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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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Lynna [10]
Kinetic energy is the energy associated with the motion of an object. It's a scalar quantity, there is no direction associated with KE and it has no components. KE =  \frac{mv ^{2} }{2} =  \frac{8 *14.3 ^{2} }{2} = 4 *204.49 = 817.96J. Therefore Kinetic energy is 817.96J.
6 0
3 years ago
A beam of protons enter the electric field of magnitude E = 0.5 N/C between a pair of parallel plates. There is a magnetic field
HACTEHA [7]

Answer:

0.217 m/s

Explanation:

The protons in the beam passes undeflected when the electric force is equal to the magnetic force:

qE = qvB

where

q is the proton's charge

E is the magnitude of the electric field

v is the speed of the protons

B is the magnitude of the magnetic field

Re-arranging the equation,

v=\frac{E}{B}

And by substituting

E = 0.5 N/C

B = 2.3 T

We find

v=\frac{0.5}{2.3}=0.217 m/s

3 0
3 years ago
WHOEVER GET IT RIGHT GETS 50 POINTS
AVprozaik [17]

Answer:

ANSWER BELOW I

                             I

                            V

Remember that w=mg where w is weight in Newtons, m is mass in kilograms, and g is gravity in

m/s2

. For example, for Earth, 445 N = 45.4 × 9.8

m/s2

:Notice that the x-axis values will be gravity in

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3 0
3 years ago
The driver accelerates a 330.0 kg snowmobile, which results in a force being exerted that speeds up the snowmobile from 6.00 m/s
just olya [345]

Answer:

(a) 5610 kgm/s

(b) 5610 Ns.

(c)  78. 64 N

Explanation:

a. Change in momentum: This can be defined as the product of the mass of a body to its change in velocity. The S.I unit of change in momentum is kgm/s.

Mathematically, change in momentum is expressed as

ΔM = mΔv......................... Equation 1

Where ΔM = change in momentum, m = mass of snowmobile, Δv = change in velocity.

Given: m = 330 kg, Δv = v₂-v₁ = 23-6 = 17 m/s.

Note: v₁ and v₂ are the initial and the final velocity of the snowmobile.

ΔM = 330(17)

ΔM = 5610 kgm/s.

(b) Impulse: This can be defined as the product and force and time. The S.I unit of impulse is Ns.

Note: From Newton's second law of motion, impulse is equal to change in momentum.

Therefore,

I = ΔM................ Equation 2

Where I = impulse of the force.

Since ΔM = 5610 kgm/s.

Therefore

I = 5610 Ns.

Thus the impulse = 5610 Ns.

(c) Force: This can be defined as the product of the mass of a body and its acceleration. The S.I unit of force is Newton (N).

F = ma ................................. Equation 3

F = force, m = mass of the body, a = acceleration

But,

a = ( v₂-v₁)/t

Where v₂ = 23.0 m/s, v₁ = 6.0 m/s t = 60.0 s.

a = (23-6)/60

a = 0.283 m/s².

Substituting the value a and m into equation 3

F = 330(0.2383)

F = 78.639 N.

F ≈ 78. 64 N

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