All categories

2 years ago

What are usability heuristics?

Physics

1 answer:

labwork [276]2 years ago

8 0

Answer:

A heuristic evaluation is a usability inspection method mainly used to identify any design issues associated with the user interface.

Explanation:

They are called “heuristics” because they are broad rules of thumb and not specific usability guidelines.

You might be interested in

What is the wavelength

melamori03 [73]

The distance between two particles that are <em>in phase</em>

4 0

3 years ago

Two brothers, Frank and Dion, are on roller skates and standing still when they push off from one another. Frank's mass is 103.4

tigry1 [53]

Around -3.5 ( might not be completely right ) Frank has a greater mass to he has a low velocity at first, Dion is around half of his weight so he doesn’t have to use as much force to get more than Frank

3 0

2 years ago

A tiny object carrying a charge of +44 μC and a second tiny charged object are initially very far apart. If it takes 21 J of wor

STatiana [176]

Answer:

The magnitude of the second charge is $\rm 1.062\times 10^{-7}\ C$ or $\rm 0.1062\ \mu C.$

Explanation:

The work done in bringing a charged particle from one point to another in the presence of some electric field is equal to the change in the electric potential energy of the charge in moving from one point to another.

The electric potential energy of some charge $q_o$ at a point in the electric field of another charge $q$ is given by the product of the amount of charge $q_o$ and electric potential at that point due to the charge $q$ .

$U = q_o\ V.$

The electric potential at that point is given by

$V = \dfrac{kq}{r}.$

where $k$ is the Coulomb's constant.

Therefore,

$U=q_o\ \dfrac{kq}{r}.$

Now, We have given two charges $q_1 = +44\ \mu C = +44\times 10^{-6}\ C$ and $q_2$ , whose value is to be found.

When the two charges are infinitely dar apart, the electric potential energy of the system is given by

$U_i = \dfrac{kq_1q_2}{\infty}=0.$

When the coordinates of position of the two charges are

$(x_1,\ y_1) = (1.00\ mm,\ 1.00\ mm).\\(x_2,\ y_2) = (1.00\ mm,\ 3.00\ mm).$

The distance between the two charges is given by

$r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(1.00-1.00)^2+(3.00-1.00)^2}=2.00\ mm = 2.00\times 10^{-3}\ m.$

The electric potential energy of the charges in this configuration is given by

$U_f = \dfrac{kq_1q_2}{r}\\=\dfrac{(8.99\times 10^9)\times (+44\times 10^{-6})\times q_2}{2.00\times 10^{-3}}\\=1.9778\times 10^8\times q_2.$

The change in the electric potential energy of the system is equal to the work done to bring the system from inifinitely far apart position to given configuration.

Therefore,

$W = U_f-U_i\\21=(1.9778\times 10^8\times q_2)-0\\\Rightarrow q_2 = \dfrac{21}{1.9778\times 10^8}\\=1.062\times 10^{-7}\ C\\=0.1062\times 10^{-6}\ C\\=0.1062\ \mu C.$