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3 years ago

What are three factors that move the blood back to the heart

1 answer:

3 0

The arteries move blood away from the heart so therefore, the veins retract the blood throughout the body which eventually goes back to the heart and repeats itself nonstop

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After which action would the concentration of a solution remain constant?(1 point)

Alika [10]

Answer:

C. Evaporating water from the container.

Explanation:

The concentration of solution changes when solvent or solute are added/removed from a solution.

5 0

3 years ago

Two boats leave the same port at the same time, with boat A traveling north at 15 knots (nautical miles per hour) and boat B tra

Mrrafil [7]

Answer:

The chance in distance is 25 knots

Explanation:

The distance between the two particles is given by:

$s^2 = (x_A - x_B)^2+(y_A - y_B)^2$ (1)

Since A is traveling north and B is traveling east we can say that their displacement vector are perpendicular and therefore (1) transformed as:

$s^2 = x_B^2+y_A^2$ (2)

Taking the differential with respect to time:

$\displaystyle{2s\frac{ds}{dt}= 2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt}}$ (3)

where $\displaystyle{\frac{dx_B}{dt}}=v_B$ and $\displaystyle{\frac{dx_A}{dt}}=v_A$ are the respective given velocities of the boats. To find $s$ and $x_B$ we make use of the given position for A, $y_A=30$ , the Pythagoras theorem and the relation between distance and velocity for a movement with constant velocity.

$\displaystyle{y_A = v_A\cdot t\rightarrow t = \frac{y_A}{v_A}=\frac{30}{15}=2 h$

with this time, we know can now calculate the distance at which B is:

$\displaystyle{x_B = v_B\cdot t= 20 \cdot 2 = 40\ nmi$

and applying Pythagoras:

$\displaystyle{s = \sqrt{x_B^2+y_A^2}=\sqrt{30^2 + 40^2}=\sqrt{2500}=50}$

Now substituting all the values in (3) and solving for $\displaystyle{\frac{ds}{dt} }$ we get:

$\displaystyle{\frac{ds}{dt} = \frac{1}{2s}(2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt})}\\\displaystyle{\frac{ds}{dt} = 25 \ knots}$

4 0

3 years ago

Underground water is being pumped into a pool whose cross section is 3 m x 4 m while water is discharged through a 0.076m-diamet

Svetllana [295]

Given:

Area of pool = 3m×4m
Diameter of orifice = 0.076m
Outlet Velocity = 6.3m/s
Accumulation velocity = 1.5cm/min

Required:

Inlet flowrate

Solution:

The problem can be solved by this general formula.

Accumulation = Inlet flowrate - Outlet flowrate
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice

First, we need to convert the units of the accumulation velocity into m/s to be consistent.

Accumulation velocity = 1.5cm/min × (1min/60s)×(1m/100cm)
Accumulation velocity = 0.00025 m/s

We then calculate the area of the pool and the area of the orifice by:

Area of pool = 3 × 4 m²
Area of pool = 12m²

Area of orifice = πd²/4 = π(0.076m)²/4
Area of orifice = 0.00454m²

Since we have all we need, we plug in the values to the general equation earlier

Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice

0.00025 m/s × 12m² = Inlet flowrate - 6.3m/s × 0.00454m²

Transposing terms,

Inlet flowrate = 0.316 m³/s

6 0

3 years ago

Crying is a .........of distress to babies.

RSB [31]

Hey there

Crying is a signal <span>of distress to babies.</span>

5 0

3 years ago

The study of the interaction of electrical and magnetic fields, and their interaction with matter, is called _____.

Dafna11 [192]

It is called Electromagnetism

8 0

3 years ago