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3 years ago

Which list of factors best supports the concept that they are "just right" for the planet earth to support life?

Physics

2 answers:

Ivanshal [37]3 years ago

8 0

Temperature, gravity, atmosphere and water

GREYUIT [131]3 years ago

7 0

Answer: temperature, gravity, atmosphere, and water

The environment of the earth favours the sustainance of life on earth. The temperature of the Earth is influenced by the composition of the atmosphere, which consists of heat-trapping gases and chemicals produce due to volcanic eruption and human activities. Temperature varies according to the seasonal variations, global warming and other variables. Temperature provides a base for the existence of life on Earth. Gravity is a force that is exerted by the Earth, it attracts objects or organisms towards itself. It facilitates a substrate to the living organism. Atmosphere is a mixture of gases, some of these gases like carbon dioxide, oxygen and nitrogen supports life in a biosphere. These gases are necessary for photosynthesis, respiration and nitrogen fixation in living organisms. Water is a colorless, shapeless substance, which acts as a solvent. It is a chief component present in all organisms living on earth. Water is required for various processes like photosynthesis, respiration, and digestion.

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Brilliant_brown [7]

Answer:

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Read 2 more answers

A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien

n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

$\phi=NBA$

Put the value into the formula

$\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2$

$\phi=7.85\times10^{-7}\ Tm^2$

We need to calculate the induced emf

Using formula of induced emf

$\epsilon=\dfrac{d\phi}{dt}$

Put the value into the formula

$\epsilon=\dfrac{7.85\times10^{-7}}{dt}$

Put the value of emf from ohm's law

$\epsilon =IR$

$IR=\dfrac{7.85\times10^{-7}}{dt}$

$Idt=\dfrac{7.85\times10^{-7}}{R}$

$Idt=\dfrac{7.85\times10^{-7}}{0.50}$

$Idt=0.00000157=1.57\times10^{-6}\ C$

We know that,

$Idt=dq$

$dq=1.57\times10^{-6}\ C$

We need to calculate the voltage across the capacitor

Using formula of charge

$dq=C dV$

$dV=\dfrac{dq}{C}$

Put the value into the formula

$dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}$

$dV=1.57\ V$

Hence, The voltage across the capacitor is 1.57 V.

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