Chromosomes from the mother control whether a child is male or female. Please select the best answer from the choices provided.
2 answers:
You might be interested in
Answer:
a) perfectly inelastic, b) collision is inelastic, c) elastic
Explanation:
In this exercise, it is asked to identify what type of shock occurs between the skater and the frisbee, for this we must define a system formed by the skater and the fribee, so that the forces during the crash have been internal and the amount of movement is preserved
Initial instant. Before the skater touches the frisbee
p₀ = M v₁ + m v₂
where M and m are the masses of the skater and frisbee, respectively
for the final moment they give us several possibilities, in all case the moment is conserved
p₀ =
case a)
Final instant. grabs the frisbee and holds it
p_{f} = (M + m) v '
p₀ = p_{f}
We can see that this shock is perfectly inelastic, it holds the fressbee
case b)
final instant.
This case is similar to the previous one, but the final speed of fresbee is zero, therefore this collision is inelastic and the kinetic energy is not conserved.
case c)
final instant. Grab the fressbee and resend it
this is an elastic Shock since the equivalent of a rebound of the fressbee, the kinetic energy is conserved.
Answer:
Explanation:
- We have to make a hollow sphere of inner radius
and outer radius
.
Then the mass of the material required to make such a sphere would be calculated as:
Total volume of the spherical shell:
And the volume of the hollow space in the sphere:
Therefore the net volume of material required to make the sphere:
- Now let the density of the of the material be
.
<u>Then the mass of the material used is:</u>
Answer:
The answer to the question
The steady state response is u₂(t) = -cos(3t + π/4)
of the form R·cos(ωt−δ) with R = , ω = 3 and δ = -π/4
Explanation:
To solve the question we note that the equation of motion is given by
m·u'' + γ·u' + k·u = F(t) where
m = mass = 2.00 kg
γ = Damping coefficient = 1
k = Spring constant = 3 N·m
F(t) = externally applied force = 27·cos(3·t)−18·sin(3·t)
Therefore we have 2·u'' + u' + 3·u = 27·cos(3·t)−18·sin(3·t)
The homogeneous equation 2·u'' + u' + 3·u is first solved as follows
2·u'' + u' + 3·u = 0 where putting the characteristic equation as
2·X² + X + 3 = 0 we have the solution given by =
This gives the general solution of the homogeneous equation as
u₁(t) =
For a particular equation of the form 2·u''+u'+3·u = 27·cos(3·t)−18·sin(3·t) which is in the form u₂(t) = A·cos(3·t) + B·sin(3·t)
Then u₂'(t) = -3·A·sin(3·t) + 3·B·cos(3·t) also u₂''(t) = -9·A·cos(3·t) - 9·B·sin(3·t) from which 2·u₂''(t)+u₂'(t)+3·u₂(t) = (3·B-15·A)·cos(3·t) + (-3·A-15·B)·sin(3·t). Comparing with the equation 27·cos(3·t)−18·sin(3·t) we have
3·B-15·A = 27
3·A +15·B = 18
Solving the above linear system of equations we have
A = -1.5, B = 1.5 and u₂(t) = A·cos(3·t) + B·sin(3·t) becomes 1.5·sin(3·t) - 1.5·cos(3·t)
u₂(t) = 1.5·(sin(3·t) - cos(3·t) = ·cos(3·t + π/4)
The general solution is then u(t) = u₁(t) + u₂(t)
however since u₁(t) = ⇒ 0 as t → ∞ the steady state response = u₂(t) =
·cos(3·t + π/4) which is of the form R·cos(ωt−δ) where
R =
ω = 3 and
δ = -π/4