Answer:
Explanation:
The mass balance is an application of conservation of mass, to the analysis of physical system. This is given in an equation form as
Input = Output + Accumulation
The conservation law that is used in this analysis of the system actually depends on the context of the problem. Nevertheless, they all revolve around conservation of mass. By conservation of mass, I mean that the fact that matter cannot disappear or be created spontaneously.
Answer:
77.88 lbm/ft³
Explanation:
Given,
Specific gravity, SG = 1.25
Density of water, ρ = 62.30 lbm/ft³
density of the fluid =
= S.G x ρ_{water}
= 62.30 x 1.25
= 77.88 lbm/ft³
Density of the fluid is equal to 77.88 lbm/ft³
The mass of the hoop is the only force which is computed by:F net = 2.8kg*9.81m/s^2 = 27.468 N
the slow masses that must be quicker are the pulley, ring, and the rolling sphere.
The mass correspondent of M the pulley is computed by torque τ = F*R = I*α = I*a/R F = M*a = I*a/R^2 --> M = I/R^2 = 21/2*m*R^2/R^2 = 1/2*m
The mass equal of the rolling sphere is computed by: the sphere revolves around the contact point with the table. So using the proposition of parallel axes, the moment of inertia of the sphere is I = 2/5*mR^2 for spin about the midpoint of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere. I = 7/5*mR^2 M = 7/5*m
the acceleration is then a = F/m = 27.468/(2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2
Answer:
gdgdrhhdhhsbsnjsjdgjdjdjsbdubjdbfn
Explanation:
nnndhdhjdjddjrjjdkdnfidjdbdjbdksjd
Answer:
The new force on the test charge is 145.02N
Explanation:
Force on a unit positive charge can be calculated using coulomb's law.
F =Kq²/r²
Where
F is the force on the charge = 321 N
K is a constant = 8.99 X10⁹ Nm²/C²
q is a test charge = 4.89 μC = 4.89 X10⁻⁶ C
r is the distance between the charge and the surface = 4.1cm
F ∝1/r²
Force on the charge is inversely proportional the square of distance between the charge and the surface.
Fr² = constant
F₁r₁² = F₂r₂²
F₂ = F₁r₁²/r₂²
If the charge is then moved 2.00 cm farther away from the surface;
The new distance becomes 4.10 cm + 2.00 cm = 6.1 cm
Therefore, r₂ = 6.1 cm, r₁ = 4.1 cm, F₁ =321 N
F₂ = (321 X4.1²)/(6.1²)
F₂ = 145.02N
Therefore, The new force on the test charge is 145.02N
