Answer:
Explanation:
Given that,
Three children of masses and their position on the merry go round
M1 = 22kg
M2 = 28kg
M3 = 33kg
They are all initially riding at the edge of the merry go round
Then, R1 = R2 = R3 = R = 1.7m
Mass of Merry go round is
M =105kg
Radius of Merry go round.
R = 1.7m
Angular velocity of Merry go round
ωi = 22 rpm
If M2 = 28 is moves to center of the merry go round then R2 = 0, what is the new angular velocity ωf
Using conservation of angular momentum
Initial angular momentum when all the children are at the edge of the merry go round is equal to the final angular momentum when the second child moves to the center of the merry go round
Then,
L(initial) = L(final)
Ii•ωi = If•ωf
So we need to find the initial and final moment of inertia
NOTE: merry go round is treated as a solid disk then I= ½MR²
I(initial)=½MR²+M1•R²+M2•R²+M3•R²
I(initial) = ½MR² + R²(M1 + M2 + M3)
I(initial) = ½ × 105 × 1.7² + 1.7²(22 + 28 + 33)
I(initial) = 151.725 + 1.7²(83)
I(initial) = 391.595 kgm²
Final moment of inertial when R2 =0
I(final)=½MR²+M1•R²+M2•R2²+M3•R²
Since R2 = 0
I(final) = ½MR²+ M1•R² + M3•R²
I(final) = ½MR² + (M1 + M3)• R²
I(final)=½ × 105 × 1.7² + ( 22 +33)•1.7²
I(final) = 151.725 + 158.95
I(final) = 310.675 kgm²
Now, applying the conservation of angular momentum
L(initial) = L(final)
Ii•ωi = If•ωf
391.595 × 22 = 310.675 × ωf
Then,
ωf = 391.595 × 22 / 310.675
ωf = 27.73 rpm
So, the final angular momentum is 27.73 revolution per minute
It is proposed that future space stations create an artificial gravity by rotating. Suppose a space station is constructed as a 1000-m-diameter cylinder that rotates about its axis. The inside surface is the deck of the space station. What rotation period will provide "normal" gravity?
To solve this problem, apply the concepts related to the centripetal acceleration as the equivalent of gravity, and the kinematic equations of linear motion that will relate the speed, the distance traveled and the period of the body to meet the needs given in the problem. Centripetal acceleration is defined as,
[tex]a_c = \frac{v^2}{r}[/tex]
Here,
v = Tangential Velocity
r = Radius
If we rearrange the equation to get the velocity we have,
[tex]v = \sqrt{a_c r}[/tex]
But at this case the centripetal acceleration must be equal to the gravitational at the Earth, then
[tex]v = \sqrt{gr}[/tex]
[tex]v = \sqrt{(9.8)(\frac{1000}{2})}[/tex]
[tex]v = 70m/s[/tex]
The perimeter of the cylinder would be given by,
[tex]\phi = 2\pi r[/tex]
[tex]\phi = 2\pi (500m)[/tex]
[tex]\phi = 3141.6m[/tex]
Therefore now related by kinematic equations of linear motion the speed with the distance traveled and the time we will have to
[tex]v = \frac{d}{t} \rightarrow \text{ But here } d = \phi[/tex]
[tex]v = \frac{\phi}{t} \rightarrow t = \frac{\phi}{t}[/tex]
[tex]t = \frac{3141.6}{70m/s}[/tex]
[tex]t = 44.9s[/tex]
Therefore the period will be 44.9s
Answer:
[tex]T \approx 44.88\,s[/tex]
Explanation:
"Normal" gravity is equal to 9.807 meters per squared second and cylinder must rotate at constant speed in order to simplify the equation of acceleration, which is in the radial direction. The centrifugal acceleration experimented by people allow them to be on the inside surface.
[tex]g = \omega^{2}\cdot R[/tex]
The angular speed required to provide "normal" gravity is:
[tex]\omega = \sqrt{\frac{g}{R} }[/tex]
[tex]\omega = \sqrt{\frac{9.807\,\frac{m}{s^{2}} }{500\,m} }[/tex]
[tex]\omega \approx 0.14\,\frac{rad}{s}[/tex]
The rotation period is:
[tex]T = \frac{2\pi}{\omega}[/tex]
[tex]T = \frac{2\pi}{0.14\,\frac{rad}{s} }[/tex]
[tex]T \approx 44.88\,s[/tex]
In Problem 1, the rake angle was changed to a neutral rake angle. If the friction angle remains the same, determine: (a) the shear plane angle (b) the chip thickness (c) the shear strain for the operation.
Answer: seen in the explanations
Explanation:
Since no values for this problem is given, I will just give the formulas to solve the problem in mechanics of orthogonal plane cutting.
a) shear plane angle will be calculated using:
tan ϕ = r Cos α / 1- r sin α
Where :
r= chip thickness ratio
α = rake angle
ϕ = shear angle.
B) chip thickness
To get the chip thickness, we must calculate the chip thickness ratio using the formula:
Rt = Lc/ L
Where:
Lc = length of chip formed
L= uncut chip length.
Use the answer to solve for chip thickness with the formula;
Tc = t/ Rt
Where :
t = depth of cut
Rt = chip thickness ratio
C) shear strain for the operation will be solved using;
ε = cot β• + tan β•
Where:
ε = shear strain
β• = orthogonal shear angle.
