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3 years ago

On a hot sunny day, while playing on the ground our body temperature increases. Explain and define the process.

Physics

1 answer:

spayn [35]3 years ago

4 0

Explanation:

Its because of the sunlight being much hoter than your body absorbs the heat of the sun.

And its also because of aerobic respiration in which your body takes in oxygen and digests gluecose to creates energy(beacuse of any physical activity like, playing and stuff). So during the proscess in which uour body creates energy(energy-andenoisine triphosphate) HEAT is also created

Thus, your body fells hot during playing and in sunlight

Hope it helps

Also see my comment pls theres another answer.

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the resistivity of gold is 2.44×10−8Ω⋅m at room temperature. A gold wire that is 0.9 mm in diameter and 14 cm long carries a cur

cricket20 [7]

Answer:

0.03605 V/m is the electric field in the gold wire.

Explanation:

Resistivity of the gold = $\rho = 2.44\times 10^{-8} \Omega.m$

Length of the gold wire = L = 14 cm = 0.14 m ( 1 cm = 0.01 m)

Diameter of the wire = d = 0.9 mm

Radius of the wire = r = 0.5 d = 0.5 × 0.9 mm = 0.45 mm = $0.45\times 0.001 m$

( 1mm = 0.001 m)

Area of the cross-section = $A=\pi r^2=\pi r^(0.45\times 0.001 m)^2$

Resistance of the wire = R

Current in the gold wire = 940 mA = 0.940 A ( 1 mA = 0.001 A)

$R=\rho\times \frac{L}{A}$

$V(voltage)=I(current)\times R(Resistance)$ ( Ohm's law)

$\frac{V}{I}=\rho\times \frac{L}{A}$

We know, Electric field is given by :

$E=\frac{dV}{dr}$

$E=\frac{V}{L}$

$E=\frac{V}{L}=\rho\times \frac{I}{A}$

$E=2.44\times 10^{-8} \Omega.m\times \frac{0.940 A}{\pi r^(0.45\times 0.001 m)^2}=0.03605 V/m$

0.03605 V/m is the electric field in the gold wire.

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