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2 years ago

How is steam foremed

2 answers:

4 0

Steam is water in the gas phase, which is formed when water boils. Steam is invisible; however, "steam" often refers to wet steam, the visible mist or aerosol of water droplets formed as this water vapour condenses.
hope this helps plz give brainliest!

bearhunter [10]2 years ago

3 0

Steam is formed by the heating of water. When water gets too hot is forms into steam and rises into the air until it becomes cold enough to fall back down to the earth.

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The overall length of a piccolo is 32.0 cm. The resonating air column vibrates as in a pipe that is open at both ends. (a) Find

lana66690 [7]

Answer:

lowest frequency = 535.93 Hz

distance between adjacent anti nodes is 4.25 cm

Explanation:

given data

length L = 32 cm = 0.32 m

to find out

frequency and distance between adjacent anti nodes

solution

we consider here speed of sound through air at room temperature 20 degree is approximately v = 343 m/s

lowest frequency will be = $\frac{v}{2L}$ ..............1

put here value in equation 1

lowest frequency will be = $\frac{343}{2(0.32)}$

lowest frequency = 535.93 Hz

and

we have given highest frequency f = 4000Hz

wavelength = $\frac{v}{f}$ ..............2

put here value

wavelength = $\frac{343}{4000}$

wavelength = 0.08575 m

so distance = $\frac{wavelength}{2}$ ..............3

distance = $\frac{0.08575}{2}$

distance = 0.0425 m

so distance between adjacent anti nodes is 4.25 cm

3 0

3 years ago

A fuel pump sends gasoline from a car's fuel tank to the engine at a rate of 5.64 10-2 kg/s. the density of the gasoline is 735

Irina18 [472]

Given:
Gasoline pumping rate, R = 5.64 x 10⁻² kg/s
Density of gasoline, D = 735 kg/m³
Radius of fuel line, r = 3.43 x 10⁻³ m

Calculate the cross sectional area of the fuel line.
A = πr² = π(3.43 x 10⁻³ m)² = 3.6961 x 10⁻⁵ m²

Let v = speed of pumping the gasoline, m/s
Then the mass flow rate is
M = AvD = (3.6961 x 10⁻⁵ m²)*(v m/s)*(735 kg/m³) = 0.027166v kg/s

The gasoline pumping rate is given as 5.64 x 10⁻² kg/s, therefore
0.027166v = 0.0564
v = 2.076 m/s

Answer: 2.076 m/s
The gasoline moves through the fuel line at 2.076 m/s.

7 0

3 years ago

Read 2 more answers

A train traveling at 6.4 m/s accelerates at 0.10 m/s 2 over a distance of 100 m. How large is the train’s final velocity?

klasskru [66]

The final velocity of the train at the end of the given distance is 7.81 m/s.

The given parameters;

initial velocity of the train, u = 6.4 m/s
acceleration of the train, a = 0.1 m/s²
distance traveled, s = 100 m

The final velocity of the train at the end of the given distance is calculated using the following kinematic equation;

v² = u² + 2as

v² = (6.4)² + (2 x 0.1 x 100)

v² = 60.96

v = √60.96

v = 7.81 m/s

Thus, the final velocity of the train at the end of the given distance is 7.81 m/s.

Learn more here:brainly.com/question/21180604

4 0

2 years ago

A 0.58 kg mass is moving horizontally with a speed of 6.0 m/s when it strikes a vertical wall. The mass rebounds with a speed of

Sunny_sXe [5.5K]

Answer:

$5.8\; {\rm kg\cdot m \cdot s^{-1}}$ .

Explanation:

If the mass of an object is $m$ and the velocity of that object is $v$ , the linear momentum of that object would be $m\, v$ .

Assume that the initial velocity of the mass is positive ( $6.0\; {\rm m\cdot s^{-1}}$ .) However, the direction of the velocity is reversed after the impact. Thus, the sign of the new velocity of the object would be negative- the opposite of that of the initial velocity. The new velocity would be $(-4.0\; {\rm m\cdot s^{-1}})$ .

Thus, the change in the velocity of the mass would be:

$\begin{aligned}& (\text{Change in Velocity}) \\ =\; & (\text{Final Velocity}) - (\text{Initial Velocity}) \\ =\; & (-4.0\; {\rm m\cdot s^{-1}}) - (6.0\; {\rm m\cdot s^{-1}}) \\ =\; & (-10\; {\rm m\cdot s^{-1})\end{aligned}$ .

The change in the linear momentum of the mass would be:

$\begin{aligned} & \text{change in momentum} \\ =\; & (\text{mass}) \times (\text{change in velocity}) \\ =\; & 0.58\; {\rm kg} \times (-10\; {\rm m\cdot s^{-1}}) \\ =\; & (-5.8\; {\rm kg \cdot m \cdot s^{-1}})\end{aligned}$ .

Thus, the magnitude of the change of the linear momentum would be $5.8\; {\rm kg \cdot m \cdot s^{-1}}$ .

7 0

2 years ago

If the atoms of one object (initially neutral) have electrons rubbed off through friction with a second object, the first object

snow_lady [41]

Positive. The 1st object loses electrons and will thus have an imbalance of charge with loss of electrons.

4 0

3 years ago