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notka56 [123]
4 years ago
8

A basket has a mass of 5.5 kg. Find the magnitude of the normal force if the basket is at rest on a ramp inclined above the hori

zontal at:
---
a) 0 degrees (horizontal surface)
b) 12 degrees
c) 25 degrees
d) 45 degrees


can you write the answer and the steps how you found the magnitude
Physics
2 answers:
storchak [24]4 years ago
8 0
Here is the correct answer of the given problem above.
Given that the basket has a mass of 5.5kg, the magnitude of the normal force if the basket is at rest on a ramp inclined above the horizontal is at 12 degrees. The solution is simple: 
<span>Fn at rest = lmgl </span>
<span>= 5.5kg (9.80N/kg) 
=</span><span> mgCos12degrees 
Hope this answer helps. </span>
miskamm [114]4 years ago
4 0

Answer:

(a) 53.9 N

(b) 52.72 N

(c) 48.85 N

(d) 38.11 N

Explanation:

The normal force is always perpendicular to the surface.

(a) the normal force = m x g = 5.5 x 9.8 = 53.9 N

(b) Normal force = m x g x CosФ = 5.5 x 9.8 x Cos 12 = 52.72 N

(c) Normal force = m x g x CosФ = 5.5 x 9.8 x Cos 25 = 48.85 N

(d) Normal force = m x g x CosФ = 5.5 x 9.8 x Cos 45 = 38.11 N

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Answer:

La presión neumática para levantar un automóvil de 17,640 newtons es 220,500 pascales.

Explanation:

Asumiendo que la presión (P), medida en pascales, tiene una distribución uniforme sobre la superficie del pistón, se calcula a partir de la siguiente expresion:

P = \frac{F}{A}

Donde:

F - Fuerza motriz, medida en newtons.

A - Área del pistón, medida en metros cuadrados.

La fuerza motriz es equivalente al peso del automóvil. El área del pistón (A), medido en metros cuadrados, es determinado por:

A=\frac{\pi}{4}\cdot D^{2}

Donde D es el diámetro del pistón, medido en metros.

Si D = 0.32\,m y F =17,640\,N, entonces la presión neumática es:

A = \frac{\pi}{4}\cdot (0.32\,m)^{2}

A \approx 0.080\,m^{2}

P = \frac{17,640\,N}{0.080\,m^{2}}

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Answer:

1270.64\ \text{J}

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m = Mass of object = \dfrac{mg}{g}

mg = Weight of object = 20 N

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

v = Final velocity = 15 m/s

u = Initial velocity = 0

d = Distance moved by the object = 150 m

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f = Force of friction

fd = Work done against friction

The force balance of the system is

\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}

The work done against friction is 1270.64\ \text{J}.

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