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A fast-food restaurant uses a conveyor belt to send the burgers through a grilling machine. if the grilling machine is 1.2 m long and the burgers require 2.7 min to cook, how fast must the conveyor belt travel?if the burgers are spaced 15 cm apart, what is the rate of burger production (in burgers/min)?

Respuesta :

length of the grilling machine is 1.2 m

time taken to cook the burger is 2.7 min = 162 s

so the speed of the machine should be like this that if must have to cook till it cross the machine

[tex]v = \frac{d}{t}[/tex]

[tex]v = \frac{1.2}{162}[/tex]

[tex]v = 7.41* 10^{-3} m/s[/tex]

now in one minute the total length of the machine that is covered is given by

[tex]L = v*t[/tex]

[tex]L = 7.41*10^{-3}* 60 = 44.4 cm[/tex]

now distance between the burgers is 15 cm

so total production rate will be

[tex]N = \frac{44.4}{15} = 3 burger/min[/tex]

so it will produce 3 burger per minute

if the burgers are spaced 15 cm apart, what is the rate of burger production (in burgers/min) the rate is N = 44.4/15 = 3burger / m in so it'll produce three burger in line with minutes are 3 burgers .

How do you calculate the price of anything?

Use the components r = d/t. Your price is 24 miles divided via way of means of 2 hours, so: r = 24 miles ÷ 2 hours = 12 miles in line with hours.The duration of the grilling device is 1.2 m.

  1. The time taken to prepare dinner the burger is 2.7 min = 162 sso the velocity of the device ought to be like this that if have to need to prepare dinner until it go the device.
  2. v = d/tv = 1.2/162v = 7.41 * 10 ^ - 3 m / snow in a single minute the overall duration of the device this is protected is given via way of means of L = vtL = 7.41 * 10 ^ - 3* 60 = 44.4cmnow distance among the burgers is 15 cmso the general manufacturing price will beN = 44.4/15 = 3burger / min.

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