Answer:
weight at height = 100 N .
Explanation:
The problem relates to variation of weight due to change in height .
Let g₀ and g₁ be acceleration due to gravity , m is mass of the object .
At the surface :
Applying Newton's law of gravitation
mg₀ = G Mm / R²
At height h from centre
mg₁ = G Mm /h²
Given mg₀ = 400 N
400 = G Mm / R²
400 = G Mm / (6400 x 10³ )²
G Mm = 400 x (6400 x 10³ )²
At height h from centre
mg₁ = 400 x (6400 x 10³ )²/ ( 2 x 6400 x 10³)²
= 400 / 4
= 100 N .
weight at height = 100 N
Answer:
False
Explanation:
When the location of the poles changes in the z-plane, the natural or resonant frequency (ω₀) changes which in turn changes the damped frequency (ωd) of the system.
As the poles of a 2nd-order discrete-time system moves away from the origin then natural frequency (ω₀) increases, which in turn increases damped oscillation frequency (ωd) of the system.
ωd = ω₀√(1 - ζ)
Where ζ is called damping ratio.
For small value of ζ
ωd ≈ ω₀
soy de texas, united states
Answer:
Speed at which the ball passes the window’s top = 10.89 m/s
Explanation:
Height of window = 3.3 m
Time took to cover window = 0.27 s
Initial velocity, u = 0m/s
We have equation of motion s = ut + 0.5at²
For the top of window (position A)
For the bottom of window (position B)
We also have
Solving
So after 1.11 seconds ball reaches at top of window,
We have equation of motion v = u + at
Speed at which the ball passes the window’s top = 10.89 m/s
