Answer:
The dose absorbed by the person's body is 52600 mrem.
Explanation:
Given that,
Radiation = 52.6 rad
We need to calculate the absorbed dose
We know that,
The equivalent dose is equal to the absorbed radiation for beta source.
So, The patient is exposed 52.6 rad of radiation.
Therefore, The absorbed radiation is equal to the exposed 52.6 rad of radiation.
So, radiation absorbed = 52.6 rad
Hence, The dose absorbed by the person's body is 52600 mrem.
Answer:
a) 3.65 seconds
b) 35.87 m/s
Explanation:
s = Displacement = 65.6 m
u = Initial velocity
v = Final velocity
t = Time taken
a = Acceleration due to gravity = 9.81 m/s² (downward direction is taken as positive and upward is taken as negative)
b) Equation of motion
Initial pop up velocity is 35.87 m/s
a)
It took 3.65 seconds to reach this height
Answer
In this question we have given,
Height of plane, h1=2000m
Height at which charge concentration is 40C, h2=3000m
Height at which charge concentration is -40C, h3=1000m
charge concentaration, q1=40C
charge concentaration, q2=-40C
let the charge concentrations at height h2 and h3 as point charges
Now we will first find the electric feild on plane due to positive charge q1=40
Here k=8.98755*10^9N.m^2/C^2
q1=40C
put values of k, q1 , h1 and h2 in equation 1
\\
V/m[/tex]
similarly electric feild due to negative charge q2=-40
\\
Total electric feild E at the aircraft is given as
...............(2)
Put values of E1 and E2 in equation2
therefore s Total electric feild E at the aircraft is E= 719004V/m