Answer:
The value is 
Explanation:
From the question we are told that
The mass of the object is 
The unstressed length of the string is 
The length of the spring when it is at equilibrium is 
The initial speed (maximum speed)of the spring when given a downward blow 
Generally the maximum speed of the spring is mathematically represented as

Here A is maximum height above the floor (i.e the maximum amplitude)
and
is the angular frequency which is mathematically represented as

So

=> 
Gnerally the length of the compression(Here an assumption that the spring was compressed to the ground by the hammer is made) by the hammer is mathematically represented as

=> 
=> 
Generally at equilibrium position the net force acting on the spring is

=> 
=> 
So

=> 
I think the correct answer from the choices listed above is the second option. For endothermic reactions, the reactants have less energy than the products. Which would mean that energy should be added to the reaction for it to proceed. Hope this answers the question.
Microwaves and radio waves are employed in radio and satellite communications while infrared waves are used in remote controls and infrared features of new phones and other electrons. However, gamma rays have far too much energy and cause damage to the body. They are not used in communication.
The answer is A.
Answer:
5.4 J.
Explanation:
Given,
mass of the object, m = 2 Kg
initial speed, u = 5 m/s
mass of another object,m' = 3 kg
initial speed of another orbit,u' = 2 m/s
KE lost after collusion = ?
Final velocity of the system
Using conservation of momentum
m u + m'u' = (m + m') V
2 x 5 + 3 x 2 = ( 2 + 3 )V
16 = 5 V
V = 3.2 m/s
Initial KE = 
= 
= 31 J
Final KE = 
Loss in KE = 31 J - 25.6 J = 5.4 J.
Answer:
Thus the time taken is calculated as 387.69 years
Solution:
As per the question:
Half life of
= 28.5 yrs
Now,
To calculate the time, t in which the 99.99% of the release in the reactor:
By using the formula:

where
N = No. of nuclei left after time t
= No. of nuclei initially started with

(Since, 100% - 99.99% = 0.01%)
Thus

Taking log on both the sides:


t = 387.69 yrs