Answer:
The speed of the wave on the longer wire is 580m/s
Explanation:
The velocity possessed by a stretched string is directly proportional to the tension in the string and inversely proportional to the mass per unit length of the string. Mathematically,
V = √T/m
Where V is the velocity of wave in the wire
T is the tension in the wire
M is the mass per unit length of the wire
Let m1 and m2 be the mass per unit length of the wires
Let T1 and T2 be their respective tensions
Since the tension and mass of the wire is the same
m1= m2= m.
T1=T2=T
Let m1 =M/l
m2 =M/4l( since the second is tour times as far apart)
V1 = 290m/s(velocity in shorter wire)
V2 is the velocity of the longer wire.
V1 = √T/(m/l)
290 = √Tl/m
290² = Tl/m... 1
V2 = √4Tl/m
V2²= 4Tl/m... 2
Dividing equation 1 by 2 we have;
290²/V2² = {Tl/m}/{4Tl/m}
290²/V2² = Tl/m × m/4Tl
290²/V2² = 1/4
Cross multiplying we have;
V2² = 290²×4
V2 = √290²×4
V2 = 290×2
V2 = 580m/s
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6.
Basically, all of these questions use F = ma
m = 80 kg
a = 1.62
Weight = F = 80*1.62 = 129.6
The closest answer is B
7.
<em><u>Step One</u></em>
Find the mass here on earth
m = F/a
F = 40 kg
a = 9.2
F = 40/9.2 = 4.34
Now take everything to the moon
F = 20 kg
a = ??
m = 4.34
a = F/m = 20/4.34 = 4.6 m/s^2
8
m = 40 kg
F = 20 N
a = ??
a = F/m = 20N/40kg = 1/2 m/s^2
Comment
All of these depend on F = m*a. None but the first one talk about vertical forces where gravity would play a part. Moving horizontally means that there is no gravitational force if there is no friction. a = 9.8 has nothing to do with the problem.
Efficiency =
(useful work you get out of it) / (work you put into it)
= (100,000 J) / (250,000 J)
= 0.4 = 40% (choice-A)
