Answer:
x = 11.23 m
Explanation:
For this interesting exercise, we must use angular kinematics, linear kinematics and the relationship between angular and linear quantities.
Let's reduce to SI system units
θ = 155 rev (2pi rad / rev) = 310π rad
α = 2.00rev / s2 (2pi rad / 1 rev) = 4π rad / s²
Let's look for the angular velocity at the time the piece is released, with starting from rest the initial angular velocity is zero (wo = 0)
w² = w₀² + 2 α θ
w =√ 2 α θ
w = √(2 4pi 310pi)
w = 156.45 rad / s
The relationship between angular and linear velocity
v = w r
v = 156.45 0.175
v = 27.38 m / s
In this part we have the linear speed and the height that it travels to reach the floor, so with the projectile launch equations we can find the time it takes to arrive
y = 
t - ½ g t²
As it leaves the highest point its speed is horizontal
y = 0 - ½ g t²
t = √ (-2y / g)
t = √ (-2 (-0.820) /9.8)
t = 0.41 s
With this time we calculate the horizontal distance, because the constant horizontal speed
x = vox t
x = 27.38 0.41
x = 11.23 m
Answer:
trigonometry (guessing)
Explanation:
ellipse: is the shape of an orbit : looks like an oval
periapsis : shortest distance between something like the moon and the planet its orbiting around like the earth
parallax is triangulation. like how gps works. looking at a star one day and then looking at it again 6 months later, an astronomer can see a difference in the viewing angle for the star. With trigonometry, the different angles yield a distance. This technique works for stars within about 400 light years of earth
https://science.howstuffworks.com/question224.htm
By comparing the intrinsic brightness to the star's apparent brightness we can calculate the distance of stars
1/r^2 rule states that the apparent brightness of a light source is proportional to the square of its distance.Jan 11, 2022
https://www.space.com/30417-parallax.html
alternative distance measurement for stars used by most astronomers is the parsec. A star with a parallax angle of 1 arcsecond has a distance of 1 parsec, or 1 parsec per arcsecond of parallax, which is about 3.26 light years
blossoms.mit.edu
.
"Changing water salinity" is the most significant challenge for organisms that live in estuaries.
<u>Answer:</u> Option D
<u>Explanation:</u>
For estuaries, alkalinity levels are usually the maximum at a river's mouth where the ocean water falls for, and the minimum upstream where freshwater falls in. Although salinity vary throughout the tidal cycle. In estuaries, salinity rates usually decrease in spring as snow melt and rain raises the freshwater flow from streams and groundwater.
It influences the chemical environments within the estuary, especially the dissolved oxygen (DO) levels in the water. The level of oxygen that would get dissolved in water or its solubility get declined when the alkalinity rises.
Answer:
The volume of copper is 2.198 ml
Explanation:
Given;
mass of copper, m = 20 g
density of copper, ρ = 9.1 g/ml
Density is given by;
Density = mass / volume
Volume = mass / density
Volume = (20 g) / (9.1 g/ml)
Volume = 2.198 ml
Therefore, the volume of copper is 2.198 ml
