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Ksju [112]
3 years ago
12

A package of aluminum foil contains 50. ft2 of foil, which weighs approximately 6.0 oz. Aluminum has a density of 2.70 g/cm3. Wh

at is the approximate thickness of the foil in millimeters? Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
SIZIF [17.4K]3 years ago
3 0

Answer:

1.36 x 10^-3 cm

Explanation:

Area = 50 ft^2 = 46451.5 cm^2

mass = 6 oz = 170.097 g

density = 2.70 g/cm^3

Let t be the thickness of foil in cm.

mass = volume x density

mass = area x thickness x density

170.097 = 46451.5 x t x 2.70

t = 1.36 x 10^-3 cm

Thus, the thickness of aluminium foil is 1.36 x 10^-3 cm.

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Using Newton's Second Law, can you explain why one of the major advancements in spaceflight was the development of strong cerami
katovenus [111]

Answer:

high density can withstand high acceleration and applied forces

Heavy metals are toxic to humans,

the clay is quite abundant and in general it is not toxic

Explanation:

The selection of materials for the construction of rockets takes into account many aspects, the technical resistance to the demands of space travel, but also the abundance of the material. Heavy metals have two very serious problems. The first one, some of them are a little scarce in nature, but the most serious problem is that almost all of them are toxic to humans, for example: lead and mercury.

On the other hand, the clay is quite abundant and in general it is not toxic to living beings.

If we use Newton's second law

          F = m a

let's use the concept of density

         rho = m / V

        m = rho V

let's substitute

         F = rho V a

From this expression we see that a material with high density can withstand high acceleration and applied forces, such as those existing in spacecraft clearance and re-entry to Earth.

Unfortunately with this law there is no criterion to select a material unless its density is high, in addition to this criterion low toxicity criteria for human beings are used,

8 0
2 years ago
What is the speed of a bobsled whose distance-time graph indicates that it traveled 124m in 26s
krek1111 [17]
It's average speed during that 26 seconds was about 4.77 m/s. Without seeing the graph, we can't tell if it was going faster or slower at any particular time during that period. All we can tell is its average for the full interval.
5 0
3 years ago
In a belly-flop diving contest, the winner is the diver who makes the biggest splash upon hitting the water. the size
ad-work [718]

The second diver have to leap to make a competitive splash by 4.08 m high.

<h3>What is potential energy?</h3>

The energy by virtue of its position is called the potential energy.

PE = mgh

where, g = 9.81 m/s²

Given is the diver jumps from a 3.00-m platform. one diver has a mass of 136 kg and simply steps off the platform. another diver has a mass of 100 kg and leaps upward from the platform.

The potential energy of the first diver must be equal to the second diver.

P.E₁ = P.E₂

m₁gh₁ = m₂gh₂

Substitute the vales, we have

136 x 3  = 100 x h₂

h₂ = ₂4.08 m

Thus, the second diver need to leap by 4.08 m high.

Learn more about potential energy.

brainly.com/question/24284560

#SPJ1

7 0
2 years ago
50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the bui
soldi70 [24.7K]

Answer:

98.13m

Explanation:

Complete question

Daniel is 50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the building

CHECK THE ATTACHMENT

From the figure, using trigonometry

Tan(θ ) = opposite/adjacent

Where Angle (θ )= 63°

Opposite= X = height of the building

Adjacent= 50 m

Then substitute the values we have

Tan(63)= X/50

1.9626= X/50

X= 1.9626 × 50

X= 98.13m

Hence, the height of the building is 98.13m

8 0
2 years ago
Monochromatic light from a distant source is incident on a slit 0.750 mm wide. On a screen 2.00 m away, the distance from the ce
jasenka [17]

Answer:

\lambda= 506.25 nm

Explanation:

Diffraction is observed when a wave is distorted by an obstacle whose dimensions are comparable to the wavelength. The simplest case corresponds to the Fraunhofer diffraction, in which the obstacle is a long, narrow slit, so we can ignore the effects of extremes.

This is a simple case, in which we can use the Fraunhofer single slit diffraction equation:

y=\frac{m \lambda D}{a}

Where:

y=Displacement\hspace{3}from\hspace{3} the\hspace{3} centerline \hspace{3}for \hspace{3}minimum\hspace{3} intensity =1.35mm\\\lambda=Light\hspace{3} wavelength \\D=Distance\hspace{3}between\hspace{3}the\hspace{3}screen\hspace{3}and\hspace{3}the\hspace{3}slit=2m\\a=width\hspace{3}of\hspace{3}the\hspace{3}slit=0.750mm\\m=Order\hspace{3}number=1

Solving for λ:

\lambda=\frac{y*a}{mD}

Replacing the data provided by the problem:

\lambda=\frac{(1.35\times 10^{-3})*(0.750\times 10^{-3})}{1*2} =5.0625\times 10^{-7}m =506.25nm

7 0
3 years ago
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