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kenny6666 [7]
2 years ago
7

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

Physics
1 answer:
Lady bird [3.3K]2 years ago
5 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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Answer:

Vertical Height = 0.784 meter, Speed back at starting point = 10 m/s

Explanation:

Given Data:

V is the overall velocity vector, Vi and Ui are its initial vertical and horizontal components

R = 10 m/s\\ Projection Angle (theta) = 30 degrees\\Vi   = 10*sin(30) = 5 m/s\\Ui  = 10*cos(30) = 8.66 m/s

To find:

Max Height h achieved

Calculation:

1) Using the 3^{rd} equation of motion, we know

2*a*s = Vf^{2}  - Vi^{2}

2) In terms of gravity g height h and  the vertical component of Velocity Vf , Vi.

3) As Vf = 0 as at maximum height the vertical component of velocity is zero maximum height achieved

2*g*h = Vf^{2}  -Vi^{2}

putting values

4) h = 0.784 m/s

5) As for the speed when it reaches back its starting point, it will have a speed similar to its launching speed, the reason being the absence of air friction (Air drag)

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If 745-nm and 660-nm light passes through two slits 0.54 mm apart, how far apart are the second-order fringes for these two wave
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Answer:

0.82 mm

Explanation:

The formula for calculation an n^{th} bright fringe from the central maxima is given as:

y_n=\frac{n \lambda D}{d}

so for the distance of the second-order fringe when wavelength \lambda_1 = 745-nm can be calculated as:

y_2 = \frac{n \lambda_1 D}{d}

where;

n = 2

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substituting the parameters in the above equation; we have:

y_2 = \frac{2(745nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}

y_2 = 0.00276 m

y_2 = 2.76 × 10 ⁻³ m

The distance of the second order fringe when the wavelength \lambda_2 = 660-nm is as follows:

y^'}_2 = \frac{2(660nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}

y^'}_2 = 1.94 × 10 ⁻³ m

So, the distance apart the two fringe can now be calculated as:

\delta y = y_2-y^{'}_2

\delta y = 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m

\delta y = 10 ⁻³ (2.76 - 1.94)

\delta y = 10 ⁻³ (0.82)

\delta y = 0.82 × 10 ⁻³ m

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\delta y = 0.82 mm

Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm

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A shaving or makeup mirror is designed to magnify your face by a factor of 1.40 when your face is placed 20.0cm in front of it
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Answer:

(a) convex mirror

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(b) The image is virtual and magnified.

(c) Let the distance of image is v.

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