Explanation:
Draw a free body diagram of the toolbox. There are two forces:
Weight force mg pulling down,
and applied force F pulling up.
Sum of forces in the y direction:
∑F = ma
F − mg = ma
45 N − 15 N = (3 kg) a
a = 10 m/s²
The answer should be B. It's possible the answer key has a mistake.
To solve this problem we need the concepts of Energy fluency and Intensity from chemical elements.
The energy fluency is given by the equation
Where
The energy fluency
c = Activity of the source
r = distance
E = electric field
In the other hand we have the equation for current in materials, which is given by
Then replacing our values we have that
We can conclude in this part that 1.3*10^7Bq is the activity coming out of the cylinder.
Now the energy fluency would be,
The uncollided flux density at the outer surface of the tank nearest the source is
(a) The magnitude of the acceleration of the electron is 5.62 x 10¹³ m/s².
(b) The speed of the electron after the given time is 4.78 x 10⁵ m/s.
<h3>
Acceleration of the electron</h3>
The acceleration of the electron is calculated as follows;
F = qE
ma = qE
a = qE/m
a = (1.6 x 10⁻¹⁹ x 320)/(9.11 x 10⁻³¹)
a = 5.62 x 10¹³ m/s²
<h3>Speed of the electron</h3>
v = at
v = 5.62 x 10¹³ m/s² x 8.50 x 10⁻⁹ s
v = 4.78 x 10⁵ m/s
Learn more about speed here: brainly.com/question/4931057
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<span>P = energy/t = 0.0025/1E-8 = 250000 W
I(ave) = P/A = 250000/(pi*0.425E-3^2) = 4.4056732E11 W/m^2
I(peak) = 2I(ave) = 8.8113463E11 W/m^2
Electric field E = sqrt(I(peak)*Z0) = 1.8219499E7 V/m, where
free-space impedance Z0 = sqrt(µ0/e0) = 376.73031 ohms</span>
