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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 positive particle has a charge of 31.7 mC and the particles are 2.80 mm apart, what is the electric field at point A located
vichka [17]

Answer:

the electric field at point A is

E = 5.5 ×10¹³N/C(-x direction)

Explanation:

given

electrostatics constant k = 9.0×10⁹

charge q = 31.7mC= 31.7×10⁻³C

distance r = 2.80mm

distance from midpoint to point A = 2.00mm

attached is the diagram of the solution, describing the position of the charge

note x = r/2, where x is the distance from midpoint of r to the particle

using Pythagoras theorem as in the attachment, x = 2.44mm= 2.44×10⁻³m

the electric field at point A is given as

vector <em>E </em>= 2E×cos θ( -x direction)

recall E =kq/x²

where k is the electrostatics constant = 1/4πε₀

where ε₀ is permittivity of free space

therefore using E =2{kq/x²}cosθ

∴cosθ = adjacent/hypotenuse

cosθ=1.40/2.44

E =2 {(9.0×10⁹ × 31.7×10⁻³) ÷ (2.44×10⁻³)²}×(1.40/2.44)(-x)

E=2{4.79×10¹³}×(0.574)(-x)

E = 2×2.75 ×10¹³N/C(-x direction)

Vector <em>E= </em>5.5 ×10¹³N/C(-x direction)

3 0
3 years ago
How many zebras automatically survive the first interaction with the lions in Generation 1?
lesya692 [45]
I think the answer is c. but I think it depends on how many zebras you have
5 0
3 years ago
How does the increasing mass effect the force of an object in motion?​
irina [24]

Answer:

<u>According </u><u>to </u><u>second </u><u>law </u><u>of </u><u>motion</u><u>,</u><u>t</u><u>he acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.</u>

<em>So </em><em>simply</em><em>,</em><em> </em><em>it </em><em>can </em><em>be </em><em>affected </em><em>due </em><em>to </em><em>increasing </em><em>force </em><em>as </em><em>there </em><em>is </em><em>close </em><em>relationship </em><em>between </em><em>momentum.</em>

Explanation:

<em>The more inertia that an object has, the more mass that it has. A more massive object has a greater tendency to resist changes in its state of motion.</em>

<em>I </em><em>hope </em><em>it </em><em>was </em><em>helpful </em><em>for </em><em>you </em><em>:</em><em>)</em>

7 0
1 year ago
A shot-putter accelerates a 7.2 kg shot from rest to 17 m/s . what work did the shot-putter do on the ball?
garri49 [273]
<span>1.0x10^3 Joules The kinetic energy a body has is expressed as the equation E = 0.5 M V^2 where E = Energy M = Mass V = Velocity Since the shot was at rest, the initial energy is 0. Let's calculate the energy that the shot has while in motion E = 0.5 * 7.2 kg * (17 m/s)^2 E = 3.6 kg * 289 m^2/s^2 E = 1040.4 kg*m^2/s^2 E = 1040.4 J So the work performed on the shot was 1040.4 Joules. Rounding the result to 2 significant figures gives 1.0x10^3 Joules</span>
6 0
3 years ago
According to the principle of faunal succession, what must be true of the fossils on a layer below? Now
Firdavs [7]
It means that they were set earlier therefor they are older.
7 0
3 years ago
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