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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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Ice floats on water.

OleMash [197]

Answer:

Yeah ice floats on water.

Observation

Example in those areas were ice is found like Antarctica ice is found on top of water.

5 0

2 years ago

If an object is to rest on an incline without slipping, then friction must equal the component of the weight of the object paral

leonid [27]

Answer:

$\theta = tan^{-1}\mu$

Explanation:

As we know that if the object is placed on the inclined plane then the force of friction on the object is counterbalanced by the component of the weight of the object along the inclined plane.

So we can say

$F_f = mgsin\theta$

now if we increase the inclination of the plane then the component of the weight weight along the inclined plane will increase and hence the friction force will also increase.

As we know that the limiting value or the maximum value of friction force at the static condition is given by

$F_f = \mu N$

$N = mg cos\theta$

so we have

$F_f = \mu (mg cos\theta)$

so we will have

$mg sin\theta = \mu (mg cos\theta)$

so now we have

$tan\theta = \mu$

so maximum possible angle of the inclined plane is

$\theta = tan^{-1}\mu$

3 0

3 years ago

The net force acting on an object equals the applied force plus the force of friction.

Georgia [21]

Answer:

False

Explanation:

The net force is equal to the applied force minus the force of friction. It is possible for friction to act in the same direction as an applied force, but that would mean there would have to be more than two forces acting on the object.

3 0

2 years ago

A bullet is fired straight up from a gun with a

melamori03 [73]

Answer: 815.51 m

Explanation:

This situation is related to projectile motion or parabolic motion, in which the initial velocity of the bullet has only y-component, since it was fired straight up. In addition, we are dealing with constant acceleration (due gravity), therefore the following equations will be useful to solve this problem:

$V=V_{o}+gt$ (1)

$V^{2}=V_{o}^{2}+2gy$ (2)

Where:

$V$ is the final velocity of the bullet

$V_{o}=152 m/s$ is the initial velocity of the bullet

$g=-9.8 m/s^{2}$ is the acceleration due gravity, always directed downwards

$t=6.9 s$ is the time

$y$ is the vertical position of the bullet at $t=6.9 s$

Let's begin by finding $V$ from (1):

$V=152 m/s-9.8 m/s^{2}(6.9 s)$ (3)

$V=84.38 m/s$ (4)

Now we have to substitute (4) in (2):

$(84.38 m/s)^{2}=(152 m/s)^{2}-2(9.8 m/s^{2})y$ (5)

Isolating $y$ :

$y=815.511 m$ This is the displacement of the bullet after 6.9 s

8 0

2 years ago

Baseball player a bunts the ball by hitting it in such a way that it acquires an initial velocity of 2.4 m/s parallel to the gro

LUCKY_DIMON [66]

Let $\mathbf r_A$ denote the position vector of the ball hit by player A. Then this vector has components

$\begin{cases}r_{Ax}=\left(2.4\,\frac{\mathrm m}{\mathrm s}\right)t\\r_{Ay}=1.2\,\mathrm m-\frac12gt^2\end{cases}$

where $g=9.8\,\dfrac{\mathrm m}{\mathrm s^2}$ is the magnitude of the acceleration due to gravity. Use the vertical component $r_{Ay}$ to find the time at which ball A reaches the ground:

$1.2\,\mathrm m-\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2=0\implies t=0.49\,\mathrm s$

The horizontal position of the ball after 0.49 seconds is

$\left(2.4\,\dfrac{\mathrm m}{\mathrm s}\right)(0.49\,\mathrm s)=12\,\mathrm m$

So player B wants to apply a velocity such that the ball travels a distance of about 12 meters from where it is hit. The position vector $\mathbf r_B$ of the ball hit by player B has

$\begin{cases}r_{Bx}=v_0t\\r_{By}=1.6\,\mathrm m-\frac12gt^2\end{cases}$

Again, we solve for the time it takes the ball to reach the ground:

$1.6\,\mathrm m-\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2=0\implies t=0.57\,\mathrm s$

After this time, we expect a horizontal displacement of 12 meters, so that $v_0$ satisfies

$v_0(0.57\,\mathrm s)=12\,\mathrm m$

$\implies v_0=21\,\dfrac{\mathrm m}{\mathrm s}$

5 0

3 years ago