A bubble of air has a volume of 16 cm3 when at a depth of 6.5 m. As the bubble rises, it expands. What is the bubble’s volume just below the surface of the water? (Note: Assume the temperature of the air in the bubble doesn’t change, in which case pressure times volume is constant.)
Final answer:
The volume of the bubble just below the surface of the water is 16 cm3.
Explanation:
To solve this problem, we can use Boyle's law, which states that the pressure times the volume of a gas is constant if the temperature remains constant. In this case, the volume of the air bubble at the bottom of the water is 16 cm3 and the depth is 6.5 m. As the bubble rises, the pressure decreases, causing the volume to increase. We can set up a proportion to solve for the volume just below the surface of the water.
Using the formula P1V1 = P2V2, where P1 is the pressure at the bottom of the water, V1 is the initial volume, P2 is the pressure just below the surface, and V2 is the volume just below the surface, we can plug in the values and solve for V2.
Since the temperature of the air in the bubble doesn't change, we can assume that the pressure times the volume is constant. Therefore, we have:
P1V1 = P2V2
Using the given values, we have:
(1 atm)(16 cm3) = (?, just below the surface) V2
To find V2, we can rearrange the equation:
V2 = (P1V1) / P2
Since the pressure just below the surface is atmospheric (1 atm), we can substitute that value in:
V2 = (1 atm)(16 cm3) / (1 atm)
V2 = 16 cm3
Therefore, the volume of the bubble just below the surface of the water is 16 cm3.
At a given instant, a 3.9-A current flows in the wires connected to a parallel-plate capacitor. Part APart complete What is the rate at which the electric field is changing between the plates if the square plates are 1.60 cm on a side? Express your answer using two significant figures.
Answer:
[tex]1.7\cdot 10^{15} V m^{-1} s^{-1}[/tex]
Explanation:
The electric field between the plates of a parallel-plate capacitor is given by
[tex]E=\frac{V}{d}[/tex] (1)
where
V is the potential difference across the capacitor
d is the separation between the plates
The potential difference can be written as
[tex]V=\frac{Q}{C}[/tex]
where
Q is the charge stored on the plates of the capacitor
C is the capacitance
So eq(1) becomes
[tex]E=\frac{Q}{Cd}[/tex] (2)
Also, the capacitance of a parallel-plate capacitor is
[tex]C=\frac{\epsilon_0 A}{d}[/tex]
where
[tex]\epsilon_0[/tex] is the vacuum permittivity
A is the area of the plates
Substituting into (2) we get
[tex]E=\frac{Q}{\epsilon_0 A}[/tex] (3)
Here we want to find the rate of change of the electric field inside the capacitor, so
[tex]\frac{dE}{dt}[/tex]
If we calculate the derivative of expression (3), we get
[tex]\frac{dE}{dt}=\frac{1}{\epsilon_0 A}\frac{dQ}{dt}[/tex]
However, [tex]\frac{dQ}{dt}[/tex] corresponds to the definition of current,
[tex]I=\frac{dQ}{dt}[/tex]
So we have
[tex]\frac{dE}{dt}=\frac{I}{\epsilon_0 A}[/tex]
In this problem we have
I = 3.9 A is the current
[tex]A=(0.0160 m)\cdot (0.0160 m)=2.56\cdot 10^{-4} m^2[/tex] is the area of the plates
Substituting,
[tex]\frac{dE}{dt}=\frac{3.9}{(8.85\cdot 10^{-12})(2.56\cdot 10^{-4})}=1.7\cdot 10^{15} V m^{-1} s^{-1}[/tex]
The rate at which the electric field is changing in the parallel-plate capacitor is approximately 1.7 x 10¹⁴ V/m·s.
Current (I) = 3.9 ASide length of square plates (L) = 1.60 cm = 0.016 m1. The area (A) of the plates:
A = L² = (0.016 m)² = 0.000256 m²2. The displacement current (I_D) is given by:
I_D = ε₀ dΦ/dtwhere
ε₀ (the permittivity of free space) ≈ 8.85 x 10⁻¹² F/mdΦ/dt = rate of change of the electric flux.3. The electric flux (Φ) is related to the electric field (E) by:
Φ = E ASince:
dΦ/dt = A dE/dt4. Substitute this into the displacement current equation:
I_D = ε₀ A dE/dtRearranging for dE/dt:
dE/dt = I / (ε₀ A)5. Plugging in the values:
dE/dt = 3.9 A / (8.85 x 10⁻¹² F/m * 0.000256 m²) dE/dt ≈ 1.73 x 10¹⁴ V/m·sThus, the rate at which the electric field is changing is approximately 1.7 x 10¹⁴ V/m·s.
In order to determine the mass moment of inertia of a flywheel of radius 600 mm, a 12-kg block is attached to a wire that is wrapped around the flywheel. The block is released and is observed to fall 3 m in 4.6 s. To eliminate bearing friction from the computation, a second block of mass 24 kg is used and is observed to fall 3 m in 3.1 s. Assuming that the moment of the couple due to friction remains constant, determine the mass moment of inertia of the flywheel.
Answer:
Explanation:
Given that,
When Mass of block is 12kg
M = 12kg
Block falls 3m in 4.6 seconds
When the mass of block is 24kg
M = 24kg
Block falls 3m in 3.1 seconds
The radius of the wheel is 600mm
R = 600mm = 0.6m
We want to find the moment of inertia of the flywheel
Taking moment about point G.
