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pantera1 [17]
3 years ago
10

Why would knowing the characteristics of circuits be important in designing electrical circuits?

Physics
2 answers:
zubka84 [21]3 years ago
8 0
Well if you didn't you could make mistakes, which would lead ,in the best case, at a fail of the circuit , or if it goes out of control it could be dangerous

for example you have to know that the wires become hot and they loose their abbilitys as connecters(the hotter it will, the more energy you lose becouse the R will be bigger)
Pachacha [2.7K]3 years ago
4 0
In designing electrical circuits, it is important to know the characteristics of circuits in order to know how to solve related problems that will<span> arise whenever there is certain parameter required.

I hope my answer has come to your help. God bless and have a nice day ahead!
</span>
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If the absolute temperature of a gas is 600 K, the temperature in degrees Celsius is: A. 705°C. B. 873°C. C. 273°C. D. 327°C
zlopas [31]

Answer:

D). 327 ^0 C

Explanation:

As we know that temperature scale is linear so we will have

\frac{^0C - 0}{100 - 0} = \frac{K - 273}{373 - 273}

now we have

\frac{^0 C - 0}{100} = \frac{K - 273}{100}

so the relation between two scales is given as

^0 C = K - 273

now we know that in kelvin scale the absolute temperature is 600 K

so now we have

T = 600 - 273 = 327 ^0 C

so correct answer is

D). 327 ^0 C

4 0
3 years ago
Sample Response: When there is less rainfall, land, which is part of the geosphere, becomes dry. When there is not enough water,
GalinKa [24]

grow alot of plants that's help to make more h2o

8 0
1 year ago
A 2.964 kilogram truck strikes a fence at 7.00 meters/second and comes to
Makovka662 [10]

10.92N

Explanation:

Given parameters:

Mass of truck = 2.964kg

Velocity of truck = 7m/s

Time taken = 1.9s

Unknown:

Average force on the car = ?

Solution:

According to newton's third law of motion "action and reaction are equal and opposite".

The force with which the truck struck the fence is the same as the force the fence acted on the truck with but in another direction.

 From newton's second law:

      Force  = mass x acceleration

      We know that acceleration is the change in velocity with time;

   acceleration = \frac{change in velocity }{time }

   Force =  mass x  \frac{change in velocity }{time }

  Force = \frac{2.964 x 7}{1.9} = 10.92N

learn more:

Newton's laws brainly.com/question/11411375

#learnwithBrainly

3 0
3 years ago
The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of
adell [148]

Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = ma_a (0.35)

1.8v^2(1.1) - 1200(9.8)(1.25) = 1200a(0.35)

1.8v^2(1.1) - 14700 = 420 a   ------- equation (1)

F_x = ma_x \\ \\ = 1.8v^2 = 1200 \ a             --------- equation (2)

Replacing equation 2 into equation 1 ; we have :

{1.1 * 1200 \ a} - 14700 = 420 a

1320 a - 14700 = 420 a

1320 a -  420 a =14700

900 a = 14700

a = 14700/900

a = 16.33 m/s²

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

5 0
2 years ago
A car with a mass of 1.1 × 103 kilograms hits a stationary truck with a mass of 2.3 × 103 kilograms from the rear end. The initi
snow_lady [41]

Answer:

The velocity of the truck after this elastic collision is 15.7 m/s            

Explanation:

It is given that,

Mass of the car, m_1=1.1\times 10^3\ kg

Mass of the truck, m_2=2.3\times 10^3\ kg

Initial velocity of the car, u_1=22\ m/s

Initial velocity of the truck, u₂ = 0

After the collision the velocity of the car is, v₁ = -11 m/s

Let v₂ is the velocity of the truck after this elastic collision. Using the conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

1.1\times 10^3\times 22+0=1.1\times 10^3\times (11)+2.3\times 10^3\times v_2    

v_2=15.7\ m/s

So, the velocity of the truck after this elastic collision is 15.7 m/s. Hence, the correct option is (c).

4 0
3 years ago
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