Answer:
for this problem, 2.5 = (5+2/2)-(5-2/2)erf (50×10-6m/2Dt)
It now becomes necessary to compute the diffusion coefficient at 750°C (1023 K) given that D0= 8.5 ×10-5m2/s and Qd= 202,100 J/mol.
we have D= D0exp( -Qd/RT)
=(8.5×105m2/s)exp(-202,100/8.31×1023)
= 4.03 ×10-15m2/s
Answer:
<h2>1.5 ohms</h2>
Explanation:
Power is expressed as P = V²/R
R = resistance
V = supplied voltage
Given P = 600W and V = 30V
R = V²/P
R = 30²/600
R = 900/600
R = 1.5ohms
magnitude of its resistance is 1.5ohms
Answer:
t = 4.17 [s]
Explanation:
We know that work is defined as the product of force by distance.
W = F*d
where:
F = force [N] (units of Newtons)
d = distance = 6.34 x 10⁴ [mm] = 63.4 [m]
In order to find the force, we must determine the weight of the box, the weight can be determined by means of the product of mass by gravitational acceleration.
w = m*g
where:
m = mass = 1.47 x 10⁴ [g] = 14.7 [kg]
g = gravity acceleration = 9.81 [m/s²]
w = 14.7*9.81
w = 144.2 [N]
Therefore the work can be calculated.
W = w*d
W = 144.2*63.4
W = 9142.72 [J] (units of Joules)
Power is now defined in physics as the relationship of work at a given time
P = W/t
where:
P = power = 2190 [W]
t = time [s]
Now clearing t, we have.
t = W/P
t = 9142.72/2190
t = 4.17 [s]
Coefficient of static friction = tan(a) = 0.4
r = 740 m
g = 9.8 m/s²
v = √(9.8 × 740 × 0.4) m/s
v ≈ 53.85908 m/s
Answer
given,
change in enthalpy = 51 kJ/mole
change in activation energy = 109 kJ/mole
when a reaction is catalysed change in enthalpy between the product and the reactant does not change it remain constant.
where as activation energy of the product and the reactant decreases.
example:
ΔH = 51 kJ/mole
E_a= 83 kJ/mole
here activation energy decrease whereas change in enthalpy remains same.
