<h3>Option B</h3><h3>The time constant of a 10 H inductor and a 200 ohm resistor connected in series is 50 millisecond</h3>
<em><u>Solution:</u></em>
Given that,
10 H inductor and a 200 ohm resistor connected in series
To find: time constant
<em><u>The time constant in seconds is given as:</u></em>
Where,
L is the inductance in henry and R is the resistance in ohms
Convert to millisecond
1 second = 1000 millisecond
0.05 second = 0.05 x 1000 = 50 millisecond
Thus time constant is 50 millisecond
Answer:
2: Condensation. 4: 1,d; 2,c; 3,a; 4,e;5,b.
Explanation:
Answer:54 kj
Explanation:P1 = P2 = 1000 kPa
1Q2 = 84 kJ
1W2 = P1 (V2 – V1)
= 1000 (0.06 – 0.03) = 30 kJ
1Q2 = 1W2 + 1U2
U2 – U1= 1Q2 – 1W2 = 84 – 30 = 54 kJ
Answer:
Explanation:
Kinetic energy gained by alpha particle
= charge x potential difference
1/2 mv² = 3.2 x 10⁻¹⁹ x 3.45 x 10⁻³
.5 x 6.68 x 10⁻²⁷ V² = 11.04 x 10⁻²²
V² = 3.3 x 10⁵
V = 5.74 x 100
= 574 m / s
Mechanical energy is conserved in respect of A , C and D .
Part B
B , C, are unknown .