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Luda [366]
2 years ago
8

Consider a venturi with a small hole drilled in the side of the throat. This hole is connected via a tube to a closed reservoir.

The purpose of the venturi is to create a vacuum in the reservoir when the venturi is placed in an airstream. (The vacuum is defined as the pressure difference below the outside ambient pressure.) The venturi has a throat-to-inlet area ratio of 0.85. Calculate the maximum vacuum obtainable in the reservoir when the venturi is placed in an airstream of 90 m/s at standard sea level conditions
Physics
1 answer:
Marina CMI [18]2 years ago
6 0

Answer:

(P_1-P_2)=1913.31 N/m^2

Explanation:

given:

\frac{A_t}{A_1}=0.85

V_1=90 m/s

γ∞=1.23 kg/m^3

solution:

since outside pressure is atm pressure vaccum can be defined by (P_1-P_2)

V_1=√2(P_1-P_2)/γ∞[\frac{A_t}{A_1}^2-1]

(P_1-P_2)=1913.31 N/m^2

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A beam of light is traveling through a medium at 200,000 km/s. It enters a different medium and speeds up to almost 250,000 km/s
shtirl [24]

Answer

From the question we can see that the in medium  1 speed of light is 200,000 Km/s where as in Medium 2 speed of light is 250,000 km/s.

We can conclude that medium 1 is denser than Medium 2.

In third medium Light speed halts so, the third medium is opaque.

We can say that

Medium 1 can be either water, glass.

Medium 2 will be gas

Medium 3 will be an opaque material.

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3 years ago
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Can atoms go bad?, not in the reversible way like ionization and isotopes but really malfunction our die.​
Elza [17]

Answer:

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Explanation:

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2 years ago
A hawk is flying horizontally at 18.0 m/s in a straight line, 230 m above the ground. A mouse it has been carrying struggles fre
Lisa [10]

Answer:

a) vd = 47.88 m/s

b) θ = 80.9°

c) t = 6.8 s

Explanation:

In the situation of the problem, you can assume that the trajectory of the hawk and the trajectory of the mouse form a rectangle triangle.

One side of the triangle is the horizontal trajectory of the hawk after 2.00s of flight, the other side of the triangle is the distance traveled by the mouse when it is falling down. And the hypotenuse is the trajectory of the hawk when it is trying to recover the mouse.

(a) In order to calculate the diving speed of the hawk, you first calculate the hypotenuse of the triangle.

One side of the triangle is c1 = (18.0m/s)(2.0s) = 36m

The other side of the triangle is c2 = 230m - 3m = 227 m

Then, the hypotenuse is:

h=\sqrt{(36m)^2+(227m)^2}=229.83m    (1)

Next, it is necessary to calculate the falling down time of the mouse, this can be done by using the following formula:

y=y_o+v_ot+\frac{1}{2}gt^2    (2)

yo: initial height = 230m

vo: initial vertical speed of the mouse = 0m/s

g: gravitational acceleration = -9.8m/s^2

y: final height of the mouse = 3 m

You replace the values of the parameters in (2) and solve for t:

3=230-4.9t^2\\\\t=\sqrt{\frac{227}{4.9}}=6.8s

The hawk traveled during 2.00 second in the horizontal trajectory, hence, the hawk needed 6.8s - 2.0s = 4.8 s to travel the distance equivalent to the hypotenuse to catch the mouse.

You use the value of h and 4.8s to find the diving speed of the hawk:

v_d=\frac{229.83m}{4.8s}=47.88\frac{m}{s}

The diving speed of the Hawk is 47.88m/s

(b) The angle is given by:

\theta=cos^{-1}(\frac{c_1}{h})=cos^{-1}(\frac{36m}{229.83m})=80.9 \°

Then angle between the horizontal and the trajectory of the Hawk when it is descending is 80.9°

(c) The mouse is falling down during 6.8 s

4 0
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Answer:

I believe the answer is C

Explanation:

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LiRa [457]
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