Then,
Clockwise moment = Anticlockwise moment
ΣM_G = Σ(M_G)_eff
M•g•R - Mf = I•α + M•a•R
Relationship between angular acceleration and linear acceleration
a = αR
α = a / R
M•g•R - Mf = I•a / R + M•a•R
Case 1, when y = 3 t = 4.6s
M = 12kg
Using equation of motion
y = ut + ½at², where u = 0m/s
3 = ½a × 4.6²
3 × 2 = 4.6²a
a = 6 / 4.6²
a = 0.284 m/s²
M•g•R - Mf = I•a / R + M•a•R
12 × 9.81 × 0.6 - Mf = I × 0.284/0.6 + 12 × 0.284 × 0.6
70.632 - Mf = 0.4726•I + 2.0448
Re arrange
0.4726•I + Mf = 70.632-2.0448
0.4726•I + Mf = 68.5832 equation 1
Second case
Case 2, when y = 3 t = 3.1s
M= 24kg
Using equation of motion
y = ut + ½at², where u = 0m/s
3 = ½a × 3.1²
3 × 2 = 3.1²a
a = 6 / 3.1²
a = 0.6243 m/s²
M•g•R - Mf = I•a / R + M•a•R
24 × 9.81 × 0.6 - Mf = I × 0.6243/0.6 + 24 × 0.6243 × 0.6
141.264 - Mf = 1.0406•I + 8.99
Re arrange
1.0406•I + Mf = 141.264 - 8.99
1.0406•I + Mf = 132.274 equation 2
Solving equation 1 and 2 simultaneously
Subtract equation 1 from 2,
Then, we have
1.0406•I - 0.4726•I = 132.274 - 68.5832
0.568•I = 63.6908
I = 63.6908 / 0.568
I = 112.13 kgm²
Astronaut Jennifer's lifeline to her spaceship comes loose and she finds herself stranded, "floating" 100 m from the mothership. She suddenly throws her 2.00-kg wrench at 20 m/s in a direction away from the ship. If she and her spacesuit have a combined mass of 200 kg, how long does it take her to coast back to her spaceship
Answer:
The time it will take the astronaut to coast back to her ship is 500 seconds
Explanation:
Mass of wrench, m₁ = 2 kg
Mass of astronaut with spacesuit, m₂ = 200 kg
Velocity of wrench, v₁ = 20 m/s
Velocity of astronaut with spacesuit v₂
Distance between the astronaut and the ship is 100 m
As linear momentum is conserved
[tex]m_1v_1+m_2v_2=0\\\\\Rightarrow v_2=\frac{m_1v_1}{m_2}\\\\\Rightarrow v_2=\frac{2\times 20}{200}\\\\\Rightarrow v_2=0.2\ m/s[/tex]
[tex]Time = \frac{ Distance}{Speed}[/tex]
[tex]Time=\frac{100}{0.2}=500\ s[/tex]
Hence, The time it will take the astronaut to coast back to her ship is 500 seconds
By conserving momentum, we find that Astronaut Jennifer will move at a velocity of 0.2 m/s after throwing the wrench. It will take her 500 seconds to travel the 100 meters back to her spaceship.
Explanation:To determine the time it takes for Astronaut Jennifer to return to her spaceship, we will use the conservation of momentum principle. Since there are no external forces acting on the system in space, the momentum before and after Jennifer throws the wrench is conserved.
Let's calculate the velocity of Jennifer after she throws the wrench using the following equation:
Momentum before = Momentum after
(mass of Jennifer) × (initial velocity of Jennifer) + (mass of wrench) × (initial velocity of wrench) = (mass of Jennifer) × (final velocity of Jennifer) + (mass of wrench) × (final velocity of wrench)
Since Jennifer and the wrench start at rest, their initial velocities are 0, so:
0 + 0 = 200 kg × (final velocity of Jennifer) + 2.00 kg × (-20 m/s)
Therefore, the final velocity of Jennifer v_j is:
200 kg × v_j = -2.00 kg × (-20 m/s)
v_j = (2.00 kg × 20 m/s) / 200 kg
v_j = 0.2 m/s
Now, we need to find out how long it takes her to cover 100 meters at this velocity:
Time = Distance / Velocity
Time = 100 m / 0.2 m/s
Time = 500 s
It would take Jennifer 500 seconds to coast back to her spaceship.
A horse shoe magnet is placed on a mass balance such that a uniform magnetic field of magnitude B runs between it from North to South. A coil of resistance R is connected to a battery which supplies a potential difference of V across the coil and is suspended such that a section of the coil of length L meters lies between it with current running from East to West. The mass balance measures a mass of M. • What is the measured change in mass due to the effect of Fmag? • What is the total measured mass of the magnets? Keep in mind the effect of Newton's third law.
