Answer:
W = 55.12 J
Explanation:
Given,
Natural length = 6 in
Force = 4 lb, stretched length = 8.4 in
We know,
F = k x
k is spring constant
4 = k (8.4-6)
k = 1.67 lb/in
Work done to stretch the spring to 10.1 in.
![W = \dfrac{k}{2}[x^2]_6^{10.1}](https://tex.z-dn.net/?f=W%20%3D%20%5Cdfrac%7Bk%7D%7B2%7D%5Bx%5E2%5D_6%5E%7B10.1%7D)
W = 55.12 J
Work done in stretching spring from 6 in to 10.1 in is equal to 55.12 J.
Answer:
8050 J
Explanation:
Given:
r = 4.6 m
I = 200 kg m²
F = 26.0 N
t = 15.0 s
First, find the angular acceleration.
∑τ = Iα
Fr = Iα
α = Fr / I
α = (26.0 N) (4.6 m) / (200 kg m²)
α = 0.598 rad/s²
Now you can find the final angular velocity, then use that to find the rotational energy:
ω = αt
ω = (0.598 rad/s²) (15.0 s)
ω = 8.97 rad/s
W = ½ I ω²
W = ½ (200 kg m²) (8.97 rad/s)²
W = 8050 J
Or you can find the angular displacement and find the work done that way:
θ = θ₀ + ω₀ t + ½ αt²
θ = ½ (0.598 rad/s²) (15.0 s)²
θ = 67.3 rad
W = τθ
W = Frθ
W = (26.0 N) (4.6 m) (67.3 rad)
W = 8050 J
The answer is reverse faults.
Answer:
order d> a = e> c> b = f
Explanation:
Pascal's law states that a change in pressure is transmitted by a liquid, all points are transmitted regardless of the form
P₁ = P₂
Using the definition of pressure
F₁ / A₁ = F₂ / A₂
F₂ = A₂ /A₁ F₁
Now we can examine the results
a) F1 = 4.0 N A1 = 0.9 m2 A2 = 1.8 m2
F₂ = 1.8 / 0.9 4
F₂a = 8 N
b) F1 = 2.0 N A1 = 0.9 m2 A2 = 0.45 m2
F₂b = 0.45 / 0.9 2
F₂b = 1 N
c) F1 2.0 N A1 = 1.8 m2 A2 = 3.6 m2
F₂c = 3.6 / 1.8 2
F₂c = 4 N
d) F1 = 4.0N A1 = 0.45 m2 A2 = 1.8 m2
F₂d = 1.8 / 0.45 4.0
F₂d = 16 m2
e) F1 = 4.0 N A1 = 0.45 m2 A2 = 0.9 m2
F₂e = 0.9 / 0.45 4
F₂e = 8 N
f) F1 = 2.0N A1 = 1.8 m2 A2 = 0.9 m2
F₂f = 0.9 / 1.8 2.0
F₂f = 1 N
Let's classify the structure from highest to lowest
F₂d> F₂a = F₂e> F₂c> F₂b = F₂f
I mean the combinations are
d> a = e> c> b = f
