The driving force behind an efficient supply chain is __________.

1 answer:

5 0

Customer satisfaction is considered to be the "driving force" in order to achieve an efficient supply chain. An efficient supply chain takes place when the organization, or the company itself, meets with the demands of the consumers to improve and provides services that satisfies the people.

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A ball of mass m and weight mg is held at rest by two strings. One string makes an angle θ with the vertical. The other is horiz

malfutka [58]

Answer:

W = M g weight of ball

T cos θ = W balancing vertical forces

T sin θ = F balancing horizontal forces

tan θ = F / W dividing equations

F = W tan θ when θ equals zero F equals zero

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1 year ago

How does the change in the sa/v ratio compare with the change in distance from the center of the cell to the nearest face?

Ivanshal [37]

<span>Answer: I'm pretty sure the SA / V ratio would get smaller. Assume that the cell is more or less spherical. SA = 4(pi)r^2, while V = (3/4)(pi)r^3. The ratio = (4(pi)r^2)/((3/4)(pi)r^3), which can be simplified to 3/r. Thus, the larger r gets, the smaller the ratio becomes.</span>

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3 years ago

Would the frequency of the angular simple harmonic motion (SHM) of the balance wheel increase or decrease if the dimensions of t

storchak [24]

Answer:

Yes the frequency of the angular simple harmonic motion (SHM) of the balance wheel increases three times if the dimensions of the balance wheel reduced to one-third of original dimensions.

Explanation:

Considering the complete question attached in figure below.

Time period for balance wheel is:

$T=2\pi\sqrt{\frac{I}{K}}$

$I=mR^{2}$

m = mass of balance wheel

R = radius of balance wheel.

Angular frequency is related to Time period as:

$\omega=\frac{2\pi}{T}\\\omega=\sqrt{\frac{K}{I}} \\\omega=\sqrt{\frac{K}{mR^{2}}$

As dimensions of new balance wheel are one-third of their original values

$R_{new}=\frac{R}{3}$

$\omega_{new}=\sqrt{\frac{K}{mR_{new}^{2}}}\\\\\omega_{new}=\sqrt{\frac{K}{m(\frac{R}{3})^{2}}}\\\\\omega_{new}={3}\sqrt{\frac{K}{mR^{2}}}\\\\\omega_{new}={3}\omega$