Answer:
(a) Measured change in mass (Δm) = BVL/Rg
(b) Total measured mass M' = M - BVL/Rg
Explanation:
Current (I) across is coil is given by the formula;
I = V/R ------------------------1
The magnetic force is given by the formula;
Fb = B*I*L -------------------2
Putting equation 1 into equation 2, we have;
Fb = B*V*L/R -------------------3
Change in mass (Δm) is given as:
Δm = Fb/g -----------------------4
Putting equation 3 into equation 4, we have;
Δm = BVL/Rg
Therefore, change in mass (Δm) = BVL/Rg
2. Since B runs from North to South and current running from East to West, then the magnetic force is directed upward.
Therefore,
Total measure mass M' = M - BVL/Rg
The roads are icy, and you observe a head-on collision on Summit, at the corner with Rhodes: a 1ton car swerves out of his lane and slides through a stop sign at 34mph straight into a 3ton SUV traveling at 13mph in the other direction. The car and the SUV crumple from the collision, and stick together. What is the final velocity, in MPH (you don't need to enter MPH in your answer) , of the SUV/car entanglement (the positive direction is the direction the car was initially going)?
Answer:
v = -1.3 mph
Explanation:
Assuming no external forces acting during the collision, total momentum must be conserved, so the following condition must be met:[tex]p_{init} = p_{final} (1)[/tex]
The initial momentum is the sum of the momenta of both vehicles, taking into account their relative velocities as they are going in opposite directions. If we take as positive the direction the car was initially going, we can write the following expression:[tex]p_{init} = m_{car} * v_{car} - m_{SUV} * v_{SUV}[/tex]
Replacing by the values of the masses of both vehicles and their speeds, we have:[tex]p_{init} = 1t*34 mph - 3t*13 mph = - 5t*mph (2)[/tex]
This must be equal to the final momentum of the car/SUV entanglement, as follows:[tex]p_{final} =( m_{car} + m_{SUV} )* v_{final} (3)[/tex]
Replacing in (3) for the masses, and equating (1) and (3), we can solve for vfinal, as follows:[tex]v_{final} = \frac{p_{init}}{(m_{car} + m_{SUV} } = \frac{-5t*mph}{4t} = -1.3 mph[/tex]
This means that the car/SUV entanglement will move in the opposite direction that the car was initially going.
Which liquid is the most viscous?
Assuming that the tungsten filament of a lightbulb is a blackbody, determine its peak wavelength if its temperature is 3 200 K.
Answer:
the peak wavelength when the temperature is 3200 K = [tex]9.05625*10^{-7} \ m[/tex]
Explanation:
Given that:
the temperature = 3200 K
By applying Wien's displacement law ,we have
[tex]\lambda _m[/tex]T = 0.2898×10⁻² m.K
The peak wavelength of the emitted radiation at this temperature is given by
[tex]\lambda _m[/tex] = [tex]\frac{0.2898*10^{-2} m.K}{3200 K}[/tex]
[tex]\lambda _m[/tex]= [tex]9.05625*10^{-7} \ m[/tex]
Hence, the peak wavelength when the temperature is 3200 K = [tex]9.05625*10^{-7} \ m[/tex]
A simple pendulum consisting of a small object of mass m attached to a string of length l has a period T.
A pendulum with what combination of object mass m and string length l will also have period T?
Any pendulum that will have the same period with mass, m, and length, L, must have the ratio of (L/g) and the ratio of its mass to force constant (m/k) must also be equal to this ratio.
For a simple pendulum, the period is given as
[tex]\bold {T = 2\pi \sqrt{\dfrac L{g}}}[/tex]
This is also given as
[tex]\bold {T = 2\pi \sqrt{\dfrac m{k}}}[/tex]
where
T = period of oscillation
m = mass of the pendulum
L = length
g = acceleration due to gravity
k = force constant
Equate these equations,
[tex]\bold {T = 2\pi \sqrt{\dfrac L{g}}} = \bold {T = 2\pi \sqrt{\dfrac m{k}}}\\\\\bold {\bold { \dfrac L{g} = \dfrac m{k}}} }\\\\\bold {\bold { \dfrac m{L} = \dfrac k{g}}} }[/tex]
So, any pendulum that will have the same period with mass, m, and length, L, must have the ratio of (L/g) and the ratio of its mass to force constant (m/k) must also be equal to this ratio.
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5. 3.4 kg of water is heated from 78 C to boiling and it is all turned to steam at
100C. How much heat did it take?
Answer:
7.7 MJ
Explanation:
Let water specific heat be c = 0.004186 J/kgC and specific latent heat of vaporization be L = 2264705 J/kg
There would be 2 kinds of heat to achieve this:
- Heat to change water temperature from 78C to boiling point:
[tex]H_1 = mc\Delta t = 3.4 * 0.004186 * (100 - 78) = 0.313 J[/tex]
- Heat to turn liquid water into steam:
[tex]H_2 = mL = 3.4 * 2264705 \approx 7.7 \times 10^6J[/tex] or 7.7 MJ
So the total heat it would take is [tex]H_1 + H_2 = 7.7 \times 10^6 + 0.313 \approx 7.7 MJ[/tex]
A physics student hurries through their lab, releasing the bob of a simple pendulum from a height, and allowing it to swing. He measures the period of the pendulum and calculates that g = 8 m/s2. What might the student have done wrong? Select all that apply. The length of the pendulum string was too long, so the equation for the period of a pendulum was no longer valid in this case. The string wasn't taut when he released the bob, causing the bob to move erratically. The student only timed one cycle, introducing a significant timing error. The angle of release was too large, so the equation for the period of a pendulum was no longer valid in this case. The mass of the bob was too large, so the equation for the period of a pendulum was no longer valid in this case. Instead of releasing the bob from rest, the student threw the bob downward.
