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Marat540 [252]
3 years ago
9

Which of the following is not a necessary condition for seeing a magnified image with a

Physics
1 answer:
Thepotemich [5.8K]3 years ago
5 0

The option stating that the observer must be placed within the focal length is false or not a necessary condition for magnification.

Explanation:

Lens are used for magnification provided the nature of the lens is known and by positioning the object at correct position. So the object should be placed in the focal point of the lens for perfect magnification, it can also be kept within the focal point or focal length. But the object should not be kept beyond the focal length of the lens.

So among the given options, the object and image should be on the same side of the lens for magnification is true. Then the nature of lens should be converging for magnification is also true. And the most important point that the object should be placed within the focal length of the lens is the most mandatory for obtaining magnification. So the option stating that the observer must be placed within the focal length is false or not a necessary condition for magnification.

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Answer:

It takes a little, be patient.

8 0
4 years ago
Two spherical asteroids have the same radius R. Asteroid 1 has mass M and asteroid 2 has mass 1.97·M. The two asteroids are rele
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Answer:

0.536\sqrt{\frac{GM}{R}}

Explanation:

We are given that

Mass of one  asteroid 1,m_1=M

Mass of asteroid 2,m_2=1.97 M

Initial distance between their centers,d=13.63 R

Radius of each asteroid=R

d'=R+R=2R

Initial velocity of both asteroids

u=0

We have to find the speed of second asteroid just before they collide.

According to law of conservation of momentum

(m_1+m_2)u=m_1v_1+m_2v_2

(M+1.97 M)\times 0=Mv_1+1.97Mv_2

Mv_1=-1.97 Mv_2

v_1=-1.97v_2

According to law of conservation of energy

Gm_1m_2(\frac{1}{d'}-\frac{1}{d})=\frac{1}{2}m_1v^2_1+\frac{1}{2}m_2v^2_2

GM(1.97M)(\frac{1}{2R}-\frac{1}{13.63R})=\frac{1}{2}M(-1.97v_2)^2+\frac{1}{2}(1.97M)v^2_2

1.97M^2G(\frac{13.63-2}{27.26R})=\frac{1}{2}Mv^2_2(3.8809+1.97)

1.97MG(\frac{11.63}{27.26 R})=\frac{1}{2}(5.8509)v^2_2

v^2_2=\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}

v_2=\sqrt{\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}}

v_2=0.536\sqrt{\frac{GM}{R}}

Hence, the speed of second asteroid =0.536\sqrt{\frac{GM}{R}}

8 0
3 years ago
8. Contrast. The energy that can be released during a nuclear fission reaction with the energy that can be released during a nuc
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beta particles are high energy electrons or positrons.  They travel further due to their small size but can be stopped by a thin barrier of plastic or wood.
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3 years ago
Elements in which family are most likely to have properties associated with
Elena-2011 [213]

Answer:  As with all metals, the alkali metals are malleable, ductile, and are good conductors of heat and electricity. The alkali metals are softer than most other metals.

Alkaline earth metals

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Explanation:

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The train is accelerated from rest at a speed of 0.11 m/s2 to reach a speed of 15.6 m/s. After reaching that speed the train mov
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Answer:

cliff

Explanation:

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