Put the waves in order from shortest to longest wavelength

A 5.0 Ω resistor is hooked up in series with a 10.0 Ω resistor followed by a 20.0 Ω resistor. The circuit is powered by a 9.0 V

yan [13]

<h2>Answer:</h2>

(a) Attached to the response as Figure 1.

(b) 35.0Ω

(c) Across 5.0Ω = 1.3V

Across 10.0Ω = 2.6Ω

Across 20.0Ω = 5.2Ω

<h2>Explanation:</h2>

(a) The labelled circuit using the correct symbols (for the resistors and battery) has been attached to this response.

(b) Since the resistors are hooked up in series, their equivalent resistance R, is found by adding the individual resistances of the resistors (R₁, R₂ and R₃). i.e

R = R₁ + R₂ + R₃ -------------------(i)

Where;

R₁ = 5.0 Ω

R₂ = 10.0 Ω

R₃ = 20.0 Ω

Substitute these values into equation (i) as follows;

∴ R = 5.0 Ω + 10.0 Ω + 20.0 Ω

∴ R = 35.0 Ω

Therefore, the equivalent resistance is ∴ R = 35.0Ω

(c) When resistors are connected in series, the same current passes through them. To get the current through each resistor;

i. First, replace the resistors by their equivalent resistor as calculated above. The diagram has been attached to this response.

ii. As seen in the diagram, the current flowing through the equivalent resistor can be calculated using Ohm's law as follows;

V = I R ------------------(ii)

Where;

V = Voltage supplied to the circuit = 9.0V

I = Current through the circuit

R = Resistance of the equivalent resistor = 35.0Ω

Substitute these values into equation (ii)

9.0 = I x 35.0

I = $\frac{9.0}{35.0}$

I = 0.26A

This is also the current flowing through each of the resistors separately.

iii. Calculate the voltage drop across

1. 5.0 Ω resistor

Applying Ohm's law from equation (ii)

V = I x R

Where;

V = voltage drop across the 5.0Ω resistor

I = current through the 5.0Ω resistor = 0.26A

R = resistance of the 5.0Ω resistor = 5.0Ω

=> V = 0.26 x 5.0

=> V = 1.3V

2. 10.0 Ω resistor

Applying Ohm's law from equation (ii)

V = I x R

Where;

V = voltage drop across the 10.0Ω resistor

I = current through the 10.0Ω resistor = 0.26A

R = resistance of the 10.0Ω resistor = 10.0Ω

=> V = 0.26 x 10.0

=> V = 2.6V

3. 20.0 Ω resistor

Applying Ohm's law from equation (ii)

V = I x R

Where;

V = voltage drop across the 20.0Ω resistor

I = current through the 20.0Ω resistor = 0.26A

R = resistance of the 20.0Ω resistor = 10.0Ω

=> V = 0.26 x 20.0

=> V = 5.2V

7 0

3 years ago

What is defined by force per unit area

sleet_krkn [62]

Explanation:

Force per unit area is called Pressure.

7 0

3 years ago

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Technician A says that low pressure smoke installed in the fuel system can be used to check for leaks. Technicians B says that n

pickupchik [31]

Answer:

A. Technician A only.B.

Explanation: The fuel system of a vehicle is made up of the fuel pump,the fuel filter,the injector or carburettor and the fuel tank. The main function of the fuel system is supply fuel to the engine of a vehicle. In order to check for leakage in the fuel system a small pressure smoke is usually installed in the fuel system.

Nitrogen at low pressure has also been used to check for leakage in the fuel system of a vehicle.

4 0

3 years ago

The phase change in which a substance changes from a gas directly to a solid is

vlabodo [156]

Condensation is the change of the substance from liquid to solid phase. Example of this is the formation of ice. Vaporization is the change of substance from liquid to gas phase. Example of this is the boiling of water. Deposition is the change of a substance from gas to solid phase. Example of this is the formation of ice on a winter day. Sublimation is the change of a substance from solid to gas phase. Example of this is dry ice. The answer is letter C.

7 0

3 years ago

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An uncharged capacitor is connected to a 21-V battery until it is fully charged, after which it is disconnected from the battery

attashe74 [19]

Answer:

The voltage bewtween the plates will be 9.5V

Explanation:

Facts:

The capacitance of a parallel plate capacitor having plate area A and plate separation d is C=ϵ0A/d.

Where ϵ0 is the permittivity of free space.

A capacitor filled with dielectric slab of dielectric constant K, will have a new capacitance C1=ϵ0kA/d

C1=K(ϵ0A/d)

C1=KC ----------- 1

Where C is the capacitance with no dielectric and C1 is the capacitance with dielectric.

The new capacitance is k times the capacitance of the capacitor without dielectric slab.

This implies that the charge storing capacity of a capacitor increases k times that of the capacitor without dielectric slab.

Given points

• The terminal voltage of the battery to which the capacitor is connected to charge V=25V

• A dielectric slab of paraffin of dielectric constant K=2 is inserted in the space between the capacitor plates after the fully charged capacitor is disconnected

The charge stored in the original capacitor Q=CV

The charge stored in the original capacitor after inserting dielectric Q1=C1V1

The law of conservation of energy states that the energy stored is constant:

i.e Charge stored in the original capacitor is same as charge stored after the dielectric is inserted.

Q = Q1

CV = C1V1

 CV = C1V1 -------2

We derived C1=KC from equation 1. Inserting this into equation 2

CV = KCV1

V1 = (CV)/KC

V/K

= 21/2.2

= 9.5

4 0

3 years ago