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A water bed weighs 1025 N, and is 1.5 m wide by 2.5 m long. How much pressure does the water bed exert if the entire lower surface of the bed makes contact with the floor?

a.

3.6 x 10^2 Pa

b.

2.7 x 10^4 Pa


c.

1.2 x 10^3 Pa

d.

2.7 x 10^2 Pa

Answer :

carlosego
By definition we have to:
 [tex]P = F / A [/tex]
 Where,
 F: Force
 A: area
 For the area we have:
 [tex]A = (w) * (l) [/tex]
 Where,
 w: width
 l: long
 Substituting values we have:
 [tex]A = (1.5) * (2.5) A = 3.75 m ^ 2[/tex]
 Substituting the values we have:
 [tex]P = 1025 / 3.75 P = 273.3333333 N / m ^ 2[/tex]
 Rewriting we have:
 [tex]P = 2.7 * 10 ^ 2 Pa [/tex]
 Answer:
 d.
 2.7 x 10 ^ 2 Pa
Louli
Answer:
D) 2.7 * 10² Pa

Explanation:
To answer this question, we will use the following equation:
pressure = [tex] \frac{force}{area} [/tex]

1- getting the area:
The water bed is in the form of a rectangle. Therefore, the area can be calculated as follows:
area = length * width
We are given that:
lenth = 2.5 m
width = 1.5 m
This means that:
area = 2.5 * 1.5 = 3.75 m²

2- getting the pressure:
We noted that:
pressure = [tex] \frac{force}{area} [/tex]

We have the force = 1025 N
We calculated the area = 3.75 m²

Substitute in the above equation to get the pressure as follows:
pressure = [tex] \frac{1025}{3.75} [/tex]
pressure = 2.7 * 10² Pa

Hope this helps :)

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