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mote1985 [20]
3 years ago
15

Which energy level of the outermost in aluminum

Physics
1 answer:
Eduardwww [97]3 years ago
8 0
The outermost energy level of an element are called the valence shell, that holds the valence electrons. they consist of  the highest energy level. In aluminum, the valence electrons are 3. <span />
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Which of the following phenomena provides evidence of electric fields in the atmosphere?
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The ground(Earth's surface) is at low potential whereas the clouds are at higher potential. Due to this potential difference, we see lightning.
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True or false
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3 years ago
The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz. Find the possible range of wavelengths in ai
taurus [48]

Answer:

The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

Explanation:

Given that,

The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz.

The speed of sound in air is 343 m/s.

To find,

The wavelength range for the corresponding frequency.

Solution,

The speed of sound is given by the following relation as :

v=f_1\lambda_1

Wavelength for f = 45 Hz is,

\lambda_1=\dfrac{v}{f_1}

\lambda_1=\dfrac{343}{45}=7.62\ m

Wavelength for f = 375 Hz is,

\lambda_2=\dfrac{v}{f_2}

\lambda_2=\dfrac{343}{375}=0.914\ m/s

So, the possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

6 0
4 years ago
After the water has boiled, the temperature of the water decreases by 22 °C.
Leviafan [203]

Explanation:

Use the formula:

e = mc \delta \theta

e is the energy released,

m is the mass of water,

c is the specific heat capacity,

δθ is the change in temperature ( 100 - 22)

4 0
3 years ago
An emf is induced by rotating a 1060 turn, 20.0 cm diameter coil in the Earth's 5.25 ✕ 10−5 T magnetic field. What average emf (
lara31 [8.8K]

Answer:

The average emf induced in the coil is 175 mV

Explanation:

Given;

number of turns of the coil, N = 1060 turns

diameter of the coil, d = 20.0 cm = 0.2 m

magnitude of the magnetic field,  B = 5.25 x 10⁻⁵ T

duration of change in field, t = 10 ms = 10 x 10⁻³ s

The average emf induced in the coil is given by;

E = N\frac{\delta \phi}{dt} \\\\E = N\frac{\delta B}{\delta t}A

where;

A is the area of the coil

A = πr²

r is the radius of the coil = 0.2 /2 = 0.1 m

A = π(0.1)² = 0.03142 m²

E = \frac{NBA}{t} \\\\E = \frac{1060*5.25*10^{-5}*0.03142}{10*10^{-3}} \\\\E = 0.175 \ V\\\\E = 175 \ mV

Therefore, the average emf induced in the coil is 175 mV

3 0
3 years ago
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