A. The formula for mean free time is:

t = V/(4π√2 r²vN)
where
N = 1×10¹⁶ molecules (per m³)
V = 1 m³
r = 111×10⁻⁷m (atomic radius of silicon)

Let's solve for v first:
v = √(3RT/M) = √(3(8.314 m³·Pa/mol·K)(25 + 273 K)/28.1 g/mol Si)
v = 16.26 m/s

t = (1 m³)/(4π√2 (111×10⁻⁷m)²(16.26 m/s)(1×10¹⁶ molecules))
<em>t = 2.81×10⁻9 s</em>

<em>Pure silicon has a high resistivity relative to copper because copper is a conductor, while silicon is a semi-conductor. </em>

3 0

3 years ago

A golf ball is hit with a golf club. While the ball flies through the air, which forces act on the ball? Neglect air resistance.

34kurt

To solve this problem it is necessary to apply the equation related to the Gravitational Force, the equation describes that

$F = \frac{GMm}{r^2}$

Where,

G = Gravitational Universal Constant

M = Mass of Earth (or Bigger star)

m = Mass of Object (or smallest star)

r = Radius

From the statement we know that once the impact is made, the golf ball is subjected to the forces that are exerted in nature. Since the air resistance, which would represent the drag force, is ignored. Only the forces related to gravity remain.

The gravitational force carries 'pushes' or 'attracts' the body towards the earth, while the speed decreases as it reaches its maximum height.

When the ball has reached its maximum height only the force of gravity begins to act on it, generating the attraction to the earth in parabolic motion.

Therefore the correct answer is B.

6 0

3 years ago

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As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arr

Natalija [7]

Answer:

a) F = 680 N, b) W = 215 .4 J , c) F = 1278.4 N

Explanation:

a) Hooke's law is

F = k x

To find the displacement (x) let's use the elastic energy equation

$K_{e}$ = ½ k x²

k = 2 $K_{e}$ / x²

k = 2 85.0 / 0.250²

k = 2720 N / m

We replace and look for elastic force

F = 2720 0.250

F = 680 N

b) The definition of work is

W = ΔEm

W = $K_{ef}$ - $K_{eo}$

W = ½ k ( $x_{f}$ ² - x₀²)

The final distance

$x_{f}$ = 0.250 +0.220

$x_{f}$ = 0.4750 m

We calculate the work

W = ½ 2720 (0.47² - 0.25²)

W = 215 .4 J

We calculate the strength

F = k $x_{f}$

F = 2720 0.470

F = 1278.4 N

6 0

3 years ago