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sweet [91]
3 years ago
13

What is the angular diameter (in arcseconds) of an object that has a linear diameter of 75 cm and a distance of 2 km?

Physics
1 answer:
charle [14.2K]3 years ago
8 0

Answer:

The angular diameter is 77.35 arc-seconds.

Explanation:

The angular diameter, as shown in the figure, is the angle x  subtended by the the diameter of the object.

Before we do the calculation, we first convert everything to meters.

The diameter of the of the object in meters is

75cm =0.75m,

and the distance to the object in meters is

2 km = 2000 m.

Now, from trigonometry we get:

tan (\frac{x}{2} )= \dfrac{radius}{length} = \dfrac{0.75/2}{2000} \\\\\dfrac{x}{2} = tan^{-1}(\dfrac{0.75/2}{2000})\\\\\dfrac{x}{2}= 0.0107\\\\\boxed{x= 0.0215^o}

and since 1 degree = 3600 arc-seconds, x in arc-seconds is

x= 0.0215*3600 \\\\\boxed{x= 77.35''}

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a) x = 14t^{2} -2t^{3}

at t = 5s

x = 14*5^{2} -2*5^{3} = 14*25 - 2*125 = 350 - 250 = 100 m

b) v = \frac{dx}{dt}

       = 28t - 6t^2

at t = 5 s

v = 28*5-6*25 = 140 - 150 = -10 m/s

c) a = \frac{dv}{dt}

     = 28 - 12t

at t = 5 s

a = 28 -12*5= 28-60= -32 m/s^2

d) At maximum positive coordinate velocity = 0

So, 0 = 28t - 6t^2

         t = \frac{28}{6} = \frac{14}{3} = 4.66 s

  At t = 4.66 s

   X = 14 * 4.66^2 - 2* 4.66^3 = 202.39  m

e) At t = 4.66 s

f) At maximum positive velocity a = 0

   0=28-12t

   t = \frac{28}{12} = \frac{7}{3} = 2.33 s

At t = 2.33 s

V = 28*2.33- 6*2.33^2= 32.67 m/s

g) t = 2.33 s

h) When particle is not moving v = 0

So 0= 28t - 6 t^2

 t = \frac{28}{6} = 4.66 seconds

At t = 4.66 s

a = 28 - 12 * 4. 66 = -27.93m/s^2

i) At t = 0s, X =0m

       t = 5s, X = 100m

So, Displacement = 100m

Velocity = \frac{Displacement}{Time}  = \frac{100}{5} = 20m/s

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What 2 key vocabulary words are used when you send an object in motion?
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Explanation:

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Explanation:

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4 years ago
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If an otherwise empty pressure cooker is filled with air of room temperature and then placed on a hot stove, what would be the m
Xelga [282]

Answer:

The magnitude of the net force F₁₂₀ on the lid when the air inside the cooker has been heated to 120 °C is \frac{135.9}{A}N

Explanation:

Here we have

Initial temperature of air T₁ = 20 °C = ‪293.15 K

Final temperature of air T₁ = 120 °C = 393.15 K

Initial pressure P₁ = 1 atm = ‪101325 Pa

Final pressure P₂ = Required

Area = A

Therefore we have for the pressure cooker, the volume is constant that is does not change

By Chales law

P₁/T₁ = P₂/T₂

P₂ = T₂×P₁/T₁ = 393.15 K× (‪101325 Pa/‪293.15 K) = ‭135,889.22 Pa

∴ P₂ = 135.88922 KPa = 135.9 kPa

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4 0
4 years ago
3–101 It is estimated that 90 percent of an iceberg’s volume is below the surface, while only 10 percent is visible above the su
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Answer:

The density of iceberg upper water and under water are 102.5 kg/m³ and 922.5 kg/m³

Explanation:

Given that,

Volume of seawater = 90 % of Volume of iceberg

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We need to calculate the density of the iceberg

Using equilibrium condition

W_{I}=F_{B}

\rho_{I}\times V_{I}\times g= \rho_{w}\timesV_{w}\times g

Put the value into the formula

\rho_{I}\times V_{I}\times g=\rho_{w}\times0.9V_{I}\times g

\rho_{I}=1025\times0.9

\rho_{I}=922.5kg/m^3

We need to calculate the density of the iceberg

Using equilibrium condition

W_{I}=F_{B}

\rho_{I}\times V_{I}\times g= \rho_{w}\timesV_{w}\times g

Put the value into the formula

\rho_{I}\times V_{I}\times g=\rho_{w}\times0.1V_{I}\times g

\rho_{I}=1025\times0.1

\rho_{I}=102.5\ kg/m^3

Hence, The density of iceberg upper water and under water are 102.5 kg/m³ and 922.5 kg/m³

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3 years ago
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