Answer:
525 Bq
Explanation:
The decay rate is directly proportional to the amount of radioisotope, so we can use the half-life equation:
A = A₀ (½)^(t / T)
A is the final amount
A₀ is the initial amount,
t is the time,
T is the half life
A = (8400 Bq) (½)^(18.0 min / 4.50 min)
A = (8400 Bq) (½)^4
A = (8400 Bq) (1/16)
A = 525 Bq
Answer:
<h3>The answer is 80 N</h3>
Explanation:
The force acting on the object can be found by using the formula
![f = \frac{w}{d} \\](https://tex.z-dn.net/?f=f%20%3D%20%20%5Cfrac%7Bw%7D%7Bd%7D%20%20%5C%5C%20)
where
d is the distance
w is the work done
We have
![f = \frac{3600}{45} \\](https://tex.z-dn.net/?f=f%20%3D%20%20%5Cfrac%7B3600%7D%7B45%7D%20%20%5C%5C%20)
We have the final answer as
<h3>80 N</h3>
Hope this helps you
Answer:
Explanation:
Given
mass of drop ![m=2 kg](https://tex.z-dn.net/?f=m%3D2%20kg)
height of fall ![h=1 m](https://tex.z-dn.net/?f=h%3D1%20m)
ball leaves the foot with a speed of 18 m/s at an angle of ![55^{\circ}](https://tex.z-dn.net/?f=55%5E%7B%5Ccirc%7D)
Velocity of ball just before the collision with the floor
![u^=2gh](https://tex.z-dn.net/?f=u%5E%3D2gh)
![u=\sqrt{2gh}](https://tex.z-dn.net/?f=u%3D%5Csqrt%7B2gh%7D)
![u=\sqrt{2\times 9.8\times 1}=4.42 m/s](https://tex.z-dn.net/?f=u%3D%5Csqrt%7B2%5Ctimes%209.8%5Ctimes%201%7D%3D4.42%20m%2Fs)
Impulse delivered in Y direction
![J_y=m(v\sin (55)-(-u))](https://tex.z-dn.net/?f=J_y%3Dm%28v%5Csin%20%2855%29-%28-u%29%29)
![J_y=2(18\sin (55)+4.42)](https://tex.z-dn.net/?f=J_y%3D2%2818%5Csin%20%2855%29%2B4.42%29)
![J_y=38.32 kg-m/s](https://tex.z-dn.net/?f=J_y%3D38.32%20kg-m%2Fs)
Impulse in x direction
![J_x=m\times v\cos (55)](https://tex.z-dn.net/?f=J_x%3Dm%5Ctimes%20v%5Ccos%20%2855%29)
![J_x=2\times 4.42\cos (55)=20.646](https://tex.z-dn.net/?f=J_x%3D2%5Ctimes%204.42%5Ccos%20%2855%29%3D20.646)
![J_{net}=\sqrt{J_x^2+J_y^2}](https://tex.z-dn.net/?f=J_%7Bnet%7D%3D%5Csqrt%7BJ_x%5E2%2BJ_y%5E2%7D)
![J_{net}=\sqrt{(38.32)^2+(20.64)^2}](https://tex.z-dn.net/?f=J_%7Bnet%7D%3D%5Csqrt%7B%2838.32%29%5E2%2B%2820.64%29%5E2%7D)
![J_{net}=43.52 kg-m/s](https://tex.z-dn.net/?f=J_%7Bnet%7D%3D43.52%20kg-m%2Fs)
at an angle of ![\tan \phi =\frac{J_y}{J_x}=\frac{38.32}{20.64}](https://tex.z-dn.net/?f=%5Ctan%20%5Cphi%20%3D%5Cfrac%7BJ_y%7D%7BJ_x%7D%3D%5Cfrac%7B38.32%7D%7B20.64%7D)
![\phi =tan^{-1}(1.856)](https://tex.z-dn.net/?f=%5Cphi%20%3Dtan%5E%7B-1%7D%281.856%29)
Fluorine (F) has higher potential energy as Neon (N) is a noble/inert gas.
Fluorine will lose another electron to gain stability.