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Evgesh-ka [11]
1 year ago
14

A 2kg mass is moving at 3m/s. What is its kinetic energy?

Physics
1 answer:
Illusion [34]1 year ago
3 0
<h2><u>KINETIC ENERGY</u></h2>

<h3>Problem:</h3>

» A 2kg mass is moving at 3m/s. What is its kinetic energy?

<h3>Answer:</h3>
  • \color{hotpink} \bold{9 \: J} \\

— — — — — — — — — —

<h3>Formula:</h3>

To calculate the velocity of a kinetic energy, we can use formula

  • \underline{ \boxed{  \tt KE =  \frac{1}{2} m{v}^{2}  } }

where,

  • v is the velocity in m/s
  • KE is the kinetic energy in J (joules)
  • m is the mass in kg

— — —

Based on the problem, the givens are:

  • KE (Kinetic energy) = ? (unknown)
  • m (mass) = 2 kg
  • v (velocity) = 3 m/s

<h3>Solution:</h3>

To get the velocity, substitute the givens in the formula above then solve.

\:   \: \tt KE =  \frac{1}{2} m{v}^{2}   \\ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \: \tt \:   KE =  \frac{1}{2} \times  2 \times {(3)}^{2}  \\ \tt  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  KE =  \frac{1}{2}  \times(2\times 9) \\ \tt  KE =  \underline{ \boxed{ \blue{ \tt9 \: J}}}

Therefore, the kinetic energy is 9 Joules.

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What is the need of force in our life​
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Two converging lenses are placed 30 cm apart. The focal length of the lens on the right is 20 cm while the focal length of the l
Masja [62]

Answer:

a)   I2 = 3 (o-10) / (o- 30) , b)   h ’/h=  3 (o-10) / o (o-30)

Explanation:

The builder's equation is

          1 / f = 1 / o + 1 / i

Where f is the focal length, or e i are the distance to the object and image, respectively

As the separation between the lenses is greater than the focal distances, we must work them individually and separately. Let's start with the leftmost lens with focal length f = 15 cm

Let's calculate the position of the image of this lens

         1 / i1 = 1 / f - 1 / o

         1 / i1 = 1/15 - 1 / o

         i1 = o 15 / (o-15)

Let's calculate the distance to the image of the second lens, for this the image of the first is the distance to the object of the second

        o2 = d-i1

We write the builder equation

       1 / f2 = 1 / o2 + 1 / i2

       1 / i2 = 1 / f2 -1 / o2

       1 / i2 = 1 / f2 - 1 / (d-i1)

       1 / i2 = 1/20 - 1 / (d-i1)            (1)

Let's evaluate the last term

      d-i1 = d - 15 o / (o-15)

      d-i1 = (d (o-15) - 15 o) / (o-15)

      d- i1 = (30 or -30 15 -15 o) / (o-15)

      d-i1 = (15 or - 450) / (o- 15)

      d-i1 = = (15 or -450) / (o-15)

replace in 1

      1 / i2 = 1/20 - (or - 15) / (15 or -450)

      1 / i2 = [(15 o-450) - (o-15) 20] / (15 or -150)

      1 / i2 = (15 or - 450 - 20 or + 300) / (15 or - 150)

      1 / i2 = (-5 or -150) / (15 or -150)

      1 / i2 = (or -30) / (3 or - 30)

      I2 = 3 (o-10) / (o- 30)

Part B

The height of the image, we use the magnification equation

     m = h ’/ h = - i / o

     h ’= - h i / o

In our case

     h ’= h i2 / o

     h ’= h 3 (o-10) / o (o-30)

If they give the distance to the object it is easier

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What is the amount of work done when a force of 10N moves a 20kg mass of 8 meters?
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Yo sup??

we should know that

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W=F.s

F=10 N

s=8 m

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Hope this helps.

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