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erik [133]
1 year ago
7

a strong wind blows over the house shown in the figure. the wind is much stronger over the house's roof than lower down, and the

house has an open chimney. a window on the ground floor is open, and so are the doors inside the house. which way will air flow through the house?
Physics
1 answer:
lana66690 [7]1 year ago
8 0

A strong wind blows over the house shown in the figure. The wind is much stronger over the house's roof than lower down, and the house has an open chimney. a window on the ground floor is open, and so are the doors inside the house.  Air flow  in the window and out the chimney .

Bernoulli's principle: Within a horizontal flow of fluid, points of higher fluid speed will have less pressure than points of slower fluid speed.

Since , the wind is much stronger over the house's roof than down , that means there will be less pressure over the house's roof than down the house . Since , gas travel from high pressure to low pressure . Hence , air flow  in the window and out the chimney .

To learn more about Bernoulli's principle here :

brainly.com/question/2249706

#SPJ4

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A car starts from rest and accelerates uniformly at 2.0 m/s2 toward the north. A second car starts from rest 4.0 s later at the
yKpoI14uk [10]

Answer:

the correct solution is 13 s

Explanation:

This is a kinematic problem, let's use accelerated rectilinear motion relationships.

For the first car it has an accelerometer of 2.0 m/s²

       x = v₀₁ t + ½ a₁ t²

The second car leaves the same point, but 4.0 seconds later

       x = v₀₂ (t-4) + ½ a₂ (t-4)²

With this form we use the same time for both cars.

The initial speeds are zero for both vehicles leave the rest, at the point where they are located has the same position

        x = ½ a₁ t²

        x = ½ a₂ (t-4)²

Let's solve

       a₁  t² = a₂ (t-4)²

      a₁/a₂ t² = t² -2 4 t + 16

      t² (1- 2.0 / 4.0) - 8 t +16

      t² 0.5 - 8 t +16 = 0

      t² -16 t + 32 = 0

Let's solve the second degree equation

     t = [16 ±√( 16² - 4 32)] / 2  

     t = ½ (16 ± 11,3)

Solutions

     t1 = 13.66 s

     t2 = 2.34 s

These are the mathematical solutions for the meeting point, but car 2 leaves after 4 seconds, so the only solution is 13.66 s

the correct solution is 13 s, if you have to select one the nearest 12s

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3 years ago
What is the mechanical advantage of a pulley?
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<span>the mechanical advantage of a pulley is 1.0

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Answer:15

Explanation:

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3 years ago
A solid sphere of radius 40.0 cm has a total positive charge of 16.2 μC uniformly distributed throughout its volume. Calculate t
Jobisdone [24]

Answer:

(a) E=0  :   0 cm from the center of the sphere

(b) E= 227.8*10³ N/C   :    10.0 cm from the center of the sphere

(c)E= 911.25*10³ N/C    :    40.0 cm from the center of the sphere

(d)E= 411.84 * 10³ N/C  :    59.5 cm from the center of the sphere

Explanation:

If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere

E=\frac{K*Q}{r^{2} } : Formula (1) To calculate the electric field in the region outside the sphere r ≥ a

E=k*\frac{Q}{a^{3} } *r :Formula (2) To calculate the electric field in the inner region of the sphere. r ≤ a

Where:

K: coulomb constant

a: sphere radius

Q:  Total sphere charge

r : Distance from the center of the sphere to the region where the electric field is calculated

Equivalences

1μC=10⁻⁶C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

Q=16.2 μC=16.2 *10⁻⁶C

a= 40 cm = 40*10⁻²m = 0.4m

Problem development

(a)Magnitude of the electric field at  0 cm :

We replace r=0 in the formula (2) , then, E=0

(b) Magnitude of the electric field at 10.0 cm from the center of the sphere

r<a , We apply the Formula (2):

E=9*10^{9} *\frac{16.2*10^{-6} }{0.4^{3} } *0.1

E= 227.8*10³ N/C

(c) Magnitude of the electric field at 40.0 cm from the center of the sphere

r=a, We apply the Formula (1) :

E=\frac{9*10^{9}*16.2*10^{-6} }{0.4^{2} }

E= 911.25*10³ N/C

(d) Magnitude of the electric field at 59.5 cm from the center of the sphere  

r>a , We apply the Formula (1) :

E=\frac{9*10^{9}*16.2*10^{-6} }{0.595^{2} }

E= 411.84 * 10³ N/C

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A. THE ENERGY THAT IS RELEASED IN A NUCLEAR REACTION.

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