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Alex73 [517]
3 years ago
4

An 11.98-gram sample of zinc metal is placed in a hot water bath and warmed to 78.4°C. It is then removed and placed into a Styr

ofoam cup containing 50.0 mL of room temperature water (T=27.0°C; density = 1.00 g/mL). The water warms to a temperature of 28.1°C. Determine the specific heat capacity of the zinc.
Physics
1 answer:
Ipatiy [6.2K]3 years ago
0 0

Answer:

383.34 J/kg.K

Explanation:

From the question,

Heat lost by zinc = heat gained by water in the Styrofoam cup.

c'm'(t₁-t₃) = cm(t₃-t₂).............. Equation 1

Where c' = specific heat capacity of zinc, m' = mass of zinc, t₁ = initial temperature of zinc, t₂ = initial temperature of water in the Styrofoam cup, t₃ = temperature of the mixture

Make c' the subject of the equation

c' = cm(t₃-t₂)/m'(t₁-t₃).............. Equation 2

given: m' = 11.98 g = 0.01198 kg, m = density of water×volume of water = (1×50) = 50 g = 0.05 kg, t₁ = 78.4°C, t₂ = 27°C,  t₃ = 28.1°C

Constant: c = 4200 J/kg.K

Substitute these values into equation 2

c' = 0.05×4200(28.1-27)/[0.01198×(78.4-28.1)]

c' = 231/0.602594

c' = 383.34 J/kg.K

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Answer: A, B, and D,

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Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the su
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Complete Question

Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the surface. Find the electric field at the following locations, with radially outward defined as the positive direction and radially inward defined as the negative direction. The permittivity of free space ????0 is 8.85×10−12 C/(V⋅m). What is the electric field

E⃗ 1 inside the cell at a distance of 3.05 μm from the center?

E⃗ 2 Just inside the surface of the cell

E⃗ 3 Just outside the surface of the cell

E⃗ 4 At a point outside the cell 3.05 μm from the surface

Answer:

E⃗ 1

      0 V/m

E⃗ 2

      0 V/m

E⃗ 3

         E_3 =  2.153 *10^{9} \  V/m

E⃗ 4

E_4 =  5.754 *10^ {8} \  V/m

Explanation:

From the question we are told that

The diameter is d =  6.53 \mu m  = 6.53*10^{-6}\  m

The charge is Q =  -.2.55 *10^{-12} \  C

The permittivity of free space is \epsilon_o  =  8.85* 10^{-12}\  C / V.m

The distance considered is d =  3.05 \mu m  =  3.05 *10^{-6} \ m

Generally the electric field inside the cell at a distance of 3.05 μm from the center is

0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just inside the surface of the cell is 0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just outside the cell is mathematically represented as

E_3 =  \frac{ k  *  |Q|}{ r^2 }

Here k is the coulomb constant with value

k  =   9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}

r is the radius of the sphere which is mathematically as

r =  \frac{d}{2} =   \frac{6.53*10^{-6}}{2}  = 3.265 *10^{-6} \  m

E_3 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [3.265 *10^{-6} ]^2 }

E_3 =  2.153 *10^{9} \  V/m

Generally the electric field at a point outside the cell 3.05 μm from the surface is mathematically represented as

E_4 =  \frac{ k  *  |Q|}{ R^2 }

Here R is mathematically represented as

R  =  3.265 *10^{-6} +  3.05 *10^{-6}

=>       R  =  6.315 *10^{-6}

So

E_4 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [ 6.315 *10^{-6} ]^2 }

E_4 =  5.754 *10^ {8} \  V/m

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The largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.

The given parameters;

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  • <em>combined mass of the two black holes = 5.20 x 10³⁰ kg</em>

The product of the two masses is calculated from Newton's law of universal gravitational as follows;

F = \frac{Gm_1m_2}{r^2} \\\\m_1m_2 = \frac{Fr^2}{G} \\\\m_1m_2 = \frac{(6.9\times 10^{25}) \times (1.5\times 10^{12})^2}{6.67\times 10^{-11}} \\\\m_1m_2 = 2.328 \times 10^{60} \ kg^2

The sum of the two masses is given as;

m₁ + m₂ = 5.2 x 10³⁰ kg

m₂ = 5.2 x 10³⁰ kg - m₁

The first mass is calculated as follows;

m₁(5.2 x 10³⁰ - m₁) = 2.328 x 10⁶⁰

5.2 x 10³⁰m₁ - m₁² = 2.328 x 10⁶⁰

m₁² - 5.2 x 10³⁰m₁  + 2.328 x 10⁶⁰ = 0

<em>solve the quadratic equation using formula method</em>;

a = 1, b =-  5.2 x 10³⁰, c = 2.328 x 10⁶⁰

m_1 = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\m_1 = \frac{-(-5.2\times 10^{20})  \ \ +/- \ \ \sqrt{(-5.2\times 10^{20})^2 - 4(1\times 2.328\times 10^{60})} }{2(1)} \\\\m_1 = 4.7 \times 10^{30} \ kg \ \ or \ \ 4.9 \times 10^{29} \ kg

The second mass is calculated as follows;

m₂ = 5.2 x 10³⁰ kg - m₁

m₂ = 5.2 x 10³⁰ kg  -  4.7 x 10³⁰ kg

m₂ = 5 x 10²⁹ kg

or

m₂ = 5.2 x 10³⁰ kg  -  4.9 x 10²⁹ kg

m₂ = 4.7 x 10³⁰ kg

Thus, the largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.

Learn more here:brainly.com/question/9373839

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