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Oksanka [162]
3 years ago
5

Why does the surface of the sun appear to have a sharp edge?

Physics
1 answer:
Anna11 [10]3 years ago
6 0
The photosphere is relatively thin compared to the other atmospheric regions.
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Law of conservation of energy states that
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A) energy cannot be created nor destroyed
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3 years ago
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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.

3 0
4 years ago
In cloud formation, water vapor turns into water droplets which get bigger and bigger until it rains.
True [87]

Answer:

a. get warmer.

Explanation:

When the water vaporous reach the upper layer of the atmosphere they get a cooler air to which they loose their temperature and condense to form clouds as a the temperature of the air increases.

It may be noted that the water looses its high amount of latent heat of vaporization to condense into water this significantly increases the temperature of the air in contact.

4 0
3 years ago
A particle is attached to a spring and is pushed so that the spring is compressed more and more. As a result, the spring exerts
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In order for particles to perform a simple harmonic motion, we must follow the law of force of the form F = -kx, where x is the displacement of the object from the equilibrium position and k is the spring constant. The force shown in <span>F = -kx is always the restoring force in the sense that the particles are pulled towards the equilibrium position.

The repulsive force felt when the charge q1 is pushed into another charge q2 of the same polarity is given by Coulomb's law
                                F = </span><span>k *q1* q2 / r^2.
</span>It is clear that Coulomb's law is an inverse-square relationship. It does not have the same mathematical form as the equation <span><span>F = -kx.</span> Thus, charged particles pushed towards another fixed charged particle of the same fixed polarity do not show a simple harmonic motion when released. Coulomb's law does not describe restoring force. When q1 is released, it just fly away from q2 and never returns.</span>
5 0
4 years ago
for a freely falling object dropped from rest, what is the acceleration at the end of the fifth second of fall?
lara31 [8.8K]

acceleration due to gravity is always 9.8 m/s/s (on earth)
3 0
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