The period of a spring-cart system motion when the system is in orbit inside the international space station will not change because in conditions of weightlessness it is not possible to measure the mass of a body from its weight and the period depends on the mass of the body.
Answer:
The direction the plane would have to fly to compensate for a wind velocity of 62.0 km/h[N] is 4.5° S of W
Explanation:
The given parameters are;
Velocity of Jet = 792 km/h
Direction of jet velocity = West
Velocity of wind = 62.0 km/h
Direction of wind velocity = North
Therefore, the jet has to have a component of 62.0 km/h South of West to compensate for the wind velocity
The direction of the plane, θ° South of West (S of W) to compensate for the wind is given as follows;

Therefore;

The direction the plane would have to fly to compensate for a wind velocity of 62.0 km/h[N] = 4.5° S of W.
Answer:
a) F = 5.14 10⁻⁸ Hz, f = 4.76 10-8 Hz, b) v = 2.29 m / s, f = 42.5 Hz
Explanation:
a)This problem has two parts.
For the calculations relative to the planet Venus, we use that the period and the frequency are related
f = 1 / T
frequency of the orbit around the Sun
Let's reduce the period to the SI system
T = 225 days (24h / 1days) (3600 s / 1h) = 1.94 10⁷ s
F = 1 / 1.94 10⁷
F = 5.14 10⁻⁸ Hz
rotation frequency
T = 243 d = 2.1 107 s
f = 1 / T
f = 1 / 2.1 107
f = 4.76 10-8 Hz
b) give the data of some marine waves
the speed of the wave can be found with kinematics
v = x / t
v = 50.0 / 21.8
v = 2.29 m / s
If the wavelength is L = 9.28m
this distance is the distance between two consecutive ridges or valleys
λ / 2 = L
λ = 2L
λ = 2 9.28
λ = 18.56 m
the speed of the wave is
v = λ f
f = v /λ
f = 2.29 / 18.56
f = 42.5 Hz
Answer:
1.024 × 10⁸ m
Explanation:
The velocity v₀ of the orbit 8RE is v₀ = 8REω where ω = angular speed.
So, ω = v₀/8RE
For the orbit with radius R for it to maintain a circular orbit and velocity 2v₀, we have
2v₀ = Rω
substituting ω = v₀/8RE into the equation, we have
2v₀ = v₀R/8RE
dividing both sides by v₀, we have
2v₀/v₀ = R/8RE
2 = R/8RE
So, R = 2 × 8RE
R = 16RE
substituting RE = 6.4 × 10⁶ m
R = 16RE
= 16 × 6.4 × 10⁶ m
= 102.4 × 10⁶ m
= 1.024 × 10⁸ m
Answer:
4/10 of L.
Explanation:
A stopped pipe is a pipe with one closed end and one opened end. it is also called a closed pipe.
The fundamental mode of a stopped pipe is also called its fundamental frequency, and is f₁=v/4L.
Where f₁=fundamental frequency, v= velocity of sound, L= Length of pipe.
The fifth harmonic of the stopped pipe f₅ =5v/4L .................(1)
For an open pipe,
Fundamental mode is also called fundamental frequency f₁₀=v/2l₀ .......... (2)
Where f₁₀ = fundamental frequency of a closed pipe, v= velocity of sound and l₀=length of the resulting open pipe.
from the question, the fundamental mode of the resulting open pipe = The fifth harmonic of the original stopped pipe.
∴ f₅=f₁₀
⇒5v/4L = v/2l₀
Equating v from both side of the equation,
⇒ 5/4L = 1/2l₀
Cross multiplying the equation,
5×2l₀ = 4L× 1
10l₀ = 4L
Dividing both side of the equation by the coefficient of l₀ i.e 10
10l₀/10 = 4L/10
∴ l₀ = 4/10(L)
∴ 4/10 of L must be cut off