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vitfil [10]
3 years ago
11

What process is used to release energy in nuclear power plants

Physics
2 answers:
Hunter-Best [27]3 years ago
6 0
Fission is the process used to release energy in nuclear power plants (:
Anika [276]3 years ago
5 0
.........Nucler fission.... .
You might be interested in
A laser beam with a frequency of 180 Hz forms an 8 m standing wave with 10 nodes.
DIA [1.3K]

Answer:33

Explanation:

F = frequency

N =  Node count

w = wave lenght

v = wave velocity

L = distance wave traveled

First find wave length of laser

w = (2/(N))*(L)

w = (2/(10))*(8)

w = 1.6

then using (w), find velocity

V =  (w)(F)

V = (1.6)*(108)

V = 288

Plug in V and the new frequency to solve for new node count

F = NV/2L

(600) = (N)*(288) / 2 * (8)

(N) = 33.33

there are 33 nodes

8 0
3 years ago
In a thin film experiment, a wedge of air is used between two glass plates. If the wavelength of the incident light in air is 48
cluponka [151]

Answer:

The thickness is  \Delta y =  2.4 *10^{-6} \  m

Explanation:

From the question we are told that

   The wavelength is  \lambda  = 480 \ nm = 480*10^{-9} \  m

    The first order of the dark  fringe is  m_1 =  16

     The second order of dark fringe considered is  m_2 = 6

Generally the condition for destructive interference is mathematically represented as

        y = \frac{m \lambda}{2}

Here y is the path difference between the central maxima(i.e the origin) and any dark fringe

So  the path difference between the 16th dark fringe and the 6th dark fringe is mathematically represented as

      y_1 - y_2 = \Delta y =  \frac{m_1 \lambda}{2} -  \frac{m_2 \lambda}{2}

=>  y_1 - y_2 = \Delta y =  \frac{16 *480*10^{-9}}{2} -  \frac{6 *480*10^{-9}}{2}

=>  y_1 - y_2 = \Delta y =  5 (480*10^{-9})

=>  \Delta y =  2.4 *10^{-6} \  m

8 0
3 years ago
Graphs help you see ______.
Anastaziya [24]
D- if i was right please mark me as a brainliest answer and a thank you
4 0
3 years ago
Read 2 more answers
Which of the following would decrease in size during the contraction of a sarcomere? The width of the I-bands The width of the A
ANEK [815]

Hi!


The correct answer would be: the width of I-bands


The sacromere is the smallest contractile unit of striated muscles. These units comprise of filaments (fibrous proteins) that, upon muscle contraction or relaxation, slide past each other. The sacromere consists of thick filaments (myosin) and thin filaments (actin).


<em>Refer to the attached picture to clearly see the structure of a sacromere.</em>


<u>When a sacromere contracts, a series of changes take place which include:</u>

<em>- Shortening of I band, and consequently the H zone</em>

<em>- The A line remains unchanged</em>

<em>- Z lines come closer to each other (and this is due to the shortening of the I bands) </em>

The only changes that take place occur in the zones/areas in the sacromere (as mentioned), not in the filaments (actin and myosin) that make the up the sacromere; hence all other options are wrong.


Hope this helps!

8 0
3 years ago
A typical nuclear fission power plant produces about 1.00 GW of electrical power. Assume the plant has an overall efficiency of
mamaluj [8]

Answer:

mass consumed by 235U each day = 2 kg

Explanation:

electrical power produced = 1 GW = 1 × 10⁹ × (6.24151 × 10¹⁸ ) eV

                                            = 6.24151× 10²¹ MeV/s

thermal energy =  0.420 * 250 = 105 MeV

\dfrac{1 GW}{150 MeV}= \dfrac{6.24151\times 10^{21}}{105}

                                      = 5.94 × 10¹⁹ fission/second

                                       =5.94 × 10¹⁹× 24 × 60 ×60)

                                      =  5.13 × 10²⁴ fission/day

mu = 235.04393 ×  1.660× 10 ⁻²⁷ = 390.1729× 10⁻²⁷ Kg

M = mu ×5.13 × 10²⁴

   = 390.1729× 10⁻²⁷ ×5.13 × 10²⁴

M   =  2 kg(approx.)

mass consumed by 235U each day = 2 kg

3 0
3 years ago
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