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2 years ago

10

A particular eco solar system has five planets in total a b c d and e the table lists the orbital periods of these planets in da

ys
will mark brainliest !!!

1 answer:

nadezda [96]2 years ago

5 0

Kepler's Law (3)

T² ≈ r³

The square of the orbital period ≈ the cube the distance to the sun

The farther away the planet is, the bigger the radius, the bigger the period

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Given that R1 is 13 Ω, R2 is 10 Ω, and R3 is 4 Ω, what is the current coming out of the 9-V battery? What is the power dissipati

Kisachek [45]

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3 years ago

An electron is pushed into an electric field where it acquires a 1-v electrical potential. suppose instead that two electrons ar

Lapatulllka [165]

1V electrical potential

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3 years ago

Read 2 more answers

umka21 [38]

Answer:

Part a)

$a = 3.68 m/s^2$

Part b)

$a = 11.8 m/s^2$

Explanation:

Part a)

For force conditions of two blocks we will have

$m_1g - T = m_1 a$

$T - m_2g = m_2 a$

now from above equations we have

$(m_1 - m_2) g = (m_1 + m_2) a$

$a = \frac{m_1 - m_2}{m_1 + m_2} g$

now we know that

$m_1 = \frac{908}{9.8} = 92.65 kg$

$m_2 = \frac{412}{9.8} = 42 kg$

now from above equation we have

$a = \frac{92.65 - 42}{92.65 + 42}(9.8)$

$a = 3.68 m/s^2$

Part b)

When heavier block is removed and F = 908 N is applied at the end of the string then we have

$F - mg = ma$

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3 years ago

4.2 mol of monatomic gas A interacts with 3.2 mol of monatomic gas B. Gas A initially has 9500 J of thermal energy, but in the p

Komok [63]

Answer:

14657.32 J

Explanation:

Given Parameters ;

Number of moles mono atomic gas A , n 1 = 4 .2 mol

Number of moles mono atomic gas B , n 2 = 3.2mol

Initial energy of gas A , K A = 9500 J

Thermal energy given by gas A to gas B , Δ K = 600 J

Gas constant R = 8.314 J / molK

Let K B be the initial energy of gas B.

Let T be the equilibrium temperature of the gas after mixing.

Then we can write the energy of gas A after mixing as

(3/2)n1RT = KA - ΔK

⟹ (3/2) x 4.2 x 8.314 x T = 9500 - 600

T = (8900 x 3 )/(2x4.2x8.314) = 382.32 K

Energy of the gas B after mixing can be written as

(3/2)n2RT = KB + ΔK

⟹ (3/2) x 3.2 x 8.314 x 382.32 = KB + 600

⟹ KB = [(3/2) x 3.2 x 8.314 x 382.32] - 600

⟹ KB = 14657.32 J

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3 years ago

The angle between the axes of two polarizing filters is 45.0^\circ45.0 ∘ . By how much does the second filter reduce the int

suter [353]

Answer

given,

angle between two polarizing filters = 45°

filter reduce intensity = ?

a) I = I₀ Cos² θ

here θ = 45⁰

$I = \dfrac{I_0}{2}$

intensity of the light is reduced by 0.500

correct answer from the given option D

b) direction of the polarization

θ = 45°

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3 years ago