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Fofino [41]
3 years ago
14

Which statement accurately compares the weight of an object on these two planets

Physics
1 answer:
wel3 years ago
5 0
"An object weighs about two times as much on Jupiter as on Neptune." Hope this helps :)
You might be interested in
THIS MARCIN
nekit [7.7K]

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’.

5 0
2 years ago
It's for #8....I know the answer is 293.2 ft I just don't know how to get it
Oksi-84 [34.3K]
Kinectic Energy=1/2(mass)(velocity)^2 so 1.2=1/2(.0012)(position/2) so it travels 4000 m. Not sure how it is 293.2 ft
7 0
2 years ago
People living near wind farms sometimes object to them because of visual pollution and what other
ddd [48]

Answer:

Wind turbines are a source of clean renewable energy, but some people who live nearby describe the shadow flicker, the audible sounds and the subaudible sound pressure levels as "annoying." They claim this nuisance negatively impacts their quality of life

4 0
2 years ago
A steel ball with mass m is suspended from the ceiling at the bottom end of a light, 17.0-m-long rope. The ball swings back and
iogann1982 [59]

Answer:

1. 18.25 m/s

2. 0 m/s

Explanation:

1.So the centripetal acceleration of the ball at this lowest point must be, taking gravity into account

a_c = \frac{T - mg}{m} = \frac{3mg - mg}{m} = 2g

The speed at this point would then be

v^2 = a_c r = 2gr = 2*9.8*17 = 333.2

v = \sqrt{333.2} = 18.25 m/s

2. Similarly, if T = mg, then the centripetal acceleration must be

a_c = \frac{T - mg}{m} = \frac{mg - mg}{m} = 0

As the ball has no centripetal acceleration, its speed must also be 0 as well.

6 0
3 years ago
What are some possible injuries that can occur as a result of anabolic steroid use?
Natali5045456 [20]
Most data on the long-term effects of anabolic steroids in humans come from case reports rather than formal epidemiological studies. Serious and life-threatening adverse effects may be underreported, especially since they may occur many years later. One review found 19 deaths in published case reports related to anabolic steroid use between 1990 and 2012; however, many steroid users also used other drugs, making it difficult to show that the anabolic steroid use caused these deaths.39 One animal study found that exposing male mice for one fifth of their lifespan to steroid doses comparable to those taken by human athletes caused a high frequency of early deaths
3 0
3 years ago
Read 2 more answers
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