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xenn [34]
3 years ago
15

According to Newton’s Second Law of Motion, an object will accelerate if you apply what kind of force?

Physics
1 answer:
Alex777 [14]3 years ago
6 0

If you got this question from a book or homework sheet, then
you should take it back and comp[lain. It's a defective question.

It's fishing for the answer "unbalanced force", but that's a misleading
and misuse of the term.  There's no such thing as a "balanced force"
or an "unbalanced force". 

A GROUP of two or more forces can be balanced ... if the forces
all cancel each other out and add up to zero ... or unbalanced ...
if they don't.  But it's never correct to apply that description to
a single force.

According to Newton's second law of motion, an object will
accelerate if the forces on it are unbalanced.

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Which of these accurately describes the products of a reaction?
stiv31 [10]
The one that accurately describes the products of a reaction is : B. new substances that are present at the end of a reaction
For example the process of photosynthesis transform CO2 and other nutrients into O2 and H2O

hope this helps
3 0
2 years ago
Read 2 more answers
The amplitude of a sound is the A. frequency of the sound. B. magnitude of displacement of a sound pressure wave. C. psychologic
AlladinOne [14]

Answer:

Option B. magnitude of displacement of a sound pressure wave

Explanation:

Amplitude is simply the maximum displacement of a wave from its mean position.

6 0
3 years ago
Read 2 more answers
Do the math: How many seconds would it take an echo sounder’s ping to make the trip from a ship to the Challenger Deep (10,994 m
Lera25 [3.4K]

Answer:

<h2>14.66secs</h2>

Explanation:

Given the formula for calculating the depth in metres expressed as

depth in meters = ½ (1500 m/sec × Echo travel time in seconds)

Given depth of the challenger = 10, 994 meters, we will substitute this given value into the formula given to calculate the time take for the echo to travel.

10, 994 = depth in meters = ½ * 1500 m/sec × Echo travel time in seconds

10,994 = 750 * Echo travel time in seconds

Dividing both sides by 750;

Echo travel time in seconds = 10,994 /750

Echo travel time in seconds ≈ 14.66secs (to two decimal places)

Therefore, it would take an echo sounder’s ping 14.66secs to make the trip from a ship to the Challenger Deep and back

6 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 <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
A piece of silicon sample has a resistivity of 0.1 ω.Cm. Its thickness is 100µm. The electron mobility is 1350cm 2 v -1sec-1. Wh
lawyer [7]

The answer is Rh = 135 cm^3 and B = 0.05185 wh/m^2


Explanation:

Resitivity of silicon = 0.1

thickness = 100um

so, I = ma

Required to find out concentration of electron , we know that

Rh = up

By putting in the values,

Rh = 1350 x 0.1

Rh = 135 cm^3


Now consider,

Rh = 1 / Rh.q

= 1 / Rh . q

= 1 / 135 x1.609 x10^-19

= 4.6037 x 10^16 / cm^3

Vh = BIRh / w

B = Vh w/ IRh

B = -70 x10^-6 x 100 x10^-6 / 1x 10^-3 x 135 x 10^-6

B = 0.05185 wh / m^2


5 0
3 years ago
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