Answer:
answer : option (b) 0.016 amp
explanation : resistance of resistor , R = 10 Ω
inductance of inductor , X_LX
L
= 20H
voltage of AC circuit , V = 120volts
frequency, ff =60Hz
so, angular frequency, \omega=2\pi fω=2πf = 2 × π × 60 = 120π rad/s
now, current , i=\frac{V}{\sqrt{R^2+\omega^2L^2}}i=
R
2
+ω
2
L
2
V
= 120/√{10² + (120π)² × 20²}
= 120/√{100 + 14400π² × 400}
after solving this we get, i = 0.016 amp
By definition we have to:
LOG (k2 / k1)=(-Ea/R)*(1/T1-1/T2)
Where,
k1 = 0.0117 s-1
K2 = 0.689 s-1
T1 = 400.0 k
T2 = 450.0 k
R is the ideal gas constant
R = 8.314 KJ / (Kmol * K)
Substituting
ln (0.0117/0.689)=-Ea/(8.314)*((1/400)-(1/450))
Clearing Ea:
Ea = 122 kJ
answer
<span> the activation energy in kilojoules for this reaction is
</span> Ea = 122 kJ
<span>
</span>
Answer:
Explanation:
We are given that
We have to find the exit temperature.
By steady energy flow equation
Substitute the values
Answer:
The acceleration of the snowball is 0.3125
Explanation:
The initial speed of the snowball up the hill, u = 0
The speed the snowball reaches, v = 5 m/s
The length of the hill, s = 40 m
The equation of motion of the snowball given the above parameters is therefore;
v² = u² + 2·a·s
Where;
a = The acceleration of the snowball
Plugging in the values, we have;
5² = 0² + 2 × a × 40
∴ 2 × 40 × a = 5² = 25
80 × a = 25
a = 25/80 = 5/16
a = The acceleration of the snowball = 5/16 m/s².
The acceleration of the snowball = 5/16 m/s² = 0.3125 m/s² .
