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RoseWind [281]
2 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]2 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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A solid sphere of radius R is placed at a height of 30 cm on a15 degree slope. It is released and rolls, without slipping, to th
photoshop1234 [79]

Answer:

The height is  h_c = 42.857

A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )

Explanation:

   From the question we are told that

           The height is h_s = 30 \ cm

            The angle of the slope is \theta = 15^o

According to the law of conservation of energy

     The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

                          mgh = \frac{1}{2} I w^2 + \frac{1}{2}mv^2

Where I is the moment of inertia which is mathematically represented as this for  a sphere

                    I = \frac{2}{5} mr^2

  The angular velocity w is mathematically represented as

                         w = \frac{v}{r}

So the equation for conservation of energy becomes

               mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2

              mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]

             mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]

            gh_s =[\frac{7}{10} ] v^2

              v^2 = \frac{10gh_s}{7}

Considering a circular hoop

   The moment of inertial is different for circle and it is mathematically represented as

             I = mr^2

Substituting this into the conservation equation above

              mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2

Where h_c is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

                 mgh_c  = mv^2

                     gh_c = v^2

                     h_c = \frac{v^2}{g}

Recall that   v^2 = \frac{10gh_s}{7}

                    h_c= \frac{\frac{10gh_s}{7} }{g}

                      = \frac{10h_s}{7}

            Substituting values

                   h_c = \frac{10(30)}{7}

                       h_c = 42.86 \ cm    

       

     

                         

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2 years ago
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Aloiza [94]

Democritus was the one who did not have experimental evidence to support his theory of the atom.

Answer: Option 4

<u>Explanation: </u>

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After this, Dalton made the specific assumptions formulating some postulates for the atomic theory with proof. Then the cathode rays tube experiments performed by Thomson lead to the formation of plum pudding models of atom.

This is followed by Rutherford’s gold foil experiment discovering the presence of nucleus inside the atoms. So, Democritus first stated but due to absence of experimental evidences, his theory of atoms were not supported at that time.

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