A bees wings move so rapidly that studying them, even seeing them, has proved difficult.
The honeybees have a rapid wing beat honeybee flaps its wings 230 times every second.
Therefore, a bee flaps its wings over 200 times in about 1ms
ummmm it might be 300... i used a calculator
sorry if it is wrong
Answer: 3.12 * 10^12 F ( 3.12 pF)
Explanation: To calculate this capacitor of two hollow, coaxial, iron cylinders, we have to determine the potental differente between them and afeter that to use C=Q/ΔV
The electric field in th eregion rinner<r<router
By using the Gaussian law
∫E*ds=Q inside/εo
E*2*π*rinner^2*L= Q /εo
E=Q/(2*π*εo*r^2)
[Vab]=\int\limits^a_b {E} \, dr
where a and b are the inner and outer radii.
Then we have:
ΔV= 2*k*(Q/L)* ln (b/a)
replacing the values and using that C=Q/ΔV
we have:
C= L/(2*k*ln(b/a)=0.17/(2*9*10^9*3.023)=3.12 pF
The spring will come to rest 4.9 m below the natural length
Explanation:
The mass-spring system will come to rest when the restoring force on the spring (pulling upward) balances the weight of the mass (pulling downward). Mathematically, this can be written as
where
k is the spring constant
x is the elongation of the spring
m is the mass
g is the acceleration of gravity
In this problem, we have:
is the mass
is the acceleration of gravity
is the spring constant
Solving the equation for x,
Therefore, the spring will come to rest 4.9 m below the natural length.
Learn more about forces:
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