The constant angular acceleration (in rad/s2) of the centrifuge is 194.02 rad/s².
<h3> Constant angular acceleration</h3>
Apply the following kinematic equation;
ωf² = ωi² - 2αθ
where;
- ωf is the final angular velocity when the centrifuge stops = 0
- ωi is the initial angular velocity
- θ is angular displacement
- α is angular acceleration
ωi = 3400 rev/min x 2π rad/rev x 1 min/60s = 356.05 rad/s
θ = 52 rev x 2π rad/rev = 326.7 rad
0 = ωi² - 2αθ
α = ωi²/2θ
α = ( 356.05²) / (2 x 326.7)
α = 194.02 rad/s²
Thus, the constant angular acceleration (in rad/s2) of the centrifuge is 194.02 rad/s².
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Explanation:
It is given that,
Force on piston, F₁ = 8800 N
Area, 
Area, 
Let F₂ is the force exerted on the second piston. Using Pascal's law as :
Pressure at piston 1 = Pressure at piston 2




So, the force exerted by a second piston is 35200 N. Hence, this is the required solution.
Answer:
Explanation:
Work: This can be defined as the product of force and distance. The unit of work is Joules (J). it can be expressed mathematically as
W = F×d
or
W =
.................................. Equation 1
Where b = upper limit, a = lower limit, Fx = expression of force.
<em>Given: a = 0 , b = 1.3 m, Fx = 4 + 15.7x - 1.5x²</em>
Substituting these values into equation 1
<em>W =
</em>
W = ᵇ[4x + 15.7x²/2-1.5x³/3 +C]ₐ
Work = upper limit - lower limit
Work = ᵃ[4x + 15.7x²/2 - 1.5x³/3 +C] - [4x + 15.7x²/2 + 1.5x³/3 +C]ᵇ............... Equation 2
Substituting the values of a and b into equation 2
Work = [4(1.3) + 15.7(1.3)²/2-1.5(1.3)³/3 + C] - [0 +C]
Work = [5.2 + 26.53 -3.29 + C] - C
Work = 28.44 J
Work done by the force = 28.44 J.
Explanation is in the file
tinyurl.com/wpazsebu
Hi there!
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Why - "radiation in the form of elementary particles emitted by an atomic nucleus, as alpha rays or gamma rays, produced by decay of radioactive substances or by nuclear fission."