the answer is the Spiral Galaxy
Incomplete question as time is missing.I have assumed some times here.The complete question is here
Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, (d) 2.00 s, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 10 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.
Explanation:
Given data
Vi=10 m/s
S=70 m
(a) t₁=0.5 s
(b) t₂=1 s
(c) t₃=1.5 s
(d) t₄=2 s
(e) t₅=2.5 s
To find
Displacement S from t₁ to t₅
Velocity V from t₁ to t₅
Solution
According to kinematic equation of motion and given information conclude that v is given by

Also get the equation of displacement

These two formula are used to find velocity as well as displacement for time t₁ to t₅
For t₁=0.5 s

For t₂
For t₃

For t₄

For t₅

The object particle move faster.
Answer:
A paper being burned is a chemical change.
Answer:
678.2 km/h and 80.54° north of east
Explanation:
From the question,
Using pythagoras theorem,
a² = b²+c²..................... Equation 1
Where a = resultant velocity
Given: b = 600 km/h, c = 100 km/h
Substitute these values into equation 1
R² = 600²+100²
R² = 360000+10000
R² = 460000
R = √460000
R = 678.2 km/h.
And the direction is
tanθ = 600/100
tanθ = 6
tanθ = 6
θ = tan⁻¹(6)
θ = 80.54°.
Hence the resultant velocity of the aircraft is 678.2 km/h and 80.54° north of east