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3 years ago

Reading Check: (Describe) how pressure changes as the velocity of a fluid increases.

Physics

1 answer:

juin [17]3 years ago

4 0

Answer:

Bernoulli's equation states mathematically that if a fluid is flowing through a tube and the tube diameter decreases, then the velocity of the fluid increases, the pressure decreases, and the mass flow (and therefore volumetric flow) remains constant so long as the air density is constan

Explanation:

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PLS HELP BEING TIMED!!!

Elodia [21]

Answer:

$K_{i}+U_{g,i} = K_{f}+U_{g,f}$

Explanation:

A closed system is a system where exists energy interactions with surroundings, but not mass interactions. If we neglect any energy interactions from boundary work, heat, electricity, magnetism and nuclear phenomena and assume that process occurs at steady state and all effects from non-conservative forces can be neglected, then the equation of energy conservation is reduce to this form:

$\Delta K +\Delta U_{g} = 0$ (1)

Where:

$\Delta K$ - Change in kinetic energy of the system, measured in joules.

$\Delta U_{g}$ - Change in gravitational potential energy of the system, measured in joules.

If we know that $\Delta K=K_{i}-K_{f}$ and $\Delta U_{g} = U_{g,i}-U_{g,f}$ , then we get the following equation:

$K_{i}+U_{g,i} = K_{f}+U_{g,f}$ (2)

Where $i$ and $f$ stands for initial and final states of each energy component.

Hence, the right answer is $K_{i}+U_{g,i} = K_{f}+U_{g,f}$

7 0

3 years ago

Read 2 more answers

Experiment: Group 1 drinks 500 mL of coffee a day.

slega [8]

Answer:

How does the drink content affect an individual's blood pressure?

Explanation:

In every experiment using the scientific method, an observation lays the foundation of that experiment. A problem must be observed, which then leads to asking a SCIENTIFIC QUESTION in order to investigate. A scientific question must include the variable being changed called INDEPENDENT VARIABLE and the variable being measured called DEPENDENT VARIABLE.

In this experimental procedure or set up,

- Group 1 drinks 500 mL of coffee a day.

- Group 2 drink 500 mL of tea a day,

- Group 3 is a control group i.e no drink

At the end of 60 days all participants

blood pressure is tested.

This set up indicates that the variable being changed (independent) is the DRINK CONTENT while the variable being measured (dependent) is the BLOOD PRESSURE. Hence, these variables serve as the template to ask a scientific question which goes thus:

HOW DOES THE DRINK CONTENT AFFECT AN INDIVIDUAL'S BLOOD PRESSURE?

This scientific question relates how the independent variable (drink) causes the dependent variable to respond (blood pressure).

3 0

3 years ago

How many different types of atoms are present in one molecule of aluminum hydroxide, Al(OH)3?

tatyana61 [14]

There are approximately 3 different types of atoms that are present in one molecule of aluminum hydroxide, AI(OH)3.

8 0

3 years ago

Read 2 more answers

A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm

Sergio [31]

Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

$F = - k * \Delta x (1)$

where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
Solving for k:

$k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)$

Assuming no friction present, total mechanical energy mus keep constant.
When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

$U = \frac{1}{2}* k* (\Delta x)^{2} (3)$

Replacing k and Δx by their values, we get:

$U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)$

When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
So, we can write the following expression:

$\frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2} = \frac{1}{2}*k*\Delta x^{2} (5)$

Replacing the right side of (5) with (4), k, m, and v by the givens, and simplifying, we can solve for Δx₁, as follows:

$\frac{1}{2}* 200N/m* \Delta x_{1} ^{2} + \frac{1}{2} * 1.8kg* (-2.0m/s)^{2} = 4J (6)$

⇒ $\frac{1}{2}* 200N/m* \Delta x_{1} ^{2} = 4J - 3.6 J = 0.4 J (7)$

⇒ $\Delta x_{1} = \sqrt{\frac{0.8J}{200N/m} } = 6.3 cm (8)$

6 0

3 years ago

IN WHAT CONDITION DO SOUND ECHO

DerKrebs [107]

Answer:

The conditions necessary for hearing the echo. The distance between the sound source and the reflecting surface must not be less than 17 metres where the time period between hearing the original sound and its echo should not be less than 0.1 of a second.

6 0

2 years ago