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rjkz [21]
3 years ago
7

A 0.50 kilogram ball is held at a height of 20 meters. What is the kinetic energy of the ball when it reaches halfway after bein

g released?
Physics
1 answer:
Svetach [21]3 years ago
5 0

While the ball is held 20 meters above the floor,

                                   Potential energy = (mass) (gravity) x (height) =

                                                        (0.5 kg) x (9.8 m/s²) x (20 m) =  98 joules.

When it's dropped and reaches 1/2 of its initial height, it has half as much
potential energy.  The rest has turned into kinetic energy  =  <em>49 joules.

</em>
============================================================<em>
</em>

Want to know what else you can figure out ?  Watch this . . . 

                                                 <u> Kinetic energy = (1/2) x (mass) x (speed)²</u>

                                                          49 joules = (1/2) x (0.5 kg) x (speed)²

Divide each side by (0.25 kg):       49 joules / 0.25 kg = speed²

Take the square root of each side:        7/0.5 = speed = <em><u>14 meters per second</u></em>

That's the speed of the ball after falling 10 meters ! 
How cool is that !

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4.96 × 10⁵ Pa

Explanation:

F = mg

F = (m_{person}+m_{stool})g\\\\F =  (4.5 + 89)*9.8\\\\F = 916.3 N

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radius, r = d/2

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total cross sectional area of the three legs, A = 3*pi*r^2

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P = / 1.847 × 10⁻³\\P = \frac{ 916.3N}{1.847\times10^-^3} \\\\P= 496032.9Pa

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Answer:

The image is formed at a ‘distance of 16.66 cm’ away from the lens as a diminished image of height 3.332 cm. The image formed is a real image.

Solution:

The given quantities are

Height of the object h = 5 cm

Object distance u = -25 cm

Focal length f = 10 cm

The object distance is the distance between the object position and the lens position. In order to find the position, size and nature of the image formed, we need to find the ‘image distance’ and ‘image height’.

The image distance is the distance between the position of convex lens and the position where the image is formed.

We know that the ‘focal length’ of a convex lens can be found using the below formula

1f=1v−1u\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

f

1

=

v

1

−

u

1

Here f is the focal length, v is the image distance which is known to us and u is the object distance.

The image height can be derived from the magnification equation, we know that

Magnification=h′h=vu\text {Magnification}=\frac{h^{\prime}}{h}=\frac{v}{u}Magnification=

h

h

′

=

u

v

Thus,

h′h=vu\frac{h^{\prime}}{h}=\frac{v}{u}

h

h

′

=

u

v

First consider the focal length equation to find the image distance and then we can find the image height from magnification relation. So,

1f=1v−1(−25)\frac{1}{f}=\frac{1}{v}-\frac{1}{(-25)}

f

1

=

v

1

−

(−25)

1

1v=1f+1(−25)=110−125\frac{1}{v}=\frac{1}{f}+\frac{1}{(-25)}=\frac{1}{10}-\frac{1}{25}

v

1

=

f

1

+

(−25)

1

=

10

1

−

25

1

1v=25−10250=15250\frac{1}{v}=\frac{25-10}{250}=\frac{15}{250}

v

1

=

250

25−10

=

250

15

v=25015=503=16.66 cmv=\frac{250}{15}=\frac{50}{3}=16.66\ \mathrm{cm}v=

15

250

=

3

50

=16.66 cm

Then using the magnification relation, we can get the image height as follows

h′5=−16.6625\frac{h^{\prime}}{5}=-\frac{16.66}{25}

5

h

′

=−

25

16.66

So, the image height will be

h′=−5×16.6625=−3.332 cmh^{\prime}=-5 \times \frac{16.66}{25}=-3.332\ \mathrm{cm}h

′

=−5×

25

16.66

=−3.332 cm

Thus the image is formed at a distance of 16.66 cm away from the lens as a diminished image of height 3.332 cm. The image formed is a ‘real image’.

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