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4 years ago

5

If two planets orbit a star, but planet B is twice as far from the star as planet A, planet A will receive ____ times the flux t

hat planet B receives.

1 answer:

Tresset [83]4 years ago

7 0

Answer:

The nearest plant (A) receives 4 times more radiation from the farthest plant

Explanation:

The energy emitted by the star is distributed on the surface of a sphere, whereby intensity received is the power emitted between the area of the sphere

I = P / A

P = I A

The area of the sphere is

A = 4π r²

Since the amount of radiation emitted by the star is constant, we can write this expression for the position of the two planets

P = I₁ A₁ = I₂ A₂

I₁ / I₂ = A₂ / A₁

Suppose index 1 corresponds to the nearest planet,

r2 = 2 r₁

I₁ / I₂ = r₁² / r₂²

I₁ / I₂ = r₁² / (2r₁)²

I₁ / I₂ = ¼

4 I₁ = I₂

The nearest plant (A) receives 4 times more radiation from the farthest plant

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A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 2.9 m and moment of inertia 900 kg⋅m

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Explanation:

It is given that,

Mass of person, m = 70 kg

Radius of merry go round, r = 2.9 m

The moment of inertia, $I_1=900\ kg.m^2$

Initial angular velocity of the platform, $\omega=0.95\ rad/s$

Part A,

Let $\omega_2$ is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :

$I_1\omega_1=I_2\omega_2$

Here, $I_2=I_1+mr^2$

$I_1\omega_1=(I_1+mr^2)\omega_2$

$900\times 0.95=(900+70\times (2.9)^2)\omega_2$

Solving the above equation, we get the value as :

$\omega_2=0.574\ rad/s$

Part B,

The initial rotational kinetic energy is given by :

$k_i=\dfrac{1}{2}I_1\omega_1^2$

$k_i=\dfrac{1}{2}\times 900\times (0.95)^2$

$k_i=406.12\ rad/s$

The final rotational kinetic energy is given by :

$k_f=\dfrac{1}{2}(I_1+mr^2)\omega_1^2$

$k_f=\dfrac{1}{2}\times (900+70\times (2.9)^2)(0.574)^2$

$k_f=245.24\ rad/s$

Hence, this is the required solution.

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3 years ago

You throw a glob of putty straight up toward the ceiling, which is 3.50 mm above the point where the putty leaves your hand. The

LuckyWell [14K]

Answer: $4.65\ m/s$

Explanation:

Given

Distance putty has to travel is 3.5 m

The initial speed of putty is 9.50 m/s

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$v^2-u^2=2as$

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So, the velocity of putty just before hitting is $4.65\ m/s$

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Your answer:

In Greek mythology, this constellation is related with the time the Olympian gods sought refuge in Egypt. Unfortunately, following their epic fighting with the Titans, peace did not closing for long, as the monster Typhon, son of the Titan Tartarus and Earth, sought revenge. Typhon was once a fearsome fire-breathing creature, taller than mountains and with palms which possessed dragons' heads in region of fingers. The Olympian gods sought to break out by way of adopting a number disguises: Zeus, a ram - Hera, a white cow, Bacchus (another model of the fable suggests Pan) a goat. As Typhon approached, Bacchus/Pan threw himself into the Nile but, in a panic, solely succeeded in altering part of his body, ending up with a goat's physique and the tail of a fish. Meanwhile, Zeus had been dismembered via Typhon, however was saved when Bacchus/Pan let out an ear-splitting yell, distracting the monster lengthy ample for an agile Hermes to gather the supreme god's limbs and cautiously fix him. In gratitude, Zeus transferred Bacchus/Pan to the heavens.

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3 years ago

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calculate the time rate of change in air density during expiration. Assume that the lung has a total volume of 6000mL, the diame

kipiarov [429]

Answer:

The time rate of change in air density during expiration is 0.01003kg/m³-s

Explanation:

Given that,

Lung total capacity V = 6000mL = 6 × 10⁻³m³

Air density p = 1.225kg/m³

diameter of the trachea is 18mm = 0.018m

Velocity v = 20cm/s = 0.20m/s

dv /dt = -100mL/s (volume rate decrease)

= 10⁻⁴m³/s

Area for trachea =

$\frac{\pi }{4} d^2\\= 0.785\times 0.018^2\\= 2.5434 \times10^-^4m^2$

0 - p × Area for trachea =

$\frac{d}{dt} (pv)=v\frac{ds}{dt} + p\frac{dv}{dt}$

$-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)$

$-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)$

⇒ $-0.623133\times10^-^4+1.225\times10^-^4=6\times10^-^3\frac{ds}{dt}$

$\frac{ds}{dt} = \frac{0.6018\times10^-^4}{6\times10^-^3} \\\\= 0.01003kg/m^3-s$

ds/dt = 0.01003kg/m³-s

Thus, the time rate of change in air density during expiration is 0.01003kg/m³-s

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3 years ago

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DanielleElmas [232]

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