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fenix001 [56]
3 years ago
6

The small piston of a hydraulic press has an area of 2.0 cm^2 and the large piston has an area of 16 cm^2. If the small piston h

as a force of 4.0 N applied to it, the output force of the large piston is ____? A.) 0.50 N B.) 2.0 N C.) 8.0 N D.) 32 N
Mathematics
1 answer:
marysya [2.9K]3 years ago
4 0

Answer:

32N

Step-by-step explanation:

Given data

A1= 2cm^2

A2= 16cm^2

F1= 4N

F2=???

Applying the formula

Presure= force/area

P=F/A

F1/A1=F2/A2

substitute

4/2= F2/16

cross multiply

F2= (16*4)/2

F2= 64/2

F2= 32N

Hence the output force is 32N

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a coyote can run up to 43 mph while a rabbit can run up to 35 mph. Write two equivalent expressions and then find how many more
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The coyote can run 48 more miles than the rabbit
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3 years ago
To the nearest tenth, what is the distance between the point (10, -11) and (-1, -5)
Eva8 [605]

Answer:

The distance will be in centimeters

Step-by-step explanation:

You will do the multiply and subtract

5 0
3 years ago
Field book of an land is given in the figure. It is divided into 4 plots . Plot I is a right triangle , plot II is an equilatera
Kazeer [188]

Answer:

Total area = 237.09 cm²

Step-by-step explanation:

Given question is incomplete; here is the complete question.

Field book of an agricultural land is given in the figure. It is divided into 4 plots. Plot I is a right triangle, plot II is an equilateral triangle, plot III is a rectangle and plot IV is a trapezium, Find the area of each plot and the total area of the field. ( use √3 =1.73)

From the figure attached,

Area of the right triangle I = \frac{1}{2}(\text{Base})\times (\text{Height})

Area of ΔADC = \frac{1}{2}(\text{CD})(\text{AD})

                        = \frac{1}{2}(\sqrt{(AC)^2-(AD)^2})(\text{AD})

                        = \frac{1}{2}(\sqrt{(13)^2-(19-7)^2} )(19-7)

                        = \frac{1}{2}(\sqrt{169-144})(12)

                        = \frac{1}{2}(5)(12)

                        = 30 cm²

Area of equilateral triangle II = \frac{\sqrt{3} }{4}(\text{Side})^2

Area of equilateral triangle II = \frac{\sqrt{3}}{4}(13)^2

                                                = \frac{(1.73)(169)}{4}

                                                = 73.0925

                                                ≈ 73.09 cm²

Area of rectangle III = Length × width

                                 = CF × CD

                                 = 7 × 5

                                 = 35 cm²

Area of trapezium EFGH = \frac{1}{2}(\text{EF}+\text{GH})(\text{FJ})

Since, GH = GJ + JK + KH

17 = \sqrt{9^{2}-x^{2}}+5+\sqrt{(15)^2-x^{2}}

12 = \sqrt{81-x^2}+\sqrt{225-x^2}

144 = (81 - x²) + (225 - x²) + 2\sqrt{(81-x^2)(225-x^2)}

144 - 306 = -2x² + 2\sqrt{(81-x^2)(225-x^2)}

-81 = -x² + \sqrt{(81-x^2)(225-x^2)}

(x² - 81)² = (81 - x²)(225 - x²)

x⁴ + 6561 - 162x² = 18225 - 306x² + x⁴

144x² - 11664 = 0

x² = 81

x = 9 cm

Now area of plot IV = \frac{1}{2}(5+17)(9)

                                = 99 cm²

Total Area of the land = 30 + 73.09 + 35 + 99

                                    = 237.09 cm²

7 0
3 years ago
Can someone pls help me :)
Diano4ka-milaya [45]

Answer: c

Step-by-step explanation:

If the population triples every three hours, you would multiply three by the number of one hour to the power of three, making three hours. This function would triple it every three hours.

4 0
3 years ago
Help help help help help
nekit [7.7K]

Answer:

C, C is the answer

5 0
3 years ago
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