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2 years ago

9

Who studied how electromagnetic waves are formed?

2 answers:

Shalnov [3]2 years ago

6 0

James Clerk Maxwell and Heinrich Hertz are two scientists who studied how electromagnetic waves are formed and how fast they travel.

HACTEHA [7]2 years ago

4 0

The two scientists that studied how electromagnetic waves are formed are James Clerk Maxwell and Heinrich Hertz

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the density of dry air at 20 degrees celsius is 1.20 g/L. What is the mass of air, in kilograms, of a room that measures 24.0m b

horsena [70]

<h2>Mass of air in a room that measures 24.0 m by 15.0 m by 4.0 m is 1728 kg.</h2>

Explanation:

Density of air = 1.20 g/L = $\frac{1.20\times 10^{-3}kg}{10^{-3}m^3}=1.2kg/m^3$

Size of room is 24.0m by 15.0 m by 4.0 m

Volume of room = 24 x 15 x 4 = 1440 m³

We know the equation

Mass = Volume x Density

Mass = 1440 x 1.2

Mass = 1728 kg

Mass of air in a room that measures 24.0 m by 15.0 m by 4.0 m is 1728 kg.

8 0

3 years ago

A revolutionary war cannon, with a mass of 2260 kg, fires a 15.5 kg ball horizontally. The cannonball has a speed of 109 m/s aft

Anton [14]

Answer:

The gain in velocity is 0.37m/s

Explanation:

We need solve this problem though the conservation of momentum. That is,

$m_1 v_1 = m_2 v_2$

$m_1=2260Kg\\m_2=15.5Kg\\v_2= 109m/s$

Using the equation to find $v_1$ ,

$v_1=\frac{m_2 v_2}{m_1}\\v_1=\frac{15.5*109}{2260}\\v_1= 0.7475$

Using the conservation of energy equation, we have,

$KE= \frac{1}{2}m*v^2$

$KE_{cball}=\frac{1}{2}(15.5)(109)^2=92077.75J$

$KE_{cannon}=\frac{1}{2}(2260)(0.7475)^2=631.39J$

$Total KE= 92077.75+13425530=92708.9J$

Now this energy over the cannonball

$KE=\frac{1}{2}m*v_2^2$

$92708.9=\frac{1}{2}15.5v_2^2$

$V_2 = 109.37m/s$

The gain in velocity is 0.37m/s

4 0

3 years ago

_____ are small buildings containing transformers and electrical equipment

Vika [28.1K]

This room is called Substation

6 0

2 years ago

Why is water not used as a liquid in glass thermometers?

Maslowich

Answer:

Water is not able to be used as a thermometer liquid because of its higher freezing point and lower boiling point than the other liquids in general. If water is used in a thermometer, it will start phase variation at 0∘C and 100∘C. This will not help in measuring temperature, beyond this range.

Explanation:

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3 0

2 years ago

Read 2 more answers

A 23.5 g piece of aluminum metal is initially at 100.0°C. It is dropped into a coffee cup-calorimeter containing 130.0 g of wate

vivado [14]

Answer: The molar heat capacity of aluminum is $25.3J/mol^0C$

Explanation:

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of water = 130.0 g

$m_2$ = mass of aluminiunm = 23.5 g

$T_{final}$ = final temperature = $26.0^oC=(273+26)K=299K$

$T_1$ = temperature of water = $23^oC=(273+23)K=296K$

$T_2$ = temperature of aluminium = $100^oC=273+100=373K$

$c_1$ = specific heat of water= $4.184J/g^0C$

$c_2$ = specific heat of aluminium= ?

Now put all the given values in equation (1), we get

$130.0\times 4.184\times (299-296)=-[23.5\times c_2\times (299-373)]$

$c_2=0.938J/g^0C$

Molar mass of Aluminium = 27 g/mol

Thus molar heat capacity = $0.938J/g^0C\times 27g/mol=25.3J/mol^0C$

5 0

3 years ago