1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
oee [108]
3 years ago
14

In a two-source circuit, one source acting alone produces 10 ma through a given branch. the other source acting alone produces 8

ma in the opposite direction through the same branch. the actual current through the branch is

Physics
1 answer:
pashok25 [27]3 years ago
6 0
Refer to the figure below.
R = resistance.

Case 1:
The voltage source is V₁ and the current is 10 mA. Therefore
V₁ = (10 mA)R

Case 2:
The voltage source is V₂ and the current is 8 mA. Therefore
V₂ = (8 mA)R

Case 3:
The voltage across the resistance is V₁ - V₂. Therefore the current I is given by
V₁ - V₂ = IR
10R - 8R = (I mA)R
2 = I
The current is 2 mA.

Answer: 2 mA

You might be interested in
In the diagram, q1= -2.60*10^-9 C and
Alekssandra [29.7K]

Answer:

The magnitude of the net electric field is:

E_{net}=90.37\: N/c

Explanation:

The electric field due to q1 is a vertical positive vector toward q1 (we will call it E1).

On the other hand, the electric field due to q2 is a horizontal positive vector toward q2(We will call it E2).

Knowing this, the <u>magnitude of the net electric</u> field will be the<u> E1 + E2. </u>

Let's find first E1 and E2.

The electric field equation is given by:

|E_{1}|=k\frac{|q_{1}|}{d_{1}^{2}}

Where:

  • k is the Coulomb constant (k = 9*10^{9} Nm²/C²)
  • q1 is the first charge
  • d1 is the distance from q1 to P

|E_{1}|=(9*10^{9})\frac{|-2.60*10^{-9}|}{0.538^{2}}

|E_{1}|=80.84\: N/C

And E2 will be:

|E_{2}|=k\frac{|q_{2}|}{d_{2}{2}}

|E_{2}|=(9*10^{9})\frac{|-8.30*10^{-9}|}{1.36^{2}}

|E_{2}|=40.39\: N/C

Finally, we need to use the  Pythagoras theorem to find the magnitude of the net electric field.

E_{net}=\sqrt{E_{1}^{2}+E_{2}^{2}}

E_{net}=\sqrt{80.84^{2}+40.39^{2}}

E_{net}=90.37\: N/c

I hope it helps you!

7 0
3 years ago
A kettle is rated at 1 kW, 220 V. Calculate the working resistance of the kettle.
Anna [14]

Explanation:

Power of electric kettle, P = 1 kW

Voltage, V = 220 V

(a) Electric power is given by the formula as follows :

P=\dfrac{V^2}{R}

R is resistance

R=\dfrac{V^2}{P}\\\\R=\dfrac{(220)^2}{10^3}\\\\R=48.4\ \Omega

(b) When connected to a 220 V supply, it takes 3 minutes for the water in the kettle to reach boiling point.

Energy supplied is given by :

E=P\times t

P is power, P=\dfrac{V^2}{R}

E=\dfrac{V^2}{R}t\\\\E=\dfrac{(220)^2}{48.4}\times 180\\\\E=180000\ J\\\\E=180\ kJ

5 0
2 years ago
What is the speed of a wave that has a frequency of 125 Hz and a wavelength of 1.25 meters?
kirill115 [55]
Using the "v = f. λ" <span>equation...
 
Your "v" or </span>velocity = 156.25 meters/second
5 0
3 years ago
PLEASE HELP show the calculation of the number of neutrons of hydrogen​
xz_007 [3.2K]
The answer is 2.3 hope this helps texted me and tell me if it’s right
7 0
2 years ago
A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the
Maurinko [17]

Answer:

(a). The decay constant is 1.55\times10^{-5}\ s^{-1}

The half life is 11.3 hr.

(b). The value of N₀ is 2.38\times10^{11}\ nuclei

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity R_{0}=10\ mCi

Time t_{1}=4\ hours

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

R=R_{0}e^{-\lambda t}

\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})

Put the value into the formula

\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})

\lambda=0.0000154\ s^{-1}

\lambda=1.55\times10^{-5}\ s^{-1}

We need to calculate the half life

Using formula of half life

T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}

Put the value into the formula

T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}

T_{\dfrac{1}{2}}=44.719\times10^{3}\ s

T_{\dfrac{1}{2}}=11.3\ hr

(b). We need to calculate the value of N₀

Using formula of N_{0}

N_{0}=\dfrac{3.70\times10^{6}}{\lambda}

Put the value into the formula

N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}

N_{0}=2.38\times10^{11}\ nuclei

(c). We need to calculate the sample's activity

Using formula of activity

R=R_{0}e^{-\lambda\times t}

Put the value intyo the formula

R=10e^{-(1.55\times10^{-5}\times30\times3600)}

R=1.87\ mCi

Hence, (a). The decay constant is 1.55\times10^{-5}\ s^{-1}

The half life is 11.3 hr.

(b). The value of N₀ is 2.38\times10^{11}\ nuclei

(c). The sample's activity is 1.87 mCi.

4 0
3 years ago
Other questions:
  • Find the current passing through a circuit consisting of a battery and one resistor. The resistor has a resistance of 2 ohms and
    11·2 answers
  • Which of the following is a metal?<br> Calcium (Ca)<br> Iron (Fe)<br> Sodium (Na)<br> all of these
    13·1 answer
  • Environmental advantage of using plant oil
    13·1 answer
  • Which data set has the largest standard deviation
    11·1 answer
  • An 80 kg skydiver is falling at terminal velocity. What is the value of air resistance acting on his body? Consider, what are th
    14·1 answer
  • How are force and motion related
    9·1 answer
  • What is gama rays an it's uses​
    15·1 answer
  • A baseball is hit with a speed of 27.0 m/s at an angle of 47.0 ∘ . It lands on the flat roof of a 10.0 m -tall nearby building.
    12·1 answer
  • An object travels back and forth along a straight line. its velocity, in centimeters per second, is given by the function v(t) =
    9·1 answer
  • Someone pls review my work
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!