Answer:
Explanation:
brooke speed, u = 9 km/h = 2.5 m/s
Jamail speed, v = 4 km/h = 1.11 m/s
Let Booke is at (0, 0) and Jamail is at 35 m. So, at time t, the position of Brooke is (2.5 t , 0 ) and Jamail is (- 1.11 t, 35)
Let l be the distance between them.
Distance at t = 5 s
L = 39.38 m
Differentiate with respect to time
Put t = 5 s
dL/dt = 0.127 m/s
Average speed = (total distance covered) / (total time to cover the distance)
= (2,742 km) / (4.33 hours)
= (2,742 / 4.33) km/hr
= 633 km/hr (rounded)
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Answer:
k = 6.72
Explanation:
K of paper = 3.7
k of air = 1
Given that charge Q on the capacitor is constant because cell is disconnected from the circuit. So
V = Q / C = 2.5
Capacity becomes C / 3.7 in air .
capacity becomes C/3.7 when paper is replaced by air .
V₁ = Q / (C/3.7)
= 3.7 Q/C
3.7 x 2.5
= 9.25 V
In the second case ,
capacitance due to new unknown dielectric k
= C/3.7 x k
= kC / 3.7 ( Capacitance in air is C/3.7 )
V ( new ) = Q / ( kC/3.7 )
= 3.7 Q/kC
.55 x 2.5 = 3.7 x( 2.5 / k )
k = 3.7 / .55
= 6.72