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Salsk061 [2.6K]
3 years ago
11

What types of simple machines are in a bicycle

Physics
1 answer:
yarga [219]3 years ago
3 0
A bike is not a simple machine. It is a compound machine. A bike is made of simple machines such as gears, a pulley, le
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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 <u>afferent arteriole</u> 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 <u>renal glomerulus</u>. 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.

<u>Nephrons</u> 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 <u>collecting duct</u> 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
QUICK ITS DUE IN 6 MINUTES. create a scenario that applies the 3 laws of motion. Explain in complete sentences how each law if d
yKpoI14uk [10]

Answer:

In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

5 0
3 years ago
A formula for the normal systolic blood pressure for a man age a , measured in mmhg, is given as p=0. 006a2−0. 02a 120. Find the
djyliett [7]

The age of a man whose normal blood pressure measures 123 mm of hg

9 years

<h3>What is Quadratic equation ?</h3>

A quadratic equation as an equation of degree 2, meaning that the highest exponent of this function is 2. The standard form of a quadratic equation is y = ax^{2} + bx + c, where a, b, and c are numbers and a cannot be 0

P(A) = 0.006 a^{2} - 0.02a + 120

123 = 0.006- 0.02a + 120

0=0.006 a^{2} - 0.02a - 3

you can use the quadratic equation  formula to solve for the man's age.

A = (-b ± (\sqrt{b^{2} - 4*a*c})  ) / (2a)

A = (0.02 ±  \sqrt{(-0.02)^{2} - 4*0.006*(-3)}/ (2*0.006)

A = (0.02 ± \sqrt{0.0076}) / 0.012

A = 9 , -5.67

Age of the man will be 9 years

To learn more about quadratic equation  here

brainly.com/question/17177510?referrer=searchResults

#SPJ4

5 0
2 years ago
You're an electrical engineer designing an alternator (the generator that charges a car's battery). Mechanical engineers specify
Julli [10]

Answer:

13.78 mT

Explanation:

The peak voltage ε = ωNAB where ω = angular speed of coil = 1500 rpm = 1500 × 2π/60 rad/s = 50π rad/s = 157.08 rad/s, N = number of turns of coil = 250, A = area of coil = πr² where r = radius of coil = 10 cm = 0.10 m,

A = π(0.1 m)² = 0.03142 m² and B = magnetic field strength

So,

B = ε/ωNA  

substituting the values of the variables into the equation given that ε = 17 V

So, B = ε/ωNA

B = 17 V/(157.08 rad/s × 250 turns × 0.03142 m²)

B = 17 V/(1233.8634 rad-turns-m²/s)

B = 0.01378 T

B = 13.78 mT

8 0
3 years ago
The force F required to compress a spring a distance x is given by F 2 F0 5 kx where k is the spring constant and F0 is the prel
IrinaVladis [17]

Answer:

a)W=8.333lbf.ft

b)W=0.0107 Btu.

Explanation:

<u>Complete question</u>

The force F required to compress a spring a distance x is given by F– F0 = kx where k is the spring constant and F0 is the preload. Determine the work required to compress a spring whose spring constant is k= 200 lbf/in a distance of one inch starting from its free length where F0 = 0 lbf. Express your answer in both lbf-ft and Btu.

Solution

Preload = F₀=0 lbf

Spring constant k= 200 lbf/in

Initial length of spring x₁=0

Final length of spring x₂= 1 in

At any point, the force during deflection of a spring is given by;

F= F₀× kx  where F₀ initial force, k is spring constant and x is the deflection from original point of the spring.

W=\int\limits^2_1 {} \, Fds \\\\\\W=\int\limits^2_1( {F_0+kx} \,) dx \\\\\\W=\int\limits^a_b {kx} \, dx ; F_0=0\\\\\\W=k\int\limits^2_1 {x} \, dx \\\\\\W=k*\frac{1}{2} (x_2^{2}-x_1^{2}  )\\\\\\W=200*\frac{1}{2} (1^2-0)\\\\\\W=100.lbf.in\\\\

Change to lbf.ft by dividing the value by 12 because 1ft=12 in

100/12 = 8.333 lbf.ft

work required to compress the spring, W=8.333lbf.ft

The work required to compress the spring in Btu will be;

1 Btu= 778 lbf.ft

?= 8.333 lbf.ft----------------cross multiply

(8.333*1)/ 778 =0.0107 Btu.

6 0
3 years ago
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