Water and baking soda can be used, too.
gravitational potential is directly proportional to the height of the object relative to a reference line and is given as
PE = mgh
where m = mass of object , g = acceleration due to
gravity and h = height of the object above the reference line .
as the skydiver falls , its height above the ground decrease and hence the gravitational potential energy of the skydiver decrease.
as per conservation of energy , total energy of the skydiver must remain constant all the time . hence the decrease in potential energy appears as increase in kinetic energy by same amount to keep the total energy constant
KE + PE = Total energy
so as the skydiver falls , it gains speed and hence the kinetic energy of skydiver increase since kinetic energy is directly proportional to the square of the speed.
when the parachute opens, the skydiver experience force in upward which tries to balance the weight of the skydiver. hence the speed of the skydiver decrease until upward force becomes equal to the downward force. hence the kinetic energy decrease just after the parachute opens
<h3>
Answer:</h3>
225 meters
<h3>
Explanation:</h3>
Acceleration is the rate of change in velocity of an object in motion.
In our case we are given;
Acceleration, a = 2.0 m/s²
Time, t = 15 s
We are required to find the length of the slope;
Assuming the student started at rest, then the initial velocity, V₀ is Zero.
<h3>Step 1: Calculate the final velocity, Vf</h3>
Using the equation of linear motion;
Vf = V₀ + at
Therefore;
Vf = 0 + (2 × 15)
= 30 m/s
Thus, the final velocity of the student is 30 m/s
<h3>Step 2: Calculate the length (displacement) of the slope </h3>
Using the other equation of linear motion;
S = 0.5 at + V₀t
We can calculate the length, S of the slope
That is;
S = (0.5 × 2 × 15² ) - (0 × 15)
= 225 m
Therefore, the length of the slope is 225 m
Answer:
α = 13.7 rad / s²
Explanation:
Let's use Newton's second law for rotational motion
∑ τ = I α
we will assume that the counterclockwise turns are positive
F₁ 0 + F₂ R₂ - F₃ R₃ = I α
give us the cylinder moment of inertia
I = ½ M R₂²
α = (F₂ R₂ - F₃ R₃) 
let's calculate
α = (24 0.22 - 13 0.10) 
2/12 0.22²
α = 13.7 rad / s²
