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3 years ago

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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The position of a particle moving along an x axis is given by x = 14.0t2 - 2.00t3, where x is in meters and t is in seconds. det

Mumz [18]

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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3 years ago

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

ad-work [718]

Answer:

push or pull

Explanation:

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3 years ago

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What is the radius of a nucleus of an atom?

Advocard [28]

Answer:

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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

Where Force = $\frac{Pressure}{Area}$ we have

Force = $F_{120}=\frac{135.9}{A}N$ .

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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

Artist 52 [7]

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

Volume of visible surface = 10 %

Density of seawater = 1025 kg/m³

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