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3 years ago

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A satellite is in a circular orbit 21000 km above the Earth’s surface; i.e., it moves on a circular path under the influence of

nothing but the Earth’s gravity. Find the speed of the satellite. The radius of the Earth is 6.37 × 106 m, and the acceleration of gravity at the satellite’s altitude is 0.532655 m/s 2. Find the time it takes to complete one orbit around the Earth.

1 answer:

mina [271]3 years ago

6 0

Answer:

(orbital speed of the satellite) V₀ = 3.818 km

Time (t) = 4.5 × 10⁴s

Explanation:

Given that:

The radius of the Earth is 6.37 × 10⁶ m; &

the acceleration of gravity at the satellite’s altitude is 0.532655 m/s

We can calculate the orbital speed of the satellite by using the formula:

Orbital Speed (V₀) = √(r × g)

radius of the orbit (r) = 21000 km + 6.37 × 10⁶ m

= (2.1 × 10⁷ + 6.37 × 10⁶) m

= 27370000

= 2.737 × 10⁷m

Orbital Speed (V₀) = √(r × g)

Orbital Speed (V₀) = √(2.737 × 10⁷ × 0.532655 )

= 3818.215

= 3.818 × 10³

= 3.818 Km

To find the time it takes to complete one orbit around the Earth; we use the formula:

Time (t) = 2 π × $\frac{r}{V_o}$

= 2 × 3.14 × $\frac{2.737*10^7}{3.818*10^3}$

= 45019.28

= 4.5 × 10 ⁴ s

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Complete Question

A radio technician measures the frequency of an AM radio transmitter. The frequency is 14603 kHz . What is the frequency in megahertz? Write your answer as a decimal.

Answer:

The value is $x = 14.6 \ MHz$

Explanation:

From the question we are told that

The frequency is $f = 14603 \ kHz = 14603 *1000 = 14603000 \ Hz$

Generally

$1 Hz \to 1.0 *10^{-6} \ MHz$

$14603000 \ Hz \to x MHz$

=> $x = \frac{14603000 * 1.0*10^{-6}}{1 }$

=> $x = 14.6 \ MHz$

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Kepler deduced this law of motion from observations of Mars. What information confirms his conclusion that the orbit of Mars is

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Kepler noticed an imaginary line drawn from a planet to the Sun and this line swept out an equal area of space in equal times, If we then draw a triangle out from the Sun to a planet’s position at one point in time, it is notice that the area doesn't change even after the planet has left the original position say like after 2 to 3days or 2hours. So to have same area of triangle means that the the planet move faster when that are closer to the sun and slowly when they are far from the sun.

This led to Kepler's law of orbital motion.

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Second Law: The radius vector from the sun to a planet sweeps equal areas in equal times.

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A 15 m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60° angle with horizontal. (a) Find t

scoray [572]

Answer:

a) F₁ = 267.3 N, N₁ = 1300 N, b) μ = 0.324

Explanation:

For this exercise we use the rotational equilibrium condition, we have a reference system is the floor and the anticlockwise rotations as positive, in the adjoint we can see a diagram of the forces

let's use subscript 1 for the ladder and 2 for the firefighter

∑ τ = 0

-W₁ x₁ - W₂ x₂ + N₁ y = 0

N₁ = $\frac{W_1 x_1 + W_2 x_2}{y}$ (1)

the center of mass of the ladder is at its geometric center,

d = L / 2 = 15/2 = 7.5 m

cos 60 = x₁ / d₁

x₁ = d₁ cos 60

x₁ = 7.5 cos 60

x₁ = 3.75 m

for the firefighter d₂ = 4 m

cos 60 = x₂ / d₂

x₂ = d₂ cos 60

x₂ = 4 cos 60 = 2 m

for the fulcrum d₃ = 15 m

sin 60 = y / d₃

y = d₃ sin 60

y = 15 sin 60

y = 13 m

we look for the Normal by substituting in equation 1

N₂ = $\frac{500 \ 3.75 \ + 800 \ 2}{13}$

N₂ = 267.3 N

now let's use the translational equilibrium relations

X axis

F₁ - N₂ = 0

F₁ = N₂

F₁ = 267.3 N

Axis y

N₁ - W₁ -W₂ = 0

N₁ = W₁ + W₂

N₁ = 500 + 800

N₁ = 1300 N

b) for this case change the firefighter's distance d₂ = 9 m

x₂ = 9 cos 60

x₂ = 4.5 m

we substitute in 1

N₂ = \frac{500 \ 3.75 \ + 800 \ 4.5}{13}

N₂ = 421.15 N

of the translational equilibrium equation on the x-axis

fr = F₁ = N₂

fr = 421.15 N

friction force has the expression

fr = μ N

in this case the reaction of the Earth to the support of the ladder is N1 = 1300N

μ = fr / N₁

μ = 421.15 / 1300

μ = 0.324

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slega [8]

Answer:

Explanation:

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The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr

MA_775_DIABLO [31]

Answer:

$q=49Q/64$

and

$x =16R/15$

Explanation:

See attached figure.

$E_{Q}=$ E due to sphere

$E_{q}=$ E due to particule

$E_{total}=E_{Q}-E_{q}=0$ (1)

according to the law of gauss and superposition Law:

$E_{Q}=E_{1}+E_{2}=E_{2}$ ; electric field due to the small sphere with r1=R/4

$E_{Q}=kq_{2}/(r_{1}^{2})=$

$q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}$

then: $E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})$ (2)

on the other hand, for the particule:

$E_{q}=kq/(r_{p}^{2})$

$r_{p}=2R-R/4=7R/4$ ⇒ $E_{q}=16kq/(49R^{2})$ (3)

We replace (2) y (3) in (1):

$E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})$

$q=49Q/64$

--------------------

if R<x<2R AND $E_{total}=E_{Q}-E_{q}=0$

$E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})$

remember that $q=49Q/64$

then:

$Q(2R-x^{2})=49/64*x^{2}$

solving:

$x_{1} =16R/15$

$x_{2} =16R$

but: R<x<2R

so : $x =16R/15$

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