The range of force exerted at the end of the rope is 285.7 N to 1,000 N.
<h3>Net horizontal force of the cylinder</h3>
The net horizontal force of the cylinder when it is at equilibrium position is determined by applying Newton's second law of motion.
∑F = 0
F - μFn = 0
F - 0.2(5,000) = 0
F - 1,000 = 0
F = 1,000 N
The strength of the applied force increases as the number of turns of the rope increases.
minimum force = total force/number of turns of rope
minimum force = 1,000/3.5
minimum force = 285.7 N
Thus, the range of force exerted at the end of the rope is 285.7 N to 1,000 N.
Learn more about Newton's second law of motion here: brainly.com/question/3999427
Answer:
the energy absorbed is 4.477 x 10⁶ J
Explanation:
mass of the liquid, m = 13 kg
initial temperature of the liquid, t₁ = 18 ⁰C
final temperature of the liquid, t₂ = 100 ⁰C
specific heat capacity of water, c = 4,200 J/kg⁰C
The energy absorbed is calculated as;
H = mcΔt
H = mc(t₂ - t₁)
H = 13 x 4,200(100 - 18)
H = 4.477 x 10⁶ J
Therefore, the energy absorbed is 4.477 x 10⁶ J
The direction of an electric current is by convention the direction in which a positive charge would move. Thus, the current in the external circuit is directed away from the positive terminal and toward the negative terminal of the battery. Electrons would actually move through the wires in the opposite direction.
Answer:
34 m/s
Explanation:
Potential energy at top = kinetic energy at bottom + work done by friction
PE = KE + W
mgh = ½ mv² + Fd
mg (d sin θ) = ½ mv² + Fd
Solving for v:
½ mv² = mg (d sin θ) − Fd
mv² = 2mg (d sin θ) − 2Fd
v² = 2g (d sin θ) − 2Fd/m
v = √(2g (d sin θ) − 2Fd/m)
Given g = 9.8 m/s², d = 150 m, θ = 28°, F = 50 N, and m = 65 kg:
v = √(2 (9.8 m/s²) (150 m sin 28°) − 2 (50 N) (150 m) / (65 kg))
v = 33.9 m/s
Rounded to two significant figures, her velocity at the bottom of the hill is 34 m/s.