Answer:
The string wasn't taut when he released the bob, causing the bob to move erratically.
The student only timed one cycle, introducing a significant timing error.
The angle of release was too large, so the equation for the period of a pendulum was no longer valid in this case.
Explanation:
The period of a simple pendulum is given by the formula
[tex]T=2\pi \sqrt{\frac{L}{g}}[/tex]
where
T is the period of oscillation
L is the length of the pendulum
g is the acceleration due to gravity
Therefore, it is possible to measure the value of [tex]g[/tex] in an experiment, by taking a pendulum, measuring its length L, and measuring its period of oscillation T. Re-arranging the equation above, we get the value of g as:
[tex]g=(\frac{2\pi}{T})^2 L[/tex]
Here the value of g measured in the experiment is [tex]8 m/s^2[/tex] instead of [tex]9.8 m/s^2[/tex]. Let's now analyze the different options:
The length of the pendulum string was too long, so the equation for the period of a pendulum was no longer valid in this case. --> FALSE. There is no constraint on the length of the pendulum.
The string wasn't taut when he released the bob, causing the bob to move erratically. --> TRUE. This is possible, as if the string is not taut, the pendulum would not start immediately its oscillation, so the period would be larger causing a smaller value measured for g.
The student only timed one cycle, introducing a significant timing error. --> TRUE. This is also impossible: in fact, we can get a more accurate measurement of the period if we measure several oscillations (let's say 10), and then we divide the total time by 10.
The angle of release was too large, so the equation for the period of a pendulum was no longer valid in this case. --> TRUE. The formula written above for the period of the pendulum is valid only for small angles.
The mass of the bob was too large, so the equation for the period of a pendulum was no longer valid in this case. --> FALSE. The equation that gives the period of the pendulum does not depend on the mass.
Instead of releasing the bob from rest, the student threw the bob downward. --> FALSE. In fact, this force would have been applied only at the very first moment, but then later the only force acting on the pendulum is the force of gravity, so the formula of the period would still be valid.
Answer:
The angle of release was too large, so the equation for the period of a pendulum was no longer valid in this case.
The string wasn't taut when he released the bob, causing the bob to move erratically.
The student only timed one cycle, introducing a significant timing error.
Instead of releasing the bob from rest, the student threw the bob downward.
Explanation:
The angle of release was too large, so the equation for the period of a pendulum was no longer valid in this case. your equation only works for angles > 20 degrees
The string wasn't taut when he released the bob, causing the bob to move erratically. this could lead to errors in either direction
The student only timed one cycle, introducing a significant timing error. unforseen events like wind could heavily influence one cycle.
Instead of releasing the bob from rest, the student threw the bob downward. this would artificially decrease your period (T) and therefor throw the other variables in your equation off. since L is a constant g would be forced to change.
Evaluate the solution:
A friend proposes a solution for the folowing problem.
A violin A string is 0.33 m long and has mass 0.30 x 10^-3 kg. It vibrates a fundamental frequency of 440 Hz (concert A). What is the tension in the string?
Proposed solution: Speed depends on the tension and string mass (v = [T/m]^1/2). Thus:
T = v^2 m = (340 m/s)^2 (0.30g) = 34.680 N.
a. Evaluate the solution and identify any errors.
b. Provide a corrected solution if you find errors.
Answer:
Tension, T = 105.09 N
Explanation:
Given that,
Length of the string, l = 0.33 m
Mass of the string, [tex]m=0.3\times 10^{-3}\ kg[/tex]
Fundamental frequency, f = 440 Hz
The expression for the speed in terms of tension is given by :
[tex]v=\sqrt{\dfrac{T}{(m/l)}}[/tex]
[tex]v^2=\dfrac{Tl}{m}\\\\T=\dfrac{v^2m}{l}\\\\T=\dfrac{(340)^2\times 0.3\times 10^{-3}}{0.33}\\\\T=105.09\ N[/tex]
So, the tension in the string is 105.09 N.
The unit of current, the ampere, is defined in terms of the force between currents. Two 1.0-meter-long sections of very long wires a distance 5.0 mm apart each carry a current of 1.0 AA.What is the force between them? (If the force between two actual wires has this value, the current is defined to be exactly 1 A.)
The force between two parallel conductors carrying identical currents of one ampere over a length of one meter and separated by a distance of 5 mm is 4 x 10^-5 Newtons.
Explanation:The force between two parallel conductors carrying a current is used to define the unit of current, the ampere. According to the definition, one ampere of current through each of two parallel conductors of infinite length, separated by one meter in empty space free of other magnetic fields, causes a force of exactly 2x10^-7 N/m on each conductor. But the wires in your question are shorter and closer together.
In this case, a simplified formula can be used which is F = (2 * k * I1 * I2 * L) / r, where k = 10^-7 N/A², I1 and I2 are the currents in the wires, L the length of the wires, and r is the distance between the wires. With I1 = I2 = 1 A, L = 1 m, and r = 5 mm = 0.005 m, the formula gives: F = (2 * 10^-7 N/A² * 1 A * 1 A * 1 m) / 0.005 m = 4 * 10^-5 N.
Therefore, the force between these two wires is 4 * 10^-5 Newtons.
