F=ma so,we have
f=300.3=900 N
but you should say 3 m/s^2
Answer:
θ=5.65°
Explanation:
Given Data
Mass m=1.5 kg
Length L=0.80 m
First spring constant k₁=35 N/m
Second spring constant k₂=56 N/m
To find
Angle θ
Solution
As the both springs take half load so apply Hooks Law:
Force= Spring Constant ×Spring stretch
F=kx
x=F/k
as
![d=x_{1}-x_{2}\\ as \\x=F/k\\so\\d=\frac{F_{1} }{k_{1}} -\frac{F_{2}}{k_{2}}\\ Where \\F=1/2mg\\d=\frac{(1/2)mg}{k_{1}} -\frac{(1/2)mg}{k_{2}}\\ d=\frac{mg}{2}(\frac{1}{k_{1}} -\frac{1}{k_{2}} )\\ And\\Sin\alpha=d/L\\\\alpha =sin^{-1}[\frac{mg}{2L}(1/k_{1}-1/k_{2})]\\\alpha =sin^{-1}[\frac{(1.5kg)(9.8m/s^{2} )}{2(0.80m)}(1/35Nm-1/56Nm) ]\\\alpha =5.65^{o}](https://tex.z-dn.net/?f=d%3Dx_%7B1%7D-x_%7B2%7D%5C%5C%20%20as%20%5C%5Cx%3DF%2Fk%5C%5Cso%5C%5Cd%3D%5Cfrac%7BF_%7B1%7D%20%7D%7Bk_%7B1%7D%7D%20-%5Cfrac%7BF_%7B2%7D%7D%7Bk_%7B2%7D%7D%5C%5C%20Where%20%5C%5CF%3D1%2F2mg%5C%5Cd%3D%5Cfrac%7B%281%2F2%29mg%7D%7Bk_%7B1%7D%7D%20-%5Cfrac%7B%281%2F2%29mg%7D%7Bk_%7B2%7D%7D%5C%5C%20d%3D%5Cfrac%7Bmg%7D%7B2%7D%28%5Cfrac%7B1%7D%7Bk_%7B1%7D%7D%20-%5Cfrac%7B1%7D%7Bk_%7B2%7D%7D%20%29%5C%5C%20And%5C%5CSin%5Calpha%3Dd%2FL%5C%5C%5C%5Calpha%20%3Dsin%5E%7B-1%7D%5B%5Cfrac%7Bmg%7D%7B2L%7D%281%2Fk_%7B1%7D-1%2Fk_%7B2%7D%29%5D%5C%5C%5Calpha%20%20%20%3Dsin%5E%7B-1%7D%5B%5Cfrac%7B%281.5kg%29%289.8m%2Fs%5E%7B2%7D%20%29%7D%7B2%280.80m%29%7D%281%2F35Nm-1%2F56Nm%29%20%5D%5C%5C%5Calpha%20%3D5.65%5E%7Bo%7D)
θ=5.65°
Answer:
what is your question?..... .
Answer:
YEss this is her sister
Explanation: got a problem
Answer:
Explanation:
initial angular velocity, ωo = 0 rad/s
angular acceleration, α = 30.5 rad/s²
time, t = 9 s
radius, r = 0.120 m
let the velocity is v after time 9 s.
Use first equation of motion for rotational motion
ω = ωo + αt
ω = 0 + 30.5 x 9
ω = 274.5 rad/s
v = rω
v = 0.120 x 274.5
v = 32.94 m/s