The question is incomplete. The complete question is :
A plate of uniform areal density
is bounded by the four curves:




where x and y are in meters. Point
has coordinates
and
. What is the moment of inertia
of the plate about the point
?
Solution :
Given :




and
,
,
.
So,

, 



![$I=2 \int_1^2 \left( \left[ (x-1)^2y+\frac{(y+2)^3}{3}\right]_{-x^2+4x-5}^{x^2+4x+6}\right) \ dx$](https://tex.z-dn.net/?f=%24I%3D2%20%5Cint_1%5E2%20%5Cleft%28%20%5Cleft%5B%20%28x-1%29%5E2y%2B%5Cfrac%7B%28y%2B2%29%5E3%7D%7B3%7D%5Cright%5D_%7B-x%5E2%2B4x-5%7D%5E%7Bx%5E2%2B4x%2B6%7D%5Cright%29%20%5C%20dx%24)



So the moment of inertia is
.
Answer:

Explanation:
As per energy conservation we know that the electrostatic potential energy of the charge system is equal to the initial kinetic energy of the alpha particle
So here we can write it as

now we know that


z = 79
here kinetic energy of the incident alpha particle is given as

now we have

now we have

Answer:
A & D
Explanation:
A single-displacement reaction is a chemical reaction whereby one element is substituted for another one in a compound and thereby generating a new element and also a new compound as products.
From the options, only options A & D fits this definition of single-displacement reactions.
For option D: Both left and hand and right hand sides each have one element and one compound. We can see that K is substituted from KBr to join Cl to form KCl and Br2 on the right hand side.
For option A: Both left and hand and right hand sides each have one element and one compound. We can see that OH is substituted from 2H2O to join Mg to form Mg(OH)2 and H2 on the right hand side.
The other options are not correct because they don't involve only and element and a compound on each side of the reaction.
A)We know the formula of the angular speed is ω = 2π / TWhere T is the time period.When second hand completes one revolution then the time taken is 60s.So T = 60sThen the angular speed of the second hand is ω= 2π / (60s) = 0.1047 rad/sb)When the minute hand completes one revolution the time taken is T = 1 hr = 3600sThen the angular speed of the minute hand is ω =(2π) / (3600s) = 0.001745 rad/sc)When the hour hand completes one revolution then the timeperiod is T = 12hrs = (12)(3600)sThen the angular speed of the hour hand is ω =(2π) / [(12)(3600)s] = 1.45444 x 10^-4 rad/s