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The force between the two sections of wire carrying a current of 1.0 A each and separated by a distance of 5.0 mm is 1x10*-9 N per meter
Explanation:The force between two 1.0-meter-long sections of very long wires carrying a current of 1.0 A each can be calculated using the definition of the ampere. According to the definition, one ampere of current through each of two parallel conductors separated by one meter in empty space causes a force of exactly 2x10-7 N/m on each conductor. In this case, the wires are separated by a distance of 5.0 mm, which is equivalent to 0.005 m. Therefore, the force between the wires is 2x10-7 N/m * 0.005 m = 1x10-9 N per meter of separation.
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A straight fin fabricated from 2024 Aluminum alloy (k=185 W/mK) has a base thickness of t=3 mm and a length of L=15 mm. Its base temperature is Tb=100oC, and it is exposed to a fluid for which T[infinity] =20oC and h=50 W/m2K. For the foregoing conditions and a fin of unit width, compare the fin heat rate, efficiency, and volume for rectangular, triangular, and parabolic profiles.
The heat rate, efficiency, and volume of a straight fin made from 2024 Aluminum alloy can be calculated using relevant formulas considering its physical dimensions, the transferred heat, and the profile of the fin. Comparisons across different profiles (rectangular, triangular, parabolic) are commonly done using numerical or graphical solution methods.
Explanation:To compare the heat rate, efficiency, and volume for a straight fin made of 2024 Aluminum alloy with a rectangular, triangular, and parabolic profile, we first need to convert all known variables into SI units. The fin has a base thickness (t) of 3 mm or 0.003 m, a length (L) of 15 mm or 0.015 m, and it's exposed to a fluid at a temperature (T infinity) of 20C. The base temperature (Tb) of the fin is 100oC, and the heat transfer coefficient (h) is 50 W/m2K.
Heat Transfer CalculationWe can estimate the heat transfer rate (Q) by applying the formula Q = hA(Tb - T infinity), where A represents the surface area of the fin which would depend on the fin's profile. For a unit width, the area of a rectangular fin is A = wt = unit width*L, for a triangular profile A = 0.5*wt, and for a parabolic profile A = (2/3)*wt.
Fin EfficiencyFin efficiency can be calculated by dividing the actual heat transferred by the fin by the maximum possible heat transfer. Since these are dependent on the fin's profile (shape), numerical or graphical solution methods are commonly used for calculations.
Fin VolumeVolume can be calculated as the product of the surface area and thickness, which again would depend on the fin's profile.
Understanding these differences among fin profiles is important in heat transfer management and finding ways to increase fin efficiency.
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Compare the fin heat rate using equation: Qfin = √(hPkA) (Tb - T∞), efficiency using equation: η = Qfin / (hAt (Tb - T∞)) , and volumes are found to be bLt, (1/2)bLt and (2/3)bL for rectangular, triangular, and parabolic profiles, respectively.
1.) Heat Transfer Analysis of a Straight Fin
To compare the fin heat rate, efficiency, and volume for different fin profiles (rectangular, triangular, and parabolic), we need to perform the following calculations:
Step 1: Identify Knowns and Convert to SI Units
Given:
Thermal conductivity, k = 185 W/mKBase thickness, t = 3 mm = 0.003 mLength, L = 15 mm = 0.015 mBase temperature, Tb = 100°CAmbient temperature, T∞ = 20°CHeat transfer coefficient, h = 50 W/m²KStep 2: Determine the Fin Parameter, m
The fin parameter, m, is calculated using the formula:
m = sqrt(hP / kA)For a rectangular fin:
P = 2(bt + L) and A = btWe adjust the perimeter and area for parabolic and triangular profiles likewise.
Step 3: Heat Transfer Rate
For each fin profile, the heat transfer rate, Qfin, is given by:
Qfin = √(hPkA) (Tb - T∞)where A is the cross-sectional area, and P is the perimeter of the respective fin.
Step 4: Fin Efficiency
Fin efficiency, η, is determined as:
η = Qfin / (hAt (Tb - T∞))where At is the total surface area of the fin.
Step 5: Compare Fin Volume
The volume, V, can be found by:
Rectangular: V = bLtTriangular: V = (1/2)bLtParabolic: V = (2/3)bLAfter calculating these values, you can compare the heat rate, efficiency, and volume between the fin profiles to determine the most efficient design.
A salad spinner is used to dry lettuce that has been recently washed. For the salad spinners to operate properly, the lettuce should be evenly distributed in the basket of the salad spinner. Otherwise, after you spin the basket and remove your hand, the salad spinner will wobble and slip on the kitchen counter.
You're curious how unbalanced your salad spinner can be before it starts to slip on the counter.
-You measure the dimensions of your salad spinner and determine that the basket has a diameter of 32.2 cm and a height of 21.0 cm.
-You place the salad spinner on a kitchen scale and determine that the mass of the salad spinner is 449 g.
-The bottom of the salad spinner is rubberized. Consulting friction coefficient tables, you estimate that the coefficient of friction between the bottom of the salad spinner and your kitchen counter is 0.83.
-You search online for information about the rotational speed of your spinner. You come across a journal article about students at Rice University who modified a salad spinner to create an inexpensive medical centrifuge to diagnose anemia in remote or low-resource areas. The article states that the maximum rotational speed of their salad spinner is 600 rpm. Your salad spinner is similar to the one featured in the article.
