Answer:
Explanation:
If a small piece of Styrofoam packing material is dropped from a height of 1.90 m above the ground and reaches a terminal speed after falling 0.400m, the Change in distance will be 1.90m - 0.400 = 1.50m
If it takes 5.4secs fo r the Styrofoam to reach the ground, the terminal velocity will be expressed as;
Vt = change in distance/time
Vt = 1.5m/5.4s
Vt = 0.28m/s
Note that the Styrofoam reaches its final velocity when the acceleration is zero.
To get the constant value B from the equation a = g-Bv
a = 0m/s²
g = 9.81m/s²
v = 0.28m/s
Substituting the parameters into the formula.
0 = 9.81-0.28B
-9.81 = -0.28B
Divide both sides by -0.28
B = -9.81/-0.28
B = 35.04
b) at t = 0sec, the initial terminal velocity is also zero.
Substituting v = 0 into the equation to get the acceleration.
a = g-Bv
a = g-B(0)
a = g
Hence the acceleration at t =0s is equal to the acceleration due to gravity which is 9.81m/s²
c) Given speed v = 0.150m/s
g = 9.81m/s²
B = 35.04
Substituting the given data into the equation a = g-Bv
a = 9.81-35.04(0.15)
a = 9.81 - 5.26
a = 4.55m/s²
Answer:
a
[tex]v = 0.28 \ m/s [/tex]
b
[tex]a = 9.8 m/s^2[/tex]
c
[tex]a = 4.55 \ m/s^2[/tex]
Explanation:
From the question we are told that
The fall height is [tex]h = 1.90 \ m[/tex]
The magnitude of the acceleration is [tex]a = g -Bv[/tex]
The height at which terminal speed is attained is [tex]h_t = 0.400 \ m[/tex]
The time taken to reach the ground from the terminal velocity height is t = 5.40 s
Generally the height which object traveled with terminal velocity is
[tex]H = h- h_t[/tex]
=> [tex]H = 1.90 - 0.400[/tex]
=> [tex]H = 1.50 \ m [/tex]
Generally the terminal velocity is mathematically represented as
[tex]v = \frac{H}{t}[/tex]
=> [tex]v = \frac{1.50}{5.40}[/tex]
=> [tex]v = 0.28 \ m/s [/tex]
Generally at terminal velocity , acceleration is zero so
[tex] g -Bv = 0 [/tex]
substituting
[tex]9.8 \ m/s^2[/tex]
=> [tex] 9.8 -B (0.28) = 0 [/tex]
=> [tex] v = 35 \ m/s [/tex]
Generally at t = 0 s the velocity v = 0 (That no motion at time zero)
So from acceleration equation
[tex]a = 9.8 -35(0)[/tex]
=> [tex]a = 9.8 m/s^2[/tex]
Generally when the speed v = 0.150 m/s
The acceleration is
[tex]a = 9.8 -35(0.150)[/tex]
=> [tex]a = 4.55 \ m/s^2[/tex]
