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Likurg_2 [28]
4 years ago
11

A comet is in an elliptical orbit around the sun. its closest approach to the sun is a distance of 4.5 1010 m (inside the orbit

of mercury), at which point its speed is 9 104 m/s. its farthest distance from the sun is far beyond the orbit of pluto. what is its speed when it is 6 1012 m from the sun?
Physics
1 answer:
barxatty [35]4 years ago
4 0

r1 = 5*10^10 m , r2 = 6*10^12 m

v1 = 9*10^4 m/s

From conservation of energy

K1 +U1 = K2 +U2

0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2

0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2

M is mass of sun = 1.98*10^30 kg

G = 6.67*10^-11 N.m^2/kg^2

0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))

v2 = 5.35*10^4 m/s

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Light of wavelength 480 nm illuminates a pair of slits separated by 0.27 mm. If a screen is place 1.7 m from the slits, determin
Dahasolnce [82]

Answer:

  Δy = 6.05 mm

Explanation:

The double slit phenomenon is described by the expression

      d sin θ = m λ                constructive interference

      d sin θ = (m + ½) λ       destructive interference

      m = 0,±1, ±2, ...

As they tell us that they measure the dark stripes, we are in a case of destructive interference, let's use trigonometry to find the sins tea

      tan θ = y / x

      y = x tan θ

In the interference experiments the measured angle is very small so we can approximate the tangent

      tan θ = sin θ / cos θ

     cos θ = 1

     tan θ = sin θ

     y = x sin θ

We substitute in the destructive interference equation

     d (y / x) = (m + ½) λ

    y = (m + ½) λ x / d

The first dark strip occurs for m = 0 and the third dark strip for m = 2. Let's find the distance for these and subtract it

 m = 0

      y₀ = (0+ ½) 480 10⁻⁹ 1.7 / 0.27 10⁻³

      y₀ = 1.511 10⁻³ m

 m = 2

     y₂ = (2 + ½) 480 10⁻⁹ 1.7 / 0.27 10⁻³

     y₂ = 7.556 10⁻³ m

The separation between these strips is Δy

     Δy = y₂-y₀

     Δy = (7.556 - 1.511) 10⁻³

     Δy = 6.045 10⁻³ m

     Δy = 6.05 mm

5 0
3 years ago
A 100100​-watt ​[W] motor ​(6060​% ​efficient) is available to raise a load 1010 meters​ [m] into the air. If the task takes 808
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Answer:

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The load is 48.93kg.

Explanation:

Given;

Power P = 100watts

Efficiency e = 60% = 0.60

distance d = 10 m

time taken t = 80 s

Acceleration due to gravity g = 9.81m/s

Workdone W can be expressed as;

W = ePt = Fd = mgd

ePt = mgd

Making m the subject of formula;

m = ePt/gd

Substituting the values;

m = (0.60×100×80)/(9.81×10)

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The load is 48.93kg.

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Answer:

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