Respuesta :
Answer:
a).[tex]I=0.3794 N/s[/tex]
b).[tex]F_{avg}=210.78 N[/tex]
c).[tex]F_{max}=315.8N[/tex]
d).[tex]v_b=0.843 m/s[/tex]
Explanation:
Given the function and the information of the motion
[tex]m_f=0.45kg[/tex]
Δ[tex]T=1.80x10^{-3}s[/tex]
[tex]f(t)=[7.01*10^5*t-3.89x10^8*t^2][/tex]
a).
[tex]I=\int\limits^a_b {f(t)} \, dt[/tex]
[tex]I=\int\limits^{1.8x10{-3}}_0 {[7.01x10^5*t -3.89x10^8]t^2} \, dt[/tex]
[tex]I=[\frac{7.01*10^5}{2}*t^2-\frac{3.89x10^8}{3}*t^3]|0, 1.8x10^{-3}[/tex]
[tex]I=0.3794 N/s[/tex]
b).
[tex]F_{avg}=\frac{I}{t}=\frac{0.379N/s}{1.8x10^{-3}s}=210.78 N[/tex]
c).
[tex]F'(t)=7.01x1^5-7.78x10^8*t[/tex]
[tex]t=9.01x10^{-4}s[/tex]
[tex]F''*(t)=-7.78x10^8[/tex]
[tex]t_{max}=0.901x10^{-3}s[/tex]
[tex]F_{max}(t_{max})=7.01x10^5*(0.901x10^{-3}s)- 3.89x10^8*(0.901x10^{-3}s)^2[/tex]
[tex]F_{max}=315.8N[/tex]
d).
[tex]I=m_b*v_b[/tex]
[tex]v_b=\frac{I}{m_b}[/tex]
[tex]v_b=\frac{0.3794N/s}{0.45kg}=0.843 m/s[/tex]
The impulse, average force, maximum force and the balls speed immediately after loosing contact are; I = 0.3794 N/s; 210.78; 316 N; 0.843 m/s
What is the force and impulse?
We are given;
Mass; m = 0.45 kg
time; t = 1.8 * 10⁻³ s
Force; F(t) = [(7.01 * 10⁵)t - (3.89 * 10⁸)t²]
A) Impulse is gotten from;
∫ [(7.01 * 10⁵)t - (3.89 * 10⁸)t²] between the boundary of 1.8 * 10⁻³ s and 0 s.
Integrating this between those boundaries gives the impulse as;
I = 0.3794 N/s
B) Average force is gotten from the formula;
F_avg = I/t
F_avg = 0.3794/(1.8 * 10⁻³)
F_avg = 210.78 N
C) Change in force with time is;
df/dt = F'(t) = (7.01 * 10⁵) - (7.78 * 10⁸)t
At F'(t) = 0, we have;
t = 9.01 * 10⁻⁴ s
Maximum force is;
F_max = [(7.01 * 10⁵)(9.01 * 10⁻⁴) - (3.89 * 10⁸)(9.01 * 10⁻⁴)²]
F_max = 316 N
D) To find the speed, we will use the formula;
v = I/m
v = 0.3794/0.45
v = 0.843 m/s
Read more about Force and Impulse at; https://brainly.com/question/20586658
