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mr_godi [17]
3 years ago
10

A heater gives off heat at a rate of 330 kj/min. what is the rate of heat output in kilocalories per hour? (1 cal 4.184 j)

Physics
1 answer:
Cerrena [4.2K]3 years ago
4 0
Answer: 4732.3135 kcal/hour

To answer this question you need to convert the kj into kcalorie then convert the min into hour. Becareful because 1 kilocalories = 1000 calories = 4.184 kjoule.
The calculation would be: 
330kj/min x (1kilocalorie/4.184kj) x 60 min/hour= 4732.3135 kcal/hour
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The electric field of a sinusoidal electromagnetic wave obeys the equation E=(375V/m)cos[(1.99×107rad/m)x+(5.97×1015rad/s)t].a.
andrey2020 [161]

Answer:

A) max = Eo = 375 V/m, B)  B = 125 10⁻⁸ T , C) f = 9.50 101⁴ Hz, D) λ = 3.158 10⁻⁷ m, E)  T = 1.05 10⁻¹⁵ s , F) invisible for humans   G)  v = c = 3 10⁸ m/s

Explanation:

The expression given for the electric field is

       E = 375 cos (1.99 107x + 5.97 105t)

The general formula for the electric field of a transverse traveler is

     E = Eo cos (kx-wt)

Where k is the wave number and w the angular velocity

A) The amplitudes as electric is

     Emax = Eo = 375 V / m

B) the electric and magnetic field are related

     E / B = c

     B = E / c

     B = 375/3 108

     B = 125 10⁻⁸ T

C) angular velocity and frequency is related

    .w = 2π f

     f = w / 2π

     f = 5.97 10¹⁵ / 2π

     f = 9.50 101⁴ Hz

D) the speed of light has the formula

      c = λ f

      λ = c / f

      λ = 3 10⁸ / 9.50 10¹⁴

      λ = 3.158 10⁻⁷ m

E) The period

     T = 1 / f

     T = 1 / 9.5 10¹⁴

     T = 1.05 10⁻¹⁵ s

F) let's reduce the wavelength nm

   λ = 3.158 10⁻⁷ m (10⁹nm / 1m)

   λ = 3.158 10²nm = 315.3 nm

The visible radiation range is between 400nm and 700nm. This radiation is ultraviolet and is invisible humans

G) All electromagnetic radiation has a speed at the speed of light (c)

      v = c = 3 10⁸ m/s

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3 years ago
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2 J / 4 C = 1/2 joule per coulomb.

That's 1/2 volt.
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Answer:

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