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3 years ago

Blood has a density of 1.06 She donates 473.176 mL how much mass does she donate

Physics

2 answers:

Tcecarenko [31]3 years ago

7 0

Answer:

the answer is 501.57

Explanation:

Degger [83]3 years ago

3 0

Density = mass / volume

So..

1.06 = mass / 473.176

Multiply 473.176 over to the other side leading to

(1.06)(473.176) = mass

mass = 501.56656 --> 501.57

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Which of the following statements is NOT a correct assumption of the classicalmodel of an ideal gas?A. The molecules are in rand

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3 years ago

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A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate

Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is $\bf{2.81 \times 10^{4}~n~C^{-1}}$ .

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and ' $\epsilon_{0}$ ' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

$C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

$C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF$

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

$&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm$

Part(c):

If we apply Gauss' law of electrostatics, then

$&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}$

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3 years ago

1. What causes velocity to change?

tresset_1 [31]

The amount of force an object has will change the velocity

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3 years ago

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A 43.9-g piece of copper (CCu= 0.385 J/g°C) at 135.0°C is plunged into 254 g of water at 39.0°C. Assuming that no heat is lost t

Semmy [17]

Answer:

$T = 40.501\,^{\textdegree}C$

Explanation:

The interaction of the piece of copper and water means that the first one need to transfer heat in order to reach a thermal equilibrium with water. Then:

$-Q_{out,Cu} = Q_{in,H_{2}O}$

After a quick substitution, the expanded expression is:

$-(43.9\,g)\cdot (0.385\,\frac{J}{g\cdot ^{\textdegree}C} )\cdot (T-135^{\textdegree}C) = (254\,g)\cdot (4.187\,\frac{J}{g\cdot ^{\textdegree}C} )\cdot (T-39\,^{\textdegree}C)$

$-16.902\,\frac{J}{^{\textdegree}C}\cdot (T-135^{\textdegree}C) = 1063.498\,\frac{J}{^{\textdegree}C} \cdot (T-39^{\textdegree}C)$

$43758,192\,J = 1080.4\,\frac{J}{^{\textdegree}C}\cdot T$

The final temperature of the system is:

$T = 40.501\,^{\textdegree}C$

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3 years ago

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