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yawa3891 [41]
3 years ago
13

There is a close relationship between magnetic forces and the generation of electric current. Explain how magnets can be used to

generate electric current and how electric current can be used to create electromagnets. For each, give an example of a device you would find around the home.
Physics
1 answer:
Snowcat [4.5K]3 years ago
3 0
Inside a generator is a magnet, some electrical wire, and a source of mechanical energy. The mechanical energy moves the wire into the magnetic field of the magnet so that the wire cuts through the magnetic lines of force. As a result, electric current is produced. Electric generators can come in all sizes
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Which competition is the “ most important soccer competition in the world “ ?
Ganezh [65]
It would be B. The World Cup

Hope this helps

Have a great day/night
8 0
3 years ago
his is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Air enters a nozzle steadi
zepelin [54]

Answer:

his is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Air enters a nozzle steadily at 2.21 kg/m3 and 40 m/s and leaves at 0.762 kg/m3 and 192 m/s. The inlet area of the nozzle is 90 cm2.

determine (a) the mass flow rate through the nozzle, and (b) the exit area of the nozzle.

a)0.7956kg/s

b)5.437 × 10⁻³m²

Explanation:

The concepts related to the change of mass flow for both entry and exit is applied

The general formula is defined by

\dot{m}=\rho A V

Where,

\dot{m} = mass flow rate\\\rho = Density\\V = Velocity

values are divided by inlet(1) and outlet(2) by

\rho_1 = 2.21kg/m^3V_1 = 40m/s

A_1 = 90*10^{-4}m^2\\\rho_2 = 0.762kg/m^3\\V_2 = 192m/s

PART A) Applying the flow equation

\dot{m} = \rho_1 A_1 V_1\\\dot{m} = (2.21)(90*10^{-4})(40)\\\dot{m} = 0.7956kg/s

PART B) For the exit area we need to arrange the equation in function of Area, that is

A_2 = \frac{\dot{m}}{\rho_2 V_2}\\A_2 = \frac{0.7956}{(0.762)(192)}\\A_2 = 5.437*10^{-3}m^2

7 0
3 years ago
Which type of wave is classified as electromagnetic?
Nimfa-mama [501]

Answer: B) Light

Explanation: light waves are Electromagnetic waves. Visible light is one of the many types of electromagnetic waves. The others are mechanical waves.

8 0
3 years ago
Read 2 more answers
Topics related to physics​
Bezzdna [24]

Answer:

There are many topics related to physics such as :

Kinematics

Dynamics

Light

Sound

7 0
2 years ago
Convert: Thermal conductivity value of 0.3 Btu/(h ft°F) to W/(m °C). Surface heat transfer coefficient value of 105 Btu/(h ft2 o
Finger [1]

Answer:

0.3 Btu/(h ft °F) = 0.5189 W/(m°C)

105 Btu/(h ft² °F) = 596.2215 W/(m²°C)

Explanation:

<u>Thermal conductivity of a substance is defined as the measure of the tendency of the substance to conduct heat.</u>

<u>The SI unit of thermal conductivity is W.m⁻¹K⁻¹ .</u>

From the question , 0.3 Btu/(h ft °F) is to be converted to W/(m°C)

Thus,

1 Btu/(h ft °F) = 1.7296 W/(m°C)

So,

0.3 Btu/(h ft °F) = 1.7296×0.3 W/(m°C) = 0.5189 W/(m°C)

Thus,

<u>0.3 Btu/(h ft °F) = 0.5189 W/(m°C)</u>

<u>Heat transfer coefficient is defined as proportionality constant between heat flux (Thermal power per unit area) and the temperature difference of the substance for that flow of heat.</u>

<u>The SI unit of thermal conductivity is W.m⁻²K⁻¹ .</u>

From the question , 105 Btu/(h ft² °F) is to be converted to W/(m²°C)

Thus,

1 Btu/(h ft² °F) = 5.6783 W/(m²°C)

So,

105 Btu/(h ft² °F) = 5.6783×105 W/(m²°C) = 596.2215 W/(m²°C)

Thus,

<u>105 Btu/(h ft² °F) = 596.2215 W/(m²°C)</u>

3 0
3 years ago
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