The force that must be exerted on the outside wheel to lift the anchor at constant speed is 6.925 x 10⁵ N.
<h3>Force exerted outside the wheel</h3>
The force exerted on the outside of the wheel can be determined by applying the principle of conservation of angular momentum as shown below.
∑τ = 0
- Let the distance traveled by the load = 1.5 m
- Let the radius of the wheel or position of the force = 0.45 m
∑τ = R(mg) - r(F)
rF = R(mg)
0.45F = 1.5(21,200 x 9.8)
F = 6.925 x 10⁵ N.
Thus, the force that must be exerted on the outside wheel to lift the anchor at constant speed is 6.925 x 10⁵ N.
Learn more about angular momentum here: brainly.com/question/7538238
Answer:
In a series circuit, the same amount of current flows through all the components placed in it. On the other hand, in parallel circuits, the components are placed in parallel with each other due to which the circuit splits the current flow.
Answer:
Explanation:
The resistance of a piece of wire is given by
where
is the resistivity of the material
L is the length of the wire
A is the cross-sectional area
In this problem, we have
L = 7.0 m
The diameter of the wire is 0.14 cm, so the radius is 0.07 cm, therefore the cross sectional area is
Therefore, the resistance is
Answer:
Centripetal force F = 0.0078 N
Explanation: Given that the
Mass M = 41 kg
Velocity V = 0.02 m/s
Radius r = 2.1 m
Centripetal force F = MV^2/r
Substitutes all the parameters into the formula
F = (41 × 0.02^2)/2.1
F = ( 41 × 0.0004 )/2.1
F = 0.0164/2.1
F = 0.0078 N
The centripetal force acting upon her is 0.0078 Newton
