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Ivenika [448]
3 years ago
7

Which type of engineer would most likely work with a renewable energy source?

Physics
2 answers:
julia-pushkina [17]3 years ago
8 0
I would say Solar Power Engineer.
babymother [125]3 years ago
8 0

Answer: C. Solar power engineer

Explanation:

A renewable source of energy is the energy that is harnessed from the renewable resources. The renewable resources are those resources which can be replenished once used. These resources are abundantly available in nature. These includes the sunlight, water, wind and forests.

Among the given options solar power engineer is the correct option. This is because of the fact that solar power is a renewable source of energy and it is the source of energy which cannot be depleted in nature. The radiant energy and heat from the sunlight can be used to generate electricity or to run the electrical devices.

The solar power engineer will construct the solar heating appliances, photovoltic cells, solar power plants, and other instruments or devices that works on the solar energy.

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The process by which you group things based on their similarities is known as classifying.
dusya [7]
True. Classifying by similarities is the basis for biological classification.
8 0
3 years ago
An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr
Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

F= mg

Where, f = restoring force

kx=mg

k=\dfrac{mg}{x}

Put the value into the formula

k=\dfrac{2.15\times9.8}{0.0895}

k=235.41\ N/m

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2

Here, v = A\omega

K.E=\dfrac{1}{2}m\times(A\omega)^2

Here, \omega=\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}kA^2

Put the value into the formula

K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2

K.E=0.06500\ J

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0
3 years ago
Is this statement true or false? The next generation of nuclear power plants being built in California and South Africa are even
alexira [117]

Answer:

True

Explanation:

Modern safer and cheaper nuclear reactors can not only meet the range of our long term energy demands, they can also fight global warming.

Modern techniques provide ways to reduce radioactive waste amount. "A closed fuel cycle may be switched on for new kinds of nuclear plants. Alternatively, the waste is chemically dissuaded to transform the reusable element into fuel. This implies that nuclear waste would not be buried.

8 0
3 years ago
Newton's First Law says that an object at rest will stay at rest, and an object in motion will stay in motion unless:
Verdich [7]

Answer: Acted on by equal forces in opposite direction

Explanation:

Newton's First Law says that every body continue in its state of rest or constant speed on a straight line unless being acted upon by an external force.

6 0
3 years ago
The central star of a planetary nebula emits ultraviolet light with wavelength 104nm. This light passes through a diffraction gr
Gala2k [10]

Answer: 31.33 degrees

Explanation:

The diffraction angles \theta_{n} when we have a slit divided into n parts are obtained by  the following equation:

dsin\theta_{n}=n\lambda   (1)

Where:

d is the width of the slit

\lambda  is the wavelength of the light

n is an integer different from zero.

Now, the first-order diffraction angle is given when n=1, hence equation (1) becomes:

dsin\theta_{1}=\lambda   (2)

Now we have to find the value of \theta_{1}:

sin\theta_{1}=\frac{\lambda}{d}  

\theta_{1}=arcsin(\frac{\lambda}{d})   (3)

We know:

\lambda=104nm=104(10)^{-9}m

In addition we are told the diffraction grating has 5000 slits per mm, this means:

d=\frac{1mm}{5000}=\frac{1(10)^{-3}m}{5000}

Substituting the known values in (3):

\theta_{1}=arcsin(\frac{104(10)^{-9}m}{\frac{1(10)^{-3}m}{5000}})

\theta_{1}=arcsin(0.52)

<u>Finally:</u>

\theta_{1}=31.33\º >>>This is the first-order diffraction angle

4 0
3 years ago
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