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GrogVix [38]
3 years ago
5

What is the potential energy of a 3kg ball that is on the ground?

Physics
1 answer:
ELEN [110]3 years ago
5 0

This is where we have to admit that gravitational potential energy is
one of those things that depends on the "frame of reference", or
'relative to what?'.

         Potential energy = (mass) x (gravity) x (<em>height</em>).

So you have to specify <em><u>height above what</u></em> .

-- With respect to the ground, the ball has zero potential energy.
(If you let go of it, it will gain zero kinetic energy as it falls to
the ground.)

-- With respect to the floor in your basement, the potential energy is

                 (3) x (9.8) x (3 meters) = 88.2 joules.

(If you let go of it, it will gain 88.2 joules of kinetic energy as it falls
to the floor of your basement.)

-- With respect to the top of that 10-meter hill over there, the potential
energy is
                    (3) x (9.8) x (-10) = -294 joules

(Its potential energy is negative. After you let go of it, you have to give it
294 joules of energy that it doesn't have now, in order to lift it to the top of
the hill <em>where it will have zero</em> potential energy.)


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Answer:

Heat transferred = 22.9 watt

Explanation:

Given that:

T_1 = 27°C = (273 + 27) K = 300 K

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speed v = 2 m/s

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The heat transferred from the plate can be calculate by using the formula:

Heat transferred = h×A ×ΔT

From the tables of properties of air, the following values where obtained.

k = 0.02476 \ W/m.k  \\ \\  \rho = 1.225 \ kg/m^3 \\ \\ \mu = 18.6 \times 10^{-6} \ Pa.s \\ \\ c_p = 1.005 \ kJ/kg

To start with the reynolds number; the formula for calculating the reynolds number can be expressed as:

reynolds number = \dfrac{\rho \times v \times x }{\mu}

reynolds number = \dfrac{1.225 \times 2 \times 0.4}{18.6 \times 10^{-6}}

reynolds number = \dfrac{0.98}{18.6 \times 10^{-6}}

reynolds number = 52688.11204

Prandtl number = \dfrac{c_p \mu}{k}

Prandtl number = \dfrac{1.005 \times 18.6 \times 10^{-6} \times 10^3}{0.02476}

Prandtl number = \dfrac{0.018693}{0.02476}

Prandtl number = 0.754963

The nusselt number for this turbulent flow over the flat plate  can be computed as follows:

Nusselt no = \dfrac{hx}{k} = 0.0296 (Re) ^{0.8} \times (Pr)^{1/3}

\dfrac{h \times 0.4}{0.02476} = 0.0296 (52688.11204) ^{0.8} \times (0.754968)^{1/3}

\dfrac{h \times 0.4}{0.02476} =161.4252008}

h  =\dfrac{161.4252008 \times 0.02476}{ 0.4}

h = 9.992 W/m.k

Recall that:

The heat transferred from the plate can be calculate by using the formula:

Heat transferred = h×A ×ΔT

Heat transferred = h\times A \times (T_2-T_1)

Heat transferred = 9.992 × (0.4 × 0.01) ×(873-300)

Heat transferred = 22.9 watt

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