Suppose you place some lettuce into one side of the basket, pump the salad spinner until the basket is spinning at top speed and then remove your hand.
What is the maximum amount of lettuce that can be in the basket such that the salad spinner does not slip when you remove your hand?
The maximum amount of lettuce that dimensions can be in the basket such that the salad spinner does not slip when you remove your hand is 62.58 kg.
What is dimensions?Dimensions are the physical measurement of an object in terms of length, width, and height. They are used in various fields such as mathematics, engineering, and architecture. Dimensions are also used to describe the shape of an object, like a cube or a sphere.
The maximum amount of lettuce that can be in the basket is determined, Where Ff is the maximum static friction force (3470.54 N) and r is the radius of the salad spinner's basket (16.1 cm).
τ = 3470.54 x 0.161 = 559.21 Nm
To calculate the maximum amount of lettuce that can be in the basket, we need to calculate the rotational inertia (I) of the salad spinner
I = mr2
Where m is the mass of the salad spinner (449 g) and r is the radius of the salad spinner's basket (16.1 cm).
I = 449 x 0.1612 = 7.51 kg m2
The rotational inertia is the measure of how difficult it is to change the angular velocity of a spinning object.
α = τ / I
Where τ is the maximum rotational torque (559.21 Nm) and I is the rotational inertia (7.51 kg m2).
α = 559.21 / 7.51 = 74.37 rad/s2
The maximum amount of lettuce that can be in the basket without causing the spinner to slip is determined by the equation m = I/t, where m is the maximum mass of lettuce and I is the rotational inertia of the spinner (7.51 kg m2).
m = I/t
Where I is the rotational inertia of the spinner (7.51 kg m2) and t is the time it takes for the spinner to reach its final angular velocity (0.12 s).
m = 7.51 / 0.12 = 62.58 kg
Therefore, the maximum amount of lettuce that can be in the basket such that the salad spinner does not slip when you remove your hand is 62.58 kg.
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Two long, straight wires are separated by a distance of 9.15 cm . One wire carries a current of 2.79 A , the other carries a current of 4.36 A .Is the force per meter exerted on the 4.36-A wire greater than, less than, or the same as the force per meter exerted on the 2.79-A wire
Answer:
The force is the same
Explanation:
The force per meter exerted between two wires carrying a current is given by the formula
[tex]\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}[/tex]
where
[tex]\mu_0[/tex] is the vacuum permeability
[tex]I_1[/tex] is the current in the 1st wire
[tex]I_2[/tex] is the current in the 2nd wire
r is the separation between the wires
In this problem
[tex]I_1=2.79 A\\I_2=4.36 A\\r = 9.15 cm = 0.0915 m[/tex]
Substituting, we find the force per unit length on the two wires:
[tex]\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(2.79)(4.36)}{2\pi (0.0915)}=2.66\cdot 10^{-5}N[/tex]
However, the formula is the same for the two wires: this means that the force per meter exerted on the two wires is the same.
The same conclusion comes out from Newton's third law of motion, which states that when an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A (action-reaction). If we apply the law to this situation, we see that the force exerted by wire 1 on wire 2 is the same as the force exerted by wire 2 on wire 1 (however the direction is opposite).
20 POINTS! TRUE OR FALSE:
If a stationary box does not move when you push against it is because of kinetic friction.
A) TRUE
B) FALSE
Answer:
False.
Explanation:
Kinetic energy means it must move
Answer: I believe that it is A) True
Disclaimer: I'm not quite sure. We learned about this recently though so possibly I'm right. Good luck though!
Explanation:
A planet is discovered orbiting the star 51 Peg with a period of four days (0.01 years). 51 Peg has the same mass as the Sun. Mercury's orbital period is 0.24 years, and Venus's is 0.62 years.
The average orbital radius of this planet is:
a) less than Mercury's.
b) between Mercury's and Venus's.
c) greater than Venus's.
Answer:
a) less than Mercury's.
Explanation:
For orbital time period of a planet , the expression is
T² = 4π² R³ / GM
T is time period , R is radius of orbit , G is universal gravitational constant , M is mass of the star or sun
T² ∝ R³
As radius of orbit increases , time period increases . The given planet is making around a star whose mass is equal that of sun so M is same as sun .
The given planet has time period equal to .01 years which is less than that of Mercury and Venus , hence its R will be less than orbit of both of them or less than mercury's .
If the opening to the harbor acts just like a single-slit which diffracts the ocean waves entering it, what is the largest angle, in degrees relative to the incident direction, that a boat in the harbor would be protected from the wave action
Answer:
The angle that the wave would be [tex]\theta = sin ^{-1}\frac{2 \lambda}{D}[/tex]
Explanation:
From the question we are told that the opening to the harbor acts just like a single-slit so a boat in the harbor that at angle equal to the second diffraction minimum would be safe and the on at angle greater than the diffraction first minimum would be slightly affected
The minimum is as a result of destructive interference
And for single-slit this is mathematically represented as
[tex]D sin \ \theta =m \lambda[/tex]
where D is the slit with
[tex]\theta[/tex] is the angle relative to the original direction of the wave
m is the order of the minimum j
[tex]\lambda[/tex] is the wavelength
Now since in the question we are told to obtain the largest angle at which the boat would be safe
And the both is safe at the angle equal to the second minimum then
The the angle is evaluated as
[tex]\theta = sin ^{-1}[\frac{m\lambda}{D} ][/tex]
Since for second minimum m= 2
The equation becomes
[tex]\theta = \frac{2 \lambda}{D}[/tex]
The diagram shows two balls before they collide.
