Answer:
q_poly = 14.55 KJ/kg
Explanation:
Given:
Initial State:
P_i = 550 KPa
T_i = 400 K
Final State:
T_f = 350 K
Constants:
R = 0.189 KJ/kgK
k = 1.289 = c_p / c_v
n = 1.2 (poly-tropic index)
Find:
Determine the heat transfer per kg in the process.
Solution:
-The heat transfer per kg of poly-tropic process is given by the expression:
q_poly = w_poly*(k - n)/(k-1)
- Evaluate w_poly:
w_poly = R*(T_f - T_i)/(1-n)
w_poly = 0.189*(350 - 400)/(1-1.2)
w_poly = 47.25 KJ/kg
-Hence,
q_poly = 47.25*(1.289 - 1.2)/(1.289-1)
q_poly = 14.55 KJ/kg
Answer: a) 3.85 days
b) 10.54 days
Explanation:-
Expression for rate law for first order kinetics is given by:
where,
k = rate constant = ?
t = time taken for decomposition = 3 days
a = let initial amount of the reactant = 100 g
a - x = amount left after decay process =
First we have to calculate the rate constant, we use the formula :
Now put all the given values in above equation, we get
a) Half-life of radon-222:
Thus half-life of radon-222 is 3.85 days.
b) Time taken for the sample to decay to 15% of its original amount:
where,
k = rate constant =
t = time taken for decomposition = ?
a = let initial amount of the reactant = 100 g
a - x = amount left after decay process =
Thus it will take 10.54 days for the sample to decay to 15% of its original amount.
<h2>~<u>Solution</u> :-</h2>
- Here, the <u>moment arm</u> is defined as follows;
The magnitude of two forces, which when acting at right angle produce resultant force of VlOkg and when acting at 60° produce resultant of Vl3 kg. These forces are D. gravitational force of attraction towards the centre of the earth. A sample of metal weighs 219 gms in air, 180 gms in water, 120 gms in an <em>unknown fluid</em>.
Answer:
false statement : b ) For the motion of a cart on an incline plane having a coefficient of kinetic friction of 0.5, the magnitude of the change in kinetic energy equals the magnitude of the change in gravitational potential energy
Explanation:
mechanical energy = potential energy + kinetic energy = constant
differentiating both side
Δ potential energy + Δ kinetic energy = 0
Δ potential energy = - Δ kinetic energy
first statement is true.
Friction is a non conservative force so inter-conversion of potential and kinetic energy is not possible in that case. In case of second option, the correct relation is as follows
change in gravitational potential energy = change in kinetic energy + work done against friction .
So given 2 nd option is incorrect.
In case of no change in gravitational energy , work done is equal to
change in kinetic energy.