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2 years ago

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Two planets, Dean and Sam, orbit the Sun. They each have with circular orbits, but orbit at different distances from the Sun. De

an orbits at a greater average distance than Sam. According to Kepler's Third Law, which planet will have a longer orbital period? Group of answer choices Dean Sam Since they both have circular orbits, they will have the same orbital periods. There isn't enough information to tell.

1 answer:

lyudmila [28]2 years ago

3 0

Answer:

The correct answer is Dean has a period greater than San

Explanation:

Kepler's third law is an application of Newton's second law where the force is the universal force of attraction for circular orbits, where it is obtained.

T² = (4π² / G M) r³

When applying this equation to our case, the planet with a greater orbit must have a greater period.

Consequently Dean must have a period greater than San which has the smallest orbit

The correct answer is Dean has a period greater than San

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Speed is a component of skill related fitness what does speed enable you to do.

Mrrafil [7]

I'd say move faster, unless it's asking something else.

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2 years ago

I NEED HELP PLEASE, THANKS! :)

mrs_skeptik [129]

Answer:

1. Largest force: C; smallest force: B; 2. ratio = 9:1

Explanation:

The formula for the force exerted between two charges is

$F=K\dfrac{ q_{1}q_{2}}{r^{2}}$

where K is the Coulomb constant.

q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.

For simplicity, let's combine Kq₁q₂ into a single constant, k.

Then, we can write

$F=\dfrac{k}{r^{2}}$

1. Net force on each particle

Let's

Call the distance between adjacent charges d.
Remember that like charges repel and unlike charges attract.

Define forces exerted to the right as positive and those to the left as negative.

(a) Force on A

$\begin{array}{rcl}F_{A} & = & F_{B} + F_{C} + F_{D}\\& = & -\dfrac{k}{d^{2}} - \dfrac{k}{(2d)^{2}} +\dfrac{k}{(3d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(-1 - \dfrac{1}{4} + \dfrac{1}{9} \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-36 - 9 + 4}{36} \right)\\\\& = & \mathbf{-\dfrac{41}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(b) Force on B

$\begin{array}{rcl}F_{B} & = & F_{A} + F_{C} + F_{D}\\& = & \dfrac{k}{d^{2}} - \dfrac{k}{d^{2}} + \dfrac{k}{(2d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1}{4} \right)\\\\& = &\mathbf{\dfrac{1}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(C) Force on C

$\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}} + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(d) Force on D

$\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}} - \dfrac{k}{(2d)^{2}} - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(e) Relative net forces

In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

$F_{A} : F_{B} : F_{C} : F_{D} = \dfrac{41}{36} : \dfrac{1}{4} : \dfrac{9}{4} : \dfrac{49}{36}\ = 41 : 9 : 81 : 49\\\\\text{C experiences the largest net force.}\\\text{B experiences the smallest net force.}\\$

2. Ratio of largest force to smallest

$\dfrac{ F_{C}}{ F_{B}} = \dfrac{81}{9} = \mathbf{9:1}\\\\\text{The ratio of the largest force to the smallest is $\large \boxed{\mathbf{9:1}}$}$

7 0

2 years ago

A man has a mass of 66 kg on the earth. What is his weight.

LUCKY_DIMON [66]

Since force = mass X acceleration, then the force he excerts on the earth (aka his weight) equals his mass times the force of gravity.

Therefore
W = (66 kg) X (9.8 m/ss)
W = 646.8 kgm/ss

kg m/ss are also known as Newtons, so your answer is...

646.8 N

8 0

2 years ago

When a second student joins the first, the height difference between the liquid levels in the right and left pistons is 40 cm .

vlada-n [284]

Answer:

m = 56.5 kg

Explanation:

Since the addition of mass on one piston caused a change in pressure head at the other. Diameter of the piston calculated is used as 0.46 m

Δm*g / Area = p * g * Δh ..... Eq1

$m = \frac{p*h*A}{g}$

$m = \frac{850 * 0.4 * pi*0.46^2}{4*9.81}\\\\m= 56.5 kg$

8 0

2 years ago

A flux density of 1.2Wb/m^2 is required in the 1 mm air gap of an electromagnet having an iron path of length 1.5 m. Calculate t

Mandarinka [93]

Answer:

The mmf required is $1.125$ × $10^{-3}$ A

Explanation:

The Magnetomotive force (mmf) is given by the formula below

$F_{M} = Hl\\$

where $F_{M}$ is the Magnetomotive force (mmf)

$H$ is the Magnetic field strength

$l$ is the magnetic length

The magnetic permeability μ is given by

μ = $B / H$

Where $B$ is the Magnetic flux density

and $H$ is the Magnetic field strength

From the question,

$B$ = 1.2Wb/m^2

μ = 1600m

From μ = $B / H$

∴ $H = B/$ μ

$H = 1.2 / 1600\\$

$H = 7.5$ × $10^{-4}$ A/m

Now, for the Magnetomotive force (mmf)

$F_{M} = Hl\\$

From the question

$l$ = 1.5 m

∴ $F_{M} = 7.5$ × $10^{-4}$ × $1.5$

$F_{M} = 1.125$ × $10^{-3} A$

Hence, The mmf required is $1.125$ × $10^{-3}$ A

3 0

2 years ago