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Masja [62]
3 years ago
12

Is your mom the love of your life

Engineering
1 answer:
hodyreva [135]3 years ago
7 0
No but i love her
if she was the love of my life that would be weied
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Can you help me? I need solution of this question.
ollegr [7]

Answer:hmmmmmmmmmm give an hour.

Explanation:

...

6 0
3 years ago
What is an advantage of using a fully integrated cloud-based data analytics platform?
sweet [91]

Answer:

It gives decision-makers the processing capacity they need to turn raw data into useful knowledge.

Explanation:

Data analysis and the associated cycles (data integration, aggregation, hoarding, and revealing) are totally or partially directed in the cloud with cloud analytics.

3 0
3 years ago
Find the phasor form of:
mixer [17]

Answer:

I=24.598\angle 50.377

Explanation:

A tension or current expressed in cosine form and with a positive sign can be converted directly into a phasor. This is done by indicating the tension and the offset angle:

Acos(10\omega t +\phi)=A\angle \phi

So:

i(t)=10cos(10t+63)+15cos(10t-42)=10\angle 63 + 15\angle42=I

You can sum the phasors simply using a calculator, however, let's do it manually:

Let's find the rectangular form of each phasor using the next formulas:

A=\sqrt{a^2+b^2} \\\phi=arctan(\frac{b}{a})

For 10\angle 63

63=arctan(\frac{b}{a} )\\\\tan(63)=\frac{b}{a} \\\\b=a*tan(63)

10=\sqrt{a^2+(a*tan(63))^2} \\\\10^2=a^2+a^2*(tan(63))^2\\\\Solving\hspace{3}for\hspace{3}a\\\\a=\sqrt{\frac{100}{1+tan(63)^2} } =4.539904997\\\\and\hspace{3}b\\b=\sqrt{100-a^2} =8.910065242

So:

Z_1=a+bj=4.539904997+8.910065242j

For 15\angle 42

42=arctan(\frac{b_2}{a_2} )\\\\tan(42)=\frac{b_2}{a_2} \\\\b_2=a_2*tan(42)

15=\sqrt{a_2^2+(a_2*tan(42))^2} \\\\15^2=a_2^2+a_2^2*(tan(42))^2\\\\Solving\hspace{3}for\hspace{3}a_2\\\\a_2=\sqrt{\frac{225}{1+tan(42)^2} } =11.14717238\\\\and\hspace{3}b_2\\\\b_2=\sqrt{225-a^2} =10.0369591

So:

Z_2=a_2+b_2j=11.14717238+10.0369591j

Hence:

Z_T=Z_1+Z_2=(4.539904997+11.14717238)+(8.910065242+10.0369591)j\\Z_T=15.68707738+18.94702434j

Finally:

I=\sqrt{15.68707738^2+18.94702434^2} \angle arctan (\frac{18.94702434}{15.68707738} )=24.598\angle 50.377

7 0
3 years ago
If a ball is dropped from a height​ (H) its velocity will increase until it hits the ground​ (assuming that aerodynamic drag due
enot [183]

Answer

Explanation:

so the velocity is 39 feet per sec so the impact is 720 cm from the ground

take 720 * 39 sq

4 0
3 years ago
A very large thin plate is centered in a gap of width 0.06 m with a different oils of unknown viscosities above and below; one v
In-s [12.5K]

Answer:

The viscosities of the oils are 0.967 Pa.s and 1.933 Pa.s

Explanation:

Assuming the two oils are Newtonian fluids.

From Newton's law of viscosity for Newtonian fluids, we know that the shear stress is proportional to the velocity gradient with the viscosity serving as the constant of proportionality.

τ = μ (∂v/∂y)

There are oils above and below the plate, so we can write this expression for the both cases.

τ₁ = μ₁ (∂v/∂y)

τ₂ = μ₂ (∂v/∂y)

dv = 0.3 m/s

dy = (0.06/2) = 0.03 m (the plate is centered in a gap of width 0.06 m)

τ₁ = μ₁ (0.3/0.03) = 10μ₁

τ₂ = μ₂ (0.3/0.03) = 10μ₂

But the shear stress on the plate is given as 29 N per square meter.

τ = 29 N/m²

But this stress is a sum of stress due to both shear stress above and below the plate

τ = τ₁ + τ₂ = 10μ₁ + 10μ₂ = 29

But it is also given that one viscosity is twice the other

μ₁ = 2μ₂

10μ₁ + 10μ₂ = 29

10(2μ₂) + 10μ₂ = 29

30μ₂ = 29

μ₂ = (29/30) = 0.967 Pa.s

μ₁ = 2μ₂ = 2 × 0.967 = 1.933 Pa.s

Hope this Helps!!!

6 0
3 years ago
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