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3 years ago

I need help please !!!

Mathematics

1 answer:

juin [17]3 years ago

6 0

Answer:

f = 12(s - 35).

Step-by-step explanation:

The number of miles per hour he was over the limit = s - 35.

So his fine f is:

f = 12(s - 35).

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1. A weight of 18,000 lb is supported on a rectangular base plate that is 9 in. wide and 2 ft long. The base plate rests on a co

slega [8]

Answer:

a) $83.3\ lb/in^2$

b) $12000\ lb/ft^2$

Step-by-step explanation:

Given : A weight of 18,000 lb is supported on a rectangular base plate that is 9 in. wide and 2 ft long. The base plate rests on a concrete slab.

To find : Determine the stress that the base plate exerts on the concrete slab. Express your answer in units of (a) lb/in². and (b) lb/ft².

Solution :

a) A rectangular base plate that is 9 in. wide and 2 ft long.

Converting ft into inches,

1 feet = 12 inch

2 feet = 2×12= 24 inch

The area of the rectangular base is $A=l\times b$

$A_1=24\times 9$

$A_1=216\ in^2$

The weight is W=18,000 lb.

The stress that the base plate exerts on the concrete slab is given by,

$S=\frac{W}{A_1}$

$S=\frac{18000\ lb}{216\ in^2}$

$S=83.3\ lb/in^2$

b) A rectangular base plate that is 9 in. wide and 2 ft long.

Converting inches into ft,

1 inch = $\frac{1}{12}\ ft$

9 feet = $\frac{9}{12}=0.75\ ft$

The area of the rectangular base is $A=l\times b$

$A_2=2\times 0.75$

$A_2=1.5\ ft^2$

The weight is W=18,000 lb.

The stress that the base plate exerts on the concrete slab is given by,

$S=\frac{W}{A_2}$

$S=\frac{18000\ lb}{1.5\ ft^2}$

$S=12000\ lb/ft^2$

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