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spayn [35]
3 years ago
6

In a circuit, three pieces of wire labeled A, B and C are joined at a common point, D. If wire B carries 1.5 mA in a direction a

way from D and C carries current 1.3 mA toward D, what does wire A carry?
a. 0.2 mA toward D
b. 0.2 mA away from D
c. 2.8 mA toward D
d. 2.8 mA away from D
e. unable to determine

Physics
1 answer:
Alex_Xolod [135]3 years ago
7 0

Answer:

correct option is a. 0.2 mA toward D

Explanation:

given data

B carries = 1.5 mA

C carries current  = 1.3 mA

solution

we take positive direction of current going away from the point D

and negative direction of current coming towards point D

so we use here kirchoff's current law   that is

iA + iB + iC = 0    ......................1

iA + 1.5 + (-1.3) = 0

iA = - 0.2 mA  

so that current in wire A is 0.2 mA towards point D

correct option is a. 0.2 mA toward D

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A can of tuna, starting from rest, is dropped off a building. If it is only being pulled down by gravity (-9.8 m/s2) and it hits
tia_tia [17]

Answer:

<h3>50.23m</h3>

Explanation:

The distance talks about how far an object has travelled.

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Newtons _ law of motion states that every action has an equal and opposite reaction​
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Two positive point charges, each of which has a charge of 2.5 × 10−9 C, are located at y = + 0.50m and y = − 0.50m. Find the mag
nadya68 [22]

Answer:

F = 147,78*10⁻⁹ [N]

Explanation:

By symmetry the Fy components of the forces acting on charge in point x = 0,7 m canceled each other, and the total force will be twice Fx ( Fx is x axis component of one of the forces .

The angle β  ( angle between the line running through one of the charges in y axis and the charge in x axis) is

tan β  =  0,5/0,7

tan β  = 0,7142    then   β = arctan 0,7142      ⇒   β = 35 ⁰

cos  β = 0,81

d = √ (0,5)²  +  (0,7)²       d1stance between charges

d = √0,25 + 0,49

d =  √0,74  m

d = 0,86  m

Now Foce between two charges  is:

F  = K* q₁*q₂/ d²      (1)

Where  K  = 9*10⁹ N*m²/C²

q₁  =  2,5* 10⁻⁹C

q₂  = 3,0*10⁻⁹C

d²  = 0,74 m²

Plugging these values in (1)

F  =  9*10⁹* 2,5* 10⁻⁹*3,0*10⁻⁹ / 0,74       [N*m²/C²]*C*C/m²

F = 91,21 * 10⁻⁹   [N]

And  Fx  =  F*cos  β

Fx  =  91,21 * 10⁻⁹ *0,81

Fx =73,89*10⁻⁹  [N]

Then total force acting on charge located at x = 0,7 m is:

F = 2* Fx

F = 2*73,89*10⁻⁹  [N]

F = 147,78*10⁻⁹ [N]

8 0
3 years ago
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