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Sati [7]
3 years ago
12

The coefficient of static friction between Teflon and scrambled eggs is about 0.04. What is the smallest angle from the horizont

al that will cause the eggs to slick across the bottom of a Teflon-coated skillet?
Physics
1 answer:
velikii [3]3 years ago
7 0

The smallest angle from the horizontal that will cause the eggs to slick across the bottom of a Teflon-coated skillet is 2.29°

<u>Explanation:</u>

Given-

Coefficient of friction, μ = 0.04

angle, Θ = ?

In order for the scrambled eggs to slide of the teflon, the force applied on the teflon due to gravity must be equal to the maximum value of the static friction force:

F = μ

mg sinΘ = μ mg cos Θ

sinΘ = μ cosΘ

μ = sinΘ / cosΘ

μ = tanΘ

0.04 = tanΘ

Θ = tan⁻¹ (0.04)

Θ = 2.29°

Therefore, the smallest angle from the horizontal that will cause the eggs to slick across the bottom of a Teflon-coated skillet is 2.29°

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1. List a similarity between magnetic force and electrical force.
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A 94 g particle undergoes SHM with an amplitude of 8.3 mm, a maximum acceleration of magnitude 7.8 x 103 m/s2, and an unknown ph
Lelechka [254]

Answer:

a) T = 6.49*10^-3 s

b) v = 8 m/s

c) E = 3 J

d) F = 733 N

e) F = 366.5 J

Explanation:

Given

Mass of particle, m = 94 g = 0.094 kg

Amplitude of the particle, A = 8.3 mm = 8.3*10^-3 m

Maximum acceleration of particle, a = 7.8*10^3 m/s²

the equation describing Simple Harmonic Motion is given as

x = A cos (wt +φ)

To fond the acceleration of this relationship, we would have to integrate. Twice, the first would be a Velocity, and the second acceleration that we need.

Velocity = dx/dt = -Aw sin(wt + φ)

Acceleration = d²x/dt = -Aw² cos(wt + φ)

From the question, we were given, magnitude of acceleration to be 7.8*10^3 m/s²

Aw² = 7.8*10^3

w² = 7.8*10^3 / A

w² = 7.8*10^3 / 8.3*10^-3

w² = 939759

w = √939759

w = 969

Recall, T = 2π/w, so that

T = (2 * 3.142) / 969

T = 6.49*10^-3 s

Maximum speed = Aw

Maximum speed = 8.3*10^-3 * 969

Maximum speed = 8.0 m/s

Total mechanical energy oscillator =

mgx + 1/2mx² =

1/2mv(max)² =

1/2 * 0.094 * 8² =

3 J

Maximum displacement

x = A cos(wt + φ)

For x to be maximum here, then cos(wt + φ) Must be equal to 1

Acceleration = d²x/dt² = -Aw²

And force = mass * acceleration

Force = 0.094 * 7.8*10^3

Force = 733 N

x = A cos(wt + φ), where cos(wt + φ) = 1/2

d²x/dt² = -Aw² * 1/2

d²x/dt² = 733 * 0.5

= 366.5 N

7 0
3 years ago
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