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3 years ago

An atom that changes so that it has an electrical charge is a(n)

Physics

1 answer:

MrMuchimi3 years ago

6 0

Answer:

An ion.

Explanation:

Atoms can gain or lose electrons and become ions, which are atoms that have a positive or negative charge, because they have unequal numbers of protons and electrons.

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Give one example of a thermodynamic cycle that does not account for the carnot efficiency.

Arturiano [62]

Thermo-Electrochemical converter (UTEC) is a thermodynamic cycle that does not account for the Carnot Efficiency.

The Carnot cycle is a hypothetical cycle that takes no account of entropy generation. It is assumed that the heat source and heat sink have perfect heat transfer. The working fluid also remains in the same phase, as opposed to the Rankine cycle, in which the fluid changes phase. A practical thermodynamic cycle, such as the Rankine cycle, would achieve at most 50% of the Carnot cycle efficiency under similar heat source and heat sink temperatures.

<h3>What is Thermo-Electrochemical converter?</h3>

In a two-cell structure, a thermo-electrochemical converter converts potential energy difference during hydrogen oxidation and reduction to heat energy.

It employs the Ericsson cycle, which is less efficient than the Carnot cycle. In a closed system, it converts heat to electrical energy. There are no external input or output devices.

This means there will be no mechanical work to be done, as well as no exhaust. As a result, Carnot efficiency is not taken into account in this cycle. Carnot efficiency is accounted for by other options such as turbine and engine.

Learn more about Thermo-Electrochemical converter here:

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2 years ago

What is magnetic flux linkage im electromagnetic series

Dominik [7]

Answer:

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2 years ago

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Solar cells convert what type of energy into electrical energy?

zysi [14]

I think the correct answer from the choices listed above is option D. Solar cells convert electromagnetic energy into electrical energy. This is because the photons from the rays of light are electromagnetic waves or particles. Hope this answers the question.

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4 years ago

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Explain how artists use monocular cues to depth perception described in the text to create an impression of three dimensions on

lana66690 [7]

Answer:

The relative size of an object serves as an important monocular cue for depth perception. It works like this: If two objects are roughly the same size, the object that looks the largest will be judged as being the closest to the observer. This applies to three-dimensional scenes as well as two-dimensional images.

Explanation:

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3 years ago

PHYSICS PLEASE HELP 80 PTS!!!!!!

VladimirAG [237]

1. B) 32 m/s

The speed of the baseball is given by

$v=\frac{d}{t}$

where

d = 48 m is the distance travelled

t = 1.5 s is the time taken

Substituting,

$v=\frac{48}{1.5}=32 m/s$

2. C) 12 m/s

The average velocity is given by

$v=\frac{d}{t}$

where

d = 60 - 0 = 60 m is the displacement

t = 5 s is the time interval

Substituting,

$v=\frac{60}{5}=12 m/s$

3. Missing diagram

4. D) velocity

In fact, velocity is defined as the ratio between the displacement ( $\Delta s$ ) and the time interval ( $\Delta t$ ) needed to achieve that change in position:

$v=\frac{\Delta s}{\Delta t}$

Similarly, acceleration is defined as the ratio between the change in velocity ( $\Delta v$ ) and the time interval ( $\Delta t$ ):

$a=\frac{\Delta v}{\Delta t}$

Comparing the two equations, we see that displacement is to velocity as velocity is to acceleration.

5. C) 5/1 cm/year

We know that the ridge moves 25 cm in 5 years, so we have:

- Displacement (d): d = 25 cm

- Time interval (t): t = 5 y

Using the equation for the velocity:

$v=\frac{d}{t}=\frac{25 cm}{5 y}=5 cm/y$

6. D

The missing graph is attached. Each graph represents a distance (on the y-axis, measured in metres) versus a time (on the x-axis, measured in seconds). Therefore, the speed in each graph can be simply calculated from the slope of the curve, since speed is the ratio between distance and time:

$v=\frac{\Delta y}{\Delta x}$

For each graph:

A. $\frac{\Delta y}{\Delta x}=\frac{8.6-2}{10}\sim 0.66$

B. $\frac{\Delta y}{\Delta x}=\frac{5.7-0}{10}\sim 0.57$

C. $\frac{\Delta y}{\Delta x}=\frac{9.6-3}{10}\sim 0.66$

D. $\frac{\Delta y}{\Delta x}=\frac{7.5-0}{5}\sim 1.5$

So we can say that the correct graph must be graph D, the closest to 1.57.

7. B) The car has come to a stop and has zero velocity

The missing graph is attached.

The graph shows the position of the car versus time. From the graph, we can see that in segment B, the position of the car does not change: this means that its velocity is zero, since the velocity is defined as the ratio between the change in position (displacement) and the time interval:

$v=\frac{\Delta s}{\Delta t}$

And since the displacement is zero, the velocity is also zero.

8. C) from 4.5 to 5.0 seconds

Missing graph is attached.

The graph represents the distance covered versus the time taken: therefore, the speed of the ball can be evaluated from the slope of the curve.

The only region in which the slope of the curve is zero is between 4.5 seconds and 5.0 seconds: therefore, this is the region where the soccer ball is not moving.

9. D) diagonal line with varying slope, from 3 to 4.

The table of data is:

Time (s) 0 1 2 3 4

Distance (m) 0 3 6 12 16

We want to know how the distance/time graph would like like. We have:

- At t = 1, the distance is $d=3$ , so the slope is $\frac{d}{t}=\frac{3}{1}=3$

- Similarly, at t = 2: $\frac{d}{t}=\frac{6}{2}=3$

- Instead, at t =3, the slope is $\frac{d}{t}=\frac{12}{3}=4$

- Similarly, at t=4, $\frac{d}{t}=\frac{16}{4}=4$

So the correct answer is D.

10. D) To move in a circle requires an acceleration and therefore a net force. This force is supplied by the slingshot.

For an object in circular motion, the velocity constantly changes (because the direction changes): this means that an acceleration is required (towards the centre of the circle), and therefore a force must be exerted (also towards the centre of the circle) to keep the object in the circular path.

As soon as this force is removed, the object is "free" to leave the circular trajectory and continue its motion following a straight path at constant velocity, according to Newton's first law.

11. B) The acceleration will become 1/4 as much.

Re-arranging Newton's second law formula,

$F=ma \rightarrow a = \frac{F}{m}$ (1)

The force exerted on the satellite is actually the force of gravity:

$F=\frac{Gm M}{d^2}$ (2)

Substituting (2) into (1),

$a=\frac{F}{m}=\frac{GM}{d^2}$

We see that there is no dependance on the mass of the satellite (m), but only on the distance. Since in this problem the distance is doubled (d'=2d), the new acceleration will be:

$a'=\frac{GM}{(2d)^2}=\frac{1}{4}(\frac{GM}{d^2})=\frac{a}{4}$