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3 years ago

An element’s atomic number is 35. how many portions would an atom this element have

Physics

1 answer:

kotykmax [81]3 years ago

4 0

35 protons are present in an element whose atomic number is 35.

<u>Explanation: </u>

In an atom of an element, the number of protons = atomic number

Number of protons = number of electrons

Mass Number = number of protons + number of neutrons

Hence, an atom with atomic number 35 will have 35 protons.

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Which list of observations is the best evidence of only a chemical change occurring?

Aleonysh [2.5K]

Answer:

Explanation:

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3 years ago

A bowling ball rolls 33\,\text m33m33, start text, m, end text with an average speed of 2.5\,\dfrac{\text m}{\text s}2.5 s m 2

asambeis [7]

The ball rolled for 13.2 s

<h3>Further explanation</h3>

Speed is scalar and no direction

$\tt avg~speed=\dfrac{total~distance}{elapsed~time}=\dfrac{\Delta x}{\Delta t}$

A bowling ball rolls 33 m, with average speed = 2.5 m/s

So elapsed time :

$\tt t=\dfrac{33~m}{2.5`m/s}=13.2~s$

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3 years ago

create a poem that incorporates those ten words. Feel free to make it as silly as you like! MINIMUM of 6 lines with a MINIMUM of

Luba_88 [7]

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8 0

3 years ago

If ice has a density of.92g/cm, what is the volume of 1.8 kilograms of ice?

steposvetlana [31]

Given:
Density = .92 g / cm³
Volume = 1.8 kg
= (1.8 * 1000) grams
= 1800 grams
Now,
$Density (d) = \frac{Mass (m)}{volume(v)}$

$volume (v)= \frac{mass(m)}{density(d)}$

$volume (v)= \frac{1800}{.92}$

$volume (v)= 1956.521739$

$volume (v)= 1956.5~ cm^{3}$

So, the volume of 1.8 kg of ice is 1956.5 cm³

5 0

3 years ago

A section of a parallel-plate air waveguide with a plate separation of 7.11 mm is constructed to be used at 15 GHz as an evanesc

adell [148]

Answer:

the required minimum length of the attenuator is 3.71 cm

Explanation:

Given the data in the question;

we know that;

$f_{c_1$ = c / 2a

where f is frequency, c is the speed of light in air and a is the plate separation distance.

we know that speed of light c = 3 × 10⁸ m/s = 3 × 10¹⁰ cm/s

plate separation distance a = 7.11 mm = 0.0711 cm

so we substitute

$f_{c_1$ = 3 × 10¹⁰ / 2( 0.0711 )

$f_{c_1$ = 3 × 10¹⁰ cm/s / 0.1422 cm

$f_{c_1$ = 21.1 GHz which is larger than 15 GHz { TEM mode is only propagated along the wavelength }

Now, we determine the minimum wavelength required

Each non propagating mode is attenuated by at least 100 dB at 15 GHz

Attenuation constant TE₁ and TM₁ expression is;

∝₁ = 2πf/c × √( ( $f_{c_1$ / f)² - 1 )

so we substitute

∝₁ = ((2π × 15)/3 × 10⁸ m/s) × √( (21.1 / 15)² - 1 )

∝₁ = 3.1079 × 10⁻⁷

∝₁ = 310.79 np/m

Now, To find the minimum wavelength, lets consider the design constraint;

20log₁₀ $e^{-\alpha _1l_{min$ = -100dB

we substitute

20log₁₀ $e^{-(310.7np/m)l_{min$ = -100dB

$l_{min$ = 3.71 cm

Therefore, the required minimum length of the attenuator is 3.71 cm

6 0

3 years ago