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jonny [76]
3 years ago
14

Traditional explosives use the energy released by chemical reactions. Nuclear weapons use the energy released by nuclear reactio

ns. Why are nuclear weapons much more destructive than traditional explosives? Question 16 options: A. Chemical reactions can release much more energy than nuclear reactions B. Chemical reactions always use more energy than they release Nuclear weapons always contain more chemicals than traditional explosive C. Chemical reactions always release more energy than they use D. Nuclear reactions can release much more energy than chemical reactions
Physics
2 answers:
lawyer [7]3 years ago
5 0
I believe the correct answer from the choices listed above is option D. Nu<span>clear weapons are much more destructive than traditional explosives because nu</span><span>clear reactions can release much more energy than chemical reactions. Hope this answers the question.</span>
vazorg [7]3 years ago
4 0

Answer: The correct answer is option D.

Explanation:

Chemical reactions are defined as the reactions in which rearrangement of one or more substances gets converted to one or more different substances known as products. Energy is released in the reactions where energy of products is less than the energy of the reactants.

Nuclear reactions are the reactions in which atomic nucleus undergoes change and releases a huge amount of energy in the process. There are two types of nuclear reactions: Nuclear fusion and nuclear fission. Both the nuclear reactions releases a huge amount of energy.

From the given information above, we conclude that the correct option is Option D.

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Which of these systems is an oscillator? A. A barrel rolling down the hill B. A child sitting on a swing a skater falling on the
Paladinen [302]

Answer:

Option B:

A child sitting on a swing.

Explanation:

When we hear the word oscillator, a good example is the pendulum bob of a grandfather clock. We can picture the motion to get a perfect understanding of its path of motion and relate it to other systems of motion in our everyday life.

An oscillator is a system that moves in such a way that it reverses its direction after a period of time. It can be seen as a "to-and-fro" motion.

From the options, a child sitting on a swing is the perfect example of an oscillating system because the child will be moving forwards and backwards, alternately reversing the direction of motion with time.

7 0
3 years ago
7. En la sala de una casa hay una gran ventana de vidrio, por la que se presenta una pérdida significativa de calor; las medidas
jonny [76]

Answer:

I DON'T SPEAK TACO BELL!

Explanation:

7 0
3 years ago
Which of the following Variable <br> is not measurable
inessss [21]
Before anyone can help you, if this is a multiple choice question you must provide the answers.<span />
4 0
3 years ago
A person skateboarding is part of a closed system that has 1,450 j of
soldier1979 [14.2K]

The gravitational potential energy of the system will decreases from 1,250 J to 625 J. Option A is corect.

<h3>What is the law of conservation of energy?</h3>

According to the Law of Conservation of Energy, energy can neither be created nor destroyed, but it can be transferred from one form to another.

The total energy is the sum of all the energies present in the system. The potential energy in a system is due to its position in the system.

TE=KE+GPE

Case 1;

1450 = 200 J+GPE

GPE=1450 -200

GPE=1250 J

Case 2;

1450 = 825 J+GPE

GPE=1450 -825

GPE=625 J

The gravitational potential energy of the system will decreases from 1,250 J to 625 J.

Hence, option A is corect.

To learn more about the law of conservation of energy, refer to brainly.com/question/2137260.

#SPJ1

6 0
2 years ago
A van has a weight of 4000 lb and center of gravity at Gv. It carries a fixed 900 lb load which has a center of gravity at Gl. I
natulia [17]

Answer:

 x = 25 / μ     [ ft]

Explanation:

To solve this exercise we can use Newton's second law.

Let's set a reference system where the x axis is parallel to the road

Y axis  

       N_B + N_A - W_van - W_load = 0

       N_B + N_A = W_van + W_load

X axis

     fr = ma

     a = fr / m

the total mass is

        m = (W_van + W_load) / g

the friction force has the expression

      fr = μ N_{total}

      fr = μy (W_van + W_load)

we substitute

      a = μ (W_van + W_load)    \frac{g}{W_van + W_load}

      a = μ g

taking the acceleration let's use the kinematic relations where the final velocity is zero

       v² = v₀² - 2 a x

       0 = v₀² -2a x

        x = \frac{v_o^2}{2a}

        x = \frac{v_o^2}{2 \mu g}

        x = \frac{40^2}{2 \ 32 \  \mu}

        x = 25 / μ     [ ft]

5 0
3 years ago
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