A playground merry-go-round of radius R = 2.20 m has a moment of inertia I = 275 kg · m2 and is rotating at 9.0 rev/min about a frictionless vertical axle. Facing the axle, a 23.0-kg child hops onto the merry-go-round and manages to sit down on the edge. What is the new angular speed of the merry-go-round? rev/min
Answers
Answer:
6.4 rpm
Explanation:
= moment of inertia of merry-go-round = 275 kgm²
m = mass of the child = 23 kg
R = radius of the merry-go-round = 2.20 m
= moment of inertia of child after jumping on merry-go-round = mR² = (23) (2.20)² = 111.32 kgm²
Total moment of inertia after child jumps is given as
=
+
= 275 + 111.32 = 386.32 kgm²
Total moment of inertia before child jumps is given as
=
= 275 kgm²
= initial angular speed = 9 rpm
= final angular speed
using conservation of angular momentum
=
(275) (9) = (386.32)
= 6.4 rpm
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Answers
Answer:
a
b
Explanation:
From the question we are told that
The radius is
The current it carries is
The magnetic flux of the coil is mathematically represented as
Where B is the magnetic field which is mathematically represented as
Where is the magnetic field with a constant value
substituting value
The area A is mathematically evaluated as
substituting values
the magnetic flux is mathematically evaluated as
The self-inductance is evaluated as
substituting values
b) 6.67x10-19hz
c) 3x108hz
d) 1.5hz
Show calculation
Answers
Answer:
1.5 x 10¹⁸hz
Explanation:
Given parameters:
Wavelength = 2 x 10⁻¹⁰m
Unknown:
Frequency = ?
Solution:
To find the frequency, use the expression below;
V = f x wavelength
V is the speed of light = 3 x 10⁸m/s
f is the frequency
Now;
Insert the parameters
3 x 10⁸ = 2 x 10⁻¹⁰ x frequency
Wavelength = = 1.5 x 10¹⁸hz
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:
The frictional torque is
Explanation:
From the question we are told that
The mass attached to one end the string is
The mass attached to the other end of the string is
The radius of the disk is
At equilibrium the tension on the string due to the first mass is mathematically represented as
substituting values
At equilibrium the tension on the string due to the mass is mathematically represented as
The frictional torque that must be exerted is mathematically represented as
substituting values
Answer:here to earn points
Explanation:
Answers
Answer:
Explanation:
The uncertainty in energy is given by
here h is plank's constant which value is and
is the time interval which is given as
So using all the parameters the smallest possible uncertainty in electrons energy is
Answers
From the coffee cup to the thermometer
The assumption behind the science of calorimetry is that the energy gained or lost by the water is equal to the energy lost or gained by the object under study. So if an attempt is being made to determine the specific heat of fusion of ice using a coffee cup calorimeter, then the assumption is that the energy gained by the ice when melting is equal to the energy lost by the surrounding water. It is assumed that there is a heat exchange between the iceand the water in the cup and that no other objects are involved in the heat exchanged. This statement could be placed in equation form as
Qice = - Qsurroundings = -Qcalorimeter
The role of the Styrofoam in a coffee cup calorimeter is that it reduces the amount of heat exchange between the water in the coffee cup and the surrounding air. The value of a lid on the coffee cup is that it also reduces the amount of heat exchange between the water and the surrounding air. The more that these other heat exchanges are reduced, the more true that the above mathematical equation will be. Any error analysis of a calorimetry experiment must take into consideration the flow of heat from system to calorimeter to other parts of the surroundings. And any design of a calorimeter experiment must give attention to reducing the exchanges of heat between the calorimeter contents and the surroundings.
The energy calculated while dealing with the calorimeter experiment are the latent heat of vaporization, latent heat of fusion and the heat required to change the temperature of the substances.
Further Explanation:
The calorimeter works on the principle of conservation of energy. The amount of heat given by one part of the system is equal to the amount of heat gained by another part provided that the calorimeter does not loss any heat to the environment.
Consider that ice is mixed with water at some temperature. Then the water being at higher temperature losses heat to the ice at lower temperature. The ice gains the heat from the water and the system reaches an equilibrium at which the solution of ice and water has the same amount of energy at a particular temperature.
The different types of energies dealt with in the calorimetry experiment are as follows:
Latent heat of fusion:
The amount of energy required by a body when it is melted from its frozen state or freezes from its melted state is termed as the latent heat of fusion.
For example:
The small amount of ice is mixed with water in a calorimeter. Here, the ice requires the latent heat of fusion that leads to the melting of ice and converts it into water.
Latent heat of vaporization:
The amount of heat required to convert one gram of liquid to vapor without raising its temperature is known as latent heat of vaporization.
For example:
The water is boiling at in a calorimeter. Here, the water requires latent heat of vaporization which leads to the vaporization of water and convert it into vapors.
Thus, the latent heat of fusion, latent heat of vaporization and the heat required to change the temperature of the substance are the energies measured with the calorimeter.
Learn more:
1. Transnational kinetic energy brainly.com/question/9078768.
2. Expansion of gas brainly.com/question/9979757.
3. Conservation of momentum brainly.com/question/9484203.
Answer Details:
Grade: College
Subject: Physics
Chapter: Heat and Energy
Keywords:
Heat, energy, calorimeter, latent heat, vaporization, fusion, experiment, temperature, melting, boiling, liquid, vapor, evaporation, condensation, freeze.