3. What will happen to the black bass and blue gill as the floor of the ponds fills with organic

What is the abbreviation for mojave desert?

vagabundo [1.1K]

There is no abbreviation

3 0

3 years ago

Two buses are driving along parallel freeways that are 5mi apart, one heading east and the other heading west. Assuming that eac

Oksanka [162]

Answer:

101.54m/h

Explanation:

Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;

Let l be the be the distance further away at which they will meet from the current points;

$l=\sqrt{13^2-5^2}=12m\\\\\frac{dl}{dt}=-(55m/h+55m/h})\\\\=-110m/h$ #The speed toward each other.

$\frac{dh}{dt}=0, \ \ \ \ h=constant\\\\h^2+l^2=b^2\\\\2h\frac{dh}{dt}+2l\frac{dl}{dt}=2b\frac{db}{dt}\\\\2\times5\times0+2\times12\times(-110)=2\times13\frac{db}{dt}\\\\\frac{db}{dt}=-101.54m/h$

Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h

4 0

2 years ago

A water wheel rotates with a period of 2.26 s. If the water wheel has a radius of l.94 m,

Dafna11 [192]

Answer:

The correct answer is B)

Explanation:

When a wheel rotates without sliding, the straight-line distance covered by the wheel's center-of-mass is exactly equal to the rotational distance covered by a point on the edge of the wheel. So given that the distances and times are same, the translational speed of the center of the wheel amounts to or becomes the same as the rotational speed of a point on the edge of the wheel.

The formula for calculating the velocity of a point on the edge of the wheel is given as

$V_{r}$ = 2π r / T

Where

π is Pi which mathematically is approximately 3.14159

T is period of time

Vr is Velocity of the point on the edge of the wheel

The answer is left in Meters/Seconds so we will work with our information as is given in the question.

Vr = (2 x 3.14159 x 1.94m)/2.26

Vr = 12.1893692/2.26

Vr = 5.39352619469

Which is approximately 5.39

Cheers!

7 0

2 years ago

A runner drank a lot of water during a race. What is the expected path of the extra filtered water molecules?

Naddika [18.5K]

Answer:

Afferent arteriole, glomerulus, nephron tubule, collecting duct

Explanation:

Blood enters the kidney through the renal artery, a thick branch from the descending aorta. In the hilum, it is divided into several branches that are distributed through the lobes of the kidney and are branching forming numerous afferent arterioles that form the glomerular clew. It is precisely the walls of these capillaries that act as ultrafilters, allowing small particles to pass through.

Blood that flows through the afferent arteriole circulates through the capillary vessels of the kidney (the true capillaries that provide the kidney with oxygen and nutrients necessary for its function). These capillaries are grouped together to form the renal vein which, in turn, pours into the inferior vena cava.

Given the function of the kidneys to eliminate waste products through urine, it is not surprising that these organs are the ones that receive the most blood per gram of weight. One way to express renal blood flow is by considering the renal fraction or fraction of cardiac output that passes through the kidneys.

The regulation of blood flow in the glomeruli is achieved by three formations: the polar bearing, the Goormaghtigh cells and the dense macula. The polar bearing consists of a thickening of the afferent arteriole wall before it enters the renal glomerulus. The arteriole loses its elastic membrane, the endothelium becomes discontinuous and the middle tunic is arranged in two layers, formed by secretory cells: these secretory cells produce Angiotensin and Erythropoietin.

Goormaghtigh cells are arranged at an angle between afferent and effector arterioles and meet in small columns. They are closely related to polar bearing cells. Between both formations is the dense macula (or Zimmerman's dense macula) that is in contact with the distal tubule and afferent arteriole just before it penetrates the glomerulus. These three formations, polar bearing, Goormaghtigh cells and dense macula form the juxtaglomerular apparatus that regulates the blood flow in the glomerulus.

Nephrons regulate water and soluble matter (especially Electrolytes) in the body, by first filtering the blood under pressure, and then reabsorbing some necessary fluid and molecules back into the blood while secreting other unnecessary molecules.

The reabsorption and secretion are achieved with the mechanisms of Cotransporte and Contratransporte established in the nephrons and associated collection ducts. Blood filtration occurs in the glomerulus, a capping of capillaries that is inside a Bowman's capsule.

Liquid flows from the nephron in the collecting duct system. This segment of the nephron is crucial to the process of water conservation by the body. In the presence of the antidiuretic hormone (ADH; also called vasopressin), these ducts become water permeable and facilitate their reabsorption, thus concentrating the urine and reducing its volume. Conversely, when the body must remove excess water, for example after drinking excess fluid, ADH production is decreased and the collecting tubule becomes less permeable to water, making the urine diluted and abundant.

6 0

3 years ago

The force required to stretch a Hooke’s-law spring varies from 0 N to 63.5 N as we stretch the spring by moving one end 5.31 cm

Alika [10]

Answer:

Force constant will be 1195.85 N/m

Work done will be 1.6859 J

Explanation:

We have given the force, F = 63.5 N

Spring is stretched by 5.31 cm

So x = 0.0531 m

Force is given , F = 63.5 N

We know that force is given by $F=kx$

So $63.5=k\times 0.0531$

k = 1195.85 N/m

Now we have to find the work done

We know that work done is given by

$W=\frac{1}{2}kx^2=\frac{1}{2}\times 1195.85\times 0.0531^2=1.6859J$

8 0

3 years ago