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Neporo4naja [7]
3 years ago
10

A common chef's knife is single-beveled. TRUE FALSE

Engineering
1 answer:
melisa1 [442]3 years ago
8 0

Answer:

True Hope this helped!

Explanation:

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What is the need to achieve population inversion​
notka56 [123]

Explanation:

Population inversion is a process of achieving more electrons in the higher energy state than the lower energy state. In order to achieve population inversion, we need to supply energy to the laser medium. The process of supplying energy to the laser medium is called pumping.

3 0
3 years ago
For the speed equation along centerline of a diffuser, calculate the fluid acceleration along the diffuser centerline as a funct
Marrrta [24]

Answer:

a = v\cdot \frac{dv}{dx}, v (x) = v_{in}\cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1  \right)\cdot x \right]^{-1}, \frac{dv}{dx} = -v_{in}\cdot \left(\frac{1}{L}\right) \cdot \left(\frac{v_{in}}{v_{out}}-1  \right) \cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}} -1 \right) \cdot x \right]^{-2}

Explanation:

Let suppose that fluid is incompressible and diffuser works at steady state. A diffuser reduces velocity at the expense of pressure, which can be modelled by using the Principle of Mass Conservation:

\dot m_{in} - \dot m_{out} = 0

\dot m_{in} = \dot m_{out}

\dot V_{in} = \dot V_{out}

v_{in} \cdot A_{in} = v_{out}\cdot A_{out}

The following relation are found:

\frac{v_{out}}{v_{in}} = \frac{A_{in}}{A_{out}}

The new relationship is determined by means of linear interpolation:

A (x) = A_{in} +\frac{A_{out}-A_{in}}{L}\cdot x

\frac{A(x)}{A_{in}} = 1 + \left(\frac{1}{L}\right)\cdot \left( \frac{A_{out}}{A_{in}}-1\right)\cdot x

After some algebraic manipulation, the following for the velocity as a function of position is obtained hereafter:

\frac{v_{in}}{v(x)} = 1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1\right) \cdot x

v(x) = \frac{v_{in}}{1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1  \right)\cdot x}

v (x) = v_{in}\cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1  \right)\cdot x \right]^{-1}

The acceleration can be calculated by using the following derivative:

a = v\cdot \frac{dv}{dx}

The derivative of the velocity in terms of position is:

\frac{dv}{dx} = -v_{in}\cdot \left(\frac{1}{L}\right) \cdot \left(\frac{v_{in}}{v_{out}}-1  \right) \cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}} -1 \right) \cdot x \right]^{-2}

The expression for acceleration is derived by replacing each variable and simplifying the resultant formula.

8 0
3 years ago
Read 2 more answers
Describe various emerging technologies and their impact on the field of robotics.
mrs_skeptik [129]

The formal procedure for taking equipment out of service and ensuring it cannot be operated until an authorized person has returned it to service is

Explanation:

3 0
3 years ago
Would anyone be able to tell me what this is? it's for a water pump​
AveGali [126]

Answer:

it is a water thing you put it on a hose so you can have different modes on how the water come out

5 0
3 years ago
In an oscillating LC circuit, L ! 25.0 mH and C ! 7.80 mF. At time t 0 the current is 9.20 mA, the charge on the capacitor is 3.
topjm [15]

Answer:

a) 926 μJ

b) 3.802 mC

c) 8.61 A

d) 0.0721

e) 3.2137

Explanation:

The energy in the inductor is

El = \frac{1}{2}*L*I^2

El = \frac{1}{2}*25*10^{-3}*(9.2*10^{-3})^2 = 1.06*10^{-6} J = 1.06 \mu J

The energy store in a capacitor is

Ec = \frac{1}{2}*C*V^2

The voltage in a capacitor is

V = Q/C

V = \frac{3.8*10^{-3}}{7.8*10^{-3}} = 0.487 V

Therefore:

Ec = \frac{1}{2}*7.8*10^{-3}*0.487^2 = 9.256*10^{-4} J = 925.6 \mu J

The total energy is Et = 925.6 + 1.1 = 926.7 μJ

At a certain point all the energy of the circuit will be in the capacitor, at this point it will have maximum charge

Ec = \frac{1}{2}*C*V^2

V = Q/C

Ec = \frac{1}{2}*C*(\frac{Q}{C})^2

Ec = \frac{1}{2}*\frac{Q^2}{C}

Q^2 = 2*Ec*C

Q = \sqrt{2*Ec*C}

Q = \sqrt{2*926*10{-6}*7.8*10^{-3}} = 3.802 * 10{-3} C = 3.802 mC

When the capacitor is completely empty all the energy will be in the inductor and current will be maximum

El = \frac{1}{2}*L*I^2

I^2 = 2*\frac{El}{L}

I = \sqrt{2*\frac{El}{L}}

I = \sqrt{2*\frac{926.7*10^{-3}}{25*10^{-3}}} = 8.61 A

At t = 0 the capacitor has a charge of 3.8 mC, the maximum charge is 3.81 mC

q = Q * cos(vt + f)

q(0) = Q * cos(v*0 + f)

3.8 = 3.81 * cos(f)

cos(f) = 3.8/3.81

f = arccos(3.8/3.81) = 0.0721

If the capacitor is discharging it is a half cycle away, so f' = f + π = 3.2137

4 0
3 years ago
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