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2 years ago

WORTH 50 POINTS!!! Please help me with questions 1,2,3,4,5,6 and 7 ASAP!

Physics

1 answer:

boyakko [2]2 years ago

6 0

This is a good example of unit conversion. Look at the units you have and the units you need to solve for.

In #1 you’re given weight in units N and area in units cm². You want to find pressure in units Pa (pascals) which is force/area, or N/m². So you just divide the force given by the area given, but be careful because the area is given in cm², whereas pressure is in m². Before you calculate pressure, make sure you convert 324cm² into m²

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Land heats up and cools down quickly because...

chubhunter [2.5K]

I would say it reflects the sun easily. That’s also how we see it :)

6 0

2 years ago

A drop of oil of volume 10m it spread out on water to make a circular firm of radius 10m calculate the tickness of the firm

Effectus [21]

Answer:

h = 3.1 cm

Explanation:

Given that,

The volume of a oil drop, V = 10 m

Radius, r = 10 m

We need to find the thickness of the film. The film is in the form of a cylinder whose volume is as follows :

$V=\pi r^2 h\\\\h=\dfrac{V}{\pi r^2}\\\\h=\dfrac{10}{\pi \times 10^2}\\\\h=0.031\ m\\\\h=3.1\ cm$

So, the thickness of the film is equal to 3.1 cm.

8 0

2 years ago

An airplane wing is designed so that the speed of the air across the top of the wing is 255 m/s when the speed of the air below

grin007 [14]

<h2>Answer:442758.96N</h2>

Explanation:

This problem is solved using Bernoulli's equation.

Let $P$ be the pressure at a point.

Let $p$ be the density fluid at a point.

Let $v$ be the velocity of fluid at a point.

Bernoulli's equation states that $P+\frac{1}{2}pv^{2}+pgh=constant$ for all points.

Lets apply the equation of a point just above the wing and to point just below the wing.

Let $p_{up}$ be the pressure of a point just above the wing.

Let $p_{do}$ be the pressure of a point just below the wing.

Since the aeroplane wing is flat,the heights of both the points are same.

$\frac{1}{2}(1.29)(255)^{2}+p_{up}= \frac{1}{2}(1.29)(199)^{2}+p_{do}$

So, $p_{up}-p_{do}=\frac{1}{2}\times 1.29\times (25424)=16398.48Pa$

Force is given by the product of pressure difference and area.

Given that area is $27ms^{2}$ .

So,lifting force is $16398.48\times 27=442758.96N$

6 0

3 years ago

Why is carbon added to iron

swat32

Answer:

it transforms it into high carbon alloy that is harder and can be sharper but is also more brittle in the process.

Explanation:

7 0

2 years ago

The eyes of some reptiles are sensitive to 850 nm light. If the minimum energy to trigger the receptor at this wavelength is 3.1

mr_godi [17]

Answer:

Minimum number of photons required is 1.35 x 10⁵

Explanation:

Given:

Wavelength of the light, λ = 850 nm = 850 x 10⁻⁹ m

Energy of one photon is given by the relation :

$E=\frac{hc}{\lambda}$ ....(1)

Here h is Planck's constant and c is speed of light.

Let N be the minimum number of photons needed for triggering receptor.

Minimum energy required for triggering receptor, E₁ = 3.15 x 10⁻¹⁴ J

According to the problem, energy of N number of photons is equal to the energy required for triggering, that is,

E₁ = N x E

Put equation (1) in the above equation.

$E_{1}=N\times\frac{hc}{\lambda}$

Substitute 3.15 x 10⁻¹⁴ J for E₁, 850 x 10⁻⁹ m for λ, 6.6 x 10⁻³⁴ J s for h and 3 x 10⁸ m/s for c in the above equation.

$3.15\times10^{-14} =N\times\frac{6.6\times10^{-34}\times3\times10^{8}}{850\times10^{-9}}$

N = 1.35 x 10⁵

8 0

2 years ago