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GalinKa [24]
2 years ago
6

Before hanging new William Morris wallpaper in her bedroom, Brenda sanded the walls lightly to smooth out some irregularities on

the surface. The sanding block weighs 1.70 N and Brenda pushes on it with a force of 3.00 N at an angle of 30.0° with respect to the vertical, and angled toward the wall. What is the coefficient of kinetic friction between the wall and the block?

Physics
1 answer:
vova2212 [387]2 years ago
3 0

Answer:

1.0193

Explanation:

The concepts that we apply here is Frictional Force with a Force on the object with a angle.

We know that Frictional Force is equal to

F_f = \mu N

The force acts downward because the moving is upward, then from the image that I attached, the forve is equal to,

Fcos30° = 3N(cos30) = 2.6

In Y we have then

1.7N + f_x = 2.6

f_x = 1.529N

For the horizontal componentes we have that

N = (3N)sin30\°

N = 1.5 N

Using the first equation we have that

F_f = \mu N

1.529N = \mu 1.5 N

Re-arrange for \mu

\mu = 1.0193

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A charged box (m=445 g, ????=+2.50 μC) is placed on a frictionless incline plane. Another charged box (????=+75.0 μC) is fixed i
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The concept required to perform this exercise is given by the coulomb law.

The force expressed according to this law is given by

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Our values are previously given, so

q= 2.5*10^{-6}C\\Q= 75*10^{-6}C\\r=0.59

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

F= \frac{(8.99 x 10^9)(2.5*10^{-6})(75*10^{-6})}{0.59^2}

F= 4.8423N

The force acting on the block are given by,

F-mgsin\theta = ma

a = \frac{F-mgsin\theta}{m}

a = \frac{4.8423-(0.445)(9.8)sin(35)}{0.445}a = 10.31m/s^2

Therefore the box is accelerated upward.

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2 years ago
Engineers who design battery-operated devices such as cell phones and MP3 players try to make them as efficient as possible. An
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efficiency= [useful energy transferred ÷ total energy supply]×100%

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Grain is pored into a silo to be stored for later use. Due to the friction between pieces of grain as they rub against eachother
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3.1×10⁻¹¹ N

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F = k q₁ q₂ / r²

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F = 3.1×10⁻¹¹

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Assume that you stay on the earth's surface. what is the ratio of the sun's gravitational force on you to the earth's gravitatio
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First, let's determine the gravitational force of the Earth exerted on you. Suppose your weight is about 60 kg. 

F = Gm₁m₂/d²
where
m₁ = 5.972×10²⁴ kg (mass of earth)
m₂ = 60 kg
d = 6,371,000 m (radius of Earth)
G = 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²

F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(5.972×10²⁴ kg)/(6,371,000 m )²
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Next, we find the gravitational force exerted by the Sun by replacing,
m₁ = 1.989 × 10³⁰<span> kg
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Thus,
d = 149.6×10⁹ m - 6,371,000 m = 1.496×10¹¹ m

Thus,
F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(1.989 × 10³⁰ kg)/(1.496×10¹¹ m)²
F = 0.356  N

Ratio = 0.356  N/589.18 N
<em>Ratio = 6.04</em>
5 0
2 years ago
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