<span>Match the basic components of a nuclear reactor with their descriptions.
1. slows down neutrons
moderator - This is the substance that slows down fast neutrons and makes them slow neutrons which are easier to capture by the atomic nuclei so that the fission reaction can continue.
2. absorb emitted neutrons
control rods - These are rods made up of a substance that easily absorbs neutrons. Their purpose is to slow down or shut down the reaction.
3. mass of unstable atoms
nuclear fuel - The entire point of a nuclear reactor is the capture the energy released by the fission of unstable atoms. So this mass of unstable atoms is the fuel for the nuclear reactor.
4. concrete and lead enclosure
shield - This is the enclosure that prevents radiation from escaping into the general environment.
5. energy transfer medium
coolant - Since the purpose of a nuclear reactor is to generate usable energy, the coolant extracts heat from the fissioning core and that heat is generally used to boil water which in turn is used to operate turbines that power electrical generators.</span>
The first thing you should know to solve this problem is the conversion of pounds to kilograms:
1lb = 0.45 Kg
We can solve this problem by a simple rule of three
1lb ---> 0.45Kg
125lb ---> x
Clearing x we have:
x = ((125) / (1)) * (0.45) = 56.25 Kg.
Answer
her mass expressed in kilograms is 56.25 Kg.
Definitely not the last 2. My bet is on the first option. If it is wrong don't hit me please...
the area bounded by the line and the axes of a velocity-time graph is equal to the displacement of an object during that particular time period
Thank you
(a) The work done by the force applied by the tractor is 79,968.47 J.
(b) The work done by the frictional force on the tractor is 55,977.93 J.
(c) The total work done by all the forces is 23,990.54 J.
<h3>
Work done by the applied force</h3>
The work done by the force applied by the tractor is calculated as follows;
W = Fd cosθ
W = (5000 x 20) x cos(36.9)
W = 79,968.47 J
<h3>Work done by frictional force</h3>
W = Ffd cosθ
W = (3500 x 20) x cos(36.9)
W = 55,977.93 J
<h3>Net work done by all the forces on the tractor</h3>
W(net) = work done by applied force - work done by friction force
W(net) = 79,968.47 J - 55,977.93 J
W(net) = 23,990.54 J
Learn more about work done here: brainly.com/question/25573309
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