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2 months ago

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A block spring system oscillates on a frictionless surface with an amplitude of 10\text{ cm}10 cm and has an energy of 2.5 \text

{ J}2.5 J. If the block is replaced by a block with twice the mass of the original block and the amplitude of the motion is 6 \text{ cm}6 cm, what is the energy of the system.

1 answer:

ValentinkaMS [3.4K]2 months ago

8 0

Answer:

The system's energy amounts to 15 J.

Explanation:

Given that,

Energy E = 2.5 J

Amplitude = 10 cm

The spring constant needs to be computed.

Using the formula for the mechanical energy of the system,

$E=\dfrac{1}{2}kA^2$

Substituting the values into the formula:

$2.5=\dfrac{1}{2}k\times(10\times10^{-2})^2$

$k=\dfrac{2.5\times2}{(10\times10^{-2})^2}$

$k=500\ N/m$

If the block is swapped out for one with double the original mass,

Amplitude = 6 cm

We need to find the energy

Using the mechanical energy formula,

$E=\dfrac{1}{2}kA^2$

Substituting into the formula:

$E=\dfrac{1}{2}\times500\times(6\times10^{-2})$

$E=15\ J$

Thus, the system’s energy is 15 J.

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Jeff's body contains about 5.46 L of blood that has a density of 1060 kg/m3. Approximately 45.0% (by mass) of the blood is cells

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Answer:

a) Blood mass is $m= 5.7876kg$

b) The count of blood cells is $N_t=1.04*10^{13}$

Explanation:

From the problem statement, we learn that

The blood volume is $V_b = 5.46 \ L = \frac{5.46}{1000} = 0.00546m^3$

The density of blood is $\rho_b = 1060 kg/m^3$

% of blood which consists of cells is = 45.0%

the % of blood that is plasma is = 55.0%

density of blood cells is $\rho_d = 1125kg/m^3$

% of cells that are white is = 1%

% of cells that are red is = 99%

The red blood cell diameter is = $7.5 \mu m = 7.5*10^{-6}m$

The red blood cell radius is $= \frac{7.5*10^{-6}}{2} = 3.75*10^{-6}m$

The mass is generally represented mathematically as

$m = \rho_b * V_b$

Substituting values

$m = 1060 * 0.00546$

$m= 5.7876kg$

Cell mass is $m_c$ = 45% of m

= 0.45 * 5,7876

= 2.60442 kg

The volume of cells is $V_c$ = $\frac{m_c}{\rho_d}$

$= \frac{2.60442}{1125}$

$= 2.315 *10^{-3} m^3$

The white blood cells volume is $V_w$ = 1% of the cells volume

$= \frac{1}{100} * 2.315*10^{-3}$

$= 2.315*10^{-5}m^3$

The volume of a single cell is $V_s = 4 \pi r^3$

$= 4*(3.142) * (3.75*10^{-6})^3$

$= 2.21*10^{-16}m^3$

The red blood cells volume is $V_r = V_c - V_w$

$=2.315*10^{-3} - 2.315*10^{-5}$

$= 2.29*10^{-3}m^3$

The total red blood cell count is $= \frac{V_r}{V_s}$

$= \frac{2.29 *10^{-3}}{2.21*10^{-16}}$

$= 1.037*10^{13}$

The total white blood cell count is $=\frac{V_w}{V_s}$

$= \frac{2.315 * 10^{-5}}{2.21*10^{-16}}$

$= 1.04*10^{11}$

The overall number of blood cells is $N_t= 1.037*10^{13} + 1.04*10^{11}$

$N_t=1.04*10^{13}$

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