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Morgarella [4.7K]
2 years ago
13

What are the products of a fusion reaction? Check all that apply. Lighter atoms energy heavier atoms a neutron a proton.

Physics
1 answer:
svet-max [94.6K]2 years ago
6 0

The process by which two or more tiny nuclei unite to generate a bigger nucleus is known as a nuclear fusion reaction. Heavier atoms are products of a fusion reaction.

<h3 /><h3>What is nuclear fusion?</h3>

The process by which two or more tiny nuclei unite to generate a bigger nucleus is known as a nuclear fusion reaction.

For example, the fusion of two hydrogen atoms produces more energy than the fusion of one helium atom, and surplus energy is expelled into space upon binding.

Hence heavier atoms are e products of a fusion reaction.

To learn more about nuclear fusion refer to the link;

brainly.com/question/14019172

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A person is using a rope to lower a 5.0-n bucket into a well with a constant speed of 2.0 m/s. What is the magnitude of the forc
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Answer:

5 N

Explanation:

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The energy that generates wind comes from what source?
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Which of the following statements is true of electromagnetic waves but NOT mechanical waves.
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2 years ago
What force must the deltoid muscle provide to keep the arm in this position?
ruslelena [56]

Answer:

Deltoid Force, F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}

Additional Information:

Some numerical information are missing from the question. However, I will derive the formula to calculate the force of the deltoid muscle. All you need to do is insert the necessary information and calculate.  

Explanation:

The deltoid muscle is the one keeping the hand arm in position. We have two torques that apply to the rotating of the arm.

1. The torque about the point in the shoulder for the deltoid muscle, T_{Deltoid}

2. The torque of the arm, T_{arm}  

Assuming the arm is just being stretched and there is no rotation going on,

                        T_{Deltoid} = 0

                       T_{arm} = 0

       ⇒           T_{Deltoid} = T_{arm}

                  r_{d}F_{d}sin\alpha_{d} = r_{a}F_{a}sin\alpha_{a}

Where,

r_{d} is radius of the deltoid

F_{d} is the force of the deltiod

\alpha_{d} is the angle of the deltiod

r_{a} is the radius of the arm

F_{a} is the force of the arm , F_{a} = mg  which is the mass of the arm and acceleration due to gravity

\alpha_{a} is the angle of the arm

The force of the deltoid muscle is,

                                 F_{d} = \frac {r_{a}F_{a}sin\alpha_{a}}{r_{d}sin\alpha_{d}}

but F_{a} = mg ,

                ∴            F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}

7 0
2 years ago
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