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VikaD [51]
3 years ago
6

शब्द के जिस रुप से उसके पुरुष जाति के होने का बोध होता है उसे कहते हैं?​

Physics
1 answer:
AVprozaik [17]3 years ago
8 0

Answer:

A

Algebra For what values of the variables must ABCD be a parallelogram?

B

23. A 2y + 2 B 24. B

(3x + 10)

(8x + 5)º

3x + 6

54°

D

С

D

A

Зу - 9 с

ly+4

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2-The amount of internal energy needed to raise the temperature of 0.25kg of water by 0.2°C is 209.3 J. How fast must a 0.25 kg
Yakvenalex [24]

Answer:

40.92 m/s

Explanation:

The computation is shown below:

Ek = 1 ÷2mv²...............................(1)

v = √(2Ek/m).......................... (2)

Here EK denotes kinetic energy

m denotes mass

v denotes velocity

Given that

m = 0.25kg and Ek = 209.3J

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Are individual organelles (cell structures) living? Explain
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All cells are living however organelles are not living there are no organisms that consist of just a single cell
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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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Random kinetic energy possessed by objects in a material at finite temperature. An object that feels hot has a lot of this.
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Internal energy or thermal energy.
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