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anastassius [24]
2 years ago
11

A roll of kitchen aluminum foil is 30 cm wide by 22 m long (if you unroll it). If the foil is 0.15 mm thick, and the specific we

ight of aluminum is 26460 N/m3, how much does the roll of aluminum foil weigh
Physics
1 answer:
Andreyy892 years ago
6 0

The weight of the aluminum foil is 26.20 N

To find the weight of the aluminum foil when it is unrolled, we need to find its volume.

The volume V = lwt where

  • l = length of aluminum foil = 22 m,
  • w = width of aluminum foil = 30 cm = 0.30 m and
  • t = thickness of aluminum foil = 0.15 mm = 0.15 × 10⁻³ m.

So, V = lwt

= 22 m × 0.30 m × 0.15 × 10⁻³ m

= 0.99 × 10⁻³ m³.

So, its weight W = ρV where

  • ρ = specific weight of aluminum = 26460 N/m³ and
  • V = volume of aluminum foil = 0.99 × 10⁻³ m³

So, W = ρV

W = 26460 N/m³ × 0.99 × 10⁻³ m³

W = 26195.4 × 10⁻³ N

W = 26.1954 N

W ≅ 26.20 N

So, the weight of the aluminum foil is 26.20 N

Learn more about weight of aluminum foil here:

brainly.com/question/9922121

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3 years ago
A force of 350N is applied to a body. If the work done is 40kJ, what is the distance through which the body moved?
Studentka2010 [4]

The distance covered by the body is 114.3 m

Explanation:

The work done by a force exerted on an object is given by

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the object

\theta is the angle between the direction of the force and of the displacement

For the object in this problem, we have

F = 350 N is the force applied

W=40 kJ = 40,000 J is the work done

\theta=0^{\circ} if we assume that the force is applied parallel to the motion of the object

Solving for d, we find the distance covered by the object:

d=\frac{W}{F cos \theta}=\frac{40,000}{(350)(cos 0)}=114.3 m

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#LearnwithBrainly

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3 years ago
What do we measure sound intensity in?
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Read 2 more answers
The charges of two particles are as follows: Q1=2 x 10 -8 C and Q2 = 3 x 10 -7 C. Find the magnitude of the force between these
Fantom [35]

Answer:

F = 3.86 x 10⁻⁶ N

Explanation:

First, we will find the distance between the two particles:

r = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2+(z_{2}-z_{1})^2}\\

where,

r = distance between the particles = ?

(x₁, y₁, z₁) = (2, 5, 1)

(x₂, y₂, z₂) = (3, 2, 3)

Therefore,

r = \sqrt{(3-2)^2+(2-5)^2+(3-1)^2}\\r = 3.741\ m\\

Now, we will calculate the magnitude of the force between the charges by using Coulomb's Law:

F = \frac{kq_{1}q_{2}}{r^2}\\

where,

F = magnitude of force = ?

k = Coulomb's Constant = 9 x 10⁹ Nm²/C²

q₁ = magnitude of first charge = 2 x 10⁻⁸ C

q₂ = magnitude of second charge = 3 x 10⁻⁷ C

r = distance between the charges = 3.741 m

Therefore,

F = \frac{(9\ x\ 10^9\ Nm^2/C^2)(2\ x\ 10^{-8}\ C)(3\ x\ 10^{-7}\ C)}{(3.741\ m)^2}\\

<u>F = 3.86 x 10⁻⁶ N</u>

5 0
3 years ago
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