F_P + F_Q = M g
F_P = M g - F_Q
Torque, or moment of force:
∑ M_P = 0
∑ M_P = M g L - F_Q · 3 L
0 = M g L - 3 F_Q L / : L
0 = M g - 3 F_Q
3 F_Q = M g
F_Q = M g /3
Finally:
F_P = M g - M g/3
F_P = 4 M g / 3
Answer:
Yes, the car driver is exceeding the given limit.
Explanation:
<u>Given:</u>
- Speed of the car, v = 38.0 m/s.
- Speed limit of the highway,

<h2><u>
Converting the speed limit from mi/h to m/s:</u></h2>
We know,
1 mi = 1.60934 km.
1 km = 1000 m.
Therefore, 1 mi = 1.60934 × 1000 m = 1609.34 m.
1 hour = 60 minutes.
1 minute = 60 seconds.
Therefore, 1 hour = 60 × 60 seconds = 3600 seconds.
Using these values,

Therefore,

Clearly,

which means, the car driver is exceeding the given speed limit.
The difference between the above velocities is that they exist in opposite direction of each other. or it can be said that they are negative vectors of each other.
Answer:
the distance in meters traveled by a point outside the rim is 157.1 m
Explanation:
Given;
radius of the disk, r = 50 cm = 0.5 m
angular speed of the disk, ω = 100 rpm
time of motion, t = 30 s
The distance in meters traveled by a point outside the rim is calculated as follows;

Therefore, the distance in meters traveled by a point outside the rim is 157.1 m
Force, newtons 3rd law of motion stated for every action there is an equal and opposite reaction