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pentagon [3]
3 years ago
7

A bird carries a 25 g oyster to a height of 11m what is the gravitational potential energy of the oyster

Physics
1 answer:
garik1379 [7]3 years ago
7 0

Given data:

Mass of oyster (m) = 25 g,

                              = 0.025 kg            (<em> since 1 Kg = 1000 g)</em>

Height of oyster (h) = 11 m,

determine gravitational potential energy (GPE) = ?

Gravitational potential energy is energy of an object possesses because of its position on the earth. The amount of gravitational potential energy of an object on earth depends on its<em> mass</em> and<em> height </em>from the earth.  

Mathematically,

        GPE = m. g. h <em>Joules</em>

<em>         Where, </em>

<em>               </em>GPE =<em> </em>gravitational potential energy,

<em>         </em>          m = mass of object in Kg,

                  g = gravitational force in m/s²,     g = 9.8 m/s²

                  h = height of the object in meters,

     Now substituting all the values in the equation GPE

           GPE = 0.025 Kg × 9.8 m/s² × 11 m

                   = 2.695 J

Gravitational potential energy of the oyster is 2.695 J


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Complete question:

A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If the switch is closed at time t = 0 s, what is the current 7.0 ms later?

Answer:

The current in the circuit 7 ms later is 0.2499 A

Explanation:

Given;

Ideal inductor, L = 45-mH

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Time constant, is given as:

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T = (45 x 10⁻³) / (60)

T = 7.5 x 10⁻⁴ s

Change in current with respect to time, is given as;

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Current in the circuit after 7 ms later:

t = 7 ms = 7 x 10⁻³ s

I(t) = I_o(1-e^{-\frac{t}{T}})\\\\I =0.25(1-e^{-\frac{7*10^{-3}}{7.5*10^{-4}}})\\\\I = 0.25(0.9999)\\\\I = 0.2499 \ A

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ωФ_Z= (200 rev/min)(1min/ 60s) => 3.333rev/s

time 't'= 4 s

angular acceleration 'αФ_Z'=?

constant angular acceleration equation is given by,

ωФ_Z= ωФ_o_z + αФ_Zt

αФ_Z= (ωФ_Z - ωФ_o_z )/t => (3.333-8.333)/4

αФ_Z= -1.25 rev/s²

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