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3 years ago

When ___ ,the lithium ions are removed from the_ and added into the ___

Engineering

1 answer:

bezimeni [28]3 years ago

8 0

Answer:

b. Discharging; anode; cathode

Explanation:

When discharging , it means the battery is producing a flow electric current, the lithium ions are released from the anode to the cathode which generates the flow of electrons from one side to another. When charging Lithium ions are released by the cathode and received by the anode.

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A direct contact heat exchanger (where the fluid mixes completely) has three inlets and one outlet. The mass flow rates of the i

lara31 [8.8K]

Answer:

Enthalpy at outlet=284.44 KJ

Explanation:

$m_1=1 Kg/s,m_2=1.5 Kg/s,m_3=22 Kg/s$

$h_1=100 KJ/Kg,h_2=120 KJ/Kg,h_3=500 KJ/Kg$

We need to Find enthalpy of outlet.

Lets take the outlet mass m and outlet enthalpy h.

So from mass conservation

$m_1+m_2+m_3=m$

m=1+1.5+2 Kg/s

m=4.5 Kg/s

Now from energy conservation

$m_1h_1+m_2h_2+m_3h_3=mh$

By putting the values

$1\times 100+1.5\times 120+2\times 500=4.5\times h$

So h=284.44 KJ

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3 years ago

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blagie [28]

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3 years ago

Consider a space shuttle weighing 100 kN. It is travelling at 310 m/s for 30 minutes. At the same time, it descends 2200 m. Cons

mixas84 [53]

Answer:

work done = 48.88 × $10^{9}$ J

Explanation:

given data

mass = 100 kN

velocity = 310 m/s

time = 30 min = 1800 s

drag force = 12 kN

descends = 2200 m

to find out

work done by the shuttle engine

solution

we know that work done here is

work done = accelerating work - drag work - descending work

put here all value

work done = ( mass ×velocity ×time - force ×velocity ×time - mass ×descends ) 10³ J

work done = ( 100 × 310 × 1800 - 12×310 ×1800 - 100 × 2200 ) 10³ J

work done = 48.88 × $10^{9}$ J

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3 years ago

Consider a resistor made of pure silicon with a cross-sectional area pf 0.5 μm2, and a length of 50 μm. What is the resistance o

lukranit [14]

Answer: 24 pA

Explanation:

As pure silicon is a semiconductor, the resistivity value is strongly dependent of temperature, as the main responsible for conductivity, the number of charge carriers (both electrons and holes) does.

Based on these considerations, we found that at room temperature, pure silicon resistivity can be approximated as 2.1. 10⁵ Ω cm.

The resistance R of a given resistor, is expressed by the following formula:

R = ρ L / A

Replacing by the values for resistivity, L and A, we have

R = 2.1. 10⁵ Ω cm. (10⁴ μm/cm). 50 μm/ 0.5 μm2

R = 2.1. 10¹¹ Ω

Assuming that we can apply Ohm´s Law, the current that would pass through this resistor for an applied voltage of 5 V, is as follows:

I = V/R = 5 V / 2.1.10¹¹ Ω = 2.38. 10⁻¹¹ A= 24 pA

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3 years ago

When _____ ,the lithium ions are removed from the_____ and added into the _____

When ___ ,the lithium ions are removed from the_ and added into the ___