Answer:
Explanation:
(a) The force of gravity is called an attractive force because it is the force (although weak) in which a planetary body or matter uses to attract an object towards itself.
(b) Yes, it does and the formula for force of gravity between any two object is
F = G
where m1 and m2 are masses of the first and second object respectively
r is the distance between the center of the two masses
G is the gravitational constant
The density of a material is constant and it is given by the ratio of the mass to the volume of the material
The mass of the liquid and the full bottle ae
The mass of the liquid is <u>450 g</u>
The mass of the filled bottle is <u>530 grams</u>
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The reason the above values are correct are as follows:
The given parameters are;
Volume of the bottle, V = Half litre
Mass of the bottle, = 80 g
The volume of liquid in the bottle when filled, V = 500 cm³
The density of the olive oil with which the bottle is filled, ρ = 0.9 g/cm³
a. Required:
To calculate the mass of oil in the bottle
Solution:
The volume of oil in the bottle when the bottle is filled, V = 500 cm³
The mass of the liquid, m = ρ × V
∴ m = 0.9 g/cm³ × 500 cm³ = 450 g
The mass of the liquid, m = <u>450 g</u>
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b. The mass of the oil in the bottle, m = grams
The mass of the full bottle, = m +
∴ = 450 g + 80 g = 530 g
The mass of the full bottle, = <u>530 grams</u>
Learn more about density here:
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