They enjoy being alone for quit small time

Physics

1 answer:

hjlf3 years ago

4 0

Yeah and? What’s the question

You might be interested in

The distance between the lenses in a compound microscope is 18 cm. The focal length of the objective is 1.5 cm. If the microscop

Blababa [14]

Answer:

The focal length of eye piece is 6.52 cm.

Explanation:

Given that,

Angular Magnification of the microscope M = -46

the distance between the lens in microscope L= 16 cm

The focal length of objective f₀ = 1.5 cm

Normal near point N = 25 cm

Have to find focal length of eye piece f ₙ =?

The angular magnification is given by

M ≈ - (L-fₙ)N/f₀fₙ

Rearranging for fₙ

fₙ =L(1 - Mf₀/N)⁺¹

=18/2.76

fₙ = 6.52 cm

The focal length of eye piece is 6.52 cm.

6 0

3 years ago

What occurs when an object travels in a curved path

babymother [125]

<span>When an object travels in a curved path, there must be a force acting toward the center of the circular trajectory. This force is called "centripetal force", and it cause an acceleration of the object, called "centripetal acceleration". The effect of this acceleration is that the velocity of the object changes in direction: however if the circular motion is uniform, the speed (=the magnitude of the velocity) does not change. In this case, the magnitude of the centripetal force is given by
</span> $F=m \frac{v^2}{r}$ <span>
where m is the mass of the object, v its velocity, and r the radius of the circular path.</span>

7 0

3 years ago

NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA

kkurt [141]

Answer:

The answer is " $q=0.0945\,C$ ".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

$U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}$

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

$U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}$

Potential energy shifts:

$= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\$

$=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J$

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

$=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5} }{ 4,228 \times10^{5}} \right ) \\\\$