3. A child decides to keep his goldfish outside in a small bowl. He has to add water every day to keep

Give two examples of workplace environments where considerations must be made with respect to the possibility of electric discha

katrin2010 [14]

Answer:

1.) Oil and gas environment

2.) Mechanical engineering and electrical environment.

Explanation:

1.) Oil and gas environment.

Electricity and electric discharge are recognized as serious workplace hazard in oil and gas companies, exposing employees to burns, fires, and explosions.

2.) Mechanical engineering and electrical environment.

Electricity or electric discharge has long been recognized as a serious workplace hazard, exposing employees to electric shock, electrocution, burns, fires, and explosions. Workers could die from electrocutions at work. This is

accounting for on-the-job fatalities and what makes these more tragic is that most of these fatalities could have been easily avoided.

4 0

3 years ago

In a polytropic process, with the exponent n = k, where k is the specific heat ratio, 1.0 kg of neon (an ideal gas) in a frictio

Elden [556K]

A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs.
Read more on Brainly.com - brainly.com/question/1581851#readmore

3 0

3 years ago

A water skier lets go of the tow rope upon leaving the end ofa

ANTONII [103]

Answer:

1.35m

Explanation:

At the highest point of the jump, the vertical speed of the skier should be 0. So the 13m/s speed is horizontal, this speed stays the same from the jumping point to the highest point. The 14m/s speed at jumping point is the combination of both vertical and horizontal speeds.

The vertical speed at the jumping point can be computed:

$v_v^2 + v_h^2 = v^2$

$v_v^2 + 13^2 = 14^2$

$v_v^2 = 196 - 169 = 27$

$v_v = \sqrt{27} = 5.2 m/s$

When the skier jumps to the its potential energy is converted to kinetic energy:

$E_p = E_k$

$mgh = mv_v^2/2$

where m is the skier mass and h is the vertical distance traveled, $v_v$ is the vertical velocity at jumping point, and h is the highest point.

Let g = 10m/s2

We can divide both sides of the equation by m:

$gh = v_v^2/2$

$h = \frac{v_v^2}{2g} = \frac{27}{2*10} = 1.35 m$

3 0

3 years ago

What does solution mean in science

slega [8]

A solution is a mixture composed of two or more substances which is always homogeneous. One substance dissolves in with the other substance making a solution.

Hope this helps!

8 0

4 years ago

Read 2 more answers

An object with charge q = −6.00×10−9 C is placed in a region of uniform electric field and is released from rest at point A. Aft

sergeinik [125]

Answer:

a) 80 V

b) The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge

Explanation:

Given :

We are given an object with charge q = -6.00 x I0^-9 C starts moving from the rest at point A, which means its kinetic energy at point A is zero ( $K_{A}$ = 0) to the point B at distance l = 0.500m where its kinetic energy is ( $K_{B}$ = 5.00 x 10^-7J) . Also, the electric potential of q at point A is VA = + 30.0 v.

Required :

(a) We are asked to find the electric potential VB

(b) We want to determine the magnitude and the direction of the electric field E.

Solution

(a) We are given the values for VA, $K_{B}$ and q so we want to find a relationship between these three parameters and VB to get the value of VB.

As we have two states, at points A and B , where the charge moved from A to B due to the applied electric field. The mechanical energy of the object is conservative during this travel, and we can apply eq(1) in this situation:

$K_{A} +U_{A} =K_{B} +U_{B}$ .........................................(1)

Where $K_{A}$ = 0 and the potential energy U of the charge is given by U = q V

where V is the electric potential. So, equation (1) will be in the form :

0+qVA= $K_{B}$ +qVB (Divide by q)

VA= $K_{B}$ /q + VB (solve for VB)

VB=VA- $K_{B}$ /q .......................................(2)

We get the relation between VB, VA and $K_{B}$ , now we can plug our values for VA, $K_{B}$ and q into equation (2) to get VB

VB=VA- $K_{B}$ /q

=30V-(5.00 x 10^-7J)/(-6.00 x I0^-9)

=80 V

(b) After we calculated VB we can use equation a to get the electric field E that applied to the charge q, where the potential difference between the two points equals the integration of the electric field multiplied by the distance l between the two points

VA-VB = $\int\limits^1_0 {E} \, dl$ ...................................(a)

$=E\int\limits^1_0 {} \, dl$

VA-VB=E $l$ (solve for E)

E= VA-VB/ $l$ ..................................(3)

Now let us plug our values for VA, Vs and l into equation (3) to get the electric field E

E= VA-VB/ $l$

=-100 N/C

The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge

5 0

3 years ago