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vivado [14]
3 years ago
15

Give real-world examples of evidence that supports the evolution of Earth in each category:

Physics
1 answer:
expeople1 [14]3 years ago
8 0

Answer:

concave shape of the waverock

Explanation:

The wave rock is formed by the weathering of the surrounding area. This helps in proving the deposition part, as the wave rock was below the ground, occurred due to deposition of rock years over the years. It is made from very tough material from its surroundings. The weathering reduces the surrounding terrain, while the bedrock remains to witness for the history.  Also, volcanic eruptions have been changing earth for a long time. (first of all, the theory that a volcanic eruption helped in making dinos go extinct.) Real world examples include: Ash and sulfur went pretty high into Earth's atmosphere because of the Tambora eruption which in turn dimmed incoming sunlight, and lowered global temperatures by about 3°F. The Mount Pinatubo eruption in the Philippines in 1991 cooled the planet by about 1°F.

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Two square air-filled parallel plates that are initially uncharged are separated by 1.2 mm, and each of them has an area of 190
julia-pushkina [17]

Answer:

5.5\cdot 10^{-11} C

Explanation:

The capacitance of the parallel-plate capacitor is given by:

C=\epsilon_0 \frac{A}{d}

where

\epsilon_0 = 8.85\cdot 10^{-12} F/m is the vacuum permittivity

A=190 mm^2 = 190 \cdot 10^{-6} m^2 is the area of the plates

d=1.2 mm = 0.0012 m is the separation between the plates

Substituting,

C=(8.85\cdot 10^{-12}) \frac{190 \cdot 10^{-6}}{0.0012}=1.4\cdot 10^{-12}F

The energy stored in the capacitor is given by

U=\frac{Q^2}{2C}

Since we know the energy

U=1.1 nJ = 1.1 \cdot 10^{-9} J

we can re-arrange the formula to find the charge, Q:

Q=\sqrt{2UC}=\sqrt{2(1.1\cdot 10^{-9} J)(1.4\cdot 10^{-12}F )}=5.5\cdot 10^{-11} C

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Two resistors, A and B, are connected in series to a 6.0 V battery. A voltmeter connected across resistor A measures a potential
mestny [16]

Answer:

Resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

Explanation:

When the two resistors are in series, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B.

Given that V₁ + V₂ = 6.0 V and V₁ = 4.0 V,

V₂ = 6.0 V - V₁ = 6.0 V - 4.0 V = 2.0 V

Also, let the current in series be I.

So, V₁ = IR₁ and V₂ = IR₂

I = V₁/R₁ and I = V₂/R₂

equating both expressions, we have

V₁/R₁ = V₂/R₂

4.0 V/R₁ = 2.0 V/R₂

dividing through by 2.0 V, we have

2/R₁ = 1/R₂

taking the reciprocal, we have

R₂ = R₁/2

R₁ = 2R₂

From the parallel connection, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B. Since it is parallel, V₁ = V₂ = V = 6.0 V

Also, V₂ = I₂R₂ where I₂ = current in resistor B = 2.0 A and R₂ = resistance of resistor B

So, R₂ = V₂/I₂

= 6.0 V/2.0 A

= 3.0 Ω

R₁ = 2R₂

= 2(3.0 Ω)

= 6.0 Ω

So, resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

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