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astraxan [27]
3 years ago
11

Suppose a horizontal wind blows with a speed of 12.6 m/s outside a large pane of plate glass with dimensions 3.00 m x 1.80 m. As

sume the density of the air to be 1.30 kg/m3. The air inside the building is at atmospheric pressure. What If? If a second skyscraper is built nearby, the airspeed can be especially high where wind passes through the narrow separation between the buildings. (c) Find the pressure with a wind speed twice as high.
Physics
1 answer:
Wewaii [24]3 years ago
6 0

Answer:

Explanation:

Given to find the pressure

v = 12.6 m/s , A = 3.0m * 1.8 m = 54 m^2

p air = 1.3 kg/m^3

F = 1/2 * p *v^2 *A

F = 1/2 *1.3 kg/m^3 * (12.6 m/s)^2 * 54m^2

F = 5572.476 N

The stress and the pressure can find across the area

α = F /A

α = 5572.476 N / 54 m^2

α = 103.194 Pa

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The CERN particle accelerator is circular with a circumference of 7.0 km.
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Answer:

a_c=2.0196\times 10^{13}\ m/s^2

F=3.37273\times 10^{-14}\ N

Explanation:

m = Mass of proton = 1.67\times 10^{-27}\ kg

v = Speed of proton = 0.5c = 0.5\times 3\times 10^8=1.5\times 10^8\ m/s

Circumference of the colider is 7 km

P=2\pi r\\\Rightarrow r=\frac{P}{2\pi}\\\Rightarrow r=\frac{7000}{2\pi}\ m

a_c=\frac{v^2}{r}\\\Rightarrow a_c=\frac{\left(1.5\times 10^8\right)^2}{\frac{7000}{2\pi}}\\\Rightarrow a_c=2.0196\times 10^{13}\ m/s^2

Centripetal acceleration is 2.0196\times 10^{13}\ m/s^2

F_c=ma_c\\\Rightarrow F_c=1.67\times 10^{-27}\times 2.0196\times 10^{13}\\\Rightarrow F=3.37273\times 10^{-14}\ N

Force on protons is 3.37273\times 10^{-14}\ N

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3 years ago
which type of lightening device would you use for each of the following needs: an economical light source in a manufacturing pla
Yakvenalex [24]
Btw only someone who is nice will answer tour question. You can't expect for explanition when the question is only worth 5 points. Not trying to be mean sorry if i am being mean
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A group of scientists have obtained some experimental results.
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What is the mass of 5 moles of gold
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The atomic number of gold is <span>197.0g Au

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Grams is used to measure mass. 

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8 0
3 years ago
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Aparticlewhosemassis2.0kgmovesinthexyplanewithaconstantspeedof3.0m/s along the direction r = i + j . What is its angular momentu
svetoff [14.1K]

Answer:

\vec{L}=-30\frac{kgm^2}{s}\hat{k}

Explanation:

In order to calculate the angular momentum of the particle you use the following formula:

\vec{L}=\vec{r}\ X\ \vec{p}       (1)

r is the position vector respect to the point (0 , 5.0), that is:

r = 0m i + 5.0m j    (2)

p is the linear momentum vector and it is given by:

\vec{p}=m\vec{v}=(2.0kg)(3.0m/s)(\hat{i+\hat{j}})=6\frac{kgm}{s}(\hat{i}+\hat{j})   (3)

the direction of p comes from the fat that the particle is moving along the i + j direction.

Then, you use the results of (2) and (3) in the equation (1) and solve for L:

\vec{L}=-30\frac{kgm^2}{s}\hat{k}

The angular momentum is -30 kgm^2/s ^k

8 0
2 years ago
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