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Sedbober [7]
2 years ago
7

A square-based shipping crate is being designed that must contain a volume of 16 ft3 . The material that is used for the base an

d the lid costs 3 dollars/ft2 , while the material used for the sides costs 2 dollars/ft2 . What are the most cost-effective dimensions of such a crate?
Physics
1 answer:
Vlada [557]2 years ago
6 0

Answer:

Explanation:

Given

volume V=16 ft^3

Suppose base is square with side L

height of crate is h

Volume V=L^2\times h

16=L^2\times h

Cost of top and bottom area c_1=3L^2

Cost of Side area c_2=4Lh\times 2=8Lh=8L\times \frac{16}{L^2}=\frac{128}{L}

Total Cost C=c_1+c_2

Total Cost C=3L^2+\frac{128}{L}

Differentiate C w.r.t Length

\frac{dC}{dL}=6L-\frac{128}{L^2}

L^3=\frac{128}{6}

L=2.75 ft

h=\frac{16}{2.75^2}=11.46 ft

Dimensions are L\times L\times h=2.75\times 2.75\times 11.46    

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Answer:

Explanation:

  • given S = distance from the first = 3.20cm = 0.032m, t = 1.30×10−8 s
  • q = 1.6 x 10_19C
  • using S = at^2/2
  • acceleration = 0.032 X 2 /(1.30×10−8)^2

a = 3.79 x 10^14m/s^2

  • From F = ma
  • F = qE
  • ma = qE

E = ma /q = 9.11 x 10^-31 x 3.79 x 10^14 / 1.6 x 10^-19

E = magnitude of this electric field. = 2156.3N/C

b) Find the speed of the electron when it strikes the second plate ; V^2 = 2as

= 2 X 3.79 x 10^14 X 0.032

= 4.92 X 10^6m/s

5 0
2 years ago
A special rocket can produce 7.66 ✕ 10^5 N of instantaneous thrust with an exhaust speed of 3.05 ✕ 103 m/s in vacuum. What mass
adelina 88 [10]

The mass of fuel the engine burn each second to produce a thrust of 7.66×10⁵ N is 2.5×10² kg/s.

<h3 /><h3>What is mass?</h3>

Mass can be defined as the quantity of matter contained in a body. The S.I unit of mass is kilogram(kg)

To calculate the mass the engine burns each seconds, we use the formula below.

Formual:

  • M = T/v............. Equation

Where:

  • M = Mass per seconds of the rocket
  • T = Thrust
  • v = Velocity

From the question,

Given:

  • T = 7.66×10⁵ N
  • v = 3.05×10³ m/s

Substitute these values into equation 1

  • M = (7.66×10⁵)/(3.05×10³)
  • M = 2.5×10² kg/s

Hence, the mass of fuel burned in each second is 2.5×10² kg/s.

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1 year ago
the displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion s
12345 [234]

The average velocity or displacement of a particle for the first time interval is <u>Δs / Δt = 6 cm/s.</u>

Solution:

As we know that displacement is calculated in centimeters and the unit of time is second.

The average velocity for the first interval [1,2] is given

Δs / Δt = s (t2) - s (t) / t2 - t1

Δs / Δt = 2sin2  π  + 3cos 2 π -  ( 2sin π + 3cos π ) / 2 - 1

Δs / Δt = 2(0) + 3(1) - 2(0) - 3 (-1) / 1

Δs / Δt = 6 cm/s

Thus the average velocity or displacement of a particle for the first time interval is Δs / Δt = 6 cm/s

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The complete question is:

The displacement of a particle moving back and forth along a line is given by the following equation s(t) = 2sin π t + 3cos π t. Estimate the instantaneous velocity of the particle when t = 1

4 0
1 year ago
Pls answerrrrr thisssss
Jobisdone [24]

The amplitude of wave-c is 1 meter.

The speed of all of the waves is (12meters/2sec)= 6 m/s.

The period of wave-a is 1/2 second.

4 0
3 years ago
Select all the answers that apply. Scientists think past changes in climate have been caused by _____.
Natasha2012 [34]
<span>C. plate tectonics....</span>
8 0
3 years ago
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