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RoseWind [281]
3 years ago
9

A ruler has a division of 1mm. You use the ruler to measure a piece of paper. You find that it is 7.4 cm wide and 8.3 cm long. W

hat is the area of the piece of paper?
Physics
1 answer:
Anastaziya [24]3 years ago
3 0

Answer:

<h2><em>6,142mm²</em></h2>

Explanation:

Given the dimension of a paper measured by a ruler as 7.4 cm wide and 8.3 cm long, the area of the paper is expressed using the area for calculating the area of a rectangle as shown;

Area of the piece of paper = Length * Width

Given length = 7.4cm

Length = 74mm (Since 10mm = 1cm)

Width = 8.3cm

Width (in mm) = 83mm

We converted to mm since the ruler used to measure has a division of 1mm.

Substituting the given values into the formula, we will have:

Area of the piece of paper = 74mm * 83mm

Area of the piece of paper = 6,142mm²

<em>Hence, the area of the piece of paper is 6,142mm²</em>

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2 years ago
As shown in the diagram, two forces act on an object. The forces have magnitudes F1 = 5.7 N and F2 = 1.9 N. What third force wil
galina1969 [7]

Answer:

Second option 6.3 N at 162° counterclockwise from  

F1->

Explanation:

Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.

For the address x we have:

-F_3sin(b) + F_1 = 0

For the address and we have:

-F_3cos(b) + F_2 = 0

The forces F_1 and F_2 are known

F_1 = 5.7\ N\\\\F_2 = 1.9\ N

We have 2 unknowns (F_3 and b) and we have 2 equations.

Now we clear F_3 from the second equation and introduce it into the first equation.

F_3 = \frac{F_2}{cos (b)}

Then

-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°

Then we find the value of F_3

F_3 = \frac{F_1}{sin(b)}\\\\F_3 =\frac{5.7}{sin(72\°)}\\\\F_3 = 6.01 N

Finally the answer is 6.3 N at 162° counterclockwise from  

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Two tugboats are moving a barge. Tugboat A pushes on the barge with a force of 3000n. Tugboat B pulls with a force of 5000 Newto
kenny6666 [7]

The net force on the barge is 8000 N

Explanation:

In order to find the net force on the badge, we have to use the rules of vector addition, since force is a vector quantity.

In this problem, we have two forces:

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Since the two forces act in the same direction, this means that we can simply add their magnitudes to find the net combined force on the barge. Therefore, we get

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and the direction is the same as the direction of the two forces.

Learn more about forces:

brainly.com/question/11179347

brainly.com/question/6268248

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