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irina1246 [14]
2 years ago
13

Walter Arfeuille of Belgium lifted a 281.5 kg load off the ground using his teeth. Suppose Arfeuille can hold just three times t

hat mass on a 30.0° slope using the same force. What is the coefficient of static friction between the load and the slope?
Physics
2 answers:
Umnica [9.8K]2 years ago
6 0
Gravitational, g = 9.81 m/s2 
Theta = 30 degrees, using the same force. 
Let u be the coefficient of static friction 
Force applied F = mg 
Static Friction Force = u3mgcos (theta) 
The gravitational Force = 3mgsin (theta) 
So the force equation = 3mgsin (theta) - (u3mgcos (theta) + mg) = 0 
So 3mgsin (theta) = (u3mgcos (theta) + mg) => 3sin (theta) = (u3cos (theta) + 1)
 => u = (3sin (30) - 1) / 3cos (30) => u = (3(1/2) - 1) / (3 x 0.866)
 => u = (1.5 - 1) / 2.59 = 0.5 / 2.59 = 0.5 / 2.598
 Coefficient of Static Friction u = 0.19.
Alecsey [184]2 years ago
5 0
Let's call
 m = 281.5Kg
 The force that Walter can make is by definition:
 F = m * g
 where,
 g: acceleration of gravity.
 Suppose Walter is at the top of the slope:
 By making a free-body diagram we have the following forces in the direction of the slope:
 Friction force = (μ3mgcosø)
 Force of gravity = (3mgsinø),
 Force exerted by Walter = mg
 We have then:
 mg + μ3mgcosø-3mgsinø = 0.
 Clearing the friction coefficient:
 μ = (3sinø-1) / (3cosø).
 Substituting the angle:
 μ = 0.19.
 answer
 the coefficient of static friction between the load and the slope is 0.19
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The intensity of a light in a surface follows the inverse square law formula which can be mathematically expressed as,
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1 year ago
Consider a capacitor made of two rectangular metal plates of length and width , with a very small gap between the plates. There
mezya [45]

Answer:

F= σ² L² /2ε₀

F = (L² ε₀/4π)   ΔV² / r⁴

Explanation:

a)  For this exercise we can use Coulomb's law

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where the negative sign indicates that the force is attractive and the value of the charge is equal to the two plates

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         C = Q / ΔV

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also the capacitance for a parallel plate capacitor is related to its shape

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we substitute

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we substitute in the force equation

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           k = 1 / 4πε₀

           F = ε₀ /4π  L² ΔV² / r⁴4

           F = L² ΔV² ε₀/ (4π r⁴)

           F = (L² ε₀/4π)   ΔV² / r⁴

b) Another way to solve the exercise is to use the relationship between the force and the electric field

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           Ф = ∫E .dA = q_{int} / ε₀

the plate have two side

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substituting in force

          F = q σ / 2ε₀

the charge total on the other plate is

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       q = σ  L²

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4 0
2 years ago
Two small aluminum spheres, each of mass 0.0250 kilograms, areseparated by 80.0 centimeters.
sineoko [7]

Answer:

Total number of electrons

N = 7.25 \times 10^{24}

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N = 5.27 \times 10^{15}

Fraction of electrons transferred is given as

f = 7.27 \times 10^{-10}

Explanation:

As we know that moles is defined as

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n = 0.93

so number of atoms of Al in each sphere is given as

N = 0.93(6.02 \times 10^{23})

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Now number of electrons in each atom is given as

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N = 13 \times (5.58 \times 10^{23})

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Also we know that force of attraction between them is given as

F= \frac{kq_1q_2}{r^2}

1.00 \times 10^4 = \frac{(9\times 10^9)q^2}{0.80^2}

q = 8.4 \times 10^{-4} C

now we have

q = Ne

8.4 \times 10^{-4} = N(1.6 \times 10^{-19}

N = \frac{(8.4 \times 10^{-4})}{1.6 \times 10^{-19}}

N = 5.27 \times 10^{15}

Fraction of electrons transferred is given as

f = \frac{5.27 \times 10^{15}}{7.25 \times 10^{24}}

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ddd [48]

Answer:

a. mass density

Explanation:

<em>Land and sea breeze that occur near the shore are due to the variation of mass density of air with change in temperature.</em>

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<em>During the day the land is warmer than the sea so the sea breeze blows and during the night the water bodies are warmer than the land so the land breeze blows.</em>

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2 years ago
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vlada-n [284]

Answer:

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W_f = -5857.8 J

Explanation:

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Work done by all forces = change in kinetic energy of the system

so here car is moving from bottom to top

so here the change in kinetic energy is total work done on the car

so here we will have

W_f + W_g = \frac{1}{2}m(v_f^2 - v_i^2)

W_f - mgH = \frac{1}{2}m(v_f^2 - v_i^2)

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W_f - (130)(9.81)(2\times 12) = \frac{1}{2}(130)(8^2 - 25^2)

W_f = 30607.2 - 36465

W_f = -5857.8 J

6 0
2 years ago
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