Answer:
Lenz's law, in electromagnetism, statement that an induced electric current flows in a direction such that the current opposes the change that induced it. This law was deduced in 1834 by the Russian physicist Heinrich Friedrich Emil Lenz (1804–65).
Answer: it depends on the mass of the pendulum or on the size of the arc through which it swings.
Explanation:
Answer:
a) 
b) 
Explanation:
Searching the missed information we have:
E: is the energy emitted in the plutonium decay = 8.40x10⁻¹³ J
m(⁴He): is the mass of the helium nucleus = 6.68x10⁻²⁷ kg
m(²³⁵U): is the mass of the helium U-235 nucleus = 3.92x10⁻²⁵ kg
a) We can find the velocities of the two nuclei by conservation of linear momentum and kinetic energy:
Linear momentum:
Since the plutonium nucleus is originally at rest, 
:
(1)
Kinetic Energy:
(2)
By entering equation (1) into (2) we have:
Solving the above equation for 
we have:
And by entering that value into equation (1):
The minus sign means that the helium-4 nucleus is moving in the opposite direction to the uranium-235 nucleus.
b) Now, the kinetic energy of each nucleus is:
For He-4:
For U-235:
I hope it helps you!
Answer:
im not really good at explaining, but i found this website url:
https://www.numerade.com/questions/a-cyclist-travels-from-point-a-to-point-b-in-10-min-during-the-first-20-min-of-her-trip-she-maintain/
same question just with the explanation
Answer:
54%
Explanation:
So, we have that the "magnitude of its displacement from equilibrium is greater than (0.66)A—''. Thus, the first step to take in answering this question is to write out the equation showing the displacement in simple harmonic motion which is = A cos w×t.
Therefore, we will have two instances t the displacement that is to say at a point 2π/w - a2 and the second point at a = a2.
Let us say that 2π/w = A, then, we have that a = A cos ^-1 (0.66)/2π. Also, we have that a2 = A/2 - A cos^- (0.66) / 2π.
The next thing to do is to calculate or determine the total length of of the required time. Thus, the total length is given as:
2a1 + ( A - 2a2) = 2A{ cos^-1 (0.66)}/ π.
Therefore, the total percentage of the period does the mass lie in these regions = 100 × {2a1 + ( A - 2a2) }/A = 2 { cos^-1 (0.66)}/ π × 100 = 54%.
Thus, the total percentage of the period does the mass lie in these regions = 54%.
