Explanation:
The given data is as follows.
mass, m = 75 g
Specific heat of water = 4.18
First, we will calculate the heat required for water is as follows.
q = 
= 
= 8464.5 J/mol
= 8.46 kJ ......... (1)
Also, it is given that 
= (20 + 273) K = 293 K and specific heat of ice is 2.108 kJ/kg K.
Now, we will calculate the heat of fusion as follows.
q = 
= 
= -46.32 kJ ......... (2)
Now, adding both equations (1) and (2) as follows.
8.46 kJ - 46.32 kJ
= -37.86 kJ
Therefore, we can conclude that energy in the form of heat (in kJ) required to change 75.0 g of liquid water at 
to ice at 
is -37.86 kJ.
Answer:
Explanation:
The centripetal acceleration is given by:
Here v is the linear speed and r is the radius of the circular motion. v is defined as the distance traveled to make one revolution (
) divided into the time takes to make one revolution, that is, the period (T).
Replacing (2) in (1) and replacing the given values:
Newtons second law.. <span>The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.</span>
Answer:
a. The total momentum of the trolleys which are at rest before the separation is zero
b. The total momentum of the trolleys after separation is zero
c. The momentum of the 2 kg trolley after separation is 12 kg·m/s
d. The momentum of the 3 kg trolley is -12 kg·m/s
e. The velocity of the 3 kg trolley = -4 m/s
Explanation:
a. The total momentum of the trolleys which are at rest before the separation is zero
b. By the principle of the conservation of linear momentum, the total momentum of the trolleys after separation = The total momentum of the trolleys before separation = 0
c. The momentum of the 2 kg trolley after separation = Mass × Velocity = 2 kg × 6 m/s = 12 kg·m/s
d. Given that the total momentum of the trolleys after separation is zero, the momentum of the 3 kg trolley is equal and opposite to the momentum of the 2 kg trolley = -12 kg·m/s
e. The momentum of the 3 kg trolley = Mass of the 3 kg Trolley × Velocity of the 3 kg trolley
∴ The momentum of the 3 kg trolley = 3 kg × Velocity of the 3 kg trolley = -12 kg·m/s
The velocity of the 3 kg trolley = -12 kg·m/s/(3 kg) = -4 m/s
Answer:
i(t) = (E/R)[1 - exp(-Rt/L)]
Explanation:
E−vR−vL=0
E− iR− Ldi/dt = 0
E− iR = Ldi/dt
Separating te variables,
dt/L = di/(E - iR)
Let x = E - iR, so dx = -Rdi and di = -dx/R substituting for x and di we have
dt/L = -dx/Rx
-Rdt/L = dx/x
interating both sides, we have
∫-Rdt/L = ∫dx/x
-Rt/L + C = ㏑x
x = exp(-Rt/L + C)
x = exp(-Rt/L)exp(C) A = exp(C) we have
x = Aexp(-Rt/L) Substituting x = E - iR we have
E - iR = Aexp(-Rt/L) when t = 0, i(0) = 0. So
E - i(0)R = Aexp(-R×0/L)
E - 0 = Aexp(0) = A × 1
E = A
So,
E - i(t)R = Eexp(-Rt/L)
i(t)R = E - Eexp(-Rt/L)
i(t)R = E(1 - exp(-Rt/L))
i(t) = (E/R)(1 - exp(-Rt/L))
