Ω = 2.81
A = 0.232
k = 29.8
x = A cos(ωt + Ф)
at t = 0:
x = A = A cos(ωt + Ф) = A cos(Ф)
Ф = 0
at t = 1.42, with Ф = 0:
x = A cos(ωt)
U = 1/2 k x² = 1/2 k [A cos(ωt)]²
Answer:
1.7333333m/s²
Explanation:
Tension of the line = the weight + force from pulling up the fish
30N = mg + ma
30 = (6)(9.8) + (6)a
10.4 = 6a
∴ a = 1.7333333m/s²
Answer:
йцу21241
Explanation:
ere taking their seats. Finn and Jan presented the progress they had made on the project since the last meeting. Everyone engaged in the subsequent discussions, asking questions and offering ideas.
Wow ! This is not simple. At first, it looks like there's not enough information, because we don't know the mass of the cars. But I"m pretty sure it turns out that we don't need to know it.
At the top of the first hill, the car's potential energy is
PE = (mass) x (gravity) x (height) .
At the bottom, the car's kinetic energy is
KE = (1/2) (mass) (speed²) .
You said that the car's speed is 70 m/s at the bottom of the hill,
and you also said that 10% of the energy will be lost on the way
down. So now, here comes the big jump. Put a comment under
my answer if you don't see where I got this equation:
KE = 0.9 PE
(1/2) (mass) (70 m/s)² = (0.9) (mass) (gravity) (height)
Divide each side by (mass):
(0.5) (4900 m²/s²) = (0.9) (9.8 m/s²) (height)
(There goes the mass. As long as the whole thing is 90% efficient,
the solution will be the same for any number of cars, loaded with
any number of passengers.)
Divide each side by (0.9):
(0.5/0.9) (4900 m²/s²) = (9.8 m/s²) (height)
Divide each side by (9.8 m/s²):
Height = (5/9)(4900 m²/s²) / (9.8 m/s²)
= (5 x 4900 m²/s²) / (9 x 9.8 m/s²)
= (24,500 / 88.2) (m²/s²) / (m/s²)
= 277-7/9 meters
(about 911 feet)
