Explanation:
It is given that,
The Displacement x of particle moving in one dimension under the action of constant force is related to the time by equation as:

Where,
x is in meters and t is in sec
We know that,
Velocity,

(a) i. t = 2 s

At t = 4 s

(b) Acceleration,

Pu t = 3 s in above equation
So,

Hence, (a) (i) v = 55 m/s (ii) v = 211 m/s and (b) 78 m/s²
Most likely they would stick together and keep moving together
Answer:
V₀ₓ = 10.94 m/s
V₀y = 18.87 m/s
Explanation:
To find the launch velocity, we use 1st equation of motion.
Vf = Vi + at
where,
Vf = Final Velocity of Ball = Launch Speed = V₀ = ?
Vi = Initial Velocity = 0 m/s (Since ball was initially at rest)
a = acceleration = 376 m/s²
t = time = 0.058 s
Therefore,
V₀ = 0 m/s + (376 m/s²)(0.058 s)
V₀ = 21.81 m/s
Now, for x-component:
V₀ₓ = V₀ Cos θ
where,
V₀ₓ = x-component of launch velocity = ?
θ = Angle with horizontal = 59.9⁰
V₀ₓ = (21.81 m/s)(Cos 59.9°)
<u>V₀ₓ = 10.94 m/s</u>
<u></u>
for y-component:
V₀ₓ = V₀ Sin θ
where,
V₀y = y-component of launch velocity = ?
θ = Angle with horizontal = 59.9⁰
V₀y = (21.81 m/s)(Sin 59.9°)
<u>V₀y = 18.87 m/s</u>
<u></u>
Velocity velocity is a vector value and needs not only the magnitude (speed) but also a direction(west)
There is no chart provided in this statement. However, the current that flows within an electrically charged item is dependent on how it is arranged. There are two kinds of current and these are, parallel and circuit currents. Current that flows in parallel current is less than when it is farther away from the electric source; current that flows in a series current flows at a constant rate regardless of how far it is from the electrical source.