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blondinia [14]
3 years ago
15

Why is physics important to everyone? Describe 5 applications of physics in everyday life.

Physics
1 answer:
Sedaia [141]3 years ago
4 0

1. Communication

   In early ages people were using letters to communicate between them. Now days instantly we can call one person from anywhere, these are based on optical fiber communication and it is an application of physics.

2. Transportation

  We can travel from place to anther place by aeroplane , these technologies are introduced after advancement in physics.

3. Construction

 Earlier people used to stay in small houses, now days apartments with many number of storeys are built by considering all the forces coming on the structure.

4. Medical Department

 The invention of many equipment increased the health of humans.

5. Weather forecasting

 The prediction of weather considering all the available factors, prediction of natural disasters all are application of physics.

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In-s [12.5K]
Vf=vi plus 2 ad
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3 years ago
When the magnetic flux through a coil of wire changes, what is generated between the ends of the coil? Which equation predicts t
True [87]

Answer:

induced electromotive force (Voltage)    E = - N dΦ / dt

Explanation:

When the magnetic flux this coil induces a current in each turn of the coil, which is why an induced electromotive force (Voltage) appears at the ends of the coil.

This phenomenon is fully explained by Faraday's law

        E = - dΦ / dt

where in the case of a coil with N turns of has

        E = - N dΦ / dt

Rl flux is the product of the normal to the area by the magnetic field, in this case the flux changes so we can assume that the area of ​​the coil is constant

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3 years ago
Individuals who suffer from a mental disorder are most likely to commit violent crimes. true or false
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I hope this helps!
~cupcake
7 0
3 years ago
Read 2 more answers
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astraxan [27]

Answer:

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7 0
3 years ago
The average distance of the planet mercury from the sun is 0.39 times the average distance of the earth from the sun. How long i
Stels [109]

Answer:

T_1=0.24y

Explanation:

Using Kepler's third law, we can relate the orbital periods of the planets and their average distances from the Sun, as follows:

(\frac{T_1}{T_2})^2=(\frac{D_1}{D_2})^3

Where T_1 and T_2 are the orbital periods of Mercury and Earth respectively. We have D_1=0.39D_2 and T_2=1y. Replacing this and solving for

T_1^2=T_2^2(\frac{D_1}{D_2})^3\\T_1^2=(1y)^2(\frac{0.39D_2}{D_2})^3\\T_1^2=1y^2(0.39)^3\\T_1^2=0.059319y^2\\T_1=0.24y

5 0
3 years ago
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