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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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Answer:

Both the third and fifth answers are correct

8 0
2 years ago
Explain the differences between a physical and chemical change, and give examples of each.
mrs_skeptik [129]
Chemical change:  a reaction/event where the chemicals/elements present before the change are NOT the same molecules present after the change.
For example, combustion reactions such as the burning of wood or rubbing alcohol are examples of chemical change. The reactants before the combustion of rubbing alcohol are CH3O (rubbing alcohol) and O2 oxygen (oxygen). The molecules present after the combustion reaction are CO2 (carbon dioxide) and H2O (water vapor).

Physical Change: a change in structure or state of a substance but, after the change the material is the same material that we started with. One example would be breaking a glass bottle by dropping it. The glass is now in smaller pieces than when we started, but the glass molecules are still glass molecules.

Hope this helped! Leave any questions you still have for me in the comments below!
4 0
3 years ago
A 30-km, 34.5-kV, 60-Hz, three-phase line has a positive-sequence series impedance z 5 0.19 1 j0.34 V/km. The load at the receiv
zmey [24]

Answer:

(a) With a short line, the A,B,C,D parameters are:

    A = 1pu    B = 1.685∠60.8°Ω    C = 0 S    D = 1 pu

(b) The sending-end voltage for 0.9 lagging power factor is 35.96 KV_{LL}

(c) The sending-end voltage for 0.9 leading power factor is 33.40 KV_{LL}

Explanation:

(a)

Considering the short transition line diagram.

Apply kirchoff's voltage law to the short transmission line.

Write the equation showing the relations between the sending end and the receiving end quantities.

Compare the line equations with the A,B,C,D parameter equations.

(b)

Determine the receiving-end current for 0.9 lagging power factor.

Determine the line-to-neutral receiving end voltage.

Determine the sending end voltage of the short transition line.

Determine the line-to-line sending end voltage which is the sending end voltage.

(c)

Determine the receiving-end current for 0.9 leading power factor.

Determine the sending-end voltage of the short transition line.

Determine the line-to-line sending end voltage which is the sending end voltage.

8 0
3 years ago
A railroad car having a mass of 17.5 Mg is coasting at 1.5 m/s on a horizontal track. At the same time another car having a mass
Leokris [45]

Answer:

Then final velocity of the coupled system of cars will be 0.58 m/s

Explanation: Let us consider the car-1 is moving towards right (say + ve direction) and car-2 is moving towards left (say in - ve direction), accordingly velocity are considered +ve and -ve.

m_{1} = 17.5 \ Mg = 17.5 \times 10^{6} \ g = 17.5 \times 10^{3} \ Kg = 17500 \ Kg

v_{1} = + \ 1.5 \ m/s

m_{2} = 12 \ Mg = 12 \times 10^{6} \ g = 12 \times 10^{3} \ Kg = 12000 \ Kg

v_{2} = - \ 0.75 \ m/s

Applying the conservation of momentum, and let the final velocity of combined system is V m/s

m_{1} \times v_{1} + m_{2} \times v_{2} = (m_{1} + m_{2}) \times V

17500 \times 1.5 \  + 12000 \times (-0.75) = 29500 \times V

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V = \frac{17250}{29500} = 0.58 m/s

Then final velocity of the coupled system of cars will be 0.58 m/s

8 0
3 years ago
Read 2 more answers
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Alla [95]

Answer:

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Explanation:

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The ball hits the floor with a speed of 2.40 m/s i.e. u = -2.40 m/s  (-ve because the ball hits the ground)

It rebounds with a speed of 1.7 m/s i.e. v = 1.7 m/s (+ve because the ball rebounds in upward direction)

We have to find the change in the ball's momentum. It is given by :

\Delta p=p_f-p_i

\Delta p=m(v-u)

\Delta p=0.275\ kg(1.7\ m/s-(-2.4\ m/s))

\Delta p=1.1275\ kg-m/s

So, the change in the momentum is 1.1275 kg-m/s

3 0
3 years ago
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