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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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One mol of a perfect, monatomic gas expands reversibly and isothermally at 300 K from a pressure of 10 atm to a pressure of 2 at
Zolol [24]

Answer:

Explanation:

Given

1 mole of perfect, monoatomic gas

initial Temperature(T_i)=300 K

P_i=10 atm

P_f=2 atm

Work done in iso-thermal process=P_iV_iln\frac{P_i}{P_f}

P_i=initial pressure

P_f=Final Pressure

W=10\times 2.463\times ln\frac{10}{2}=39.64 J

Since it is a iso-thermal process therefore q=w

Therefore q=39.64 J

(b)if the gas expands by the same amount again isotherm-ally and irreversibly

work done is=P\Delta V

V_1=\frac{RT_1}{P_1}=\frac{1\times 0.0821\times 300}{10}=2.463 L

V_2=\frac{RT_2}{P_2}=\frac{1\times 0.0821\times 300}{2}=12.315 L

\Delta W=1\times (12.315-2.463)=9.852 J

\Delta q=\Delta W=9.852 J

\Delta U=0

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4 years ago
Choose the true statement concerning the mall or subatomic particles:
solniwko [45]
C is true, and just one of those has as much mass as about 1,840 electrons.
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What is the sharpness of the light and the amount of voltage when you put the magnet through the lower number of coils first vs
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A vehicle moving with a uniform a acceleration of 2m/s and has a velocity of 4m/s at a certain time .
Evgesh-ka [11]

Answer:

1 second later the vehicle's velocity will be:

v(1)= 6\,\,\frac{m}{s} \\

5 seconds later the vehicle's velocity will be:

v(5)=14\,\,\frac{m}{s}

Explanation:

Recall the formula for the velocity of an object under constant accelerated motion (with acceleration "a"):

v(t)=v_0+a\,t

Therefore, in this case v_0=4\,\,\frac{m}{s}  and a=2\,\,\frac{m}{s^2}

so we can estimate the velocity of the vehicle at different times just by replacing the requested "t" in the expression:

v(t)=v_0+a\,t\\v(t)=4+2\,\,t\\v(1)=4+2\,(1) = 6\,\,\frac{m}{s} \\v(5)=4+2\,(5)=14\,\,\frac{m}{s}

3 0
3 years ago
If a rock is thrown upward on the planet Mars with a velocity of 15 m/s, its height above the ground (in meters) after t seconds
s2008m [1.1K]

(a) The velocity (in m/s) of the rock after 1 second is 11.28 m/s.

(b) The velocity of the rock after 2 seconds is 7.56 m/s.

(c) The time for the block to hit the surface is 4.03.

(d) The velocity of the block at the maximum height is 0.

<h3>Velocity of the rock</h3>

The velocity of the rock is determined as shown below;

Height of the rock after 1 second; H(t) = 15(1) - 1.86(1)² = 13.14 m

v² = u² - 2gh

where;

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v² = (15)² - 2(3.72)(13.14)

v² = 127.23

v = √127.23

v = 11.28 m/s

<h3>Velocity of the rock when t = 2 second</h3>

v = dh/dt

v = 15 - 3.72t

v(2) = 15 - 3.72(2)

v(2) = 7.56 m/s

<h3>Time for the rock to reach maximum height</h3>

dh/dt = 0

15 - 3.72t = 0

t = 4.03 s

<h3>Velocity of the rock when it hits the surface</h3>

v = u - gt

v = 15 - 3.72(4.03)

v = 0

Learn more about velocity at maximum height here: brainly.com/question/14638187

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