This occurs when a person comes into contact with hazard
1 answer:
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Answer:
(A). The resulting nucleus is Fr.
(B). The resulting nucleus is Po.
(C). The resulting nucleus is Ne
(D). The resulting nucleus is Tc.
Explanation:
Given that,
A nucleus undergoes a nuclear decay.
(A). In alpha decay,
We know that,
When the nucleus emit alpha particle then atomic mass of particle reduce by 4 and atomic number reduce by 2.
We need to calculate the resulting nucleus
Using given data
The resulting nucleus is Fr.
(B). In beta-minus decay,
We know that,
When the nucleus emit beta- minus particle then atomic mass of particle is same and atomic number increase by 1.
We need to calculate the resulting nucleus
Using given data
The resulting nucleus is Po.
(C). In beta-plus decay,
We know that,
When the nucleus emit beta- plus particle then atomic mass of particle is same and atomic number decrease by 1.
We need to calculate the resulting nucleus
Using given data
The resulting nucleus is Ne.
(D). In gamma decay,
We know that,
When the nucleus emit gamma particle then atomic mass and atomic number of particle is same.
We need to calculate the resulting nucleus
Using given data
The resulting nucleus is Tc.
Hence, (A). The resulting nucleus is Fr.
(B). The resulting nucleus is Po.
(C). The resulting nucleus is Ne
(D). The resulting nucleus is Tc.
Answer:
Diagrams in pictures
Explanation:
Using energy I can get
m g h = 1/2 m v^2
So the velocity at the end of the ramp is the squareroot of two times the initial height of the box times the gravity constant.
(H= 1,3m sin25)
V=2,32m/s
V= a t
And
X= v t +1/2 a t^2
Knowing v=2,32 m/s and x= 1,3 m
I can get
a= 6,21m/s2
F= m a
I can get the force of the box when it collides with the spring
F= 12, 425 N
The force the spring makes on the box then is
F = -12,425N = -k d
Then the spring's constant is k= 51,75N/m
To make the two diagrams I need the functions of time when the box slows down
I use the same two equations
V= a t
And
X= v t + 1/2 a t^2
Being now 2,32 my initial velocity and 0 my final velocity, and my distance 0,24 m.
I get there the time t=0,0689 seconds and the acceleration a= -33,67 m/s2 (negative because it's slowing down).
Then,
V(t)= - 33,67 m/s2 t for time between 0 and 0,689 sec
X(t)= 2,32 m/s t + 1/2 33,67 m/s2 t^2.
for time between 0 and 0,689 sec
Diagrams and equations are in the pictures
