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notsponge [240]
3 years ago
5

Early American rockets used an RC circuit to set the time for the rocket to begin re-entry after launch (true story). Assume the

capacitor is initially charged, and then discharge begins at launch, t = 0 (this was when the umbilical connector was disconnected). Further assume the resistor in the RC circuit is 1k Ω, and the control circuit begins re-entry when the voltage drops ~63% or to ~37% of full-scale (i.e. 1 time constant), and re-entry is desired to commence at 5 minutes (i.e. 300 seconds after launch). Calculate the capacitor value required.
Engineering
1 answer:
koban [17]3 years ago
3 0

Answer:

C = 0.3 F

Explanation:

The expression for a voltage in a capacitor with an initial voltage V₀, as a function of time, is given by the following equation:

V= Vo*e^{-t/RC}

When V = =.37*V₀, we have the following expression:

\frac{V}{Vo} = e^{-t/RC} = 0.37

Applying ln to both sides, we have:

\frac{-t}{RC} = ln 0.37 = -1

⇒ t = R*C

if t= 300 s, and R = 10³ Ω, we can solve for C as follows:

C = \frac{t}{R} = \frac{3e2 s}{1e3 ohms} = 0.3 F

So, the required value for C is 0.3 F.

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Metals:

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Applications: Body of the vehicles, structures in the skyscrapers, cooking pots.

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Properties: low density, recyclable, poor heat and electrical conductors, plastic deformation;

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8 0
3 years ago
Considering only (110), (1 1 0), (101), and (10 1 ) as the possible slip planes, calculate the stress at which a BCC single crys
vredina [299]

Solution :

i. Slip plane (1 1 0)

Slip direction -- [1 1 1]

Applied stress direction = ( 1 0 0 ]

τ = 50 MPa    ( Here slip direction must be perpendicular to slip plane)

τ = σ cos Φ cos λ

$\cos \phi = \frac{(1,0,0) \cdot (1,1,0)}{1 \times \sqrt2}$

       $=\frac{1}{\sqrt2 }$

$\cos \lambda = \frac{(1,0,0) \cdot (1,-1,1)}{1 \times \sqrt3}$

       $=\frac{1}{\sqrt3 }$

  τ = σ cos Φ cos λ

∴ $50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3} $

  σ = 122.47 MPa

ii. Slip plane  --- (1 1 0)

   Slip direction -- [1 1 1]

  $\cos \phi = \frac{(1, 0, 0) \cdot (1, -1, 0)}{1 \times \sqrt2} =\frac{1}{\sqrt2}$

   $\cos \lambda = \frac{(1, 0, 0) \cdot (1, 1, -1)}{1 \times \sqrt3} =\frac{1}{\sqrt3}$

 τ = σ cos Φ cos λ

∴ $50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3} $

  σ = 122.47 MPa

iii. Slip plane  --- (1 0 1)

    Slip direction --- [1 1 1]

$\cos \phi = \frac{(1, 0, 0) \cdot (1, 0, 1)}{1 \times \sqrt2} =\frac{1}{\sqrt2}$

   $\cos \lambda = \frac{(1, 0, 0) \cdot (1, 1, -1)}{1 \times \sqrt3} =\frac{1}{\sqrt3}$

τ = σ cos Φ cos λ

∴ $50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3} $

  σ = 122.47 MPa

iv. Slip plane -- (1 0 1)

    Slip direction  ---- [1 1 1]

$\cos \phi = \frac{(1, 0, 0) \cdot (1, 0, -1)}{1 \times \sqrt2}=\frac{1}{\sqrt2}$

$\cos \lambda = \frac{(1, 0, 0) \cdot (1, -1, 1)}{1 \times \sqrt3} =\frac{1}{\sqrt3}$

τ = σ cos Φ cos λ

∴ $50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3} $

  σ = 122.47 MPa

∴ (1, 0, -1). (1, -1, 1) = 1 + 0 - 1 = 0

3 0
2 years ago
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