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RideAnS [48]
3 years ago
14

Name two types of battery chargers that are used in mechanics

Engineering
1 answer:
tekilochka [14]3 years ago
7 0
Primary batteries
Secondary batteries
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g Consider a thin opaque, horizontal plate with an electrical heater on its backside. The front end is exposed to ambient air th
xxTIMURxx [149]

Answer:

The electrical power is 96.5 W/m^2

Explanation:

The energy balance is:

Ein-Eout=0

qe+\alpha sGs+\alpha skyGsky-EEb(Ts)-qc=0

if:

Gsky=oTsky^4

Eb=oTs^4

qc=h(Ts-Tα)

\alpha s=\frac{\int\limits^\alpha _0 {\alpha l Gl} \, dl }{\int\limits^\alpha _0 {Gl} \, dl }

\alpha s=\frac{\int\limits^\alpha _0 {\alpha lEl(l,5800 } \, dl }{\int\limits^\alpha _0 {El(l,5800)} \, dl }

if Gl≈El(l,5800)

\alpha s=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))

lt= 2*5800=11600 um-K, at this value, F=0.941

\alpha s=(0.8*0.941)+0.3(1-0.941)=0.77

The hemispherical emissivity is equal to:

E=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))

lt=2*333=666 K, at this value, F=0

E=0+(1-0.7)(1)=0.3

The hemispherical absorptivity is equal to:

qe=EoTs^{4}+h(Ts-T\alpha  )-\alpha sGs-\alpha oTsky^{4}=(0.3*5.67x10^{-8}*333^{4})+10(60-20)-(0.77-600)-(0.3*5.67x10^{-8}*233^{4})=96.5 W/m^{2}

3 0
3 years ago
The base class Pet has attributes name and age. The derived class Dog inherits attributes from the base class Pet class and incl
Nonamiya [84]

Answer:

Explanation:

class Pet:

   def __init__(self):

       self.name = ''

       self.age = 0

   def print_info(self):

       print('Pet Information:')

       print('   Name:', self.name)

       print('   Age:', self.age)

class Dog(Pet):

   def __init__(self):

       Pet.__init__(self)

       self.breed = ''

def main():

   my_pet = Pet()

   my_dog = Dog()

   pet_name = input()

   pet_age = int(input())

   dog_name = input()

   dog_age = int(input())

   dog_breed = input()

   my_pet.name = pet_name

   my_pet.age = pet_age

   my_pet.print_info()

   my_dog.name = dog_name

   my_dog.age = dog_age

   my_dog.breed = dog_breed

   my_dog.print_info()

   print('   Breed:', my_dog.breed)

main()

3 0
3 years ago
A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.0 mm (0.20 in
Sonja [21]

The load is 17156 N.

<u>Explanation:</u>

First compute the flexural strength from:  

σ = FL / πR^{3}

   = 3000 \times (40 \times 10^-3) / π (5 \times 10^-3)^3

σ = 305 \times 10^6 N / m^2.

We can now determine the load using:

F = 2σd^3 / 3L

  = 2(305 \times 10^6) (15 \times 10^-3)^3 / 3(40 \times 10^-3)

F = 17156 N.  

7 0
3 years ago
A 20.0 µF capacitor is charged to a potential difference of 800 V. The terminals of the charged capacitor are then connected to
Sergeu [11.5K]

Answer:

a) Q_initial = 16 * 10^-3 C

b) V_1 = V_2 =  (16/3) * 10^2 V

c)  E = 64/15 J

d)  dE = 32/15 J of decrease

Explanation:

Given:

- Capacitor 1, C_1 = 20.0 uF

- Capacitor 2, C_2 = 10.0 uF

- Charged with P.d V = 800 V

Find:

a) the original charge of the system,

(b) the final potential difference across each capacitor

(c) the final energy of the system

(d) the decrease in energy when the capacitors are connected.

Solution:

a)

- The initial charge in the circuit is the one carried by the first charged capacitor.

                           Q_initial = C_1*V

                           Q_initial = 20*10^-6 * 800

                           Q_initial = 16 * 10^-3 C

b)

- After charging the other capacitor, we know that the total charge is conserved among two capacitor:

                          Q_initial = Q_1 + Q_2

- We also know that potential difference across two capacitor is also same.

                          V_1 = V_2 = Q_1 / C_1 = Q_2 / C_2

- Using the two equations and solve for charge Q_2:

                          Q_2 = Q_1*C_2/C_1

                          Q_2 = Q_1*10/20 = 0.5*Q_1

- using conservation of charge:

                          Q_initial = 1.5*Q_1

                          Q_1 = 16*10^-3 / 1.5 = 10.67*10^-3 C

- Hence the Voltage across each capacitor is:

                          V_2 = V_1 = Q_1 / C_1  

                                            = 10.67*10^-3 / 20*10^-6

                                            = (16/3) * 10^2 V

c)

- The energy in the system is:

                          E = 0.5*C_eq*V^2

Where, C_eq is the equivalent capacitance of paralle circuit.

                           E = 0.5*(20+10)*10^-6 *((16/3) * 10^2)^2

                          E = 64/15 J

d)

- The decrease in energy of the capacitors is:

                           dE = E_initial - E_final

Where, E_initial is due to charging of the C_1 only:

                          dE = 0.5*10^-6*20*800^2 - (64/15)

                          dE = 32/5 - 64/15 = 32/15 J

5 0
3 years ago
A sewage lagoon that has a surface area of 10 ha and a depth of 1 m is receiving 8,640 m^3 /d of sewage containing 100 mg/L of b
Marysya12 [62]

Answer: Coefficient= 0.35 per day

Explanation:

To find the bio degradation reaction rate coefficient, we have

k= \frac{(Cin)(Qin)-(Cout)(Qout)}{(Clagoon)V}

Here, the C lagoon= 20 mg/L

Q in= Q out= 8640 m³/d

C in= 100 mg/L

C out= 20 mg/L

V= 10 ha* 1* 10

V= 10⁵ m³

So, k= \frac{8640*100-8640*20}{20*10^5}

k= 0.35 per day

6 0
3 years ago
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