`V = k((P)/(T))^(0.33)` where k is constant. It is an,

For a perfectly crystalline solid C_(p.m) = aT^(3) , where a is constant. If C_(p.m) is 0.42 J/K - mol at 10K, molar entropy at 10K is

For a perfectly crystalline solid C_(p.m) = aT^(3) , where a is constant. If C_(p.m) is 0.42 J/K - mol at 10K, molar entropy at 10K is

The temperature T of a cooling object drops at a rate proportional to the difference T-S, where S is the constant temperature of the surrounding medium. Thus, (dT)/(dt)=-k(T-S) , where k(>0) is constant and t is the time. Solve the differntial equation if it is given that T(0)=150.

For a perfectly crystaline solid C_(p.m)=aT^(3) , where a is constant. If C_(p.m) is 0.42J//K-mol at 10 K , molar entropy at 10 K is:

The acceleration of a particle travelling along a straight line is a =(k)/(v) , where k is constant (k gt 0) . If at t=0,V=V_(0) then speed of particle at t = (V_(0)^(2))/(2) .

The acceleration of a particle travelling along a straight line is a =(k)/(v) , where k is constant (k gt 0) . If at t=0,V=V_(0) then speed of particle at t = (V_(0)^(2))/(2) .

If f(xy)=f(x)f(y), then f(t) may be of the form (a) t+k( b) ct+K(C)t^(k)+C(d)t^(k) where k is a constant.

Let I be the purchase value of an equipment and V(t) be the value after it has been used for t years. The value V(t) depreciates at a rate given by differential equation (d V(t)/(dt)=-k(T-t) , where k"">""0 is a constant and T is the total life in years of the equipment. Then the scrap value V(T) of the equipment is : (1) I-(k T^2)/2 (2) I-(k(T-t)^2)/2 (3) e^(-k T) (4) T^2-1/k