The amount left of a radioactive sample amount N0 if the decay constant is 0.00125 seconds and the time is 180 seconds is 0.7999 N.
<h3>What is half-life?</h3>
The time it takes for half of the original population of radioactive atoms to decay is called the half-life. The relationship between the half-life T1/2 and the decay constant is given by T1/2 = 0.693/λ.
- N=N0e−λt
- given λ = 0.00125 seconds
- t = 180 seconds
- Now putting values.
- N=N0e−λt = 0.799
- N= 0.7999.
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Answer:
(a) Workdone = -27601.9J
(b) Average required power = 1314.4W
Explanation:
Mass of hoop,m =40kg
Radius of hoop, r=0.810m
Initial angular velocity Winitial=438rev/min
Wfinal=0
t= 21.0s
Rotation inertia of the hoop around its central axis I= mr²
I= 40 ×0.810²
I=26.24kg.m²
The change in kinetic energy =K. E final - K. E initail
Change in K. E =1/2I(Wfinal² -Winitial²)
Change in K. E = 1/2 ×26.24[0-(438×2π/60)²]
Change in K. E= -27601.9J
(a) Change in Kinetic energy = Workdone
W= 27601.9J( since work is done on hook)
(b) average required power = W/t
=27601.9/21 =1314.4W
Answer:
9.916\times 10^{-4}T
Explanation:
Diameter of toroid is 15 cm =0.15 m
So length of the toroid is 
Number of turns of the toriod is given as 555
Current through the toroid is 0.670 A
Magnetic field 
Answer: 
Explanation:
Given
Magnitude of charge is 
Force experienced is 
Electric field intensity is the electrostatic force per unit charge
Thus, the electric field intensity is
True. It would be false if the statement was "trunk rotation is the most common <em>static</em> flexibility assessment."
So, you're answer should be "true". Hope that helped!
