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1 year ago

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rolls of foil are 304 mm wide and 0.020 mm thick. (the density of foil is 2.7 g/cm3 .) what maximum length of foil can be made f

rom 1.09 kg of foil?

1 answer:

Rainbow [258]1 year ago

5 0

The maximum length of foil will be 6639.8635 cm or 66.3986 m

To find the maximum length of the foil, the given datas are:

Width of roll of foil = 304 mm

Height or thickness = 0.020 mm

Density of foil = 2.7 g/cm³

Mass of foil = 1.09 Kg

<h3>What is Density?</h3>

Density is the substance's mass per unit volume. It can be also described by measuring how the particles in the substances tightly packed together.

d = m/ v

v = length (l) × width (w) × height (h)

First of we will convert the Kg into gram and mm into cm.

For 1 Kg = 1000 g

1.09 × 1000 = 1090 g

For 1 cm = 10 mm

304 / 10 = 30.4 cm

0.020 / 10 = 0.0020 cm

Now we will put the values in formula:

d = m/ l× h× w

l = m / d × h× w

l = 1090 g / 2.7 g/cm³× 30.4 cm × 0.0020 cm

l = 1090 g/ 0.16416 g/cm

l = 6639.8635 cm or 66.3986 m

The maximum length of foil will be 6639.8635 cm or 66.3986 m

Learn more about the density,

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How to find density of a gas.

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Answer:

Subtract the mass of the cylinder from the mass of the cylinder when it contains gasoline. This is the mass of the gasoline. Divide this figure by the volume, 100 ml, to get the density.

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3 years ago

the frequency of a beam of uv light is 1.0 ×10 ^15hz what is the energy in one quantum of this light express it in ev?

kap26 [50]

Answer:

4.14 eV

Explanation:

f = 1.0 ×10^15 Hz

h= 6.63×10^-34 J s ( this is called PLANCK 'S CONSTANT)

ENEGY = E = ?

E = hf ( THIS IS FORMULA FOR ENERGY OF ONE QUANTA OR ONE PHOTON )

E= 6.63×10^-34×1.0 ×10^15

E = 6.63×10^-19 J

As 1eV = 1.6×10^-19 J so changing energy in eV from joules we will divide energy by 1.6×10^-19

hence E in eV = 6.63×10^-19/(1.6×10^-19)

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7 0

3 years ago

A 1 036-kg satellite orbits the Earth at a constant altitude of 98-km. (a) How much energy must be added to the system to move t

Veronika [31]

Answer:

a) The Energy added should be 484.438 MJ

b) The Kinetic Energy change is -484.438 MJ

c) The Potential Energy change is 968.907 MJ

Explanation:

Let 'm' be the mass of the satellite , 'M'(6× $10^{24}$ be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67× $10^{-11}$ N/m) be the universal constant of gravitation.

We know that the orbital velocity(v) for a satellite -

v= $\sqrt{\frac{Gmm}{R+h} }$ [(R+h) is the distance of the satellite from the center of the earth ]

Total Energy(E) = Kinetic Energy(KE) + Potential Energy(PE)

For initial conditions ,

h = $h_{i}$ = 98 km = 98000 m

∴Initial Energy ( $E_{i}$ ) = $\frac{1}{2}$ m $v^{2}$ + $\frac{-GMm}{(R+h_{i} )}$

Substituting v= $\sqrt{\frac{GMm}{R+h_{i} } }$ in the above equation and simplifying we get,

$E_{i}$ = $\frac{-GMm}{2(R+h_{i}) }$

Similarly for final condition,

h= $h_{f}$ = 198km = 198000 m

∴Final Energy( $E_{f}$ ) = $\frac{-GMm}{2(R+h_{f}) }$

a) The energy that should be added should be the difference in the energy of initial and final states -

∴ ΔE = $E_{f}$ - $E_{i}$

= $\frac{GMm}{2}$ ( $\frac{1}{R+h_{i} }$ - $\frac{1}{R+h_{f} }$ )

Substituting ,

M = 6 × $10^{24}$ kg

m = 1036 kg

G = 6.67 × $10^{-11}$

R = 6400000 m

$h_{i}$ = 98000 m

$h_{f}$ = 198000 m

We get ,

ΔE = 484.438 MJ

b) Change in Kinetic Energy (ΔKE) = $\frac{1}{2}$ m[ $v_{f} ^{2}$ - $v_{i} ^{2}$ ]

= $\frac{GMm}{2}$ [ $\frac{1} {R+h_{f} }$ - $\frac{1} {R+h_{i} }$ ]

= -ΔE

= - 484.438 MJ

c) Change in Potential Energy (ΔPE) = GMm[ $\frac{1}{R+h_{i} }$ - $\frac{1}{R+h_{f} }$ ]

= 2ΔE

= 968.907 MJ

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