Answer:
The y-axis should be labelled as W in Newtons (kg·m/s²)
Explanation:
The given data is presented here as follows;
Mass (kg)
Newtons (kg·m/s²)
3.2
31.381
4.6
45.1111
6.1
59.821
7.4
72.569
9
89.241
10.4
101.989
10.9
106.892
From the table, it can be seen that there is a nearly linear relationship between the amount of Newtons and the mass, as the slope of the data has a relatively constant slope
Therefore, the data can be said to be a function of Weight in Newtons to the mass in kilograms such that the weight depends on the mass as follows;
W(m) in Newtons = Mass, m in kg × g
Where;
g is the constant of proportionality
Therefore, the y-axis component which is the dependent variable is the function, W(m) = Weight of the body while the x-axis component which is the independent variable is the mass. m
The graph of the data is created with Microsoft Excel give the slope which is the constant of proportionality, g = 9.8379, which is the acceleration due to gravity g ≈ 9.8 m/s²
We therefore label the y-axis as W in Newtons (kg·m/s²)
The Correct Answer is <u>D.Infrared/</u> <em>INFARED has a lower frequency than visible light/</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 .
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