Answer:
Without this slack, a locomotive might simply sit still and spin its wheels. The loose coupling enables a longer time for the entire train to gain momentum, requiring less force of the locomotive wheels against the track. In this way, the overall required impulse is broken into a series of smaller impulses. (This loose coupling can be very important for braking as well).
Explanation:
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
26 m/s for fifteen seconds. distance = rate times time, so distance = 26 m/s * 15 seconds. this gives you distance = 390 meters.
Answer:
d = 69 .57 meter
Explanation:
First case
Speed of car ( v ) = 20.5 mi/h = 9.164 M/S
distance ( d ) = 11.6 meter ( m = mass of the car )
Work done = 0.5 m v² = 0.5 * 9.164² * m J = 41.99 m J
Force = ( workdone /distance ) = ( 41.99 m / 11.6 ) = 3.619 m N
Second case
v = 50.2 mi/h = 22.44135 m/s
d = ?
Work done = 0.5 * 22.44² * m J = 251.7768 * m J
Since the braking force remains the same .
3.619 m = ( 251.7768 m / d )
d = 69 .57 meter
