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Sergio039 [100]
3 years ago
15

A non-licensed person may be the SOLE owner of a civil, electrical, or mechanical engineering business under which of the follow

ing conditions
Physics
2 answers:
insens350 [35]3 years ago
8 0

Answer:

hi your question lacks the options here is the complete question

A non-licensed person may be the SOLE owner of a civil, electrical, or mechanical engineering business under which of the following conditions

(a) A currently licensed California Civil, Electrical, or Mechanical Engineer, as appropriate, is in charge of the engineering practice of the business.

(b) All engineering plans and specifications are prepared by an appropriately licensed engineer.

(c) The name and business contain only the name of the licensed engineer.

(d) None. No provisions exist.

Answer: B

Explanation:

Generally a non-licensed person is not allowed to be a sole owner of an engineering business because it requires engineering plans and specifications to be made and carried out hence a certified/licensed  Engineer alone can do that.

from the conditions given the only condition suitable for a non-licensed person to be the sole owner of an engineering business is that the engineering plans and specifications are prepared by an appropriately licensed engineer either working for the engineering business or contracted .

ra1l [238]3 years ago
4 0

Answer:

No.

Explanation:

The "guide to Engineering and land surveying" for professional engineers and land surveyors by the California board reviews that an unlicensed person cannot be a sole owner of an engineering business, unless there is partnership with a licensed engineer.

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A dinner plate falls vertically to the floor and breaks up into three pieces, which slide horizontally along the floor. Immediat
Olegator [25]

Answer:

Explanation:

There will be conservation of momentum along horizontal plane because no force acts along horizontal plane.

momentum of first piece = .320 kg x 2 m/s

= 0.64 kg m/s along x -axis.

momentum of second piece = .355 kg x 1.5 m/s

= 0.5325 kg m/s along y- axis .

Let the velocity of third piece be v and it is making angle of θ with x -axis .

Horizontal component of its velocity = .100 kg x v cosθ = .1 v cosθ

vertical  component of its velocity = .100 kg x v sinθ = .1 v sinθ

For making total momentum in the plane zero

.1 v cosθ = 0.64 kg m/s

.1 v sinθ = 0.5325 kg m/s

Dividing

Tanθ = .5325 / .64 = .83

θ = 40⁰.

The angle will be actually 180 + 40 = 220 ⁰ from positive x -axis.

6 0
2 years ago
Read 2 more answers
The route followed by a hiker consists of three displacement vectors, X, Y and Z. Vector X is along a measured trail and is 1430
poizon [28]

Answer:

  • magnitude : 1635.43 m
  • Angle: 130°28'20'' north of east

Explanation:

First, we will find the Cartesian Representation of the \vec{X} and \vec{Y} vectors. We can do this, using the formula

\vec{A}= | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )

where | \vec{A} | its the magnitude of the vector and θ the angle. For  \vec{X} we have:

\vec{X}= 1430 m \ ( \ cos( 42 \°) \ , \ sin (42 \°) \ )

\vec{X}= ( \ 1062.70 m \ , \ 956.86 m \ )

where the unit vector \hat{i} points east, and \hat{j} points north. Now, the \vec{Y} will be:

\vec{Y}= - 2200 m \hat{j} = ( \ 0 \ , \ - 2200 m \ )

Now, taking the sum:

\vec{X} + \vec{Y} + \vec{Z} = 0

This is

\vec{Z} = - \vec{X} - \vec{Y}

(Z_x , Z_y) = - ( \ 1062.70 m \ , \ 956.86 m \ ) - ( \ 0 \ , \ - 2200 m \ )

(Z_x , Z_y) = ( \ - 1062.70 m \ ,  \ 2200 m \ - \ 956.86 m \ )

(Z_x , Z_y) = ( \ - 1062.70 m \ ,  \ 1243.14 m\ )

Now, for the magnitude, we just have to take its length:

|\vec{Z}| = \sqrt{Z_x^2 + Z_y^2}

|\vec{Z}| = \sqrt{(- 1062.70 m)^2 + (1243.14 m)^2}

|\vec{Z}| = 1635.43 m

For its angle, as the vector lays in the second quadrant, we can use:

\theta = 180\° - arctan(\frac{1243.14 m}{ - 1062.70 m})

\theta = 180\° - arctan( -1.1720)

\theta = 180\° - 45\°31'40''

\theta = 130\°28'20''

5 0
3 years ago
What area must the plates of a capacitor be if they have a charge of 5.7uC and an electric field of 3.1 kV/mm between them? O 0.
kirill [66]

Answer:

Area of the plates of a capacitor, A = 0.208 m²

Explanation:

It is given that,

Charge on the parallel plate capacitor, q = 5.7\ \mu C=5.7\times 10^{-6}\ C

Electric field, E = 3.1 kV/mm = 3100000 V/m

The electric field of a parallel plates capacitor is given by :

E=\dfrac{q}{A\epsilon_o}

A=\dfrac{q}{E\epsilon_o}

A=\dfrac{5.7\times 10^{-6}\ C}{3100000\ V/m\times 8.85\times 10^{-12}\ F/m}

A = 0.208 m²

So, the area of the plates of a capacitor is 0.208 m². Hence, this is the required solution.

8 0
3 years ago
A red laser from the physics lab is marked as producing 632.8-nm light. When light from this laser falls on two closely spaced s
BARSIC [14]

Answer:

The wavelength is  \lambda_R  =  649 *10^{-9}\ m

Explanation:

From the question we are told that

   The wavelength of the red laser is  \lambda_r  =  632.8 \ nm =  632.8 *10^{-9}\ m

    The spacing between  the fringe is  y_r  =  6.00\ mm =  6.00*10^{-3}  \  m

   The spacing between  the fringe for smaller laser point  is  y_R   = 6.19 \ mm =  6.19 *10^{-3} \  m

      Generally the spacing between  the fringe is mathematically represented as

       y  =  \frac{D *  \lambda  }{d}

Here  D is the distance to the screen

    and  d is the distance of the slit separation

Now for both laser red light light and  small laser  point  D and  d are same for this experiment

So

         \frac{y_r}{\lambda_r}  =  \frac{D}{d}

=>      \frac{y_r}{\lambda_r}  = \frac{y_R}{\lambda_R}

Where \lambda_R  is the wavelength produced by the small laser pointer

  So

           \frac{6.0 *10^{-3}}{ 632.8*10^{-9}}  = \frac{ 6.15 *10^{-9}}{\lambda_R}

=>       \lambda_R  =  649 *10^{-9}\ m

7 0
3 years ago
PLEASE HELP ME WITH THIS ONE QUESTION
PolarNik [594]

OPTION C is the correct answer.

6 0
2 years ago
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