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3 years ago

A series circuit is composed of two resistors:

Physics

1 answer:

WITCHER [35]3 years ago

7 0

<h3>Answer:</h3>

1.3 Amps

<h3>Explanation:</h3>

<u>We are given;</u>

A circuit with resistors, R1 and R2

R1 = 7 Ω

R2 = 11 Ω

Voltage = 24 V

We are required to calculate the current in the circuit.

<h3>Step 1: We need to find the effective resistance.</h3>

When resistors are arranged in series, the effective resistance is calculated by;

Rt = R₁ + R₂ + R₃ + ..........Rₙ

Therefore;

Total resistance = 7 + 11

= 18 Ω

<h3>Step 2: Calculate the current in the circuit</h3>

From the ohm's law;

V = IR

Rearranging the formula;

I = V/R

Thus;

I = 24 V ÷ 18 Ω

= 1.333 Amps

= 1.3 Amps

Thus, the current in the circuit is 1.3 Amps

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Electromagnetic waves polarized in the x direction propagate towards an ideal polarizer whose axis is aligned with the y-axis. H

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Answer:

there is ZERO TRANSMISSION through the polarizer.

Explanation:

Given the data in the question;

waves was polarized in the x direction propagate towards an ideal polarizer whose axis is aligned with the y-axis.

so θ = 90° { aligned with the y-axis }

now, we know that for polarized light going at angle θ to the polarizer;

I = I₀ × cos²θ

Where I₀ is the intensity of unpolarized light.

I is the intensity after polarization.

so we substitute the value of θ

I = I₀ × cos²( 90° )

I = I₀ × 0

I = 0

Therefore, there is ZERO TRANSMISSION through the polarizer.

6 0

2 years ago

Where an electric field line crosses an equipotential surface, the angle between the field line and the equipotential is

n200080 [17]

Answer:

90 degree

Explanation:

Electric field line is vertical to the electric field line.

4 0

2 years ago

If a true heading of 135° results in a ground track of 130° and a true airspeed of 135 knots results in a groundspeed of 140 kno

vladimir2022 [97]

Answer:

245.1° and 13 knots

Explanation:

The given parameters are;

The true heading = 135°

The resultant ground track = 130°

The true airspeed = 135 knots

The ground speed = 140 knots

Given that the true airspeed the ground speed and the wind direction and magnitude form a triangle, we have;

From cosine rule, we have;

a² = b² + c² - 2×b×c×cos(A)

Where

a = The magnitude of the wind speed in knot

b = The true airspeed = 135 knots

c = The ground speed = 140 knots

A = The angle in between the true heading and the resultant ground track heading = 5°

Which gives;

a² = 135² + 140² - 2×135×140×cos(5 degrees) = 168.84 knots²

a = √168.84 = 12.9934 ≈ 13 knots

We have;

135 × sin(135 degrees) - 140× sin(130 degrees) = -11.7868

135 × cos(135 degrees) - 140× cos(130 degrees) = -5.469

Tan(θ) = -11.8/-5.5 = 2.155

θ = tan⁻¹(2.155) = 65.108°

Given that the wind is moving in opposite direction (slowing down the airplane, we add 180°, to get

Therefore, the angle direction = 180 + 65.108 = 245.1

Therefore, we have;

245.1° and 13 knots

3 0

3 years ago

For the equilibrium reaction N2O4(g) ⇀↽ 2 NO2(g) taking place inside a sealed container fitted with a piston, predict the effect

Butoxors [25]

Answer:

4. The equilibrium will shift to favor formation of NO2(g)

Explanation:

According to La Chatalier's Principle which states that when an equilibrium system undergoes changes either in temperature, volume or concentration; there will be in a change in the system in order to reach equilibrium.

From the above equation,

N2O4(g) ⇀↽ 2 NO2(g)

From the above reaction, there are 2 moles of gaseous product on the left and 1 mole of gaseous reactant.

Therefore, there are more moles of gases in the left hand side than the right hand side.

Because a decrease in volume favors the direction that produces fewer moles, an increase in volume will therefore shift this system towards the side with more moles of gases that is, more products are formed hence, this system will shift to right and produce more moles of products i.e more NO2(g) formed.

6 0

2 years ago

Susan's 10.0kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30? above the f

elena-s [515]

Answer:

The speed after being pulled is 2.4123m/s

Explanation:

The work realize by the tension and the friction is equal to the change in the kinetic energy, so:

$W_T+W_F=K_f-K_i$ (1)

Where:

$W_T=T*x*cos(0)=32N*3.2m*cos(30)=88.6810J\\W_F=F_r*x*cos(180)=-0.190*mg*x =-0.190*10kg*9.8m/s^{2}*3.2m=59.584J\\ K_i=0\\K_f=\frac{1}{2}*m*v_f^{2}=5v_f^{2}$

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.

Additionally, the kinetic energy is equal to $\frac{1}{2}mv^{2}$ , so if the initial velocity $v_i$ is equal to zero, the initial kinetic energy $K_i$ is equal to zero.

Then, replacing the values on the equation and solving for $v_f$ , we get:

$W_T+W_F=K_f-K_i\\88.6810-59.5840=5v_f^{2}\\29.097=5v_f^{2}$

$\frac{29.097}{5}=v_f^{2}\\\sqrt{5.8194}=v_f\\2.4123=v_f$

So, the speed after being pulled 3.2m is 2.4123 m/s

8 0

2 years ago