Answer:
Net force on the wagon is 200 N
Explanation:
As we know by Newton's II law that net force on the system of mass is given as product of mass and acceleration
Here we know that
mass = 100 kg
a = 2 m/s/s
now we have



You may have a cold if you do not feel well, depends on the symptoms
Explanation:
The protons determines the atomic number which helps in identifying an atom of an element
Answer:
570 N
Explanation:
Draw a free body diagram on the rider. There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.
The rider is moving at constant speed, so acceleration is 0.
Sum of the forces in the x direction:
∑F = ma
F cos 30° - T cos 15° = 0
F = T cos 15° / cos 30°
Sum of the forces in the y direction:
∑F = ma
F sin 30° - W - T sin 15° = 0
W = F sin 30° - T sin 15°
Substituting:
W = (T cos 15° / cos 30°) sin 30° - T sin 15°
W = T cos 15° tan 30° - T sin 15°
W = T (cos 15° tan 30° - sin 15°)
Given T = 1900 N:
W = 1900 (cos 15° tan 30° - sin 15°)
W = 570 N
The rider weighs 570 N (which is about the same as 130 lb).
In this graph, what is the displacement of the particle in the last two seconds?of the particle in the last two seconds?
<span>0.2 meters
2 meters
4 meters
6 meters</span>
In this graph, the displacement of the particle in the last two seconds is 2 meters.