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3 years ago

¿Cuál es la aceleración centrípeta de un móvil que recorre una

Physics

1 answer:

Darina [25.2K]3 years ago

4 0

Answer:

a) a = 4.57 m/s², b) a = 6.48 m / s² , c) a = 1.42 m / s²,d) r = 82.3 m

Explanation:

The centripetal acceleration is the acceleration responsible for the change of direction of the acceleration vector and occurs in circular movements, the expression is

a = v² / r

let's apply this precaution to our cases

a) let's calculate

a = 8²/14

a = 4.57 m/s²

b) an automobile at v = 65 km / h (1000 m / 1km) (1 h / 3600 s) =18,055 m/s

let's reduce feet to meters

1 ft = 0.3048 m

r = 165 ft (0.3048 m / 1 ft) = 50.292 m

a = 18,055 2 / 50,292

a = 6.48 m / s²

c) we calculate

a = 1.25²2 / 1.1

a = 1.42 m / s²

d) we look for the radius

a = v² / r

r = v² / a

we reduce

v = 80 km / h (1000 m / 1km) (1h / 3600s) = 22.22 ms

r = 22.22²/6

r = 82.3 m

e) the cenripeta acceleration is used to take the curves on the highway,

Used in centrifuges to separate compounds

It is used in the games of the park of atraccio

Used in CD players and computer hard drives

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Bill is farsighted and has a near point located 121 cm from his eyes. Anne is also farsighted, but her near point is 74.0 cm fro

Arada [10]

Answer:

Explanation:

The lens equation is

1 / f = 1 / di + 1 / do

Where

f is focal length

di is the image distance

do is the object distance

Both Annie and Billy use a glass whose near point is 25cm

Then, the object distance is

do = 25 - 2 = 23cm

The have the same object distance.

Let find the vocal length of bills eye

Given that,

Bill near point is 121cm and distance of the glass from the eye is 2cm

Then,

Image distance of bill is

di_B = -(121-2) = -119cm

object distance do = 23cm

Then,

1 / f_B = 1 / di_B + 1 / do

1 / f_B = -1 / 119 + 1 / 23

1 / f_B = -119^-1 + 23^-1

1 / f_B = 0.0351

Then, f_B = 28.51 cm

Also, let find Annie focal length

Given that,

Annie near point is 74 cm and distance of the glass from the eye is 2cm

Then,

Image distance of Annie is

di_A = -(74-2) = -72cm

object distance do = 23cm

Then,

1 / f_A = 1 / di_A + 1 / do

1 / f_A = -1 / 72 + 1 / 23

1 / f_A = -72^-1 + 23^-1

1 / f_A = 0.02959

Then, f_A = 33.8 cm

Distance of object from the lens when Annie uses Billy glass

Then,

1 / f_B = 1 / di_A + 1 / do

1 / 28.51 = -1 / 72 + 1 / do

28.51^-1 = -72^-1 + do^-1

do^-1 = 28.51^-1 + 72^-1

do^-1 = 0.048964

do = 20.42 cm

Then, the object location from the eye will be, the eye is 2cm from the glass. Then,

do_A = 20.42 + 2 = 22.42cm

do_A = 22.42 cm

Distance of object from the lens when Billy uses Annie glass

Then,

1 / f_A = 1 / di_B + 1 / do

1 / 33.8 = -1 / 119 + 1 / do

33.8^-1 = -119^-1 + do^-1

do^-1 = 33.8^-1 + 119^-1

do^-1 = 0.03799

do = 26.32 cm

Then, the object location from the eye will be, the eye is 2cm from the glass. Then,

do_B = 26.32 + 2 = 28.32 cm

do_B = 28.32 cm

7 0

3 years ago

It is not just lab chemicals but also common household substances that can be classified as acidic, basic, and neutral materials

Montano1993 [528]

Acidic: coffee, orange juice , soft drink, vinegar

Basic: Antacid, Baking soda, Detergent, Lye, Milk of Magnesia

Neutral: water, Sugar, Table salt.

Explanation:

Acidic substances have high concentration of hydrogen ions and have pH value less than 7. Basic substances have high concentration of hydroxide ions and have pH value more than 7. Neutral substance have pH value 7. The concentration oh hydrogen ions and hydroxide ions is equal.

Salt is formed when a acid reacts with base in a neutralization reaction. Different types of salt are formed depending upon the concentration of hydrogen and hydroxide ions. Neutral salt is formed when concentration of hydrogen ion is equal to concentration of hydroxide ions.

Coffee, orange juice , soft drink, vinegar are acidic substances. Antacid, Baking soda, Detergent, Lye, Milk of Magnesia. water, Sugar, Table salt are neutral substances.

6 0

3 years ago

Blue light of wavelength λ passes through a single slit of width d and forms a diffraction pattern on a screen. If we replace th

ololo11 [35]

Answer:

We can retain the original diffraction pattern if we change the slit width to d) 2d.

Explanation:

The diffraction pattern of a single slit has a bright central maximum and dimmer maxima on either side. We will retain the original diffraction pattern on a screen if the relative spacing of the minimum or maximum of intensity remains the same when changing the wavelength and the slit width simultaneously.

Using the following parameters: y for the distance from the center of the bright maximum to a place of minimum intensity, m for the order of the minimum, λ for the wavelength, D for the distance from the slit to the screen where we see the pattern and d for the slit width. The distance from the center to a minimum of intensity can be calculated with:

$y\approx\frac{m\lambda D}{d}$

From the above expression we see that if we replace the blue light of wavelength λ by red light of wavelength 2λ in order to retain the original diffraction pattern we need to change the slit width to 2d:

$y\approx\frac{m\lambda D}{d} =\frac{m2\lambda D}{2d}$

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3 years ago

Which of these is not a form of matter? a. air b. milk c. rock d. light

OLga [1]

D. Light as light is a wave

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3 years ago

What are isotopes. Please answer soon

sweet-ann [11.9K]

each of two or more forms of the same element that contain equal numbers of protons but different numbers of neutrons in their nuclei, and hence differ in relative atomic mass but not in chemical properties; in particular, a radioactive form of an element.

5 0

3 years ago