Here in nuclear reaction we can say that sum of neutrons and protons in reactant side and product side will be same always
Here mass number on the product side is given to us
so sum of mass number is given as
now on the reactant side also the number must be same
now we will have
Now number of protons on product side is given as
now we also know that atomic number of Fe is 26
so now we will have
now the equation is given as
Answer: The work done in J is 324
Explanation:
To calculate the amount of work done for an isothermal process is given by the equation:
W = amount of work done = ?
P = pressure = 732 torr = 0.96 atm (760torr =1atm)
= initial volume = 5.68 L
= final volume = 2.35 L
Putting values in above equation, we get:
To convert this into joules, we use the conversion factor:
So, 
The positive sign indicates the work is done on the system
Hence, the work done for the given process is 324 J
The resultant force of both forces is 15.62 N.
<h3 /><h3>What is resultant?</h3>
The Resultant of forces is a single force obtained when two or more forces are combined.
To calculate the resultant of the force, we use the formula below.
Formula:
- R = √[a²+b²-2abcos∅]..................... Equation 1
Where:
- R = Resultant of the forces.
- ∅ = Angle between both forces
From the question,
Given:
Substitute these values into equation 1
- R = √[8²+10²-2×8×10cos120°]
- R = √[64+100-160cos120°]
- R =√ [164-160(-0.5)]
- R = √[164+80]
- R = √(244)
- R = 15.62 N
Hence, the resultant force of both forces is 15.62 N.
Learn more about resultant force here: brainly.com/question/25239010
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Answer:
559.5 N
Explanation:
Applying,
v² = u²+2gs............. Equation 1
Where v = final velocity,
From the question,
Given: s = 5.10 m, u = 0 m/s ( from rest)
Constant: 9.8 m/s²
Therefore,
v² = 0²+2×9.8×5.1
v² = 99.96
v = √(99.96)
v = 9.99 m/s
As the diver eneters the water,
u = 9.99 m/s, v = 0 m/s
Given: t = 1.34 s
Apply
a = (v-u)/t
a = 9.99/1.34
a = -7.46 m/s²
F = ma.............. Equation 2
Where F = force, m = mass
Given: m = 75 kg, a = -7.46 m/s²,
F = 75(-7.46)
F = -559.5 N
Hence the average force exerted on the diver is 559.5 N
