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4 years ago

If 1 foot is 30.28 centimeters.how many cm is 130 feet

Physics

1 answer:

uranmaximum [27]4 years ago

6 0

Answer:3,936.4

Explanation:

If 1 foot is 30.28 centimeters multiply 30.28 by 130

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Protons and neutrons both have positive charge, and electrons have negative charge. true or false

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Explanation:

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3 years ago

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3 years ago

An electric heater used to boil small amounts of water consists of a 15-Ω coil that is immersed directly in the water. It operat

Allisa [31]

Answer:

t = 1120 seconds

Explanation:

Law of Conservation of Energy: It states that energy can neither be created or destroyed but it can be transformed from one form to another

The heater converts electric energy to heat energy.

According to the law of conservation of Energy,

Heat supplied by the heater = heat gained by the water.

Heat supplied by the heater = (V²/R)t, Where V = voltage, R = Resistance, t = Time.

Heat gained by the water = cm(θ₂ -θ₁)

Where c = specific heat capacity of water, m = mass of water, θ₂= final temperature of water, θ₁= initial temperature of water.

∴ (V²/R)t = cm(θ₂ -θ₁)

Making t the subject of the equation,

t = cm(θ₂ -θ₁) ×R/V²

Where c= 4200 Jkg/K, m=0.80 kg, R = 15 Ω, V = 60 V, θ₁= 20°C, θ₂ = the normal boiling point of water = 100°C

∴ t = (4200×0.8)(100 -20)×15/60²

t = (3360×80×15)/3600

t = 4032000/3600 = 1120 seconds

∴ The time required by the heater to raise the temperature of water to normal boiling point is 1120 seconds.

3 0

3 years ago

You need to design a 60.0-Hz ac generator that has a maximum emf of 5200 V. The generator is to contain a 180-turn coil that has

OlgaM077 [116]

Answer:

0.082 T

Explanation:

Given:

Frequency of the generator (f) = 60.0 Hz

Maximum emf of the generator (E) = 5200 V

Number of turns (N) = 180

Area per turn (A) = 0.94 m²

Magnetic field magnitude (B) = ?

We know that, the angular frequency of the generator is given as:

$\omega=2\pi f$

Plug in the value of 'f' and solve for 'ω'. This gives,

$\omega=2\pi\times 60.0=120\pi\ rad/s$

Now, the maximum emf of the generator is given by the formula:

$E=NAB\omega$

Rewriting in terms of magnetic field, 'B', we get:

$B=\frac{E}{NA \omega}$

Plug in the given values and solve for 'B'. This gives,

$B=\frac{5200\ V}{180\times 0.94\ m^2\times 120\pi\ rad/s}\\\\B=\frac{5200\ V}{63786.897}\\\\B=0.082\ T$

Therefore, the magnitude of the magnetic field in which the coil rotates is 0.082 T.

8 0

4 years ago