Answer:
v_f = 24.3 m / s
Explanation:
A) In this exercise there is no friction so energy is conserved.
Starting point. On the roof of the building
Em₀ = K + U = ½ m v₀² + m g y₀
Final point. On the floor
Em_f = K = ½ m v_f²
Emo = Em_g
½ m v₀² + m g y₀ = ½ m v_f²
v_f² = v₀² + 2 g y₀
let's calculate
v_f = √(10² + 2 9.8 25)
v_f = 24.3 m / s
Well, for the distance traveled, the car goes from the northernmost point to the southernmost point. So, it travels half of the circle's circumference = 4.7/2 = 2.35 km.
For the displacement, by going from the northernmost point to the southernmost point, the car basically just travels the diameter of circle.
So, using the formula: Circumference = 2πr = <span>πd
Hence, the d = C/</span>π = 4.7/<span>π = 1.49605... = 1.5 km (2 significant figures)
Therefore, displacement = 1.5 km</span>
Answer:
a = 3.27 m/s²
F = 32.7 N
Explanation:
Draw a free body diagram. There are three forces:
Weight force mg pulling straight down.
Normal force N pushing perpendicular to the slope.
Friction force F pushing parallel up the slope.
Sum of forces in the parallel direction:
∑F = ma
mg sin θ − F = ma
Sum of torques about the cylinder's axis:
∑τ = Iα
Fr = ½ mr²α
F = ½ mrα
Since the cylinder rolls without slipping, a = αr. Substituting:
F = ½ ma
Two equations, two unknowns (a and F). Substituting the second equation into the first:
mg sin θ − ½ ma = ma
Multiply both sides by 2/m:
2g sin θ − a = 2a
Solve for a:
2g sin θ = 3a
a = ⅔ g sin θ
a = ⅔ (9.8 m/s²) (sin 30°)
a = 3.27 m/s²
Solving for F:
F = ½ ma
F = ½ (20 kg) (3.27 m/s²)
F = 32.7 N
Answer: I think the current produced by different types of magnets
Explanation:
Answer:
distance must be = 330 × 5/2
= 330×2.5
=825m