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3 years ago

Where are the physics in football

Physics

1 answer:

sasho [114]3 years ago

5 0

Answer:

The branch of physics that is most relevant to football is mechanics, the study of motion and its causes.

Explanation:

When the ball leaves the punter's foot, it is moving with a given velocity (speed plus angle of direction) depending upon the force with which he kicks the ball. The ball moves in two directions, horizontally and vertically. Because the ball was launched at an angle, the velocity is divided into two pieces: a horizontal component and a vertical component.

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I need to choose a theme for my physics assignment My experiment is finding g

Kobotan [32]

<h3>Question:</h3>

How to find g (acceleration due to gravity)

<h3>Solution:</h3>

We know,

Acceleration due to gravity (g)

$= \frac{GM}{ {R}^{2} }$

where, G = Gravitational constant

$= 6.67 \times {10}^{11} N {m}^{2}/k {g}^{2} \\$

M = Mass of the earth

$= 6 \times {10}^{24} \: kg$

R = Radius of the earth

$= 6.4 \times {10}^{6} m$

Putting these values of G, M and R in the above formula, we get

$g \: = \: \frac{6.67 \times {10}^{11} N {m}^{2}/k {g}^{2} \times \: 6 \times {10}^{24} \: kg }{(6.4 \times {10}^{6}m {)}^{2} } \\ = 9.8m/ {s}^{2}$

So, the value of acceleration due to gravity is

$9.8m/s ^{2}$

Hope it helps.

Do comment if you have any query.

5 0

3 years ago

The nuclei of large atoms, such as uranium, with 9292 protons, can be modeled as spherically symmetric spheres of charge. The ra

Scrat [10]

Answer:

Part 1 $E = 2.42 * 10^{21} N/C$

Part 2 $E = 1.3 * 10^{13} N/C$

Part 3 $E = 0$

Explanation:

Given

Number of protons $= 92$

Radius of nucleus $r_n = 7.4 * 10^{-15} m$

Distance of the electrons $r_1 = 1.0 * 10^ {-10} m$

Part 1

Electric field produced by just outside its surface

$E = \frac{q}{4\pi*E_0* r_n^2 } \\E = \frac{9 * 10^ 9 * 92 * 1.6 * 10^{-19}}{(7.4* 10^{-15})^2} \\E = 2.42 * 10^{21} N/C$

Part 2

Electric field produced by just outside its surface

$E = \frac{q}{4\pi*E_0* r_n^2 } \\E = \frac{9 * 10^ 9 * 92 * 1.6 * 10^{-19}}{(1* 10^{-15})^2} \\E = 1.3 * 10^{13} N/C$

Part 3

The net electric field inside a uniform shell of negative charge is zero because the electric flux lines cancel out each other

hence, the solution is

Part 1 $E = 2.42 * 10^{21} N/C$

Part 2 $E = 1.3 * 10^{13} N/C$

Part 3 $E = 0$

7 0

3 years ago

What is the resolution of three skeleton key?

Arisa [49]

It depends because it’s might be lolilolololol 21212132

3 0

3 years ago

4. Heat is added to an ideal gas and the gas expands. In such a process the temperature

Bumek [7]

Answer:

is high as 100 degrees c

Explanation:

due to high heat gas expands fast than normal

4 0

3 years ago

The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371

arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

$q=m\times c\times (T_2-T_1)$

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = $4.18J/g.K$

$T_1$ = initial temperature of water = 293.0 K

$T_2$ = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

$q=250g\times 4.18J/g.K\times (371.2-293.0)K$

$q=81719J$

Now we have to calculate the enthalpy of combustion of octane.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

$\Delta H=\frac{-81719J}{0.015mole}$

$\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol$

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0

4 years ago