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BARSIC [14]
3 years ago
5

¿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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A solenoidal coil with 26 turns of wire is wound tightly around another coil with 350 turns. The inner solenoid is 20.0 cm long
noname [10]

Answer:

Part a)

\phi = 2.76 \times 10^{-7} T m^2

Part B)

M = 5.52 \times 10^{-5} H

Part C)

EMF = 0.1 V/s

Explanation:

Part a)

Magnetic field due to a long ideal solenoid is given by

B = \mu_0 n i

n = number of turns per unit length

n = \frac{N}{L}

n = \frac{350}{0.20}

n = 1750 turn/m

now we know that magnetic field due to solenoid is

B = (4\pi \times 10^{-7})(1750)(0.100)

B = 2.2 \times 10^{-4} T

Now magnetic flux due to this magnetic field is given by

\phi = B.A

\phi = (2.2 \times 10^{-4})(\pi r^2)

\phi = (2.2 \times 10^{-4})(\pi(0.02)^2)

\phi = 2.76 \times 10^{-7} T m^2

Part B)

Now for mutual inductance we know that

\phi_{total} = M i

\phi_{total} = N\phi

\phi_{total} = 20(2.76 \times 10^{-4})

\phi_{total} = 5.52 \times 10^{-6}

now we have

M = \frac{5.52 \times 10^{-6}}{0.100}

M = 5.52 \times 10^{-5} H

Part C)

As we know that induced EMF is given as

EMF = M \frac{di}{dt}

EMF = 5.52 \times 10^{-5} (1800)

EMF = 0.1 V/s

3 0
3 years ago
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