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rewona [7]
2 years ago
13

A pendulum clock gives correct time at 20°c how many sec/day will it gain or loose when the temperature fall to 5°c? cofficient

of linear expansion of pendulum is 0.0002°c-1​
Physics
1 answer:
rewona [7]2 years ago
5 0

Answer:

129.6 seconds

Explanation:

Given that :

α = 0.0002°c-1​

θ1 = 20°C

θ2 = 5°C

Time t = one day ; Converting to seconds ; number of seconds in a day ; (24 * 60 * 60) = 86400 seconds

Let dT= change in time

Using the relation :

dT = 0.5* α * dθ * t

dθ = (20 - 5) = 15°C

dT = 0.5 * 0.0002 * 15 * 86400

dT = 129.6 seconds

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Unlike a roller coaster, the seats in a Ferris wheel swivel so that the rider is always seated upright. An 80-ft-diameter Ferris
telo118 [61]

Answer:

a

   F_A  =425.42 \ N

b

  F_A_H  = 358.58 \ N

Explanation:

From the question we are told that

    The diameter of the Ferris wheel is  d  =  80 \ ft =  \frac{80}{3.281} =  24.383

    The  period of the Ferris wheel is  T  =  24 \ s

     The  mass of the passenger is  m_g  =  40 \ kg

The  apparent weight of the passenger at the lowest point is mathematically represented as

           F_A_L  =  F_c  + W

Where  F_c is the centripetal force on the passenger,  which is mathematically represented as

         F_c  =m *  r *  w^2

Where w is the angular velocity which is mathematically represented as

         w =  \frac{2* \pi   }{T}

substituting values

         w =  \frac{2* 3.142 }{24}

         w =  0.2618 \ rad/s

and  r  is the radius which is evaluated as r =  \frac{d}{2}

   substituting values

         r =  \frac{24.383}{2}

         r = 12.19 \ ft

So

          F_c  = 40 * 12.19* (0.2618)^2

          F_c  =  33.42 \ N

W is the weight which is mathematically represented as

           W =  40 * 9.8

           W =  392 \ N

So

         F_A    =  33.42 + 392

         F_A  =425.42 \ N

The  apparent weight of the passenger at the highest point is mathematically represented as

          F_A_H  =  W- F_c

substituting values

         F_A_H  = 392 -  33.42

         F_A_H  = 358.58 \ N

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3 years ago
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7 0
3 years ago
Objects with masses of 135 kg and a 435 kg are separated by 0.500 m. Find the net gravitational force exerted by these objects o
IrinaVladis [17]

Answer:

The net gravitational force on the mass is 1.27\times 10^{-5}

Explanation:

We have by Newton's law of gravity the force of attraction between masses m_{1},m_{2}

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Applying vales we get

Force of attraction between 135 kg mass and 38 kg mass is

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Thus the net force on mass 38.0 kg is F_{2}-F_{1}=1.76\times 10^{-5}-5.47\times 10^{-6}\\\\F_{2}-F_{1}=1.27\times 10^{-5}

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