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3 years ago

The enthalpy of fusion of ice is 330,000 J/kg. If I have a 1 kg sample, how much energy is needed to melt all of it?

Physics

2 answers:

EleoNora [17]3 years ago

8 0

Answer:

C. 330,000 J

Explanation:

Latent heat = mass × specific latent heat

q = mL

Given m = 1 kg and L = 330,000 J/kg:

q = (1 kg) (330,000 J/kg)

q = 330,000 J

Anestetic [448]3 years ago

6 0

Answer:

C. 330,000 J

Explanation:

The enthalpy of fusion of ice is 330,000 J/kg. If someone has a 1 kg sample, they will need 330,000 J energy in order to melt all of it.

(1 kg) (330,000 J/kg)

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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

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Answer:

a) v = 2,9992 10⁸ m / s , b) Eo = 375 V / m , B = 1.25 10⁻⁶ T,

c) λ = 3,157 10⁻⁷ m, f = 9.50 10¹⁴ Hz , T = 1.05 10⁻¹⁵ s , UV

Explanation:

In this problem they give us the equation of the traveling wave

E = 375 cos [1.99 10⁷ x + 5.97 10¹⁵ t]

a) what the wave velocity

all waves must meet

v = λ f

In this case, because of an electromagnetic wave, the speed must be the speed of light.

k = 2π / λ

λ = 2π / k

λ = 2π / 1.99 10⁷

λ = 3,157 10⁻⁷ m

w = 2π f

f = w / 2 π

f = 5.97 10¹⁵ / 2π

f = 9.50 10¹⁴ Hz

the wave speed is

v = 3,157 10⁻⁷ 9.50 10¹⁴

v = 2,9992 10⁸ m / s

b) The electric field is

Eo = 375 V / m

to find the magnetic field we use

E / B = c

B = E / c

B = 375 / 2,9992 10⁸

B = 1.25 10⁻⁶ T

c) The period is

T = 1 / f

T = 1 / 9.50 10¹⁴

T = 1.05 10⁻¹⁵ s

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λ = 3,157 10-7 m (109 nm / 1m) = 315.7 nm

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3 years ago

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Answer:

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Answer:

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2 years ago

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Answer:

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A = π*r^2

= π*(0.0021)^2 = 13.85*10^-6 m^2

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n= 8.46*10^28

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