<span><span>Use the periodic table and your knowledge of isotopes to complete these statements.
When polonium-210 emits an alpha particle, the child isotope has an atomic mass of </span><span> ⇒ 206</span>.</span>
<span><span>I-131 undergoes beta-minus decay. The chemical symbol for the new element is </span><span> ⇒ Xe</span>.</span>
<span><span>Fluorine-18 undergoes beta-plus decay. The child isotope has an atomic mass of </span><span> ⇒ 18</span>.</span>
Answer:
0.24 kgm²
Explanation:
= length of the bat = 81.3 cm = 0.813 m
= mass of the bat = 0.96 kg
= distance of the center of mass of bat from the axis of rotation = 55.9 cm = 0.559 m
= Period of oscillation = 1.35 sec
= moment of inertia of the bat
Period of oscillation is given as


= 0.24 kgm²
Answer:
Electric field, E = 45.19 N/C
Explanation:
It is given that,
Surface charge density of first surface, 
Surface charge density of second surface, 
The electric field at a point between the two surfaces is given by :



E = 45.19 N/C
So, the magnitude of the electric field at a point between the two surfaces is 45.19 N/C. Hence, this is the required solution.
The speed of the ball is always zero and the acceleration is always -g when it reaches the top of its motion. This is because when the ball is free, only gravity acts on it which is always downwards, hence g is the net acceleration and it is always negative. However the velocity does not direction change instantly, negative acceleration first slows down the ball with a positive velocity, until that point the ball keeps moving up, then the ball velocity becomes zero just before changing direction and becoming negative after which the ball will now go down along gravity. Hence the ball velocity is zero at the top (neither going up nor down). Mathematically this can be seen as velocity is the integration of acceleration.
<span>v =Integral(F/m)dtv=Integral(F/m)dt
=â«50(6t2â’4t+3)/5dt=â«05(6t2â’4t+3)/5dt
=1/5â—(2t3â’2t2+3t)|(0,5)=1/5â—(2t3â’2t2+3t)|(0,5)
=1/5â—(250â’50+15)=1/5â—(250â’50+15)
=215/5=215/5
=43m/s.</span>