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3 years ago

Find focal length of a concave mirror having 40 cm radius of curvature

Physics

1 answer:

nignag [31]3 years ago

5 0

Answer:

The focal length of the given spherical mirror is 20cm.

Explanation:

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Meg goes swimming on a hot afternoon. When she comes out of the pool, her foot senses that the prevement is unbearably hot. Supp

mart [117]

1.Record her observation with the time it was hot.
2. Gather info about the pavement and its surroundings. Find out what it's made of and what its temp. is at different times of the day.
3. Come up with a hypothesis about why it is hot.
4. Design an experiment to test the hypothesis. If she thinks the Sun is responsible (which she should b smart enough to know), keep it covered during the day time and check it's temp.
5. Come up with a conclusion. If her hypothesis is not supported, design a new experiment or gather more info.

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.

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

$=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\$

$\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C$

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0

3 years ago

Competition in the Los Angeles Flower District results in

stiks02 [169]

Competition in the Los Angeles Flower District results in better quality flowers.

<h3>Why quality is the standard in flower competition?</h3>

Competition results in better quality flowers because in the competition, best quality of flowers will be selected as a winner so the competitors produces best quality of flowers in order to claim the prize so we can conclude that Competition in the Los Angeles Flower District results in better quality flowers

Learn more about competition here: brainly.com/question/25605883

8 0

2 years ago

9. a. Determine the mass of a football which has a weight of 0.80 N on a planet where the gravitational field strength is 2 N/kg

krek1111 [17]

Answer:

m = 0.4 [kg]

Explanation:

Weight is considered as a force and this is equal to the product of mass by gravitational acceleration.

$W=m*g\\$