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pshichka [43]
3 years ago
14

How do batteries store energy and release electricity on demand.

Engineering
1 answer:
iren [92.7K]3 years ago
7 0

Answer:

Batteries work by storing negative photons on one side then transfering them to the positive side and everything thats inbetween

Explanation:

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Nonshielded cable with a 1.5-inch diameter should have a minimum bending radius of
Flura [38]

Answer:

c is the answer because we have to double the number

6 0
2 years ago
What type of slab and beam used in construction of space neddle​
Sav [38]

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5 0
3 years ago
In a Rankine cycle, superheated steam that enters the turbine at 1273.15 K and 1.8 MPa is then expanded to a vapor at 0.1 MPa. W
GrogVix [38]

Answer:

The shaft work generated per kilogram is -1.3 \frac{MJ}{kg}

Explanation:

Given:

Temperature T = 1273.15 K

Initial Pressure P_{1} = 1.8 MPa

Final pressure P_{2} = 0.1 MPa

From the table superheated,

h_{i} = 4635 \frac{K J}{Kg} and  h_{f} = 2706.54 \frac{K J}{Kg}

Work done by shaft is,

 W = h_{f} - h_{i}

 W = 2706.54 - 4635

 W = -1928.46 \frac{kJ}{kg}

But here efficiency is 0.56,

So work generated per kg is,

Work = 0.56 \times(- 1928.46)

Work = -1.3 \frac{MJ}{kg}

Therefore, the shaft work generated per kilogram is -1.3 \frac{MJ}{kg}

6 0
3 years ago
Select four geometric shapes that are commonly used by engineers in their designs.
stealth61 [152]

Answer:

Arc ,sphere ,line ,prism

Explanation:

  • Arc is used for semicircular or short term projects
  • Sphere is used for design of spherical surfaces or projects
  • Prism is used for creating buliding designs
  • Lines are especially used for creating roads, flyover design
3 0
2 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
3 years ago
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