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Liono4ka [1.6K]
3 years ago
5

What is the level of incidence

Physics
1 answer:
-Dominant- [34]3 years ago
5 0

Answer:

The incidence rate is typically expressed as the number of cases per person-year of observation. Only new cases are considered when computing the incidence rate, while cases that were diagnosed earlier are excluded. The “population at risk” measure is usually obtained from census data.

Explanation:

The incidence rate is typically expressed as the number of cases per person-year of observation. Only new cases are considered when computing the incidence rate, while cases that were diagnosed earlier are excluded. The “population at risk” measure is usually obtained from census data.

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An organ pipe open at both ends is 1.5 m long. A second organ pipe that is closed at one end and open at the other is 0.75 m lon
telo118 [61]

Answer:

A. 110Hz,220Hz, 330 Hz

Explanation:

for organ open at open both ends;

the length of the organ for the fundamental frequency, L = A---->N + N----->A

A---->N  = λ /4 and N----->A = λ /4

L = λ /4 + λ /4 = λ /2

L = \frac{\lambda}{2} \\\\\lambda = 2L

λ  = 2 x 1.5m = 3.0 m

Wave equation is given by;

V = Fλ

Where;

V is the speed of sound

F is the frequency of the wave

F = V/ λ

F₀ = V / 2L

Where;

F₀  is the fundamental frequency

F₀ = 330 / 2(1.5)

F₀ = 330 / 3

F₀ = 110 Hz

the length of the organ for the first overtone, L = A---->N + N----->A + A----->N +  N----->A

L = 4λ /4

L = λ

λ = 1.5 m

F₁ = 330 / 1.5

F₁ = 220 Hz

Thus, F₁ = 2F₀

For open organ at one end

the length of the organ for the fundamental frequency, L = N------A

L = λ /4

λ = 4L

F₀ = V/4L

F₀ = 330 / (4 x 0.75)

F₀ = 110 Hz

the length of the organ for the first overtone, L = N-----N + N-----A

L = λ/2 + λ / 4

L = 3λ /4

F₁ = 3F₀

F₁ = 3 x 110

F₁ = 330 Hz

Thus the fundamental frequency for both organs is 110 Hz,

The first overtone for the organ open at both ends is 220 Hz

The first overtone for the organ open at one end is 330 Hz

The correct option is "A. 110Hz,220Hz, 330 Hz"

6 0
3 years ago
How much work done when .0080 C is moved through a potential difference of 1.5 V? Use W = qV. A.
grin007 [14]

Answer:

0.012 J

Explanation:

We are given:

q = 0.0080C

Potential difference =  1.5V

W=qV

Substituting the values into the equation:

W=0.0080*1.5= 0.012J

8 0
3 years ago
Does the construction of a high building slow down the rotation of the earth
omeli [17]
No, the building's size in comparison to the earth would have no change or change so increadibly miniscule, like if you were told to spin slowly and an and was placed on top of your head
5 0
3 years ago
Suppose a rock is hoisted up with a rope. The rock experiences 10 N of tension force upward and 10 N of gravitational force down
nika2105 [10]
False. Since the forces are pulling in equal and opposite directions, the net force is 0.
5 0
3 years ago
Read 2 more answers
Water flows on the inside of a steel pipe with an ID of 2.5 cm. The wall thickness is 2 mm, and the convection coefficient on th
Hatshy [7]

Answer:

U = 11.67 W/m² °C

Explanation:

Inner diameter, d = 2.5 cm = 0.025 m

Thickness of the wall, t = 2 mm = 0.002 m

thus, the outer diameter, D = d + 2t = 0.025 + 2 x 0.002 = 0.029 m

h_i =500 W/m^2.^oC

h_0 =12 W/m^2.^oC

Now, based on outside convection heat transfer

we have

\frac{1}{UA_o}=\frac{1}{h_iA_i}+\frac{1}{h_oA_o}

on rearranging, we get

\frac{1}{U}=\frac{1}{h_i}\frac{A_o}{A_i}+\frac{1}{h_o}

where, A_i\ \textup{and}\ A_o are the perimeter respective to inner and and outer diameter

on substituting the values, we get

\frac{1}{U}=\frac{1}{500}\frac{0.029^2}{0.025}+\frac{1}{12}

or

U = 11.67 W/m² °C

6 0
3 years ago
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