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3 years ago

Which of the following graphs best represents the reading on the scale

Physics

1 answer:

melisa1 [442]3 years ago

5 0

The first figure matches the description

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A quarterback throws a football with an angle of elevation 55Â° and speed 60 ft/s. Find the horizontal and vertical components o

EastWind [94]

Answer:

The horizontal component of the velocity vector is;

vh = 34.4 ft/s

The vertical component of the velocity vector is;

vy = 49.1 ft/s

Explanation:

Given;

Velocity of football v = 60 ft/s

Angle of elevation ∅ = 55°

The horizontal component of the velocity vector is;

vh = vcos∅

Substituting the values;

vh = 60cos55°

vh = 34.41458618106 ft/s

vh = 34.4 ft/s

The vertical component of the velocity vector is;

vy = vsin∅

Substituting the values;

vy = 60sin55°

vy = 49.14912265733 ft/s

vy = 49.1 ft/s

7 0

3 years ago

(a) Calculate the number of free electrons per cubic meter for gold assuming that there are 1.5 free electrons per gold atom. Th

KonstantinChe [14]

Answer:

Part A:

$n=8.85*10^{28}m^{-3}$

Part B:

$Electron Mobility=3.03*10^{-3} m^2/V$

Explanation:

Part A:

To calculate the number of free electrons n we use the following formula::

n=1.5N-Au

Where N-Au is number of gold atoms per cubic meter

$N-Au=\frac{Density*Avogadro Number}{atomic weight}$

$Density = 19.32g/cm^3$

$Avogadro Number=6.02*10^{23} atoms/mol$

$Atomic weight=196.97g/mol$

So:

$n=1.5*\frac{Density*Avogadro Number}{atomic weight}$

$n=1.5*\frac{19.32*6.02*10^{23}}{196.97}$

$n=8.85*10^{28}m^{-3}$

Part B:

$Electron Mobility=\frac{Elec-conductivity}{n * charge on electron}$

n is calculated above which is 8.85*10^{28}m^{-3}

Charge on electron=1.602*10^{-19}

Elec- Conductivity= 4.3*10^{7}

$Electron Mobility=\frac{4.3*10^{7}}{ 8.85*10^{28} * 1.602*10^{-19}}$

$Electron Mobility=3.03*10^{-3} m^2/V$

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4 years ago

Paul made a diagram to compare and contrast radio waves that are used for communication and for broadcasting. Which labels belon

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Y: Travels at the speed of light

Z: May use MF and VHF waves"

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