Answers
Answer:
i would tell you but im a hobbit
Explanation:
biden 2020
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Coach ulcer paces the sidelines. Sarting at the 30 yd. line (A), he moves to the 10 yd. line (B), back to the 50 yd. line (C) and finally to the 20yd. Line (D) in 200 seconds. Determine his average speed and velocity.
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Answers
it's 9 squared divided by 6
1. Which terms (if any) are positive?
2. Which terms (if any) are negative?
3. Which terms (if any) are zero?
b) Determine the energy output by the athlete in SI unit.
c) Determine his metabolic power in SI unit.
d) Another day he performs the same task in 1.2 s.
1. Is the metabolic energy that he expends more, less, or the same?
2. Is his metabolic power more, less, or the same?
Answers
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 .
Final answer:
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.
Explanation:
To apply the energy equation to the system, we need to determine whether each term is positive, negative, or zero:
- Positive terms:
- ΔUg - the change in gravitational potential energy is positive as the barbell is lifted vertically from the ground.
- ΔUs - the change in elastic potential energy is positive if there is any stretch or compression in the system.
- ΔK - the change in kinetic energy is negative as the barbell is lifted at a constant speed, so there is no change in velocity.
- ΔEch - the change in chemical potential energy is negative if the athlete is not ingesting any food or drinks during the exercise.
- ΔEth - the change in thermal energy is zero if there is no heat transfer in the system.
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.
Learn more about Energy equation here:
#SPJ11
Answers
Answer:
Improvement in observational, and exploratory technology
Rapid increase in knowledge
International collaboration
Explanation:
Our knowledge of the solar system has increased greatly in the past few years due to to some factors which are listed below.
Improvement in observational, and exploratory technology: In recent years, developments in technology has led to the invention of advanced observational instruments and probes, that are used to study the solar system. Also more exploratory units are now developed to go out into the solar system and gather useful data which is then further processed to yield more results about our solar system.
Rapid increase in knowledge: The past few years has seen an increased number of theories proposed to explain phenomena in the solar system. Some of these theories have been seen to be accurate under experimentation, leading to newer and fresher insights into our solar system. Also, new experiments and research are carried out, all these leading to an exponential growth in our knowledge of the solar system.
International Collaboration: The sharing of knowledge by scientists all over has led to a better, quick understanding of the solar system. Also, scientists from different countries, working together on different experiment and data sharing regarding our solar system now allows our knowledge of the solar system to deepen faster.
Answers
Answer:
The magnetic field at the center of the solenoid is approximately 0.0117 T
Explanation:
Given;
length of the solenoid, L = 15 cm = 0.15 m
number of turns of the solenoid, N = 350 turns
current in the solenoid, I = 4.0 A
The magnetic field at the center of the solenoid is given by;
Therefore, the magnetic field at the center of the solenoid is approximately 0.0117 T.
Answers
Answer:
Work done will be equal to 3186.396 J
Explanation:
We have mass m = 76.2 kg
Initial velocity u = 5 m/sec
Final velocity v = 10.4 m/sec
We have to find the work done
From work energy theorem work done is equal to change in kinetic energy
w = 3168.396 J
So work done will be equal to 3186.396 J
b. With what speed does an electron exit the electron gun if its entry speed is close to zero? [Note: ignore relativity]
c. If the capacitance of the plates is 1 nF, how much charge is stored on each plate? How many extra electrons does the cathode have?
d. If you wanted to push an electron from the anode to the cathode, how much work would you have to do?
Answers
Answer:
A. 2.083 MV/m from anode to cathode.
B. 93648278.15 m/s
C. 2.5x10^-5 C and there are about 1.56x10^14 electrons
D. 4x10^-15 Joules
Explanation:
Voltage V across plate is 25 kV = 25x10^3 V
Distance apart x = 1.2 cm = 1.2x10^-2 m
A. Electric field strength is the potential difference per unit distance
E = V/x = 25x10^3/1.2x10^-2 = 2083333.3 V/m
= 2.083 MV/m
B. Energy of electron is electron charge times the voltage across
i.e eV
Charge on electron = 1.6x10^-19 C
Energy of electron = 1.6x10^-19 x 25x10^3 = 4x10^-15 Joules
Mass of electron m is 9.12x10^-31 kg
Kinetic energy of electron = 0.5mv^2
Where v is the speed
4x10^-15 = 0.5 x 9.12x10^-31 x v^2
v^2 = 8.77x10^15
v = 93648278.15 m/s
C. From Q = CV
Q = charge
C = capacitance = 1 nF 1x10^-9 F
V = voltage = 25x10^3 V
Q = 1x10^-9 x 25x10^3 = 2.5x10^-5 C
Total number of electrons = Q/e
= 2.5x10^-5/1.6x10^-19 = 1.56x10^14 electrons
D. To push electron from cathode to anode, I'll have to do a work of about
4x10^-15 Joules