ANOTHER RUNNING DOG
Explanation:
In the given question it is to find a suitable reference point to describe the motion of dog. Here I could suggest that it is better to compare the dog with another running dog to create the relative speed difference to get a reliable motion variation.
Because the motion of dog is in the linear with respect to the another dog and to the acceleration produced by the dog in the required interval is easy to calculate with respect to another dog which is already in motion.
Hence, I suggest that Motion of dog can be analysed better by analyse the motion variation of dog with another dog running.
Energy can be one answer! There are many, but energy is a main one.
Rolling friction .<span> the force that slows down the movement of a rolling object</span>
sliding friction.
Sliding friction : The opposing force that comes into play when
one body is actually sliding over the surface of the other body
is called sliding friction. e.g. A flat block is moving over a
horizontal table.
Kinetic or dynamic friction: If the applied force is increased further
and sets the body in motion, the friction opposing the motion is called
kinetic friction
Answer:
As given that the car maintains a constant speed v as it traverses the hill and valley where both the valley and hill have a radius of curvature R.
(i) At point C, the normal force acting on the car is largest because the centripetal force is up. gravity is down and normal force is up. net force is up so magnitude of normal force must be greater than the car's weight.
(ii) At point A, the normal force acting on the car is smallest because the centripetal force is down. gravity is down and normal force is up. net force is up so magnitude of normal force must be less than car's weight.
(iii) At point C, the driver will feel heaviest because the driver's apparent weight is the normal force on her body.
(iv) At point A, the driver will feel the lightest.
(v)The car can go that much fast without losing contact with the road at A can be determined as follow:
Fn=0 - lose contact with road
Fg= mv²/r
mg=mv²/r
v=sqrt (gr)
Answer:
<h2>FUNDAMENTAL UNITS INVOLVED ARE : NEWTON AND SECOND .</h2>
<h2>FORMULA OF PRESSURE = </h2>
<h2>P=F/A </h2>
