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sdas [7]
3 years ago
8

Finderkeepers,ismadeormodel

Physics
1 answer:
ankoles [38]3 years ago
7 0

Answer:

model lol

Explanation:

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15 g of lead (specific heat = 0.128 J/g Celsius) at 120 Degrees Celsius is placed on a glacier of ice at 0 degrees Celsius. How
Anastaziya [24]

This can be solve using heat balance. The amount of heat the lead released is equal to the amount of heat the ice absorb.

The heat release by the leas = m1*Cp*deltaT

Where m1 is the mass of lead

Cp is the specific heat of lead

Delta is the change of temperature

Heat absorbed by the ice = m2Lf

Where m2 is the mass of ice

Lf is the latent heat of fusion of ice

 

M1*Cp*deltaT = m2Lf

(15g) ( 0.128 j/g C) ( 120 -0 C) = m2(333/g)

Solving for m2

<span>M2 = 0.69 kg of ice melts</span>

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3 years ago
Two people are talking at a distance of 3.0 m from where you are, and
tigry1 [53]

Answer:

The sound intensity that the other student measure, I₂ is expected to be;

A) 6.2 × 10⁻⁸ W/m²

Explanation:

The measure of sound intensity is given by the following formula;

I = \dfrac{P}{4 \cdot \pi \cdot R^2}

\dfrac{I_2}{I_1} = \dfrac{R_1^2}{R_2^2}

Where;

I = The intensity

R = The radius

P = The power of the sound

Whereby we have;

The distance of the two people talking, R₁ = 3.0 m

The measure of the sound intensity, I₁ = 1.1 × 10⁻⁷ W/m² (from an online source)

The distance of the other student from the two people talking, R₂ = 4.0 m

Therefore, the estimate of the sound intensity, I₂, is given as follows;

{I_2} = \dfrac{R_1^2}{R_2^2} \times {I_1}

{I_2} = \dfrac{(3.0 \, m)^2}{(4.0 \, m)^2} \times 1.1 \times 10^{-7} \ W/m^2 = 6.1875 \times 10^{-8} \ W/m^2

I₂ = 6.1875 × 10⁻⁸ W/m²

∴ The sound intensity that the other student measure, I₂ ≈ 6.2 × 10⁻⁸ W/m²

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