Answer:
Maximum speed of the rod, v = 10.34 m/s
Explanation:
It is given that,
Voltage of the battery, V = 2.7 V
The magnetic field perpendicularly into the plane of the paper is, B = 0.9 T
Length of the rod between the rails, l = 0.29 m
Due to the motion of the rails inside the magnetic field, an emf will induced in it which is given by :
v is the speed attained by the rod
v = 10.34 m/s
So, the maximum speed attained by the rod after the switch is closed is 10.34 m/s. Hence, this is the required solution.
Answer:
6.26 m/s
Explanation:
Pretty slow.... the PE (Potential Energy) at 2m will be converted to KE (Kinetic Energy) at the bottom of the track (neglecting friction)
PE = KE
mgh = 1/2 mv^2 divide both sides of the equation by 'm'
gh = 1/2 v^2 multiply both sides by 2
2 gh = v^2 take sqrt of both sides
v = sqrt ( 2gh) = sqrt ( 2*9.81*2) = 6.26 m/s
A peak = A Rms x Sq root 2
Therefore 3.6 x sq root of 2
A peak = 5.09
<h2>
Answer: The half-life of beryllium-15 is 400 times greater than the half-life of beryllium-13.</h2>
Explanation:
The half-life of a radioactive isotope refers to its decay period, which is the average lifetime of an atom before it disintegrates.
In this case, we are given the half life of two elements:
beryllium-13:
beryllium-15:
As we can see, the half-life of beryllium-15 is greater than the half-life of beryllium-13, but how great?
We can find it out by the following expression:
Where is the amount we want to find:
Finally:
Therefore:
The half-life of beryllium-15 is <u>400 times greater than</u> the half-life of beryllium-13.