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umka21 [38]
3 years ago
5

n a distant solar system, a planet of mass 5.0 x 1024 kg orbits a sun of mass 3.0 x 1030 kg at a constant distance of 2.0 x1011

m. How many earth days does it take for the planet ot execute one complete orbit about the sun
Physics
1 answer:
Anna35 [415]3 years ago
4 0

Answer:

F = M2 ω^2 R       centripetal force of sun on planet

ω = (F / (M2 R))^1/2 = 2 pi f = 2 pi / P        where P is the period

P = 2 pi (M2 * R / F)^1/2

F = G M1 M2 / R^2        gravitational force on planet

P = 2 pi {R^3 / (G M1)]^1/2

P = 6.28 [(2.0E11)^3 / (6.67E-11 * 3.0E30)]^1/2

P = 6.28 (8 / 20)^1/2 E7 = 3.9E7 sec

1 yr = 3600 * 24 * 365 = 3.15E7 sec

P = 3.9 / 3.2 = 1.2 years

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According to the theory of plate tectonics, the lithosphere is separated into sections that are called tectonic plates. These pl
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4 0
3 years ago
In lecture, it was discussed measurements of mass, temperature, time, volume and length. which of these is not an si base unit (
photoshop1234 [79]

From the options provided in the question, the measurement which is not an SI base unit is volume.

<h3>What is SI base unit?</h3>

This is referred to as the standard and fundamental unit of measurement of various quantities or variables which is defined arbitrarily and not by combinations of other units.

Volume is a quantity which is derived from the combination of lengths in a  three-dimensional manner which is why the formula is length× breadth×height and the unit is cm³. This is gotten from the combination of the unit of length which is cm.

This is therefore the reason why volume was chosen as the most appropriate choice.

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6 0
1 year ago
At what position or positions on the x-axis is the electric field zero?
ElenaW [278]

Answer:

The electric field will be zero at x = ± ∞.

Explanation:

Suppose, A -2.0 nC charge and a +2.0 nC charge are located on the x-axis at x = -1.0 cm and x = +1.0 cm respectively.

We know that,

The electric field is

E=\dfrac{kq}{r^2}

The electric field vector due to charge one

\vec{E_{1}}=\dfrac{kq_{1}}{r_{1}^2}(\hat{x})

The electric field vector due to charge second

\vec{E_{2}}=\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})

We need to calculate the electric field

Using formula of net electric field

\vec{E}=\vec{E_{1}}+\vec{E_{2}}

\vec{E_{1}}+\vec{E_{2}}=0

Put the value into the formula

\dfrac{kq_{1}}{r_{1}^2}(\hat{x})+\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})=0

\dfrac{kq_{1}}{r_{1}^2}(\hat{x})=\dfrac{kq_{2}}{r_{2}^2}(\hat{x})

(\dfrac{r_{2}}{r_{1}})^2=\dfrac{q_{2}}{q_{1}}

\dfrac{r_{2}}{r_{1}}=\sqrt{\dfrac{q_{2}}{q_{1}}}

Put the value into the formula

\dfrac{2.0+x}{x}=\pm\sqrt{\dfrac{2.0}{2.0}}

2.0+x=x

If x = ∞, then the equation is be satisfied.

Hence, The electric field will be zero at x = ± ∞.

4 0
3 years ago
B. Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the
goblinko [34]

We have that the instantaneous velocity of the shuttlecock when it hits the ground is

V_{int}=\sqrt{U^2+19.6H}

From the question we are told

Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the

shuttlecock when it hits the ground? Show your work below.

Generally the equation for acceleration  is mathematically given as

a=v \frac{dv}{dx}\\\\\Therefore\\\\\2ah=v^2-u^2

Where

acceleration is still -9.81 m/s2,

Hence,

V^2-U^2=2(-9.81)*-H

Therefore

V_{int}=\sqrt{U^2+19.6H}

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