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14 days ago

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Water, of density 1000 kg/m3, is flowing in a drainage channel of rectangular cross-section. The width of the channel is 15 m, t

he depth of the water is 8.0 m and the speed of the flow is 2.5 m/s. At what rate is water flowing in this channel?

1 answer:

ValentinkaMS [2.4K]14 days ago

8 0

Answer:

The flow rate of water is (300000kg/s) = (300000l/s)

Explanation:

To compute the volume of moving fluid per second in the channel, we consider the channel's section, the water depth, and the fluid velocity:

Volume flow rate = 15m × 8m × (2.5m/s) = 300 m³/s

To find the mass or liters of water flowing per second, multiply the volume of circulating fluid by the water's density:

Flow rate of water = (300m³/s) × (1000kg/m³) = (300000kg/s) = (300000l/s)

It is important to note that 1kg of water is approximately equivalent to 1 liter.

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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

ValentinkaMS [2425]

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 $N_t= 1.037*10^{13} + 1.04*10^{11}$

$N_t=1.04*10^{13}$

6 0

19 hours ago

Read 2 more answers

State two advantages of a lead-acid accumulator over a leclanche cell

kicyunya [2264]

Respuesta:

Se pueden recargar.

Poseen una vida útil mayor.

Son reutilizables.

Explicación:

Las baterías de plomo-ácido y las pilas Leclanché son tipos de células electroquímicas.

Las pilas Leclanché son células primarias. Estas células producen reacciones químicas que generan corriente eléctrica de manera irreversible.

Por su parte, las baterías de plomo-ácido son células secundarias que permiten la reversibilidad de la corriente eléctrica generada.

Esto hace que las baterías de plomo-ácido sean reutilizables, con mayor durabilidad y un tiempo de vida más prolongado.

5 0

1 month ago

A closed system of mass 10 kg undergoes a process during which there is energy transfer by work from the system of 0.147 kJ per

Softa [2035]

Result: -50.005 kJ

Details:

Provided Data

mass of the system = 10 kg

work done = 0.147 kJ/kg

Elevation change $(\Delta h)=-50 m$

initial speed $(v_1)=15 m/s$

Final Speed $(v_2)=30 m/s$

Specific internal Energy $(\Delta U)=-5 kJ/kg$

according to the first Law of thermodynamics

$Q-W=\Delta H$

$\Delta H=\Delta KE+ \Delta PE +\Delta U$

where KE represents kinetic energy

PE indicates potential energy

U denotes internal Energy

$\Delta H=\frac{m(v_2^2-v_1^2)}{2}+mg(\Delta h)+\Delta U$

$Q=W+\Delta KE+ \Delta PE +\Delta U$

$Q=0.147\times 10+\frac{10\cdot (30^2-15^2)}{2}+10\cdot 9.81\cdot (-5)-5\times 10$

Q = 1.47 + 3.375 - 4.850 - 50

Q = -50.005 kJ

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1 month ago

While looking at bromine (Br) on the periodic table, a student needs to find another element with very similar chemical properti

Sav [2230]

Response: Numerous elements can be found, all situated within the same vertical column as bromine.

Explanation:

Elements are organized by their atomic numbers on the periodic table. Those in the same vertical column (known as groups) exhibit the same valence electron configurations, resulting in similar chemical characteristics. Consequently, there are numerous elements sharing analogous chemical properties grouped with Bromine.

7 0

1 month ago

Read 2 more answers

An archer fires an arrow, which produces a muffled "thwok" as it hits a target. If the archer hears the "thwok" exactly 1 s afte

Ostrovityanka [2208]

Answer:

35.79 meters

Explanation:

We have an archer, and there is a target. Denote the distance between them as d.

The bowman releases the arrow, which travels the distance d at a velocity of 40 m/s until it hits the target. We establish the equation as:

$v_{arrow} * t_{arrow} = d\\ \\40 \frac{m}{s} * t_{arrow} = d$

Right after this, the arrow produces a muffled noise, traveling the same distance d at a speed of 340 m/s in time $t_{sound}$ . Thus, we can derive:

$v_{sound} * t_{sound} = d\\ \\340 \frac{m}{s} * t_{sound} = d$ .

Consequently, the sound reaches the archer, precisely 1 second post-firing the bow, resulting in:

$t_{arrow} + t _{sound} = 1 s$ .

Using this relationship in the distance formula for sound allows us to write:

$340 \frac{m}{s} * t_{sound} = d \\ \\ 340 \frac{m}{s} * (1 s- t_{arrow}) = d$ .

Substituting the value of d from the first equation yields:

$40 \frac{m}{s} * t_{arrow} = d \\ 40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * (1 s- t_{arrow})$ .

Now, after some calculations, we can proceed further:

$40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * 1 s - 340 \frac{m}{s} * t_{arrow} \\ \\ 40 \frac{m}{s} * t_{arrow} + 340 \frac{m}{s} * t_{arrow} = 340 m \\ \\ 380 \frac{m}{s} * t_{arrow} = 340 m \\ \\ t_{arrow} = \frac{340 m}{380 \frac{m}{s}} \\ \\ t_{arrow} = 0.8947 s$ .

Finally, the value is inserted into the initial equation:

$40 \frac{m}{s} * t_{arrow} = d$

$40 \frac{m}{s} * 340/380 s = 35,79 s = d$

6 0

1 day ago