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°
Electroreception is limited to aquatic environments because on here is the resistivity of the medium is low enough for electric currents to be generated as the result of electric fields of biological origin. In air, the resistivity of the environment is so high that electric fields from biological sources do not generate a significant electric current. Electroreceptor are found in a number of species of fish, and in at least one species of mammal, the Duck-Billed platypus.
Answer:
Magnetic energy stored in the inductor when all of the energy in the circuit is in the inductor = 0.049 mJ
If all the energy is then transferred into the capacitor, the voltage drop across the capacitor = 0.00572 V = 0.01 V (expressed to the hundredths value)
Explanation:
In an RLC circuit with maximum current of 7mA = 0.007 A
When all of the energy is stored in the inductor, maximum current will flow through it,
Hence E = (1/2) LI²
L = inductance of the inductor = 2 H
E = (1/2) (2)(0.007²) = 0.000049 J = 0.049 mJ
When all the energy in the circuit is in the capacitor, this energy will be equal to the energy calculated above.
And for a capacitor, energy is given as
E = (1/2) CV²
E = 0.000049 J, C = 3 F, V = ?
0.000049 = (1/2)(3)(V²)
V = 0.00572 V = 0.01 V
I think it might be a gravitational pull
Answer:
This question is incomplete
Explanation:
The question is incomplete. However, to determine the time (in seconds) it took a worker to hit the ground from an elevated point. The speed the worker was coming with to the ground and the distance between the elevated point and the ground will have to be considered. Thus the formula to be used here will be
Speed (in meter per second) = distance (in meters) ÷ time (in seconds)
time (in seconds) = distance (in meters) ÷ speed (in meter per seconds)
