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Mumz [18]
3 years ago
9

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

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A rectangular piece of metal is 30 in longer than it is wide. Squares with sides 6 in long are cut from the four corners and the
Alexandra [31]

Answer:

length=53 in

width= 23 in

Step-by-step explanation:

First, we know that each corner has now a square of 6 in long, so the heigth of the box will be of this size (as you can see in the picture attached). Then, the length and the width of the box would be the same as the piece, but with 12 in less. Therefore the equations to solve the problem are:

h=6 in\\w=x-12\\l=x+30-12\\l=x+18\\\\V: volume of the box\\V=6*(x-12)*(x+18)\\2706=6x^{2}+36x-1296\\0=6x^{2}+36x-4002\\\\

We factorize the 6 and we get:

0=x^{2}+6x-667\\

Finally we solve for x, and we get the initial dimensions of the piece of metal:

x=23l=30+x=30+23=53 in\\w=x= 23 in

7 0
3 years ago
The value of x is? What is the value of x in this problem?
snow_lady [41]

Answer:

x=96°

Step-by-step explanation:

For the upper left angle, we know that they are supplementary, so we can find the first angle of the triangle by

180-130=50

For the bottom right angle, we know that they are supplementary, so we can find that angle by

180-134=46

As a triangle has 180 degrees, we can subtract the values of the other two angles to find the final upper right angle.

180-50-46=84

Lastly, The upper right angle and x are supplementary so we can find x by subtracting the angle from 180

180-84=x\\x=96

5 0
3 years ago
Read 2 more answers
1. Find the value of x.<br><br> A. 90<br> B. 80<br> C. 60<br> D. 45
tia_tia [17]

Answer:

A 90

Step-by-step explanation:

multiple ways to prove this.

e.g. since the angle between the two lines from the center of the circle to the 2 tangent touching points is 90 degrees (that is the meaning of these 90 degrees here as the angle of the circle segment defined by the 2 tangent touching points and the circle center), the tangents have the same "behavior" as tan and cot = the tangents at the norm circle at 0 and 90 degrees. they hit each other outside of the circle again at 90 degrees.

another way

imagine the two right triangles of the tangents crossing point to the circle center and the tangent/circle touching points.

the Hypotenuse of each triangle is cutting the 90 degree angle of the circle segment exactly in half (due to the symmetry principle). so the angle between radius side and Hypotenuse is 90/2 = 45 degrees.

that means also the angle of such a right triangle at the tangent crossing point is 45 degrees (as the sum of all angles in a triangle must be 180, we have the remainder of 180 - 90 - 45 = 45 degrees).

the angles of both right triangles at that point are the same, and so we can add 45+45 = 90 degrees for the total angle at the tangent crossing point.

8 0
3 years ago
Integral <br><img src="https://tex.z-dn.net/?f=x%5E%7B3%7D%20%20-%207x%5E%7B2%7D%20%2B%205x%20%2B%2040%20%5Cdiv%20x%5E%7B2%7D%20
solong [7]

Answer:

ext^{4}-7x^{2}+3x+401dvx^{2}i-8ext

Step-by-step explanation:

6 0
3 years ago
I have no idea how to do this, it is due in two days. Hopefully someone sees this before then.
elena-14-01-66 [18.8K]

Hello,

m\ \widehat{ABC}=x\\m\ \widehat{BAC}=2*x\\\\So:\\ x+2x=90^o\\x=30^o\\

cos(30^o)=\dfrac{\sqrt{3} }{2} \\

In the triangle ABC,

cos(30^o)=\frac{BC}{BA} \\\\BA=\dfrac{cos(30^o)}{BC} \\\\BA=\frac{\dfrac{\sqrt{3} }{2} }{24} =16*\sqrt{3} \\\\

sin(30^o)=\dfrac{1 }{2} =\dfrac{AC}{AB} \\\\AC=\dfrac{1}{2} *16\sqrt{3} =8\sqrt{3}

In the triangle ACB,

cos(30^o)=\dfrac{AC}{AL} \\\\AL=\dfrac{8\sqrt{3} *2}{\sqrt{3} } =16\\

6 0
2 years ago
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