Cumulus belongs to vertical clouds and status to low
Answer:
y = 4 Sin (2πt)
Explanation:
Amplitude, A = 4
frequency, f = 1
Wave function is given by
y = A sinωt
where, ω is angular frequency
ω = 2 π f = 2 π x 1 = 2π
So, the desired wave function
y = 4 Sin (2πt)
<span>A. How could energy become the matter present today? </span>
Answer:
85.8 m/s
Explanation:
We know that the length of the circular path, L the plane travels is
L = rθ where r = radius of path and θ = angle covered
Now,its speed , v = dL/dt = drθ/dt = rdθ/dt + θdr/dt
where dθ/dt = ω = angular speed = v'/r where v' = maximum speed of plane and r = radius of circular path
Now, from θ = θ₀ + ωt where θ₀ = 0 rad, ω = angular speed and t = time,
θ = θ₀ + ωt = 0 + ωt = ωt
So, v = rdθ/dt + θdr/dt
v = rω + ωtdr/dt
v = (r + tdr/dt)ω
v = (r + tdr/dt)v'/r
v = v' + tv'/r(dr/dt)
v = v'[1 + t(dr/dt)/r]
Given that v' = 110 m/s, t = 33.0s, r = 120 m and dr/dt = rate at which line is shortened = -0.80 m/s (negative since it is decreasing)
So, v = 110 m/s[1 + 33.0 s(-0.80 m/s)/120 m]
v = 110 m/s[1 + 11.0 s(-0.80 m/s)/40 m]
v = 110 m/s[1 + 11.0 s(-0.02/s)]
v = 110 m/s[1 - 0.22]
v = 110 m/s(0.78)
v = 85.8 m/s
Answer:
1110 N
Explanation:
First, find the acceleration.
Given:
Δx = 300 m
v₀ = 85.5 km/h = 23.75 m/s
v = 0 m/s
Find: a
v² = v₀² + 2aΔx
(0 m/s)² = (23.75 m/s)² + 2a (300 m)
a = -0.94 m/s²
Find the force:
F = ma
F = (1180 kg) (-0.94 m/s²)
F = -1110 N
The magnitude of the force is 1110 N.
______