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-Dominant- [34]
2 years ago
6

One of the dangers of tornados and hurricanes is the rapid drop in air pressure that is associated with such storms. Assume that

the air pressure inside of a sealed house is 1.02 atm when a hurricane hits. The hurricane rapidly decreases the external air pressure to.910 atm. What net force (directed outwards) is exerted on a 2.02-msquare window on the house
Physics
1 answer:
Ierofanga [76]2 years ago
4 0

Answer:

Force = 22508.86 N

Explanation:

Given:

Area of the window, A = 2.02 m²

Air pressure inside the house, P₁ = 1.02 atm

External air pressure, P₂ = 0.910 atm

Now, The difference in the pressure, ΔP = P₁ - P₂ = 1.02 - 0.910 = 0.11 atm

or

ΔP = 0.11 atm  ×  (101300 Pa / 1 atm )

ΔP = 11143 Pascals

Now,

the Force is given as:

Force = Pressure × area

on substituting the values, we get

Force = 11143 Pa × 2.02 m²

or

Force = 22508.86 N

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Natasha2012 [34]

Answer:

air

Explanation:

The car is being slowed down by air.

5 0
2 years ago
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Changes that occur in the urinary system with aging include all of the following, EXCEPT
Sladkaya [172]

Answer: c. increased sensitivity to ADH

Explanation:

a. a decline in the number of functional nephrons: With aging the loss of nephron occurs that can be detected by the age related decrease in the glomerular filteration rate.

b. a reduction in the GFR (glomerular filtration rate): The GFR tend to decline in older age even though there is no disease. These people are required to check with the GFR in future.

d. problems with the micturition reflex: With aging people experience problem of bladder control. This leads to leakage or incontinence of urine or urinary retention that is inability to empty the bladder.

e. loss of sphincter muscle tone: With age the sphincter tone may diminish. This results in loss of control and storage capacity. The rectal muscles or sphincter muscles get loose which lead to passage of stool before reaching the washroom.

6 0
2 years ago
Lilli suggests that they explore the simulation starting with varying only a single parameter in order to understand the role of
mrs_skeptik [129]

Answer:

B.

Explanation:

One of the ways to address this issue is through the options given by the statement. The concepts related to the continuity equation and the Bernoulli equation.

Through these two equations it is possible to observe the behavior of the fluid, specifically the velocity at a constant height.

By definition the equation of continuity is,

A_1V_1=A_2V_2

In the problem A_2 is 2A_1, then

A_1V_1=2A_1V_2

V_2 = \frac{V_1}{2}

<em>Here we can conclude that by means of the continuity when increasing the Area, a decrease will be obtained - in the diminished times in the area - in the speed.</em>

For the particular case of Bernoulli we have to

P_1 + \frac{1}{2}\rho V_1^2 = P_2 +\frac{1}{2}\rho V_2^2

P_2-P_1 = \frac{1}{2} \rho (V_1^2-V_2^2)

For the previous definition we can now replace,

P_2-P_1 = \frac{1}{2} \rho (V_1^2-(\frac{V_1}{2})^2)

\Delta P =  \frac{3}{8} \rho V_1^2

<em>Expressed from Bernoulli's equation we can identify that the greater the change that exists in pressure, fluid velocity will tend to decrease</em>

The correct answer is B: "If we increase A2 then by the continuity equation the speed of the fluid should decrease. Bernoulli's equation then shows that if the velocity of the fluid decreases (at constant height conditions) then the pressure of the fluid should increase"

4 0
2 years ago
The helicopter in the drawing is moving horizontally to the right at a constant velocity. The weight of the helicopter is W=4900
Sedaia [141]

Answer:

(a) The magnitude of the lift force is 52144.71 N, approximately.

(b) The magnitude of the air resistance force opposing the movement is 17834.54 N, approximately.

Explanation:

Since the helicopter is moving horizontally at a constant velocity, we can assume that the net force acting on it is zero, then

(a) in the vertical direction we have

L\cos(20\deg)-W=0\\L=\frac{W}{\cos(20\deg)}=\frac{49000 N}{\cos(20\deg)}\approx \mathbf{52144.71 N}.

(b) Now horizontally,

L\sin(20\deg)-R=0\\R=L\sin(20\deg)=52144.71 N\times \sin(20\deg) \approx \mathbf{17834.54 N}.

3 0
2 years ago
A 15-g bullet moving at 300 m/s passes through a 2.0 cm thick sheet of foam plastic and emerges with a speed of 90 m/s. Let's as
Shkiper50 [21]

Answer:

Explanation:

a) Change in momentum, Δp = mΔv = m(v - u) = (15 * 10-3) * (90 - 300) = -3.15 kg-m2

b) Acceleration of the bullet, a = (v2 - u2) / 2s = (902 - 3002) / (2 * 0.02) = -2047500 m/s2

So, the bullet is in contact with the plastic for the time,  \bigtriangleupt = \frac{(v - u)}{a} =\frac{(90 - 300)}{(-2047500)} = 1.03 \times 10^{-4} s

c) Average force, F_{avg} =\frac{\bigtriangleup p}{\bigtriangleup t} =\frac{(-3.15)}{(1.03 \times 10^{-4})} = 30.6 kN

8 0
2 years ago
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