Explanation:
A) Use Hooke's law to find the spring constant.
F = kx
40 N = k (0.4 m)
k = 100 N/m
B) Period of a spring-mass system is:
T = 2π √(m / k)
T = 2π √(2.6 kg / 100 N/m)
T = 1 s
Frequency is the inverse of period.
f = 1 / T
f = 1 Hz
Answer:ummm ok
Explanation:I really don’t get it but ok
Answer:
a) v = √(v₀² + 2g h), b) Δt = 2 v₀ / g
Explanation:
For this exercise we will use the mathematical expressions, where the directional towards at is considered positive.
The velocity of each ball is
ball 1. thrown upwards vo is positive
v² = v₀² - 2 g (y-y₀)
in this case the height y is zero and the height i = h
v = √(v₀² + 2g h)
ball 2 thrown down, in this case vo is negative
v = √(v₀² + 2g h)
The times to get to the ground
ball 1
v = v₀ - g t₁
t₁ =
ball 2
v = -v₀ - g t₂
t₂ = - \frac{v_{o} + v }{ g}
From the previous part, we saw that the speeds of the two balls are the same when reaching the ground, so the time difference is
Δt = t₂ -t₁
Δt =
Δt = 2 v₀ / g
Gravitational force = G · (mass₁) · (mass₂) / (distance)
(distance²) = G · (mass₁) · (mass₂) / (Gravitational force)
G = 6.67 x 10⁻¹¹ n-m² / kg² (the "gravitational constant")
Distance² = (6.67 x 10⁻¹¹ n-m² / kg²) (28,500 kg) (2.2 x 10⁸ kg) / (39 N)
Distance² = (6.67 · 28,500 · 2.2 x 10⁻³ N-m²) / (39N)
Distance² = (418.209 N-m²) / (39N)
Distance² = 10.72 m²
<em>Distance = 3.275 meters</em>
An absurd scenario, but that's by golly what the math says with the numbers provided. I guess it's a teeny tiny planet orbiting 3.275 meters outside a teeny tiny black hole.