Explanation:
Load=800N
Effort=200N
1. Mechanical Advantage = LOAD/EFFORT
= 800N/200N
= 4
2 Velocity Ratio = no. Of pulleys =5
3. Efficiency = Mechanical advantage / velocity ratio × 100%
= (4/5)×100%
=80%
4. output work= load×load distance
= 800N × 5m
= 4 × 1000J
5. Efficiency = (output work/input work) ×100%
Or, 80% = (4000J/input work) ×100%
Or, 80%/100% = 4000J/inputwork
Or, 4/5 = 4000J/inputwork
Or, input work =4000J × 5/4
Input work = 5×1000J
I hope it helped! ;-)
Answer:
<em>Because </em><em>of </em><em>the </em><em>given </em><em>stranded</em><em> </em><em>wires </em><em>is </em><em>that </em><em>it's </em><em>thinner </em><em>there </em><em>are </em><em>even </em><em>more </em><em>air </em><em>gaps </em><em>and </em><em>a </em><em>greater </em><em>surface</em><em> </em><em>area </em><em>in </em><em>the </em><em>individual</em><em> </em><em>stranded</em><em> wires</em><em> </em><em>then </em><em>therefore </em><em>it </em><em>carries </em><em>less </em><em>current </em><em>than </em><em>similar </em><em>solid </em><em>wires </em><em>can </em><em>with</em><em> </em><em>each</em><em> </em><em>type </em><em>of </em><em>wire </em><em>,</em><em> insulations</em><em> </em><em>technologies </em><em>can </em><em>greatly</em><em> </em><em>assist </em><em> </em><em>in </em><em>reducing</em><em> </em><em>power </em><em>dissipation</em><em>.</em>
Answer:
Explanation:
Mass of the losing player with its all equipment is given as
M = 86 kg
Net force applied on him by another player is given as
F = 780 N
also we know that acceleration of the losing player is given as
now by Newton's 2nd law we will have
Through Shannon's Theorem, we can calculate the capacity of the communications channel using the value of its bandwidth and signal-to-noise ratio. The capacity, C, can be expressed as
C = B × log₂(1 + S/N)
where B is the bandwidth of the channel and S/N is its signal-to-noise ratio.
Since the given SN ratio is in decibels, we must first express it as a ratio with no units as
SN (in decibels) = 10 × log (S/N)
30 = 10log(S/N)
log(S/N) = 3
S/N = 10³ = 1000
Now that we have S/N, we can solve for its capacity (in bits per second) as
C = 4000 × log₂(1 + 1000)
C = 39868.91 bps
Thus, the maximum capacity of the channel is 39868 bps or 40 kbps.
Answer: 40 kbps
