It would be 429 because in the question it said “ if she was swimming downstream of HALF of her maximum speed, how far will she swim for 13 minutes” so first you would divide 66 by 2 and get 33 then you would multiply 33 by 13 to get you’re answer.
Answer:
the frequency is the fundamental and distance is L = ¼ λ
Explanation:
This problem is a phenomenon of resonance between the frequency of the tuning fork and the tube with one end open and the other end closed, in this case at the closed end you have a node and the open end a belly, so the wavelength is the basis is
λ = 4 L
In this case L = 19.4 cm = 0.194 m
let's use the relationship between wave speed and wavelength frequency and
v = λ f
where the frequency is f = 440 Hz
v = 4 L f
let's calculate
v = 4 0.194 440
v = 341.44 m / s
so the frequency is the fundamental and distance is
L = ¼ λ
Answer:
L= 1 m, ΔL = 0.0074 m
Explanation:
A clock is a simple pendulum with angular velocity
w = √ g / L
Angular velocity is related to frequency and period.
w = 2π f = 2π / T
We replace
2π / T = √ g / L
T = 2π √L / g
We will use the value of g = 9.8 m / s², the initial length of the pendulum, in general it is 1 m (L = 1m)
With this length the average time period is
T = 2π √1 / 9.8
T = 2.0 s
They indicate that the error accumulated in a day is 15 s, let's use a rule of proportions to find the error is a swing
t = 1 day (24h / 1day) (3600s / 1h) = 86400 s
e= Δt = 15 (2/86400) = 3.5 104 s
The time the clock measures is
T ’= To - e
T’= 2.0 -0.00035
T’= 1.99965 s
Let's look for the length of the pendulum to challenge time (t ’)
L’= T’² g / 4π²
L’= 1.99965 2 9.8 / 4π²
L ’= 0.9926 m
Therefore the amount that should adjust the length is
ΔL = L - L’
ΔL = 1.00 - 0.9926
ΔL = 0.0074 m
Answer:
he performed 100 push-ups, 100 sit-ups, and 100 squats, and ran 10 kilometers each day for over a year.
Explanation:
crazy right
