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irakobra [83]
2 years ago
11

La resistencia total de un circuito en paralelo se calcula con la expresión: 1/ Rt = 1/ R1 + 1 / R2 + … + 1 / Rn

Physics
1 answer:
vlabodo [156]2 years ago
8 0

Answer:

correct

Explanation:

in \: parallel

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The conservation of energy principle applies to a. all systems. b. open systems. c. closed systems. d. Ideal gas systems.
Ainat [17]

Answer:closed systems

Explanation:

A closed system is one in which matter does not enter or leave the system but there is exchange of energy between the system and its environment. In a closed system, the principle of energy conservation applies. The principle of energy conservation states that energy can neither be created nor destroyed but is converted from one form to another. An example of a closed system is a reaction vessel whose lid is closed.

5 0
3 years ago
Read 2 more answers
Which of the following would have the most momentum?
Anarel [89]
I think it should be D as momentum is the product of mass and velocity...
4 0
2 years ago
Can someone help me in this one :)
djyliett [7]
I believe it is “runs on parallel circuits”! my bad if incorrect
3 0
3 years ago
A baseball has mass 0.145 kg.
Vesnalui [34]

Answer:

a).p=15.67kg*m/s

b). F=7.83N

Explanation:

change in momentum is the subtraction from "after momentum" of the "before momentum"  as momentum is a vector quantity this is a vector subtraction.

initial momentum of ball

m1v1 = (0.145)(44.0) = 6.38 kg-m/s

after momentum

m1v2 = (0.145)(64.0) = 9.28 kg-m/s

since these momentums are 180° opposite one must be called negative so their difference

6.38+9.28= 15.67 kg-m/s

change in momentum = 15.67 kg-m/s ANS a1

Impulse = change in momenutm = 15.67 ANS a2

b)

Impulse =Favg*t

I=(2.00)Favg

Impulse from (a2) = 15.67  

Favg = 15.67/2.00 =

F= 7.83 N

3 0
3 years ago
A finite line of charge with linear charge density λ = 3.35 × 10-6 C/m, and length L = 0.808 m is located along the x axis (from
Andrews [41]

Answer: magnitude = 169.66N/C

direction = 8.6859°

Explanation:

Given from the question, we have that;

the Length of line of charge L = 0.808 m

Linear charge density λ = 3.35 × 10⁻⁶ C/m

charge q = -7.32 × 10⁻⁷ C

Coulombs force constant K = 1/(4π ε0) = 8.99 × 10⁹ N·m²/C².

NB. The picture uploaded gives a diagrammatic description of the problem.

From Pythagoras theorem we have,

tan Θ = 3.75 / (10.7-1.56)  

Θ = 22.3076 °

recall that the Electric field at point P due to the finite wire is;

È = Kλ (L / b(L+b)) Î .............. (1)

where È rep the electric field.

from equation 1, we have that  

È = 8.99 × 10⁹

where È rep the electric field.

from equation 1, we have that  

È = 8.99 × 10⁹ × 3.35 × 10⁻⁶ (0.808/ 10.7(10.7 – 0.808))

È = 0.229905 × 10³ N/C Î

recall also that the Electric field at point -P due to -q is;

È  = (8.99 × 10⁹ × 7.32 × 10⁻⁷) / ((3.75)² + (10.75-1.56)²) = 0.6742 × 10² N/C

where E = -E₁cosθÎ  + E₁sinθĴ

E = - 0.62446 ×10²Î   + 0.2562 ×10²Ĵ

The Resultant Electric charge Er is given as;

Er = 1.6771 ×10²Î + 0.2562 ×10²

Er =  [√(1.6771)² + (0.2562)² ] × 10² = 169.66 N/C

∴ Magnitude = 169.66 N/C

Having gotten the magnitude, let us find the direction;

⇒ Direction = tan Ф = 0.25621/1.6771 = 8.6859°

Direction = 8.6859°

cheers i hope this helps

7 0
3 years ago
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