1. What are the two quantities involved when calculating MOMENTUM?
1 answer:
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Answer:
0.767m
Explanation:
We are given that the time interval between each droplet is equal.
We are also given that the fourth drop is just dripping from the shower when the first hits the floor.
If they fall at the same time interval and we know that the distance between the shower head and floor are the same, they must therefore fall at the same velocity.
The distance between each drop has to be the same given that they fall at equal time intervals.
Let this distance be x.
We can then partition the entire height of the system into three parts (as shown in the diagram).
Hence, we can say that:
x + x + x = 2.3m
3x = 2.3m
=> x = 2.3/3 = 0.767m
Therefore, at the time the first drop hits the floor, the third drop is only 0.767 m below the shower head.
Answer:
v = (-4.44 i^ + 6.66 j^ ) m/s, a_average =( 0 i^ -2π j^) m/s²
Explanation:
The expression left corresponds to an oscillatory movement (MAS), the speed is defined by
v = dr / dt
the function of position
r = 2 cos πt i^ + 3 sin πt j^
let us note that it is a movement in two dimensions
let's perform the derivative
v = -2π sin πt i^ + 3π cos πt j^
we evaluate this expression for t = 0.25 s, remember that the angle is in radians
v = -2π sin (π 0.25) i^ + 3π cos (π 0.25) j^
v = (-4.44 i^ + 6.66 j^ ) m/s
To calculate the mean acceleration we use the expression
a = () / Δt
indicates that the time is the first 3 s
we look for the initial velocity t = 0 s
v₀ = 0 i ^ + 3π j ^
we look for the fine velocity, t = 3 s
v_f = - 2π sin (π 3) + 3π cos (π 3) j ^
v_f = 0 i ^ - 3π j ^
we calculate the average acceleration
Δt = (3 -0) = 3 s
a_average = (0-0) / 3 i ^ + (-3π - 3π) / 3
a_average = (0 i ^ -2π j ^ ) m/s²