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lord [1]
3 years ago
15

a circular cylinder and isused to maintain a water depth of 4 m. That is, when the water depth exceeds 4 m, thegate opens slight

ly and lets the water flow under it. Determine the weight of the gate permeter of length
Physics
1 answer:
stiv31 [10]3 years ago
3 0

Answer:

  W / A = 39200 kg / m²

Explanation:

For this problem let's use the equilibrium equation of / newton

           F = W

Where F is the force of the door and W the weight of water

         W = mg

We use the concept of density

        ρ = m / V

        m = ρ V

The volume of the water column is

          V = A h

We replace

         W = ρ A h g

On the other side the cylinder cover has a pressure

          P = F / A

          F = P A

We match the two equations

       P A = ρ A h g

        P = ρ g h

        P = 39200 Pa

The weight of the water column is

       W  = 1000 9.8 4 A

       W / A = 39200 kg / m²

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<u>Explanation</u>:

Given data,

charge = 5.5 x 10¹² C

k =9.00 x 10⁹

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V= kQ / r

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V=  49.5 x 10⁻³/ 5.00×10⁻²

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3 years ago
In a given reversible process, the temperature of an ideal gas is kept constant as the gas is compressed to a smaller volume. Wh
sesenic [268]

Answer:

B. The gas must release heat to its surroundings.

Explanation:

Since the process is reversible , it must be slow . Since temperature is kept constant so the internal energy is constant , Since volume is decreased , pressure must be increased. Since work is done on the gas ( gas is compressed )  , so heat must be released by the gas so that its internal energy remains constant.

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In a 200-turn automobile alternator, the magnetic flux in each turn is ΦB = 2.50 10-4 cos ωt, where ΦB is in webers, ω is the an
n200080 [17]

Answer:

10.63952sin(209.43951t)

10.63952 V

Explanation:

N_t = Number of turns = 200

\phi_B = Magnetic flux = 2.5\times 10^{-4}cos(\omega t)

\omega_e = Engine angular speed = 1\times 10^{3}\ rpm

Alternator angular speed is given by

N_a=2\times \omega_e\\\Rightarrow N_a=2\times 1\times 10^{3}\\\Rightarrow N_a=2\times 10^{3}\ rpm

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Induced emf is given by

\epsilon=-N_t\dfrac{d\phi_B}{dt}\\\Rightarrow \epsilon=-200\dfrac{d}{dt}2.54\times 10^{-4}cos(209.43951 t)\\\Rightarrow \epsilon=200\times 2.54\times 10^{-4}\times 209.43951 sin(209.43951t)\\\Rightarrow \epsilon=10.63952sin(209.43951t)

The function is 10.63952sin(209.43951t)

The induced maximum emf is 10.63952 V

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Answer:

b          d                   a

Explanation:

3 0
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