Answer:
The graph should have velocity (v) on the y-axis and radius (r) on the x-axis. It will have a straight, horizontal line that goes across the graph.
Explanation:
Shown above is the formula for Kinetic Energy in rotational terms. I'm new to brain.ly so I couldn't insert the omega symbol, sorry about that. Omega can be replaced with 
. Moment of Inertia (I) can be replaced with 
.
The equation becomes 
.
The r's cancel out, making the different radii negligible, causing a straight horizontal line.
The primary additive colors are red, green, and blue, which means that any color can be constructed from a linear superposition of these colors. According to this RGB (Red, Green, Blue) refers to the system for representing colors on a computer display. It is not possible <span>that someone could have a color photograph that cannot be represented using full 24-bit color. Every color photograph can be represented using the RGB.</span>
Answer:
y = 54.9 m
Explanation:
For this exercise we can use the relationship between the work of the friction force and mechanical energy.
Let's look for work
W = -fr d
The negative sign is because Lafourcade rubs always opposes the movement
On the inclined part, of Newton's second law
Y Axis
N - W cos θ = 0
The equation for the force of friction is
fr = μ N
fr = μ mg cos θ
We replace at work
W = - μ m g cos θ d
Mechanical energy in the lower part of the embankment
Em₀ = K = ½ m v²
The mechanical energy in the highest part, where it stopped
= U = m g y
W = ΔEm = - Em₀
- μ m g d cos θ = m g y - ½ m v²
Distance d and height (y) are related by trigonometry
sin θ = y / d
y = d sin θ
- μ m g d cos θ = m g d sin θ - ½ m v²
We calculate the distance traveled
d (g syn θ + μ g cos θ) = ½ v²
d = v²/2 g (sintea + myy cos tee)
d = 9.8 12.6 2/2 9.8 (sin16 + 0.128 cos 16)
d = 1555.85 /7.8145
d = 199.1 m
Let's use trigonometry to find the height
sin 16 = y / d
y = d sin 16
y = 199.1 sin 16
y = 54.9 m
E = ½KA^2 is the mechanical energy of any oscillator. It is the sum of elastic potential energy and
kinetic energy. When amplitude A
decreases by 3%, then
(E2-E1)/E1 = {½K(A2^2/A1^2) }/ ½K(A1^2)
= {(A2^2 – A1^2) / (A1^2)}
= 97^2 – 100^2/100^2
= 5.91% of the mechanical energy is lost each cycle.
<em>projectile can only follow the straight line path when it is launched upward straightly so the correct option is <u>90 degree with respect to horizontal x -axis ..:)</u></em>
