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Yakvenalex [24]
3 years ago
11

How does nuclear fusion produce energy

Physics
1 answer:
prisoha [69]3 years ago
6 0
The fusion reaction that is easiest to accomplish is the reaction between two hydrogen isotopes: deuterium, extracted from water and tritium, produced during the fusion reaction through contact with lithium. When deuterium and tritium nuclei fuse, they form a helium nucleus, a neutron and a lot of energy.
You might be interested in
Transform boundaries are classified under which type of fault?
cluponka [151]

Answer:

Strike-slip fault

Explanation:

Transform boundaries play the role of connecting the other plate boundary segments.

When the plates are rubbed against each other, they result in enormous amount of stresses which leads to the breaking of the part of a rock causing earthquakes. Places of occurrence of these breaks are termed as faults.

Strike slip faults results from compression which takes place horizontally, but but in this the rock displacement  releases energy and takes place in a horizontal direction which is parallel to the force of compression.

8 0
3 years ago
Saturated steam at 125 kpa is compressed adiabatically in a centrifugal compressor to 700 kpa at the rate of 2.5 kg⋅s−1. the com
Tpy6a [65]
M° = 2.5 kg/sec
For saturated steam tables
at p₁ = 125Kpa
hg = h₁ = 2685.2 KJ/kg
SQ = s₁ = 7.2847 KJ/kg-k
for isotopic compression
S₁ = S₂ = 7.2847 KJ/kg-k
at 700Kpa steam with S = 7.2847
h₂ 3051.3 KJ/kg
Compressor efficiency
h =  0.78
0.78 = h₂ - h₁/h₂-h₁
0.78 = h₂-h₁ → 0.78 = 3051.3 - 2685.2/h₂ - 2685.2
h₂ = 3154.6KJ/kg
at 700Kpa with 3154.6 KJ/kg
enthalpy gives
entropy S₂ = 7.4586 KJ/kg-k
Work = m(h₂ - h₁) = 2.5(3154.6 - 2685.2
W = 1173.5KW
5 0
3 years ago
A 49.0 kg wheel, essentially a thin hoop with radius 0.730 m, is rotating at 114 rev/min. It must be brought to a stop in 22.0 s
belka [17]

Explanation:

Mass of the wheel, m = 49 kg

Radius of the hoop, r = 0.73 m

Initial angular speed of the wheel, \omega_i=114\ rev/min = 11.93\ rad/s

Final angular speed of the wheel, \omega_f=0

Time, t = 22 s

(a) If I is the moment of inertia of the hoop. It is equal to,

I=mr^2

I=49\times (0.73)^2

I=26.11\ kg-m^2

We know that the work done is equal to change in kinetic energy.

W=\Delta E

W=\dfrac{1}{2}I(\omega_f^2-\omega_i^2)

W=-\dfrac{1}{2}\times 26.11\times (11.93^2)

W = -1858.05 Joules

(b) Let P is the average power. It is given by :

P=\dfrac{W}{t}

P=\dfrac{1858.05\ J}{22\ s}

P =84.45 watts

Hence, this is the required solution.

4 0
3 years ago
19 dm expressed in millimeters
alina1380 [7]
1900 millimeters thats wht i got
6 0
1 year ago
What is needed to give a larger boulder a large acceleration?
Ronch [10]
More force needs to be applied
7 0
3 years ago
Read 2 more answers
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