed î¸=30â above the horizontal). the force of gravity fg=mg (directly down, where g=9.8m/s2). the normal force n (directly up). the force of static friction fs (directly left, opposing any potential motion). if we want to find the size of the force necessary to just barely overcome static friction (in which case fs=îľsn), we use the condition that the sum of the forces in both directions must be 0. using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as: fcosî¸â’îľsn=0 fsinî¸+nâ’mg=0 in order to find the magnitude of force f, we have to solve a system of two equations with both f and the normal force n unknown. use the methods we have learned to find an expression for f in terms of m, g, î¸, and îľs (no n).

Discuss the difference between thermal energy and heat. (terrance) terrance, egolf, r., congdon, donald. physical science student text, 5th edition. bju press, 2014. vitalbook file.