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enyata [817]
3 years ago
9

For this problem, we assume that we are on planet-i. the radius of this planet is r =4200 km, the gravitational acceleration at

the surface is gi =9.1 m/s 2 , and the gravitational constant g = 6.67 × x10−11 n m2 /kg2 in si units. the mass of planet-i is not given. not all the quantities given here will be used. suppose a cannon ball of mass m = 5130 kg is projected vertically upward from the surface of this planet. it rises to a maximum height h =12306 km above the surface of the planet. caution: here the gravitational acceleration decreases as the cannon ball travels away from planet-i. r h determine the kinetic energy (in joules) of the cannon ball immediately after it is fired off. answer in units of j.
Physics
1 answer:
Minchanka [31]3 years ago
3 0
The expression commonly used for potential gravitational energy is just simplification. It is actually just the first term in Taylor expansion of the real expression. 
In general, the potential energy of gravitational field is defined as:
U=-G \frac{mM}{r}
Where G is universal gravitational constant, and r is the distance between the objects centers of mass. Negative sign represents the bound state.
Since we are not given the mass of the planet we have to calculate it.
F_g=G\frac{mM}{r_p^2}\\ mg=G\frac{mM}{r_p^2}\\ g=G\frac{M}{r_p^2}
This formula can be used for any planet. It gives you the gravitational acceleration on the planet's surface. We can use it to calculate the planet's mass:
g=G\frac{M}{r_p^2}\\ M=\frac{gr_p^2}{G}=2.41\cdot 10^{24}kg
Now we can calculate the potential energy of that cannonball when it reaches its maximum height.
U=-G \frac{mM}{r}\\ U=-G \frac{mM}{r_p+h}
When we plug in the numbers we get:
U=-4.99\cdot 10^{10} J
The potential energy has to be equal to the kinetic energy.
E_k=4.99\cdot 10^{10} J

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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
Devon was watching an old television show about people stranded on an island. One of the characters wanted to make an electric g
navik [9.2K]

D. a and c

Explanation:

To make a generator out of a bicycle, Devon will need two magnets and coil of wire.

There is a close knit relationship between magnetism and electricity. Electricity can be generated by a wire changing flux in the presence of a magnetic field.

  • The magnets provide magnetic field for the process.
  • The coil of wire is placed in between the magnetic field generated by the magnets.
  • The bicycle pedal is connected to the set up to turn the wire in between the magnetic fields.
  • This way, electricity is produced.

Learn more:

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5 0
3 years ago
a system of four particles moves along one dimension. the center of mass of the system is at rest, and the particles do not inte
Zinaida [17]

The velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

The given parameters;

m1 = 1.45 kg, v1(t) = (6.09m/s) + (0.299m/s^2) × t

m2 = 2.81 kg, v2(t) = (7.83m/s) + (0.357m/s^2) × t

m3 = 3.89 kg, v3(t) = (8.09m/s) + (0.405m/s^2) × t

m4 = 5.03kg

The velocity of the center mass of the particles is calculated as;

McmVcm = m1v1 + m2v2 +m3v3+m4v4

Vcm= m1v1 + m2v2 +m3v3 +m4v4/ Mcm

0 = m1v1 + m2v2 +m3v3 +m4v4/ Mcm

m1v1 + m2v2 +m3v3+m4v4 = 0

m4v4 = -(m1v1 + m2v2 +m3v3)

v4 =-(m1v1 + m2v2 +m3v3)/ m4

The velocity of particle 1 at time, t = 2.83 s;

vi = 6.09 + 0.299× 2.83

v1 = 6.94 m/s

The velocity of particle 3 at time, t = 2.83 s;

v2 = 7.83 + 0.357 × 2.83

v2 = 8.84 m/s

The velocity of particle 3 at time, t = 2.83 s;

v3 = 8.09 + 0.405 × 2.83

v3 = 9.24 m/s

The velocity of particle 3 at time, t = 2.83 s;

v4 = - (m1v1 + m2v2 + m3v3)/m4

v4 = -(1.45×6.94 + 2.81×8.84 + 3.89×9.24)/5.03

v4 = -14.4 m/s

Thus, the velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

Learn more about Velocity here:

brainly.com/question/18084516

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5 0
1 year ago
Find the electric potential VP at point P. [Hint: To input a natural logarithm into the answer box, simply type the letters "ln"
Alik [6]

Answer:

After finding the electric potential VP at point P = Q/Чπϵ₀L ㏑(1+\frac{L}{d})

Explanation:

I believe it is a part C question.  

The derivative of V and P  will be directly proportional to the differential dq and the inverse of Чπϵ₀δ........

Please find detailed solution in the attached picture as i believe that is the answer to the part C question you are seeking for.

8 0
3 years ago
Consider a situation in which you are moving two point charges such that the potential energy between them decreases. (NOTE: ign
Oxana [17]

Answer: Option A

Explanation:

The potential energy decreases in the case when the charges are opposite and they attract each other.

In this case there is no external energy required in order to put the charges together.

This is so because the charges are opposite and they will attract each other. Yes, the only condition should be that the charges should be alike.

Example: a negative charge and a positive charge.

5 0
3 years ago
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