1. Physical size of Russia compared to other countries, despite a lack of visible borders from space.
2. Part of Russia's outline would likely be obscured by the various clouds and objects in the stratosphere; this would allow the astronaut to view potential cloud and weather patterns on earth.
3. An astronaut could see outlines of Russia's geography such as mountain ranges.
Hope that it helps :)
Question:
The water molecules now in your body were once part of a molecular cloud. Only about onemillionth of the mass of a molecular cloud is in the form of water molecules, and the mass density of such a cloud is roughly 2.0×10−21 g/cm^3.
Estimate the volume of a piece of molecular cloud that has the same amount of water as your body.
Answer:
The volume of cloud that has the same density as the amount of water in our body is 1.4×10²⁵ cm³
Explanation:
Here, we have mass density of cloud = 2.0×10⁻²¹ g/cm^3
Density = Mass/Volume
Volume = Mass/Density = If the mass is 40 kg and the body is made up of 70% by mass of water, we have
28 kg water = 28000 g
Therefore the Volume = 28 kg/ 2.0×10⁻²¹ g/cm^3 = 1.4×10¹⁹ m³ = 1.4×10²⁵ cm³.
Therefore, the volume of cloud that has the same density as the amount of water in our body = 1.4×10²⁵ cm³.
Answer:
775.48 W
Explanation:
given,
diameter of disk = 0.6 cm
length of the disk = 0.4 m
T₁ = 450 K T₂ = 450 K T₃ = 300 K
= 1.33
now,
the value of view factor (F₁₂)corresponding to 1.33
F₁₂ = 0.265
F₁₃ = 1 - 0.265 = 0.735
now,
net rate of radiation heat transfer from the disk to the environment:
= 2 F₁₃ A₁ σ (T₁⁴ - T₃⁴)
= 2 x 0.735 x π x (0.3)² x (5.67 x 10⁻⁸ W/m²) (450⁴ - 300⁴)
= 775.48 W
Net radiation heat transfer from the disks to the environment = 775.48 W
Answer:
a = √ (a_t² + a_c²)
a_t = dv / dt
, a_c = v² / r
Explanation:
In a two-dimensional movement, the acceleration can have two components, one in each axis of the movement, so the acceleration can be written as the components of the acceleration in each axis.
a = aₓ i ^ + a_y j ^
Another very common way of expressing acceleration is by creating a reference system with a parallel axis and a perpendicular axis. The axis called parallel is in the radial direction and the perpendicular axis is perpendicular to the movement, therefore the acceleration remains
a = √ (a_t² + a_c²)
where the tangential acceleration is
a_t = dv / dt
the centripetal acceleration is
a_c = v² / r
The equation to be used here is the trajectory of a projectile as written below:
y = xtanθ +/- gx²/2v²(cosθ)²
where
y is the vertical distance
x is the horizontal distance
θ is the angle of trajectory or launch angle
g is 9.81 m/s²
v is the initial velcity
Since the angle is below horizontal, let's use the minus equation. Substituting the values:
- 0.8 m = xtan15° - (9.81 m/s²)x²/2(4.8 m/s)²(cos15°)²
Solving for x,
x = 2.549 m
However, we only take half of this distance because it was specified that the distance asked before bouncing. Hence, the horizontal distance is equal to 1.27 m.
