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2 years ago

Where is the location potential energy is converted to kinetic energy? Please answer asap!!!!!

Physics

1 answer:

kolezko [41]2 years ago

7 0

Potential energy is the energy that is possible to be realized, based on position, while the kinetic energy is the energy that is present and due to motion

The location where the potential energy is converted into kinetic energy is the point (2)

Reason:

The mechanical energy, M.E., of the roller coaster, is given as follows;

M.E. = P.E. + K.E.

Where;

P.E. = The potential energy = m·g·h

K.E. = The kinetic energy = $\dfrac{1}{2} \cdot m \cdot v^2$

Where;

v = The velocity

h = The height

m = The mass

g = Acceleration due to gravity

∴ M.E. = m·g·h + $\dfrac{1}{2} \cdot m \cdot v^2$

The mechanical energy, M.E., is constant, therefore;

When the height, h = 0, we have, the kinetic energy is maximum, given that the mechanical energy is entirely kinetic energy

Therefore;

The locations where potential energy is converted to kinetic energy are lowest locations, where the height, h, is minimum, such as the point (2)

Learn more about potential and kinetic energy here:

brainly.com/question/18963960

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Chapter 16, Problem 63. A person is standing in a room at 18 ◦C. The exposed surface area and skin temperature of the person are

Juliette [100K]

Answer: $Q=248.011 W$

Explanation:

Given

Temperature of Room $T_{\infty }=18^{\circ}\approx 291 K$

Area of Person $A=1.7 m^2$

Temperature of skin $T=32^{\circ}\approx 305 K$

Heat transfer coefficient $h=5 W/m^2.k$

Emissivity of the skin and clothes $\epsilon =0.9$

$\Delta T=32-18=14^{\circ}$

Total rate of heat transfer=heat Transfer due to Radiation +heat transfer through convection

Heat transfer due radiation $Q_1=\epsilon \sigma A(T^4-T_{\infty }^4 )$

where $\sigma =stefan-boltzman\ constant$

$Q_1=0.9\times 5.687\times 10^{-8}\times 1.7\times (305^4-291^4)$

$Q_1=129.01 W$

Heat Transfer due to convection is given by

$Q_2=hA(\Delta T)$

$Q_2=5\times 1.7\times 14=119 W$

$Q=Q_1+Q_2$

$Q=129.01+119=248.011 W$

7 0

3 years ago

Julie drives 100 mi to Grandmother's house. On the way to Grandmother's, Julie drives half the distance at 30.0 mph and half the

salantis [7]

Answer:

Explanation:

Given

Distance to grandmother's house=100 mi

it is given that during return trip Julie spend equal time driving with speed 30 mph and 70 mph

Let Julie travel x mi with 30 mph and 100-x with 70 mph

$\frac{x}{30}=\frac{100-x}{70}$

x=30 mi

Therefore

Julie's Average speed on the way to Grandmother's house $=\frac{100}{\frac{50}{30}+\frac{50}{70}}$

=42 mph

On return trip

$=\frac{100}{2\frac{30}{30}}=50 mph$

6 0

3 years ago

It says find the slope for each line I'm stuck on number one can you help me

Allushta [10]

$\text{slope}=\frac{y_2-y_1}{x_2-x_1}$

(-2, -1)(3, 1)

Therefore,

$\text{slope}=\frac{1-(-1)}{3-(-2)}=\frac{1+1}{3+2}=\frac{2}{5}$

7 0

1 year ago

please help! find magnitude and direction (the counterclockwise angle with the +x axis) of a vector that is equal to a + c

-BARSIC- [3]

Answer:

Option (2)

Explanation:

From the figure attached,

Horizontal component, $A_x=A\text{Sin}37$

$A_x=12[\text{Sin}(37)]$

= 7.22 m

Vertical component, $A_y=A[\text{Cos}(37)]$

= 9.58 m

Similarly, Horizontal component of vector C,

$C_x$ = C[Cos(60)]

= 6[Cos(60)]

= $\frac{6}{2}$

= 3 m

$C_y=6[\text{Sin}(60)]$

= 5.20 m

Resultant Horizontal component of the vectors A + C,

$R_x=7.22-3=4.22$ m

$R_y=9.58-5.20$ = 4.38 m

Now magnitude of the resultant will be,

From ΔOBC,

$R=\sqrt{(R_x)^{2}+(R_y)^2}$

= $\sqrt{(4.22)^2+(4.38)^2}$

= $\sqrt{17.81+19.18}$

= 6.1 m

Direction of the resultant will be towards vector A.

tan(∠COB) = $\frac{\text{CB}}{\text{OB}}$

= $\frac{R_y}{R_x}$

= $\frac{4.38}{4.22}$

m∠COB = $\text{tan}^{-1}(1.04)$

= 46°

Therefore, magnitude of the resultant vector will be 6.1 m and direction will be 46°.

Option (2) will be the answer.

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3 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