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MariettaO [177]
4 years ago
5

The capacity of a battery is 1800 mAh and its OCV is 3.9 V. a) Two batteries are placed in series. What is the combined battery

OCV and when charged at 2C what is its capacity? b) Two batteries are placed in parallel. What is the combined OCV and capacity?
Engineering
1 answer:
Lynna [10]4 years ago
3 0

Answer:

capacity  = 0.555 mAh

capacity  = 3600 mAh

Explanation:

given data

battery = 1800 mAh

OCV = 3.9 V

solution

we get here capacity when it is in series

so here Q = 2C  

capacity  = 2 × ampere × second   ...............1

put here value and we get

and 1 Ah = 3600 C

capacity  = \frac{2}{3600}

capacity  = 0.555 mAh

and

when it is in parallel than capacity will be

capacity = Q1 +Q2   ...............2

capacity  = 1800 + 1800

capacity  = 3600 mAh

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Answer:

K=C+273.15

Explanation:

Kelvin's climbing represents the <em>absolute temperature</em>. Temperature is a measure of the molecular kinetic energy of translation. If the molecules move quickly, with the same energy as in the walls of the container, which makes us feel like "heat". If the molecules do not move, the temperature is zero. 0 K.

The Celsius scale has an <em>artificial zero</em>, defined in the solidification temperature of the water. It is very useful to talk about the weather, and about some simpler technical matters. But it is artificial.

3 0
4 years ago
Write a program that asks the user to input a vector of integers of arbitrary length. Then, using a for-end loop the program exa
ELEN [110]

Answer:

%Program prompts user to input vector

v = input('Enter the input vector: ');

%Program shows the value that user entered

fprintf('The input vector:\n ')

disp(v)

%Loop for checking all array elements

for i = 1 : length(v)

   %check if the element is a positive number

   if v(i) > 0

       %double the element

       v(i) = v(i) * 2;

   %else the element is negative number.

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       v(i) = v(i) * 3;

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4 0
4 years ago
A compressor operates at steady state with Refrigerant 134a as the working fluid. The refrigerant enters at 0.24 MPa, 0°C, with
marissa [1.9K]

Answer:

POWER INPUT = 82.989 KW

Explanation:

For pressure P =0.24 MPa and T_1 = 0 DEGREE, Enthalapy _1 = 248.89 kJ/kg

For pressure P =1  MPa and T_1 = 50 DEGREE, Enthalapy _1 = 280.19 kJ/kg

Heat loss Q = 0.05w

Inlet diameter = 3 cm

exit diamter = 1.5  cm

volume of tank will be v = area * velocity

velocity at inlet= \frac{0.64\60 m/s}{ \frac{\pi}{4} (3\times 10^{-2})^2} = 15.09  m/s

velocity at outlet= \frac{0.64\60 m/s}{ \frac{\pi}{4} (1.5\times 10^{-2})^2} = 60.36  m/s

steady flow energy equation

E_{IN} = E_{OUT}

h_1 + \frac{v_1^2}{2g} +wc = h_2 + \frac{v_2^2}{2g} + 0.05wc

248.89 + \frac{15.09^2}{2} + wc = 280.18 + \frac{60.36^2}{2} + 0.05 wc

solving wc = 1830.64  kJ/kg

wc in KWH

we know thatwc = \dot m wc

       \dot m = 4.25 kg/m3 \times (0.64/60) m^3/s

\dot m = 0.04533  kg/s

wc = 0.04533 \times 1830.64 = 82.989 kW

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