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KiRa [710]
3 years ago
13

Which nuclear emission has the greatest penetrating power

Physics
1 answer:
velikii [3]3 years ago
3 0

Answer:

<em><u>Gamma γ radiations has the highest penetrating power,</u></em> of all the nuclear emissions.

Explanation:

The three major types of nuclear emissions are <em>alpha (α), beta (β)</em> and <em>gamma </em>radiations. Of all three, gamma (Υ) radiations have the highest penetrating power (only shielded by lead and concrete), owing to its lack of mass and charge.

It also has a low ionization power and so does not interact well with matter. IT has vast application in cancer treatment and sterilization mechanism.

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What unit can be used to measure the mass of a grain of salt
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milligrams

Explanation:

The best unit to measure the mass of a grain of salt is the milligram.

This is about; a thousandth of a gram.

  A grain of salt is a very small particle size that can still be visible with the eye.

It has a very low and small mass.

For substances like this, we use the milligram:

             1000milligram  = 1g

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Radio devices are used for communication in space. Why
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What principle of light’s behavior can be used to make a far away object appear closer and also make a small object appear large
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What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor
const2013 [10]

Complete Question

An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.

What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.

Answer:

The torque is  \tau = 34.3 \  N\cdot m

Explanation:

From the question we are told that

   The mass of the steel ball is  m  =  3.0 \  kg

    The length of arm is  l =  70 \ cm  = 0.7 \  m

    The mass of the arm is m_a  = 4.0 \  kg

Given that the arm of the athlete is uniform them the distance from the shoulder to the center of gravity of the arm is mathematically represented as

       r = \frac{l}{2}

=>    r = \frac{ 0.7}{2}  

=>    r = 0.35 \ m  

Generally the magnitude of torque about the athlete shoulder is mathematically represented as

      \tau =  m_a * g * r  + m * g *  L

=>    \tau =  4 * 9.8 * 0.35 + 3 * 9.8 *  0.70

=>    \tau = 34.3 \  N\cdot m

5 0
2 years ago
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