At one end of a production line, Isaac puts peaches into cans on a conveyor belt. He fills 18 cans per minute. As soon as Isaac
has filled 670 cans on the conveyor belt, Albert starts putting lids on the cans at the other end of the production line. As Albert finishes closing each can, he moves up along side the conveyor belt to meet the next can. Albert puts lids on 50 cans per minute. (Assume that the conveyor belt initially had no cans on it, and that Isaac stays in place as he continues to fill cans.) How long does it take for Albert to reach Isaac (measured from when Albert starts putting on lids)?
1 answer:
Answer: Long answer...
<em>Explanation:</em> As soon as Isaac has filled 670 cans on the conveyor belt, Albert starts putting lids on the cans at the other end of the production line. As Albert finishes closing each can, he moves up along side the conveyor belt to meet the next can. Albert puts lids on 50 cans per minute.
You might be interested in
Answer:
5.2 m
Explanation:
from the question we are given the following
depth of pool (d) = 3.2 m
height of laser above the pool (h) = 1 m
point of entry of laser beam from edge of water (l) = 2.5 m
we first have to calculate the angle at which the laser beam enters the water (∝),
tan ∝ = \frac{1}{2.2}
∝ = 24.44 degrees
from the attached diagram, the angle with the normal (i) = 90 - 24.4 = 65.56 degrees
lets assume it is a red laser which has a refractive index of 1.331 in water, and with this we can find the angle of refraction (r) using the formula below
refractive index = \frac{sin i}{sin r}
1.331 = \frac{sin 65.56}{sin r}
r = 43.16 degrees
we can get the distance (x) from tan r = \frac{x}{3.2}
tan 43.16 = \frac{x}{3.2}
x = 3 m
To get the total distance we need to add the value of x to 2.2 m
total distance = 3 + 2.2 = 5.2 m
Complete Question
The complete question is shown on the first uploaded image
Answer:
Explanation:
From the question we are told
The amplitude of the lateral force is
The frequency is
The mass of the bridge per unit length is
The length of the central span is
The oscillation amplitude of the section considered at the time considered is
The time taken for the undriven oscillation to decay to of its original value is t = 6T
Generally the mass of the section considered is mathematically represented as
=>
=>
Generally the oscillation amplitude of the section after a time period t is mathematically represented as
Here b is the damping constant and the is the amplitude of the section when it was undriven
So from the question
=>
=>
=>
=>
Generally the amplitude of the section considered is mathematically represented as
=>
=>
=>
=>
