The concept required to solve this problem is hydrostatic pressure. From the theory and assuming that the density of water on that planet is equal to that of the earth 
we can mathematically define the pressure as
Where,
= Density
h = Height
g = Gravitational acceleration
Rearranging the equation based on gravity
The mathematical problem gives us values such as:
Replacing we have,
Therefore the gravitational acceleration on the planet's surface is 
Answer:
d. Mass is the amount of matter in an object.
Explanation:
This is the exact definition of the word mass. My mom is a science teacher.
Acceleration is the rate at which velocity changes.
Answer:
Lacks colored blood - Scorpion
Soft, unsegmented body - Octopus
A vertebrate - Bird
Lacks antennae - Starfish
Explanation:
Scorpion belongs to the Phylum Arthropoda of the Kingdom Animalia. All the organisms of this phylum lack coloured blood.
Octopus comes under the Phylum Molusca of the Kingdom Animalia. Soft, unsegmented body is the property of the organisms of this phylum.
Bird belongs to the Vertebrata class of the Phylum Chordata of Kingdom Animalia.
Starfish belongs to Echinodermata Phylum of the Kingdom Animalia. The organisms of this category do not posses antennae.
Answer:
1.696 nm
Explanation:
For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,
dsinθ = mλ = (1)λ = λ
dsinθ = λ
sinθ = λ/d.
Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m
From trig ratios 1 + cot²θ = cosec²θ
1 + (1/tan²θ) = 1/(sin²θ)
substituting the values of sinθ and tanθ we have
1 + (D/w)² = (d/λ)²
(D/w)² = (d/λ)² - 1
(w/D)² = 1/[(d/λ)² - 1]
(w/D) = 1/√[(d/λ)² - 1]
w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹ = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.
w is also the distance from the center to the other principal maximum on the other side.
So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm
So, the minimum width of the screen must be 1.696 nm
