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Anon25 [30]

<span>A transverse wave is characterized by peaks and dips.</span>

6 0

4 years ago

Read 2 more answers

As the phase angle is decreased, the power factor is

Darina [25.2K]

The answer is increased. A power factor of one or "unity power factor" is the aim of any electric utility business from the time when if the power factor is less than one, they have to give more current to the user for an assumed amount of power use. In so doing, they gain more line harms.

7 0

3 years ago

What is the difference between transformation of energy and transfer of energy

Svetllana [295]

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In energy transfer, happens from one object to another without changing its form. For example, when you fill a pot of water and sit it on the stove, the water continues to heat until you turn it off. While in a energy transformation, the energy changes forms such as Chemical, kinetic.

6 0

4 years ago

a 4050 N sled is being pushed with a force of 575 N. If it encounters a frictional force of 385 N… determine the net force

creativ13 [48]

According to the solving the net force is:

= 3,860.

<h3>Why is net force important?</h3>

The net force is the total of all the forces operating on an object, according to its definition. Mass can be accelerated by net force. Whether a body is at rest or in motion, another force is at work on it. When a system has a lot of forces acting on it, the phrase "net force" is employed.

<h3>What's the net force equation?</h3>

When a body is moving and multiple forces, such as the normal force, frictional force, and gravitational force, are acting on it, the combined effect is known as net force. As a result, is the net force formula. FNet equals Fa, Fg, Ff, and FN.

<h3>According to the given data:</h3>

4050 N sled

force of 575 N.

encounters a frictional force of 385 N

So the net force will be:

(Net)F = F₁ + F₂ + F₃

So,

(Net)F = F₁ + F₂ - F₃(frictional force)

(Net)F = 4050 + 385 - 575

= 3,860

According to the solving the net force is:

= 3,860.

To know more about net force visit:

brainly.com/question/18109210

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8 0

1 year ago

Consult Multiple-Concept Example 15 to review the concepts on which this problem depends. Water flowing out of a horizontal pipe

kiruha [24]

Answer:

The pressure of the water in the pipe is 129554 Pa.

Explanation:

<em>There are wrongly written values on the proposal, the atmospheric pressure must be 101105 Pa, and the density of water 1001.03 kg/m3, those values are the ones that make sense with the known ones.</em>

We start usign the continuity equation, and always considering point 1 a point inside the pipe and point 2 a point in the nozzle:

$A_1v_1=A_2v_2$

We want $v_2$ , and take into account that the areas are circular:

$v_2=\frac{A_1v_1}{A_2}=\frac{\pi r_1^2 v_1}{\pi r_2^2}=\frac{r_1^2 v_1}{r_2^2}$

Substituting values we have (we don't need to convert the cm because they cancel out between them anyway):

$v_2=\frac{r_1^2 v_1}{r_2^2}=\frac{(1.8cm)^2 (0.56m/s)}{(0.49cm)^2}=7.56m/s$

For determining the absolute pressure of the water in the pipe we use the Bernoulli equation:

$P_1+\frac{\rho v_1^2}{2}+\rho gh_1=P_2+\frac{\rho v_2^2}{2}+\rho gh_2$

Since the tube is horizontal $h_1=h_2$ and those terms cancel out, so the pressure of the water in the pipe will be:

$P_1=P_2+\frac{\rho v_2^2}{2}-\frac{\rho v_1^2}{2}=P_2+\frac{\rho (v_2^2-v_1^2)}{2}$

And substituting for the values we have, considering the pressure in the nozzle is the atmosphere pressure since it is exposed to it we obtain:

$P_1=101105 Pa+\frac{1001.03Kg/m^3 ((7.56m/s)^2-(0.56m/2)^2)}{2}=129554 Pa$

3 0

3 years ago