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Travka
2 months ago
11

A person is trying to judge whether a picture (mass = 1.10 kg) is properly positioned by temporarily pressing it against a wall.

The pressing force is perpendicular to the wall. The coefficient of static friction between the picture and the wall is 0.660. What is the minimum amount of pressing force that must be used?
Physics
1 answer:
kicyunya [3.2K]2 months ago
7 0

Respuesta:

16.3 N

Explicación:

El coeficiente depende del material de la pared y del estado del objeto. Habrá dos fuerzas en acción.

P = fuerza debida a la presión

F= fuerza por fricción

Entonces F1= μP

y F2= mg (con el que cuelga la imagen)

por lo tanto F1= F2

     μP = mg

       P=  mg/μ

     Sustituyendo todos los valores: m= 1.1 kg   μ=.660       g=9.8m/s²

      P= ( 1.1 × 9.8 ) /.660

      P=  16.3 N

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Response:

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- Another 2 N weight at the 50 cm position produces a moment of 100 Ncm also clockwise.

- The weight of the 1 N stick located at its center of mass (50 cm) has a moment of 50 Ncm, clockwise.

- A 3 N weight at the 60 cm position generates a moment of 180 Ncm, clockwise.

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Ostrovityanka [3204]

Result:

1.60 g

Elucidation:

Based on the attached document:

we can infer that:

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The segment corresponds to a quarter of a circle with radius r,

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Then the radius (r) can be calculated as:

r = \frac{4x}{2 \pi}\\\\r = \frac{4*40}{2 \pi}

r = 25.5 m

Centripetal acceleration can be expressed as:

a = \frac{v^2}{r}

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a = \frac{(20 \ m/s^2)}{25.5 \ m}

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The acceleration magnitude suffered by your passengers in relation to the acceleration due to gravity can be calculated using:

a' = \frac{a}{g} g

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6 0
1 month ago
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inna [3103]

The peak wavelength for Betelgeuse is 828 nm

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