All categories

3 years ago

The force on an area 16 m by 12 m is 24,000 N. What is the pressure per square meter?

Mathematics

2 answers:

Dmitrij [34]3 years ago

6 0

Recall that
Pressure = Force ÷ Area

In this context we have,
Pressure = 24,000 ÷ (16 × 12)
= 24,000 ÷ 192
= 125 Pa

marysya [2.9K]3 years ago

5 0

Answer:

125 N/m^2

Step-by-step explanation:

To calculate the the pressure(P) we need area(A) and force(F) because pressure is a function of force and area:

$P=F/A$ $(N/m^2)$

We can calculate the Area as:

$A=16*12=192$ m^2

Therefore we can calculate pressure per square meter:

$P=24000*(192)=4608000$ N/m^2

You might be interested in

Given: MNFD is a trapezoid. MF ⊥ ND , NK ⊥ MD , NK = h, MN = FD. Find: Area of MNFD

kodGreya [7K]

Solved ... you owe me :)
A = h^2

5 0

2 years ago

F(x)=-10x-3 ; compressed horizontally by a factor of 5/8

notka56 [123]

Answer:

yes?

Step-by-step explanation:

7 0

2 years ago

Halla la tasa de variación de cada funcion en el intervalo [-4,3] e indica si es positiva , negativa o nula A) f(x)=x2-2x+4 B) f

masya89 [10]

Answer:

$A) \hspace{3}Rate\hspace{3}of\hspace{3}change=-5\hspace{3}Negative\\\\B)\hspace{3}Rate\hspace{3}of\hspace{3}change=-21\hspace{3}Negative$

Step-by-step explanation:

Given a function $f(x)$ , we called the rate of change to the number that represents the increase or decrease that the function experiences when increasing the independent variable from one value " $x_1$ " to another " $x_2$ ".

The rate of change of $f(x)$ between $x_1$ and $x_2$ can be calculated as follows:

$Rate\hspace{3}of\hspace{3}change=f(x_2)-f(x_1)$

For:

$f(x)=x^2-2x+4$

Let's find $f(x_1)$ and $f(x_2)$ , where:

$[x_1,x_2]=[-4,3]$

$f(x_1)=f(-4)=(-4)^2-2(4)+4=16-8+4=12\\f(x_2)=f(3)=(3)^2-2(3)+4=9-6+4=7$

So:

$Rate\hspace{3}of\hspace{3}change =7-12=-5\hspace{3}Negative$

And for:

$f(x)-3x+2$

Let's find $f(x_1)$ and $f(x_2)$ , where:

$[x_1,x_2]=[-4,3]$

$f(x_1)=f(-4)=-3(-4)+2=12+2=14\\f(x_2)=f(3)=-3(3)+2=-9+2=-7$

So:

$Rate\hspace{3}of\hspace{3}change =-7-14=-21\hspace{3}Negative$

<em>Translation:</em>

Dada una función $f(x)$ , llamábamos tasa de variación al número que representa el aumento o disminución que experimenta la función al aumentar la variable independiente de un valor " $x_1$ " a otro " $x_2$ ".

La tasa de variación de $f(x)$ entre $x_1$ y $x_2$ , puede ser calculada de la siguiente forma:

$Tasa\hspace{3}de\hspace{3}variacion=f(x_2)-f(x_1)$

Para:

$f(x)=x^2-2x+4$

Encontremos $f(x_1)$ y $f(x_2)$ , donde:

$[x_1,x_2]=[-4,3]$

$f(x_1)=f(-4)=-3(-4)+2=12+2=14\\f(x_2)=f(3)=-3(3)+2=-9+2=-7$

Entonces:

$Tasa\hspace{3}de\hspace{3}variacion =7-12=-5\hspace{3}Negativa$

Y para:

$f(x)-3x+2$

Encontremos $f(x_1)$ y $f(x_2)$ , donde:

$[x_1,x_2]=[-4,3]$

$f(x_1)=f(-4)=-3(-4)+2=12+2=14\\f(x_2)=f(3)=-3(3)+2=-9+2=-7$

Entonces:

$Tasa\hspace{3}de\hspace{3}variacion=-7-14=-21\hspace{3}Negativa$

8 0

3 years ago

Help!!!!!?!!!!!!!?!!?!!!!

Andreas93 [3]

Answer:

yellow

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

A random sample of 16 one-kilogram sugar packets is obtained and the actual weights of the packets are measured. The sample mean

elena55 [62]

Answer:

The 99% two-sided confidence interval for the average sugar packet weight is between 0.882 kg and 1.224 kg.

Step-by-step explanation:

We are in posession of the sample's standard deviation, so we use the student's t-distribution to find the confidence interval.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 16 - 1 = 15

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 35 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.905([tex]t_{995}$ ). So we have T = 2.9467

The margin of error is:

M = T*s = 2.9467*0.058 = 0.171

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 1.053 - 0.171 = 0.882kg

The upper end of the interval is the sample mean added to M. So it is 1.053 + 0.171 = 1.224 kg.

The 99% two-sided confidence interval for the average sugar packet weight is between 0.882 kg and 1.224 kg.

3 0

3 years ago