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3 years ago

An unknown material has a mass

Physics

2 answers:

atroni [7]3 years ago

4 0

Answer: 1896.55J/kg°C

Explanation:

The quantity of Heat Energy (Q) required to heat a material depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since,

Q = 1320 joules

Mass of material = 5.61kg

C = ? (let unknown value be Z)

Φ = 0.124°C

Then, Q = MCΦ

1320J = 5.61kg x Z x 0.124°C

1320J = 0.696kg°C x Z

Z = (1320J / 0.696kg°C)

Z = 1896.55 J/kg°C

Thus, the specific heat of the material is 1896.55J/kg°C

torisob [31]3 years ago

3 0

Answer:

1897.53J/Kg°C

Explanation:

Data obtained from the question include:

M (mass of material) = 5.61 kg

ΔT ( change in temperature) = 0.124°C

Q ( heat) = 1320 J

C (specific heat capacity of the material) =?

Using the formula Q = MCΔT, the specific heat capacity of the material can be obtained as follow:

Q = MCΔT

C = Q / MΔT

C = 1320J/ (5.61 kg x 0.124°C)

C = 1897.53J/Kg°C

Therefore, the specific heat capacity of the material is 1897.53J/Kg°C

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Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.

elena-s [515]

Answer:

$g=13.42\frac{m}{s^2}$

Explanation:

1) Notation and info given

$\rho_{center}=13000 \frac{kg}{m^3}$ represent the density at the center of the planet

$\rho_{surface}=2100 \frac{kg}{m^3}$ represent the densisty at the surface of the planet

r represent the radius

$r_{earth}=6.371x10^{6}m$ represent the radius of the Earth

2) Solution to the problem

So we can use a model to describe the density as function of the radius

$r=0, \rho(0)=\rho_{center}=13000 \frac{kg}{m^3}$

$r=6.371x10^{6}m, \rho(6.371x10^{6}m)=\rho_{surface}=2100 \frac{kg}{m^3}$

So we can create a linear model in the for y=b+mx, where the intercept b= $\rho_{center}=13000 \frac{kg}{m^3}$ and the slope would be given by $m=\frac{y_2-y_1}{x_2-x_1}=\frac{\rho_{surface}-\rho_{center}}{r_{earth}-0}$

So then our linear model would be

$\rho (r)=\rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r$

Since the goal for the problem is find the gravitational acceleration we need to begin finding the total mass of the planet, and for this we can use a finite element and spherical coordinates. The volume for the differential element would be $dV=r^2 sin\theta d\phi d\theta dr$ .