Answer:
Mg or your weight.
Explanation:
When your velocity is constant, the net force acting on you is 0. That means the upwards force of air resistance must fully balance the downwards force of gravity on you, which is Mg.
Answer:
Static Friction - acts on objects when they are resting on a surface
Sliding Friction - friction that acts on objects when they are sliding over a surface
Rolling Friction - friction that acts on objects when they are rolling over a surface
Fluid Friction - friction that acts on objects that are moving through a fluid
Explanation:
Examples of static include papers on a tabletop, towel hanging on a rack, bookmark in a book
, car parked on a hill.
Example of sliding include sledding, pushing an object across a surface, rubbing one's hands together, a car sliding on ice.
Examples of rolling include truck tires, ball bearings, bike wheels, and car tires.
Examples of fluid include water pushing against a swimmer's body as they move through it , the movement of your coffee as you stir it with a spoon, sucking water through a straw, submarine moving through water.
Answer:
He needs 1.53 seconds to stop the car.
Explanation:
Let the mass of the car is 1500 kg
Speed of the car, v = 20.5 m/s
He will not push the car with a force greater than,
The impulse delivered to the object is given by the change in momentum as :
So, he needs 1.53 seconds to stop the car. Hence, this is the required solution.
Answer:
The total frictional force is 358.0 newtons
Explanation:
Power is the amount of average work (W) an object does on a period of time (Δt):
Remember average work is average force (F) times displacement (Δs):
but displacement over time is average speed , then:
(1)
That is, the power of the car is the force the engine does times the speed of the car. As the question states, if the car is at constant velocity then the power developed is used to overcome the frictional forces exerted by the air and the road, that is by Newton's first law, the force the motor of the car does is equal the force of frictional forces. So, to find the frictional forces we only have to solve (1) for F:
Knowing that 1hp is 746W then 30hp=22380W and 1 mile = 1609m then 140 mph = 225308 = , then: