We will apply the concepts related to Newton's second law. At the same time we will convert everything to the system of international units.

The values of the velocities are,


We know that the acceleration is equivalent to the change of the speed in a certain time therefore



Now applying the Newton's second law we have,



Therefore the approximate magnitude is 8516.36N
Answer:
0.739
Explanation:
If we treat the four tire as single body then
W ( weight of the tyre ) = mass × acceleration due to gravity (g)
the body has a tangential acceleration = dv/dt = 5.22 m/s², also the body has centripetal acceleration to the center = v² / r
where v is speed 25.6 m/s and r is the radius of the circle
centripetal acceleration = (25.6 m/s)² / 130 = 5.041 m/s²
net acceleration of the body = √ (tangential acceleration² + centripetal acceleration²) = √ (5.22² + 5.041²) = 7.2567 m/s²
coefficient of static friction between the tires and the road = frictional force / force of normal
frictional force = m × net acceleration / m×g
where force of normal = weight of the body in opposite direction
coefficient of static friction = (7.2567 × m) / (9.81 × m)
coefficient of static friction = 0.739
<h2>
Answer:</h2>
Answer to this question is (A)
<h2>
Explanation</h2>
A ball bouncing on the floor is not the example of simple harmonic motion. SHM is the special kind of to and fro motion in which a particle oscillate about its mean position in a straight line. The acceleration of the particle is always directed towards its mean position and is directly proportional to its displacement from its mean position.
In case of a ball bouncing on the ground, the motion of the ball is not SHM, as neither it’s a to and fro motion nor the acceleration is proportional to its displacement from its mean position.
Answer:
B - Divide Speed and Distance
Explanation:
Hope it helps!!
:D