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denis-greek [22]

3 years ago

9

As increases, the humidity of that area will . temperature; also increase dew; also increase rainfall; alwa

ys increase temperature; never increase

1 answer:

Nady [450]3 years ago

4 0

Always increase temperature

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A violin with string length 32 cm and string density 1.5 g/cm resonates in its fundamental with the first overtone of a 2.0-m or

love history [14]

Answer:

T=1022.42 N

Explanation:

Given that

l = 32 cm ,μ = 1.5 g/cm

L =2 m ,V= 344 m/s

The pipe is closed so n= 3 ,for first over tone

$f=\dfrac{nV}{4L}$

$f=\dfrac{3\times 344}{4\times 2}$

f= 129 Hz

The tension in the string given as

T = f²(4l²) μ

Now by putting the values

T = f²(4l²) μ

T = 129² x (4 x 0.32²) x 1.5 x 10⁻³ x 100

T=1022.42 N

6 0

3 years ago

Since the particles of liquids are able to slide past each other into different locations they have ______.

amid [387]

C Bc it’s the right answers not wrong

7 0

3 years ago

Use the periodic table from the lesson to answer the following question.

kodGreya [7K]

Answer:

18.99u or 19u

Explanation:

Mass number=No of protons+No of neutrons

$\\ \sf\longmapsto 9+10$

$\\ \sf\longmapsto 19u$

Fluorine has atomic number=9
The Electronic configuration is given by

$\\ \sf\longmapsto 1s^22s^22p^5$

7 0

3 years ago

Read 2 more answers

When holes are drilled through the wall of a water tower, water will spurt out the greatest horozontal distance from the hole cl

Mnenie [13.5K]

Bottom of the water tower

Explanation:

When holes are drilled through the wall of a water tower, water will spurt out the greatest horizontal distance from the hole closest to the bottom of the water tower.

This is because the near the bottom of the tower is the greatest compared to other parts of the tower.

The pressure increases with depth.
As the depth of water increases, the column of water above exerts more pressure on the one below.
Therefore, water will spurt out more from the bottom of the tower compared to the top of the tower.

Learn more:

Pressure brainly.com/question/4868239

#learnwithBrainly

3 0

3 years ago

Air at 30°C and 2MPa flows at steady state in a horizontal pipeline with a velocity of 25 m/s. It passes through a throttle valv

erastovalidia [21]

Answer:

$V_2=159.9\ m/s$

$T_2=290.6K$

Explanation:

At initial condition

P=2 MPa

T=30°C

V=25 m/s

At final condition

P=0.3 MPa

Now from first law for open system

$h_1+\dfrac{V_1^2}{2000}=h_2+\dfrac{V_2^2}{2000}$

We know that for air

h= 1.010 x T KJ/kg

$1.01\times 303+\dfrac{25^2}{2000}=1.01\times T+\dfrac{V_2^2}{2000}$ -----1

Now from mass balance

$\rho_1A_1V_1=\rho_2A_2V_2$

We also know that

$\rho=\dfrac{P}{RT}$

$\dfrac{P_1}{RT_1}V_1=\dfrac{P_2}{RT_2}V_2$

$\dfrac{2}{R\times 303}\times 25=\dfrac{0.3}{RT_2}V_2$

$T_2=1.81V_2$ ----------2

Now from equation 1 and 2

$V_2^2+3673.749V-612916.49=0$

So we can say that

$V_2=159.9\ m/s$

This is the outlet velocity.

Now by putting the values in equation 2

$T_2=1.81\times 159.9$

$T_2=290.6K$

This is the outlet temperature.

4 0

4 years ago