Explanation:
Given that,
Initial speed of the car, u = 88 km/h = 24.44 m/s
Reaction time, t = 2 s
Distance covered during this time,
(a) Acceleration,
We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :
s = 74.66 meters
s = 74.66 + 48.88 = 123.54 meters
(b) Acceleration,
s = 37.33 meters
s = 37.33 + 48.88 = 86.21 meters
Hence, this is the required solution.
Answer:
Acceleration,
Explanation:
Initial speed of the skater, u = 8.4 m/s
Final speed of the skater, v = 6.5 m/s
It hits a 5.7 m wide patch of rough ice, s = 5.7 m
We need to find the acceleration on the rough ice. The third equation of motion gives the relationship between the speed and the distance covered. Mathematically, it is given by :
So, the acceleration on the rough ice and negative sign shows deceleration.
Answer:
F = 520 N
Explanation:
For this exercise the rotational equilibrium equation should be used
Σ τ = 0
Let's set a reference system with the origin at the back of the refrigerator and the counterclockwise rotation as positive. On the x-axis it is horizontal directed outward, eg the horizontal y-axis directed to the side and the z-axis vertical
Torque is
τ = F x r
the bold indicate vectors, we analyze each force
the applied force is horizontal along the -x axis, the arm (perpendicular distance) is directed in the z axis,
The weight of the body is the vertical direction of the z-axis, so the arm is on the x-axis
-F z + W x = 0
F z = W x
F = W
The exercise indicates the point of application of the force z = 1.5 m and the weight is placed in the center of mass of the body x = 0.6 m, we are assuming that the force is applied in the wide center of the refrigerator
let's calculate
F = 1300 0.6 / 1.5
F = 520 N