<span>virtual, upright, and magnified</span>
Answer:
Velocity = 0.309 m/s
Along negative x axis
Explanation:
A pulse moving to the right along the x axis is represented by the wave function
y(x,t) = 2/ (x - 3t)² + 1
At t =0
y(x,0) = 2/ ((x - 3(0))² + 1)
=2 / (x² + 1)
At t = 1
y(x,t) = 2/ ((x - 3(1))² + 1)
= 2 /(( x - 3)² + 1)
At t = 2
y(x,t) = 2/ ((x - 3(2))² + 1)
= 2 /(( x - 6)² + 1)
For the pulse with expression y(x,t) = 4.5
²
The Velocity is
V = 2.7 / 8.73
= 0.309 m/s
Answer: (a) Z-score are 1 and -1.2 for northern and southern regions, respectively.
Explanation: <u>Z-score</u> is how many standard deviations a data is from the population mean or how far a data point is from the mean.
The z-score is calculated by the following:

where
x is the data point
μ is population mean
σ is standard deviation
For the <u>northern</u> <u>region</u> birds:
μ = 10, σ = 3, x = 13

z = 1
The z-score for birds living in the northern region is 1, which means it is 1 standard deviation <em>above the mean</em>.
For the southern region:
μ = 16, σ = 2.5, x = 13

z = -1.2
The z-score for southern living birds is -1.2, meaning it is 1.2 standard deviations <em>below the mean</em>.
Answer:
Explanation:
(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)
ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .
ΔK = 0
ΔUg is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .
ΔUg = positive
ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .
ΔUs = 0
ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence
ΔEth = negative .
b )
work done by athlete
= 400 x 2 = 800 J
energy output = 800 J
c )
It is 25% of metabolic energy output of his body
so metalic energy output of body
= 4x 800 J .
3200 J
power = energy output / time
= 3200 / 1.6
= 2000 W .
d )
1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .
2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .