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Elden [556K]
3 years ago
7

A fountain sends a stream of water straight up into the air to a maximum height of 5.60 m. The effective area of the pipe feedin

g the fountain is 4.77 x 10-4 m2. Neglecting air resistance and any viscous effects, determine how many gallons per minute are being used by the fountain. (1 gal = 3.79 x 10-3 m3)
Physics
1 answer:
kifflom [539]3 years ago
5 0

Answer:It increases confidence because the more times you conduct the same experiment over and over should either prove your hypothesis right and wrong and eliminate any random occurrences that might affect your results.

Read more on Brainly.com - brainly.com/question/1492542#readmore

Explanation:

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A. Alternating current
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2 years ago
What category of the electromagnetic spectrum has a wavepength of 108m​
vodomira [7]

Answer: Radio Waves

Explanation:

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3 years ago
Which is a correct statement of what occurs at a turbine during electricity production?
jarptica [38.1K]

Answer:

The turbine is rotated and rotates the generator to produce electricity.

Explanation:

Within a turbine enters the superheated steam which is at high pressure and high temperature, this steam is previously formed in the boiler when the steam enters the turbine hits each one of the blades of the turbine making it rotate at a given speed, the turbine shaft is coupled to the shaft of an electric generator and thus generates electricity.

It is also important to say that when the steam comes out of the turbine comes out at low pressure, this way the internal operating process is carried out within the turbine.

4 0
3 years ago
How would doubling the mass of Earth affect the gravity we experience?​
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The strength of the gravitational force between two objects depends on two factors, mass and distance. the force of gravity the masses exert on each other. If one of the masses is doubled, the force of gravity between the objects is doubled. increases, the force of gravity decreases.
5 0
3 years ago
Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m
Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}.....(I)

tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}....(II)

Put the value of \omega=0.628\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}

A=0.0198

From equation (II)

tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}

d=0.0023

Put the value of \omega=3.14\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}

A=0.0203

From equation (II)

tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}

d=0.0120

Put the value of \omega=6.28\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}

A=0.0209

From equation (II)

tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}

d=0.0257

Put the value of \omega=9.42\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}

A=0.0217

From equation (II)

tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}

d=0.0413

Hence, This is the required solution.

5 0
3 years ago
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