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AnnZ [28]
3 years ago
8

What type of figure is this

Engineering
1 answer:
Vlad1618 [11]3 years ago
8 0
The answer is Ur moms figure
You might be interested in
If i build thing a and thing a builds thing b did i build thing b
postnew [5]

Answer:

<h3>Yes</h3>

Explanation:

If you build thing "a" and thing "a" builds thing "b" you <u>indirectly</u> build thing "b".

3 0
3 years ago
Read 2 more answers
A battery with an f.e.m. of 12 V and negligible internal resistance is connected to a resistor of 545 How much energy is dissipa
Alenkasestr [34]

Answer:

When are resistors in series? Resistors are in series whenever the flow of charge, called the current, must flow through devices sequentially. For example, if current flows through a person holding a screwdriver and into the Earth, then  

R

1

 in Figure 1(a) could be the resistance of the screwdriver’s shaft,  

R

2

 the resistance of its handle,  

R

3

 the person’s body resistance, and  

R

4

 the resistance of her shoes.

Figure 2 shows resistors in series connected to a voltage source. It seems reasonable that the total resistance is the sum of the individual resistances, considering that the current has to pass through each resistor in sequence. (This fact would be an advantage to a person wishing to avoid an electrical shock, who could reduce the current by wearing high-resistance rubber-soled shoes. It could be a disadvantage if one of the resistances were a faulty high-resistance cord to an appliance that would reduce the operating current.)

Two electrical circuits are compared. The first one has three resistors, R sub one, R sub two, and R sub three, connected in series with a voltage source V to form a closed circuit. The first circuit is equivalent to the second circuit, which has a single resistor R sub s connected to a voltage source V. Both circuits carry a current I, which starts from the positive end of the voltage source and moves in a clockwise direction around the circuit.

Figure 2. Three resistors connected in series to a battery (left) and the equivalent single or series resistance (right).

To verify that resistances in series do indeed add, let us consider the loss of electrical power, called a voltage drop, in each resistor in Figure 2.

According to Ohm’s law, the voltage drop,  

V

, across a resistor when a current flows through it is calculated using the equation  

V

=

I

R

, where  

I

 equals the current in amps (A) and  

R

 is the resistance in ohms  

(

Ω

)

. Another way to think of this is that  

V

 is the voltage necessary to make a current  

I

 flow through a resistance  

R

.

So the voltage drop across  

R

1

 is  

V

1

=

I

R

1

, that across  

R

2

 is  

V

2

=

I

R

2

, and that across  

R

3

 is  

V

3

=

I

R

3

. The sum of these voltages equals the voltage output of the source; that is,

V

=

V

1

+

V

2

+

V

3

.

 

This equation is based on the conservation of energy and conservation of charge. Electrical potential energy can be described by the equation  

P

E

=

q

V

, where  

q

 is the electric charge and  

V

 is the voltage. Thus the energy supplied by the source is  

q

V

, while that dissipated by the resistors is

q

V

1

+

q

V

2

+

q

V

3

.

Explanation:

8 0
3 years ago
Why is the lubrication system of an internal combustion engine equipped with an oil filter?
Ierofanga [76]

Answer:

to filter out any impurities such as metal shavings in the oil

6 0
3 years ago
An induced-draft cooling tower cools 90,000 gallons per minute of water from 84 to 68oF. Air at 14.61 psia, 70oF dry bulb and 60
belka [17]

Answer:

a. V = 109.64 × 10⁵ ft/min

b. Mw = 654519.54 kg/hr

Explanation:

Given Parameters

mass flow rate of water, Mw = 90000g/min = 6607.33 kg/s

inlet temperature of water, T1 = 84 F = 28.89 C

outlet temperature of water, T2 = 68 F = 20 C

specific heat capacity of water, c = 4.18kJ/kgK

rate of heat remover from water, Qw is given by

Qw = 6607.33[28.89 - 20] * 4.18

Qw = 245529.545kw

For air, inlet condition

DBT = 70 F              hi = 43.43 kJ/kg

WBT = 60 F             wi = 0.00874 kJ/kg

                                u1 = 0.8445 m/kg

oulet condition,

DBT = 70 F        RH = 100.1

h1 = 83.504kJ/kg

Wo = 0.222kJ/kg

check the attached file for complete solution

3 0
3 years ago
How can I solve 23.5 million Nona meters to millimeters using no calculator because I have to show my work
katrin2010 [14]

Express it in standard form and apply the basic indices laws to simplify

6 0
3 years ago
Read 2 more answers
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