Force = Work/distance
Force = 150/10
= 15 Newtons
Force = 15 Newtons
Therefore, 15 newtons of force is applied to the body when 150 joules of work
is done in displacing the body through a distance of 10m in the direction of the force.
Answer : The partial pressure of
is, 67.009 atm
Solution : Given,
Partial pressure of
at equilibrium = 30.6 atm
Partial pressure of
at equilibrium = 13.9 atm
Equilibrium constant = 
The given balanced equilibrium reaction is,

The expression of
will be,

Now put all the values of partial pressure, we get


Therefore, the partial pressure of
is, 67.009 atm
Then report it and it might be taken down
Answer:
<em>The period of the motion will still be equal to T.</em>
<em></em>
Explanation:
for a system with mass = M
attached to a massless spring.
If the system is set in motion with an amplitude (distance from equilibrium position) A
and has period T
The equation for the period T is given as

where k is the spring constant
If the amplitude is doubled, the distance from equilibrium position to the displacement is doubled.
Increasing the amplitude also increases the restoring force. An increase in the restoring force means the mass is now accelerated to cover more distance in the same period, so the restoring force cancels the effect of the increase in amplitude. Hence, <em>increasing the amplitude has no effect on the period of the mass and spring system.</em>