Answer:
Part a)

Part b)

Since the distance of other building is 15 m so YES it can make it to other building
Part c)

direction of velocity is given as
![[tex]\theta = 26.35 degree](https://tex.z-dn.net/?f=%5Btex%5D%5Ctheta%20%3D%2026.35%20degree)
Explanation:
Part a)
acceleration due to gravity on this planet is 3/4 times the gravity on earth
So the acceleration due to gravity on this new planet is given as


now the vertical displacement covered by the canister is given as

now by kinematics we have



Part b)
Horizontal speed of the canister is given as

now the distance moved by it



Since the distance of other building is 15 m so YES it can make it to other building
Part c)
Final velocity in X direction will remains the same

final velocity in Y direction



now magnitude of velocity is given as



direction of velocity is given as


![[tex]\theta = 26.35 degree](https://tex.z-dn.net/?f=%5Btex%5D%5Ctheta%20%3D%2026.35%20degree)
Convection, because it is the process of heat transfer from one location to the next by the movement of fluids. The moving fluid carries energy within it.
The formula that can be used to obtain a convergent series is given by;
a/1 – r.
<h3>What is a series?</h3>
In mathematics, we define a series as sum of numbers which could be convergent or divergent. In a convergent series, the summation of the numbers approaches a given value.
The formula that can be used to obtain a convergent series is given by;
a/1 – r.
Learn more about a convergent series:brainly.com/question/15415793?
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An example would be 2 types of motion. It could be rectilinear or projectile motion. There are various equations for each type. Since you don't want me to tell you the answer, I could just express it in words. Then, it will be up to you to translate into mathematical equations.
For rectilinear motion, the distance traveled is equal to the initial velocity times the time, plus one-half of the acceleration times the square of the time. For projectile motion, the maximum distance is equal to the square of the initial velocity multiplied with the square of the sine of the launch angle, all over twice the gravity.
Answer:
12.0 meters
Explanation:
Given:
v₀ = 0 m/s
a₁ = 0.281 m/s²
t₁ = 5.44 s
a₂ = 1.43 m/s²
t₂ = 2.42 s
Find: x
First, find the velocity reached at the end of the first acceleration.
v = at + v₀
v = (0.281 m/s²) (5.44 s) + 0 m/s
v = 1.53 m/s
Next, find the position reached at the end of the first acceleration.
x = x₀ + v₀ t + ½ at²
x = 0 m + (0 m/s) (5.44 s) + ½ (0.281 m/s²) (5.44 s)²
x = 4.16 m
Finally, find the position reached at the end of the second acceleration.
x = x₀ + v₀ t + ½ at²
x = 4.16 m + (1.53 m/s) (2.42 s) + ½ (1.43 m/s²) (2.42 s)²
x = 12.0 m