We could determine the acceleration using this formula
Joe accelerated from 35 m/s to 50 m/s in 50 seconds, plug in the numbers
a=
a =
a =
a = 0.3
The acceleration is 0.3 m/s²
The energy added here is potential energy since it is moving upward 180 meters in a gravitational field. This is then turned into KE when it rolls down. 2524N x 180m = 454,320J
Answer:
Yes. Towards the center. 8210 N.
Explanation:
Let's first investigate the free-body diagram of the car. The weight of the car has two components: x-direction: towards the center of the curve and y-direction: towards the ground. Note that the ground is not perpendicular to the surface of the Earth is inclined 16 degrees.
In order to find whether the car slides off the road, we should use Newton's Second Law in the direction of x: F = ma.
The net force is equal to
Note that 95 km/h is equal to 26.3 m/s.
This is the centripetal force and equal to the x-component of the applied force.
As can be seen from above, the two forces are not equal to each other. This means that a friction force is needed towards the center of the curve.
The amount of the friction force should be
Qualitatively, on a banked curve, a car is thrown off the road if it is moving fast. However, if the road has enough friction, then the car stays on the road and move safely. Since the car intends to slide off the road, then the static friction between the tires and the road must be towards the center in order to keep the car in the road.
Answer:
Γ = r x F vector notation for torque
Doubling r (the length) will double the torque applied
2kg objects in the world of a hunting group