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3 years ago

Describe two rights and two responsibilities that you have as an employee

Physics

2 answers:

Lubov Fominskaja [6]3 years ago

7 0

Answer:

Explanation:

According to me as an employee two rights include:

Right to have safe occupational condition.
Right to get appropriate money for your work.

Two responsibilities include:

Responsibility to do your work honestly.
Responsibility to respecting your co-workers and maintaining decorum.

For example, worker working in a industry have the right to get appropriate amount of holidays to refresh themselves.

Pachacha [2.7K]3 years ago

5 0

As an employee you have the right to receive money but the responsability to work as your advisor ask you for
as an employee you have the right to work safety but the responsability to do all is accept ,and respectul by that way you can save your life and other by the same time

i can give you a last exemple

as an employee you have the right to have holyday break because you also need to be with your family but the responsability to stay in work when you have to

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A girl rides her scooter to school, a total distance of 4.5km. She has to slow down twice to cross busy streets, but overall the

Whitepunk [10]

Answer:

6.93 km/h

Explanation:

To calculate her average speed, we need the "speed" formula, which is:

average speed = distance / time

You plug in your numbers and it will give you the answer.

Speed = 4.5km/0.65hr

= 6.923 km/h

8 0

3 years ago

Read 2 more answers

"Which of the following statements about electrons is not true?

gogolik [260]

Answer:

B) Within an atom, an electron can have only particular energies.

Explanation:

As we know that electrons have energy but apart from electrons we know that protons and neutrons inside the nucleus of atom will also have energy in them.

rest all the statements are true as we have

A) Electrons orbit the nucleus rather like planets orbiting the Sun.

TRUE, because electrons can move in stationary orbit around the nucleus

C) Electrons can jump between energy levels in an atom only if they receive or give up an amount of energy equal to the difference in energy between the energy levels.

Difference amount of energy is lost or absorbed by the electron in form of photons

D) An electron has a negative electrical charge.

Charge of an electron is given as $-1.6 \times 10^{-19} C$

E) Electrons have very little mass compared to protons or neutrons

Mass of an electron is given as

$m_e = 9.11 \times 10^{-31} kg$

mass of proton or neutron

$m_p = 1.67 \times 10^{-27} kg$

7 0

3 years ago

The ratio of the focal length of a lens to it's diameter is the

user100 [1]

Answer:

The f-ratio describes the relationship between the lens diameter and the focal length and is calculated by dividing the focal length by the diameter of the lens. For example, if a lens were to have a focal length of 50mm and a diameter of 10mm, then the f-ratio would be 50mm/10mm=5 or otherwise referred to as f5.

Explanation:

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3 years ago

Read 2 more answers

Consider the sections of two circuits illustrated above. Select True or False for all statements. After connecting c and d to a

Basile [38]

Answer:

Follows are the solution to the given question:

Explanation:

In this question the missing file of the circuit is not be which is defined in attached file please find it.

In Option 1, this statement is true because the current is on $R_3 \ and \ R_4$ , that is the same.

In option 2, this statement is false because $Rcd=R_3+R_4$ therefore it implies that Rcd is always larger then $R_3$ .

In option 3, this statement is true because the voltage of $R_1 \ and \ R_2$ is always equal.

In option 4, this statement is true because $Rab= \frac{1}{(\frac{1}{R1}+\frac{1}{R2})}=\frac{R1R2}{(R1+R2)}. \frac{R2}{(R1+R2)}$ is always smaller then 1 therefore, $R_1 \times (\frac{R2}{(R1+R2)})$ is always equal to R1.

5 0

3 years ago

The drawing shows a large cube (mass = 21.0 kg) being accelerated across a horizontal frictionless surface by a horizontal force

MaRussiya [10]

Answer:

The blocks must be pushed with a force higher than 359 Newtons horizontally in order to accomplish this friction levitation feat.

Explanation:

The first step in resolving any physics problem is to draw the given scenario (if possible), see the attached image to have an idea of the objects and forces involved.

The large cube in red is being pushed from the left by a force $\vec{P}$ whose value is to be found. That cube has its own weight $\vec{w}_1=m_1\vec{g}$ , and it is associated with the force of gravity which points downward. Newton's third law stipulates that the response from the floor is an upward pointing force on the cube, and it's called the normal force $\vec{N}_1$ .

A second cube is being pushed by the first, and since the force $\vec{P}$ is strong enough it is able to keep such block suspended as if it were glued to the first one, due to friction. As in the larger cube, the smaller one has a weight $\vec{w}_2=m_2\vec{g}$ pointing downwards, but the normal force in this block doesn't point upwards since its 'floor' isn't below it, but in its side, therefore the normal force directs it to the right as it is shown in the picture. Normal forces are perpendicular to the surface they contact. The final force is the friction between both cubes, that sets a resistance of one moving parallel the other. In this case, the weight of the block its the force pointing parallel to the contact surface, so the friction opposes that force, and thus points upwards. Friction forces can be set as $Fr=\mu~N$ , where $\mu$ is the coefficient of static friction between the cubes.

Now that all forces involved are identified, the following step is to apply Newton's second law and add all the forces for each block that point in the same line, and set it as equal its mass multiplied by its acceleration. The condition over the smaller box is the relevant one so its the first one to be analyzed.

In the vertical component: $\Sigma F^2_y=Fr-w_2=m_2 a_y$ Since the idea is that it doesn't slips downwards, the vertical acceleration should be set to zero $a_y=0$ , and making explicit the other forces: $\mu N_2-m_2g=0\quad\Rightarrow (0.710)N_2-(4.5)(10)=0\quad\Rightarrow N_2=(4.5)(10)/(0.710)\approx 63.38$ [N]. In the last equation gravity's acceleration was rounded to 10 [m/s $^2$ ].

In its horizontal component: $\Sigma F^2_x=N_2=m_2 a_x$ , this time the horizontal acceleration is not zero, because it is constantly being pushed. However, the value of the normal force and the mass of the block are known, so its horizontal acceleration can be determined: $63.38=(4.5) a_x \quad \Rightarrow a_x=(63.38)/(4.5)\approx 14.08$ [m/s $^2$ ]. Notice that this acceleration is higher than the one of gravity, and it is understandable since you should be able to push it harder than gravity in order for it to not slip.

Now the attention is switched to the larger cube. The vertical forces are not relevant here, since the normal force balances its weight so that there isn't vertical acceleration. The unknown force comes up in the horizontal forces analysis: $\Sigma F_x=P=m a_x$ , since the force $\vec{P}$ is not only pushing the first block but both, the mass involved in this equation is the combined masses of the blocks, the acceleration is the same for both blocks since they move together; $P=(21.0+4.5) 14.08\approx 359.04$ [N]. The resulting force is quite high but not impossible to make by a human being, this indicates that this feat of friction suspension is difficult but feasable.

4 0

3 years ago