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3 years ago

D. 'g' vanishes at centre of

Physics

1 answer:

guapka [62]3 years ago

6 0

Answer:

a. 0.8 cm

Explanation:

The distance of the object from the lens, u = 1 cm

The magnification of the lens, m = 5

The focus of a lens formula is given as follows;

$f = \dfrac{1}{\dfrac{1}{v} + \dfrac{1}{v} }$

The magnification of the lens, m = -v/u

Where;

v = The distance of the image from the lens

Therefore, we have;

v = m × u

∴ v = 5 × 1 cm = 5 cm (on the other side of the lens)

From which we get;

$f = \dfrac{1}{\dfrac{1}{1} + \dfrac{1}{5} } = \dfrac{5}{6} \approx 0.8$

The focal length ≈ 0.8 cm

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What are the starting materials for nuclear fission

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3 years ago

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What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.

valkas [14]

<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $a$ of its orbit.

This Law is originally expressed as follows:

$T^{2}=\frac{4\pi^{2}}{GM}a^{3}$ (1)

Where;

$G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$M=6.39(10)^{23}kg$ is the mass of Mars

$a=3.4(10)^{6}m$ is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

$T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}$ (2)

$T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}$ (3)

$T=\sqrt{11581157.44 s^{2}}$ (4)

Finally:

$T=3403.1099s=56.718min$ This is the orbital period of a spacecraft in a low orbit near the surface of mars

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3 years ago

A point charge Q is placed at the center of a conducting spherical shell (inner radius a, outer radius b). What is the electric

sp2606 [1]

Answer:

a) $E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C$

b)E=0

c) $E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C$

Explanation:

Given that

A point charge Q is placed at the center of a conducting spherical shell .Due to this - Q charge will induce on the inner sphere surface and +Q will induce on the outer sphere surface .

a) r < a

At a radius r ,from gauss theorem

$E.ds=\dfrac{q_i}{\varepsilon _o}$