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4 years ago

Predict the number of valence electrons for each element based on its location in the periodic table of elements.

Physics

1 answer:

o-na [289]4 years ago

6 0

Answer:

Barium=2

Lead=4

Bismuth=5

Potassium=1

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A sinusoidal voltage Δv = (100 V) sin (170t) is applied to a series RLC circuit with L = 40 mH, C = 130 μF, and R = 50 Ω.

mixer [17]

Answer:

See attached file

Explanation:

5 0

3 years ago

7) HELP ASAP PLZ PLZ HELP ME I don't know if I am right so if I am not right can you plz help me

CaHeK987 [17]

I believe that the answer is B

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3 years ago

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When you push a box with 20N of force what force does the box apply back on you ?

Elden [556K]

Newton’s 3rd Law: for every action there is an equal but opposite reaction.
Assuming that the box is motionless, we can say the force is -20N

3 0

3 years ago

A bicyclist initially at rest, begins pedaling and gained speed steadily for 4.80s during which she covers 37.0m What was her fi

lidiya [134]

Answer:

The final speed of the bicyclist is 15.44 m/s.

Explanation:

Given;

initial velocity, u = 0

time of motion, t = 4.8 s

distance covered, d = 37.0 m

The acceleration of the bicyclist is calculated as;

d = ut + ¹/₂at²

37 = 0 + ¹/₂(4.8)²a

37 = 11.52a

a = 37 / 11.5

a = 3.22 m/s²

The final speed of the bicyclist is given as;

v² = u² + 2ad

v² = 0 + 2(3.22)(37)

v² = 238.28

v = √238.28

v = 15.44 m/s

Therefore, the final speed of the bicyclist is 15.44 m/s.

4 0

3 years ago

Determine the ratio β = v/c for each of the following.

nlexa [21]

Answer:

a) $\beta = 1.111\times 10^{-7}$ , b) $\beta = 9\times 10^{-7}$ , c) $\beta = 3.087\times 10^{-6}$ , d) $\beta = 2.5\times 10^{-5}$ , e) $\beta = 0.5$ , f) $\beta = 0.877$

Explanation:

From relativist physics we know that $c$ is the symbol for the speed of light, which equal to approximately 300000 kilometers per second. (300000000 meters per second).

a) A car traveling 120 kilometers per hour:

At first we convert the car speed into meters per second:

$v = \left(120\,\frac{km}{h} \right)\times \left(1000\,\frac{m}{km} \right)\times \left(\frac{1}{3600}\,\frac{h}{s} \right)$

$v = 33.333\,\frac{m}{s}$

The ratio $\beta$ is now calculated: ( $v = 33.333\,\frac{m}{s}$ , $c = 3\times 10^{8}\,\frac{m}{s}$ )

$\beta = \frac{33.333\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 1.111\times 10^{-7}$

b) A commercial jet airliner traveling 270 meters per second:

The ratio $\beta$ is now calculated: ( $v = 270\,\frac{m}{s}$ , $c = 3\times 10^{8}\,\frac{m}{s}$ )

$\beta = \frac{270\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 9\times 10^{-7}$

c) A supersonic airplane traveling Mach 2.7:

At first we get the speed of the supersonic airplane from Mach's formula:

$v = Ma\cdot v_{s}$

Where:

$Ma$ - Mach number, dimensionless.

$v_{s}$ - Speed of sound in air, measured in meters per second.

If we know that $Ma = 2.7$ and $v_{s} = 343\,\frac{m}{s}$ , then the speed of the supersonic airplane is:

$v = 2.7\cdot \left(343\,\frac{m}{s} \right)$

$v = 926.1\,\frac{m}{s}$

The ratio $\beta$ is now calculated: ( $v = 926.1\,\frac{m}{s}$ , $c = 3\times 10^{8}\,\frac{m}{s}$ )

$\beta = \frac{926.1\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 3.087\times 10^{-6}$

d) The space shuttle, travelling 27000 kilometers per hour:

At first we convert the space shuttle speed into meters per second:

$v = \left(27000\,\frac{km}{h} \right)\times \left(1000\,\frac{m}{km} \right)\times \left(\frac{1}{3600}\,\frac{h}{s} \right)$

$v = 7500\,\frac{m}{s}$

The ratio $\beta$ is now calculated: ( $v = 7500\,\frac{m}{s}$ , $c = 3\times 10^{8}\,\frac{m}{s}$ )

$\beta = \frac{7500\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 2.5\times 10^{-5}$

e) An electron traveling 30 centimeters in 2 nanoseconds:

If we assume that electron travels at constant velocity, then speed is obtained as follows:

$v = \frac{d}{t}$

Where:

$v$ - Speed, measured in meters per second.

$d$ - Travelled distance, measured in meters.

$t$ - Time, measured in seconds.

If we know that $d = 0.3\,m$ and $t = 2\times 10^{-9}\,s$ , then speed of the electron is:

$v = \frac{0.3\,m}{2\times 10^{-9}\,s}$

$v = 1.50\times 10^{8}\,\frac{m}{s}$

The ratio $\beta$ is now calculated: ( $v = 1.5\times 10^{8}\,\frac{m}{s}$ , $c = 3\times 10^{8}\,\frac{m}{s}$ )

$\beta = \frac{1.5\times 10^{8}\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 0.5$

f) A proton traveling across a nucleus (10⁻¹⁴ meters) in 0.38 × 10⁻²² seconds:

If we assume that proton travels at constant velocity, then speed is obtained as follows:

$v = \frac{d}{t}$

Where:

$v$ - Speed, measured in meters per second.

$d$ - Travelled distance, measured in meters.

$t$ - Time, measured in seconds.

If we know that $d = 10^{-14}\,m$ and $t = 0.38\times 10^{-22}\,s$ , then speed of the electron is:

$v = \frac{10^{-14}\,m}{0.38\times 10^{-22}\,s}$

$v = 2.632\times 10^{8}\,\frac{m}{s}$

The ratio $\beta$ is now calculated: ( $v = 2.632\times 10^{8}\,\frac{m}{s}$ , $c = 3\times 10^{8}\,\frac{m}{s}$ )

$\beta = \frac{2.632\times 10^{8}\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 0.877$

4 0

3 years ago