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weeeeeb [17]
4 years ago
11

Consider the binding energy of two stable nuclei, one with 60 nucleons and one with 200 nucleons. a. Is the total binding energy

of the nucleus with 200 nucleons more than, less than, or equal to the total binding energy of the nucleus with 60 nucleons. Justify your reasoning.
Physics
1 answer:
sashaice [31]4 years ago
4 0

Answer:

The total binding energy of the nucleus with 200 nucleons more than the total binding energy of the nucleus with 60 nucleons

Explanation:

Binding energy can be given by the formula:

Binding energy = Binding energy per nucleon * Number of nucleons

This means that if the binding energy per nucleon = x MeV

Where x is a positive real number

The nucleus with 60 nucleons will have Binding energy = 60x MeV

The nucleus with 200 nucleons will have binding energy = 200x MeV

For a +ve x, 200x > 60x

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A copper wire has radius 0.800 mm and carries current I at 20.0°C. A silver wire with radius 0.500 mm carries the same current a
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Answer:

Ecu/Eag = 0.46

Explanation:

E = PI/A

Ecu = Pcu × I/A

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Ecu/Eag = 0.026875I/π × π/0.0588I = 0.46

7 0
4 years ago
Cabbage juice turns pink in solutions with pH below 7 and green in solutions with pH above 7. You drop some cabbage juice in a s
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State the Newton's first law of motion.​
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Explanation:

  • Meaning an object at rest will not move an less someone else applies a force to it
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3 years ago
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A 2.2 kg object is whirled in a vertical circle whose radius is 1.0 m. If the time of one revolution is 0.97 s,
strojnjashka [21]

Answer:

The tension in the string at the top  =  71 N

The tension in the string at the top  =  1.1 \times 10^{-2} N

Explanation:

a) at the top  

F_{cent} = F_g +F_T ----------------------(1)

Where,

F_{cent} is the centripetal force

F_g is the gravitational force

F_T  force due to tension

From (1)

F_T = F_{cent} - F_g

F_T = \frac{ 4\pi^2 Rm}{T^2} -mg

where

R is the radius

m is the mass

T is the time taken for one revolution

g is the acceleration due to gravity

On Substituting the values

F_T = \frac{4 (3.14)^2 (1.0)}{(0.97)^2} -(2.2 \times9.8)

F_T= 70.7259N

F_T =71N

b) at the bottom

On Substituting the values

F_{cent} = -F_g +F_T

F_T = F_{cent}- F_g

F_T = \frac{ \pi^2 Rm}{T^2} +mg

F_T = \frac{4 (3.14)^2 (1.0)}{(0.97)^2} +(2.2 \times9.8)

F_T  = 113.8899N

F_T = 1.1 \times 10^{-2}

8 0
3 years ago
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