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 .
Positive, negative, and zero terms in the energy equation. Calculation of energy output and metabolic power. Comparison of metabolic energy and power for different time durations.
To apply the energy equation to the system, we need to determine whether each term is positive, negative, or zero:
To determine the energy output by the athlete, we can calculate the work done on the barbell using the formula W = ΔUg. In this case, the work done is equal to the change in gravitational potential energy, which is equal to mgh. Thus, W = 400 N × 2.0 m = 800 J. So the energy output by the athlete is 800 J.
The metabolic power can be calculated using the equation P = W / t, where P is the power, W is the work done, and t is the time taken. Substituting the given values, P = 800 J / 1.6 s = 500 W. Therefore, the metabolic power of the athlete is 500 W. If the task is performed in a faster time, the metabolic energy expended will be the same. However, the metabolic power will be greater as the work is done in less time.
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Answer:
The position of the arrows will not be on the target i.e. outside the bull's eye, neither will they be close to one another (widely scattered).
Explanation:
Accuracy refers to the closeness of a measurement to an actual or accepted value while precision refers to the closeness of measurements to one another.
Using archery as an illustration of precision and accuracy, measurements (arrows) that are neither accurate not precise are those arrows that will be far away or outside the bull's eye region (target) of the board and also far apart from one another.
In a nutshell, the arrows will be distant from the bull's eye or target (not accurate) and also distant from one another (not precise).
The average wavelength of radio waves ranges from roughly two millimeters to more than 150 kilometers. The wavelengths of radio waves are the longest in the electromagnetic spectrum
It can be understood in terms of the distance between any two similar successive points across any wave for example wavelength can be calculated by measuring the distance between any two successive crests.
It is the total length of the wave for which it completes one cycle.
The wavelength is inversely proportional to the frequency of the wave as from the following relation.
C = νλ
They also have the lowest frequencies, ranging from around 4,000 cycles per second, or 3 kilohertz, to roughly 280 billion hertz, or 280 gigahertz.
The wavelengths of radio waves are the longest in the electromagnetic spectrum, ranging from roughly two millimeters to more than 150 kilometers.
To learn more about wavelength from here, refer to the link given below;
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Answer:
Radio waves have frequencies as high as 300 gigahertz(GHz)to as low as 30 hertz(Hz).At 300 GHz the corresponding wavelength is 1mm and at 30Hz is 10,000 km
Answer
3.340J
Explanation;
Using the relation. Energy stored in capacitor = U = 7.72 J
U =(1/2)CV^2
C =(eo)A/d
C*d=(eo)A=constant
C2d2=C1d1
C2=C1d1/d2
The separation between the plates is 3.30mm . The separation is decreased to 1.45 mm.
Initial separation between the plates =d1= 3.30mm .
Final separation = d2 = 1.45 mm
(A) if the capacitor was disconnected from the potential source before the separation of the plates was changed, charge 'q' remains same
Energy=U =(1/2)q^2/C
U2C2 = U1C1
U2 =U1C1 /C2
U2 =U1d2/d1
Final energy = Uf = initial energy *d2/d1
Final energy = Uf =7.72*1.45/3.30
(A) Final energy = Uf = 3.340J
Answer:
Check attachment for better understanding
Explanation:
Given that,
Current in wire I =2.2A
Capacitor plate dimension is 2cm by 2cm
s=2cm=2/100 = 0.02m
Rate at which electric field Is changing dE/dt?
The current in the wires must also be the displacement current in the capacitor. We find the rate at which the electric field is changing from
ID = ε0•A•dE/dt
Where ε0 is a constant
ε0= 8.85×10^-12C²/Nm²
Area of the square plate is
A =s² =0.02² = 0.0004m²
Then,
Make dE/dt the subject of formula
dE/dt = ID/ε0A
dE/dt = 2.2 / (8.85×10^-12 ×4×10^-4)
dE/dt = 6.215×10^14 V/m-s
Or
dE/dt = 6.215×10^14 N/C.s
The rate at which the electric field is changing between the plates is 6.215×10^14 N/C.s
Answer:
0.557 s
Explanation:
Given:
v₀ = 5.46 m/s
v = 0 m/s
a = -9.8 m/s²
Find: t
v = at + v₀
0 m/s = (-9.8 m/s²) t + 5.46 m/s
t = 0.557 s