In this case the rubber raft has horizontal and vertical motion.
Considering vertical motion first.
We have displacement
, u = Initial velocity, t = time taken, a = acceleration.
In vertical motion
s = 1960 m, u = 0 m/s, a = 9.81 

So raft will take 20 seconds to reach ground.
Now considering horizontal motion of raft
u = 109 m/s, t = 20 s, a = 0
So 
So shipwreck was 2180 meter far away from the plane when the raft was dropped.
We know that the Delta E + W(Work done by non-conservative
forces) = 0 (change of energy)
In here, the non-conservative force is the friction force
where f = uN (u =kinetic friction coefficient)
W= f x d = uNd ; N=mg
Delta E = 1/2 mV^2 -1/2mVi^2
umgd + 1/2mV^2 - 1/2mVi^2 = 0 (cancel out the m term)
This will then give us:
1/2Vi^2-ugd = 1/2V^2
V^2 = Vi^2 - 2ugd
So plugging in our values, will give us:
V= Sqrt (5.6^2 -2.3^2)
=sqrt (26.07)
= 5.11 m/s
The answer is letter B. XD
Answer:
4
Explanation:
In order for the current to continue flowing through the circuit (and for the bulbs to continue shining), there must be a closed path containing the battery where current can flow. Let's see the effect of removing each bulb on the circuit:
- 1: when removing bulb 1 only, the current can still flow through the path battery-bulb 3- bulb 4
- 2: when removing bulb 2 only, the current can still flow through the path battery-bulb 3- bulb 4
- 3: when removing bulb 3 only, the current can still flow through the path battery-bulb 1-bulb 2- bulb 4
- 4: when removing bulb 4 only, the current can no longer flow. In fact, there is no closed path that contains the battery now, so the current will not flow and all the bulbs will stop shining.