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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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A 58.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 140 m/s from the top of a cliff
strojnjashka [21]

(a) 6.43\cdot 10^5 J

The total mechanical energy of the projectile at the beginning is the sum of the initial kinetic energy (K) and potential energy (U):

E=K+U

The initial kinetic energy is:

K=\frac{1}{2}mv^2

where m = 58.0 kg is the mass of the projectile and v=140 m/s is the initial speed. Substituting,

K=\frac{1}{2}(58 kg)(140 m/s)^2=5.68\cdot 10^5 J

The initial potential energy is given by

U=mgh

where g=9.8 m/s^2 is the gravitational acceleration and h=132 m is the height of the cliff. Substituting,

U=(58.0 kg)(9.8 m/s^2)(132 m)=7.5\cdot 10^4 J

So, the initial mechanical energy is

E=K+U=5.68\cdot 10^5 J+7.5\cdot 10^4 J=6.43\cdot 10^5 J

(b) -1.67 \cdot 10^5 J

We need to calculate the total mechanical energy of the projectile when it reaches its maximum height of y=336 m, where it is travelling at a speed of v=99.2 m/s.

The kinetic energy is

K=\frac{1}{2}(58 kg)(99.2 m/s)^2=2.85\cdot 10^5 J

while the potential energy is

U=(58.0 kg)(9.8 m/s^2)(336 m)=1.91\cdot 10^5 J

So, the mechanical energy is

E=K+U=2.85\cdot 10^5 J+1.91 \cdot 10^5 J=4.76\cdot 10^5 J

And the work done by friction is equal to the difference between the initial mechanical energy of the projectile, and the new mechanical energy:

W=E_f-E_i=4.76\cdot 10^5 J-6.43\cdot 10^5 J=-1.67 \cdot 10^5 J

And the work is negative because air friction is opposite to the direction of motion of the projectile.

(c) 88.1 m/s

The work done by air friction when the projectile goes down is one and a half times (which means 1.5 times) the work done when it is going up, so:

W=(1.5)(-1.67\cdot 10^5 J)=-2.51\cdot 10^5 J

When the projectile hits the ground, its potential energy is zero, because the heigth is zero: h=0, U=0. So, the projectile has only kinetic energy:

E = K

The final mechanical energy of the projectile will be the mechanical energy at the point of maximum height plus the work done by friction:

E_f = E_h + W=4.76\cdot 10^5 J +(-2.51\cdot 10^5 J)=2.25\cdot 10^5 J

And this is only kinetic energy:

E=K=\frac{1}{2}mv^2

So, we can solve to find the final speed:

v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(2.25\cdot 10^5 J)}{58 kg}}=88.1 m/s

4 0
3 years ago
If the system is operated on mars, through what distance would the 18.0-kg mass have to fall to give the same amount of kinetic
-Dominant- [34]
The previous part of the exercise says:
"<span>Engineers are designing a system by which a falling mass m imparts kinetic energy to a rotating uniform drum to which it is attached by thin, very light wire wrapped around the rim of the drum. There is no appreciable friction in the axle of the drum, and everything starts from rest. This system is being tested on Earth, but it is to be used on Mars, where the acceleration due to gravity is 3.71 m/s². In the Earth tests, when m is set to 18.0 kg and allowed to fall through 5.50 m, it gives 300.0 J of kinetic energy to the drum."

Since Kearth = Kmars, we have, for conservation of energy, that also the potential energies must be equal:
Uearth = Umars

which means:
m </span>· gearth · hearth = m · gmars <span>· hmars

we can solve for hmars:
hmars = (gearth / gmars) </span>· hearth
           = (9.8 / 3.71) · 5.50
           = 14.53m

Therefore, the correct answer will be: the mass would have to fall from an height of 14.53m.

5 0
3 years ago
A change in the gene pool of a population due to chance is _____.
Vinvika [58]
Gene flow
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8 0
3 years ago
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The tungsten filament of a light bulb has an operating temperature of about 2 100 K. If the emitting area of the filament is 1.0
Murljashka [212]

Answer:

75 W

Explanation:

T = temperature of the filament = 2100 K

A = Emitting area of the filament = 1 cm² = 10⁻⁴ m²

e = Emissivity = 0.68

\sigma = Stefan's constant = 5.67 x 10⁻⁸ Wm⁻²K⁻⁴

Using Stefan's law, Power output of the light bulb is given as

P = \sigma e AT^{4} \\P = (5.67\times10^{-8}) (0.68) (10^{-4}) (2100)^{4}\\P = 75 W

5 0
3 years ago
In your own words explain how energy is involved in physical changes and in chemical changes
olasank [31]

Answer:

All adjustments in issue include changes in energy. Energy is either delivered or consumed. The energy is often in the form of heat, however it might be in the form of sound or light.

Explanation:

this is in my own words btw:)

8 0
3 years ago
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