I didn't understand the following matrix manipulation, as shown in the attached photo. How was it done?
1 answer:
The picture I included shows how to to a 2x2 and 2x1 matrix multiplication if that’s what you’re confused about.
If you’re wondering what you can learn from the example. The product matrices on the left are the same except the terms in top matrix each have an added term. So the difference between the two sets of matrices result in the top having two extra terms after the multiplication.
Let me know if you still need more.
If you’re wondering what you can learn from the example. The product matrices on the left are the same except the terms in top matrix each have an added term. So the difference between the two sets of matrices result in the top having two extra terms after the multiplication.
Let me know if you still need more.
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Answer:
F=1.47 KN
Explanation:
Given that
Diameter of plate = 25 cm
Height of pool h = 3 m
We know that force can be given as
F= P x A
P=ρ x g x h
Now by putting the values
P=1000 x 10 x 3
P= 30 KPa
F= 30 x 0.049 KN
F=1.47 KN
So the force on the plate will be 1.47 KN.
Answer:
a₁= 1.98 m/s² : magnitud of the normal acceleration
a₂=0.75 m/s² : magnitud of the tangential acceleration
Explanation:
Formulas for uniformly accelerated circular motion
a₁=ω²*r : normal acceleration Formula (1)
a₂=α*r: normal acceleration Formula (2)
ωf²=ω₀²+2*α*θ Formula (3)
ω : angular velocity
α : angular acceleration
r : radius
ωf= final angular velocity
ω₀ : initial angular velocity
θ : angular position theta
r : radius
Data
r =0.4 m
ω₀= 1 rad/s
α=0.3 *θ , θ= 2π
α=0.3 *2π= 0,6π rad/s²
Magnitudes of the normal and tangential components of acceleration of a point P on the rim of the disk when theta has completed one revolution.
We calculate ωf with formula 3:
ωf²= 1² + 2*0.6π*2π =1+2.4π ²= 24.687
ωf= =4.97 rad/s
a₁=ω²*r = 4.97²*0.4 = 1.98 m/s²
a₂=α*r = 0,6π * 0.4 = 0.75 m/s²