Ask question

Ask question

All categories

3 years ago

7

does the amount of work to required to move a charge in an electric field depend on the path taken or just the potential differe

nce between the end points?

1 answer:

Romashka [77]3 years ago

4 0

Answer: No

Explanation:

The work done is not path-dependent but only dependent on the final and initial position of the charge, that is the difference between the end points.

You might be interested in

A 3 kg bowling ball is thrown onto a mattress. If it takes 0.3 seconds to stop the ball using a force of 24 N, what was the init

enyata [817]

Answer:

-24 m/s

Explanation:

mass of the bowling ball = 3 kg

time (t) = 0.3 seconds

Force = 24 N

initial velocity u = ???

We know that;

Force = mass × acceleration (a)

So;

24 = 3 × a

a = 24/3

a = 8 m/s²

Also;

From equation of motion; acceleration is given by the relation;

$a =\frac{v-u}{t}$

if v = 0

then ;

$8 = \frac{0-u}{0.3}$

24 = 0- u

u = -24 m/s

Thus; the initial velocity of the bowling ball when it first touched the mattress = -24 m/s

7 0

3 years ago

What is one advantage of doing a field experiment instead of a laboratory experiment

Flauer [41]

It mimics the real world accurately

Explanation:

Experiments conducted in the field clearly presents the real world at it is to the scientist. Hardly can any part be controlled precisely and this gives a near to perfect scenario.

In the laboratory, for example, an organism is isolated from its environment and might not fully display its natural instinct and physiological capabilities.
Most laboratory set up are driven towards a model instead of real life settings.
The laboratory is more controlled and less varied and might truly represent the real world. It will only portray a part of the real world and series of further tests might have to be carried out to have a better model.

Learn more:

Experiment brainly.com/question/5096428

#learnwithBrainly

5 0

3 years ago

6. The image to the right shows a moment of inertia

Trava [24]

The moment of inertia is the rotational analog of mass, and it is given by

the product of mass and the square of the distance from the axis.

The moment of inertia changes as the position of the weight is changed, which indicates that; statement is incorrect

Reasons:

The weight on each arm that have adjustable positions can be considered as point masses.

The moment of inertia of a point mass is <em>I</em> = m·r²

Where;

m = The mass of the weight

r = The distance (position) from the center to which the weight is adjusted

Therefore;

The moment of inertia, <em>I </em>∝ r²

Which gives;

Doubling the distance from the center of rotation, increases the moment of inertia by factor of 4.

Therefore, the statement contradicts the relationship between the radius of rotation and moment of inertia.

Learn more about moment of inertia here:

brainly.com/question/4454769

7 0

2 years ago

What change in entropy occurs when a 0.15 kg ice cube at -18 °C is transformed into steam at 120 °c 4.

Studentka2010 [4]

<u>Answer:</u> The change in entropy of the given process is 1324.8 J/K

<u>Explanation:</u>

The processes involved in the given problem are:

$1.)H_2O(s)(-18^oC,255K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\3.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\4.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\5.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(120^oC,393K)$

Pressure is taken as constant.

To calculate the entropy change for same phase at different temperature, we use the equation:

$\Delta S=m\times C_{p,m}\times \ln (\frac{T_2}{T_1})$ .......(1)

where,

$\Delta S$ = Entropy change

$C_{p,m}$ = specific heat capacity of medium

m = mass of ice = 0.15 kg = 150 g (Conversion factor: 1 kg = 1000 g)

$T_2$ = final temperature

$T_1$ = initial temperature

To calculate the entropy change for different phase at same temperature, we use the equation:

$\Delta S=m\times \frac{\Delta H_{f,v}}{T}$ .......(2)

where,

$\Delta S$ = Entropy change

m = mass of ice

$\Delta H_{f,v}$ = enthalpy of fusion of vaporization

T = temperature of the system

Calculating the entropy change for each process:

<u>For process 1:</u>

We are given:

$m=150g\\C_{p,s}=2.06J/gK\\T_1=255K\\T_2=273K$

Putting values in equation 1, we get:

$\Delta S_1=150g\times 2.06J/g.K\times \ln(\frac{273K}{255K})\\\\\Delta S_1=21.1J/K$

<u>For process 2:</u>

We are given:

$m=150g\\\Delta H_{fusion}=334.16J/g\\T=273K$

Putting values in equation 2, we get:

$\Delta S_2=\frac{150g\times 334.16J/g}{273K}\\\\\Delta S_2=183.6J/K$

<u>For process 3:</u>

We are given:

$m=150g\\C_{p,l}=4.184J/gK\\T_1=273K\\T_2=373K$

Putting values in equation 1, we get:

$\Delta S_3=150g\times 4.184J/g.K\times \ln(\frac{373K}{273K})\\\\\Delta S_3=195.9J/K$

<u>For process 4:</u>

We are given:

$m=150g\\\Delta H_{vaporization}=2259J/g\\T=373K$

Putting values in equation 2, we get:

$\Delta S_2=\frac{150g\times 2259J/g}{373K}\\\\\Delta S_2=908.4J/K$

<u>For process 5:</u>

We are given:

$m=150g\\C_{p,g}=2.02J/gK\\T_1=373K\\T_2=393K$

Putting values in equation 1, we get:

$\Delta S_5=150g\times 2.02J/g.K\times \ln(\frac{393K}{373K})\\\\\Delta S_5=15.8J/K$

Total entropy change for the process = $\Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4+\Delta S_5$

Total entropy change for the process = $[21.1+183.6+195.9+908.4+15.8]J/K=1324.8J/K$

Hence, the change in entropy of the given process is 1324.8 J/K

4 0

3 years ago

Sound waves can be modeled by the equation of the form y=20sin(3t+theta). Determine what type of interference results when sound

Mariulka [41]

Answer:

go to the link quizzlet it will give you tha answer

Explanation:

4 0

2 years ago