The area of the top and bottom:
2πr²
Cost for top and bottom:
2πr² x 0.02
= 0.04πr²
Area for side:
2πrh
Cost for side:
2πrh x 0.01
= 0.02πrh
Total cost:
C = 0.04πr² + 0.02πrh
We know that the volume of the can is:
V = πr²h
h = 500/πr²
Substituting this into the cost equation to get a cost function of radius:
C(r) = 0.04πr² + 0.02πr(500/πr²)
C(r) = 0.04πr² + 10/r
Now, we differentiate with respect to r and equate to 0 to obtain the minimum value:
0 = 0.08πr - 10/r²
10/r² = 0.08πr
r³ = 125/π
r = 3.41 cm
Answer:
Check the explanation
Explanation:
Their is no data provided for the mass and length of pendulum in the picture.but it is very easy to check kinetic energy.KE is mv^2/2, m is the pendulum Bob and v is time dependent ,equation of displacement of SHM is given as x(t)=Asin(wt+∆) where ∆ is the phase angle now v=dX/dt _v=Awcos(wt+∆) and KE=
Thus w angular frequency of oscillation is√(g/l)
g acceleration due to gravity and l length of pendulum.so KE is same for pendulum having same mass and length otherwise KE expression with time will vary for all other cases.now check if all experiment pendulum have same mass and length otherwise KE will not be same.You can now easily verify.
Penn Foster Students: less than the angle of refraction
Increase in wavelength will cause decrease in frequency. When wavelength doubles, the frequency of the wave will reduce to half of its value.
<h3>
WAVE SPEED</h3>
The wave speed is the product of wave frequency and its wavelength.
v = f⋅λ
For example, if a wave has a frequency of 120 Hz and a wavelength of 5m, it would have a speed of 600 m/s.
if the wavelength is increased to 10m but the speed stays the same at 600 m/s. That is,
λ = 10m
v = 600 m/s
v = f⋅λ
Substitute f and v into the formula
600 = 10f
f = 600 / 10
f = 60 m
The frequency of the wave is reduced to half of its value.
Learn more about wave speed here: brainly.com/question/2847127
So power is considered as the rate of doing work. Base on the problem given, my analysis is that the machine who finish the work faster is machine C. Therefore, in order to finish the same amount of work in a short period of time you are going to expend the most power. My answer is Machine C.