2 balls with grey arrows pointing to them from the outside. The left ball has below it m subscript 1 = 0.6 kilograms v subscript 1 = 0.5 meters per second. The right ball has below it m subscript 2 = 0.5 kilograms v subscript 2 = negative 0.2 meters per second.
What is the momentum of the system after the collision?
Answer:
B.
Explanation:
I'm taking the test right now.
The final momentum of the system is 0.4kgm/s
According to the law conservation of momentum, the momentum of the system before the collision and after the collision remains consevred.
if [tex]m_{1}[/tex] is the mass of one ball and its initial and final velocities be [tex]v_{1}[/tex] and [tex]v_{1f}[/tex] respectively, and
if [tex]m_{2}[/tex] is the mass of one ball and its initial and final velocities be [tex]v_{2}[/tex] and [tex]v_{2f}[/tex] respectively
then, [tex]m_{1}v_{1} + m_{2}v_{2}=m_{1}v_{1f} + m_{2}v_{2f}[/tex]
Momentum after collision = [tex]m_{1}v_{1} + m_{2}v_{2}[/tex][tex]=0.6*0.5+0.5*0.2[/tex]
Momentum after collision = 0.4 kgm/s
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A parallel-plate capacitor is charged until it carries charge + q +q on one plate and charge − q −q on the other plate. The capacitor is then disconnected from the power supply and isolated. What is the direction of the magnetic field that surrounds the charged capacitor? The magnetic field is directed counterclockwise as viewed from the plate with positive charge + q +q toward the plate with negative charge − q −q. There is no magnetic field. The magnetic field is directed counterclockwise as viewed from the plate with negative charge − q −q toward the plate with positive charge + q +q. The magnetic field is directed toward the plate with positive charge + q +q inside the capacitor and toward the plate with negative charge − q −q outside of the capacitor. The magnetic field is directed toward the plate with negative charge − q −q inside the capacitor and toward the plate with positive charge + q +q outside of the capacitor.
Answer:
There is no magnetic field.
Explanation:
Since th capacitor is charged and isolated, magnetic field doesn't exist.
For magnetic field to exist, there must be flow of charge.
A 63.0 \mu F capacitor is connected to a generator operating at a low frequency. The rms voltage of the generator is 4.00 V and is constant. A fuse in series with the capacitor has negligible resistance and will burn out when the rms current reaches 15.0 A. As the generator frequency is increased, at what frequency will the fuse burn out
Answer:
[tex]f=9.5\ KHz[/tex]
Explanation:
AC Circuit
When connected to an AC circuit, the capacitor acts as an impedance of module
[tex]\displaystyle Z=\frac{1}{wC}[/tex]
Where w is the angular frequency of the power source and C is the capacitance.
If the capacitor is the only element connected to a circuit, then the Ohm's law establishes that
[tex]V=Z.I[/tex]
Where V and I are the rms voltage and current respectively. Replacing the value of Z, we have
[tex]\displaystyle V=\frac{I}{wC}[/tex]
Solving for w
[tex]\displaystyle w=\frac{I}{VC}[/tex]
The question provides us the following values
[tex]C=63\ \mu F=63\cdot 10^{-6}\ F[/tex]
[tex]V=4\ Volt[/tex]
[tex]I=15\ A[/tex]
Plugging in the values
[tex]\displaystyle w=\frac{15}{4\cdot 63\cdot 10^{-6}}[/tex]
[tex]w=59523.81\ rad/s[/tex]
Since
[tex]w=2\pi f[/tex]
Then
[tex]\displaystyle f=\frac{59523.81}{2\pi}=9473.51\ Hz[/tex]
[tex]f=9.5\ KHz[/tex]
Final answer:
The fuse in the AC circuit with a 63.0 µF capacitor will burn out when the generator frequency is increased to approximately 1,001 Hz.
Explanation:
To find at what frequency the fuse will burn out in the circuit with a 63.0 µF capacitor and a 4.00 V rms voltage generator, we will need to calculate the capacitive reactance (XC) and then use the relationship between current, voltage, and reactance for an AC circuit.
The capacitive reactance is given by the formula XC = 1 / (2πfC), where f is the frequency in hertz (Hz), and C is the capacitance in farads. The rms current I in the circuit is given by the rms voltage V divided by XC:
I = V / XC
Setting I to the maximum allowable current of 15.0 A, we have:
15.0 A = 4.00 V / (1 / (2πf × 63.0 × 10⁻⁶ F))
Solving for f, we get:
f = 1 / (2π × 63.0 × 10⁻⁶ F × (4.00 V / 15.0 A))
f ≈ 1,001 Hz
Therefore, the fuse will burn out when the generator frequency is increased to approximately 1,001 Hz.
The simplest form of a traveling electromagnetic wave is a plane wave. For a wave traveling in the x direction whose electric field is in the y direction, the electric and magnetic fields are given by the picture shown above. This wave is linearly polarized in the y direction.
1. In these formulas, it is useful to understand which variables are parameters that specify the nature of the wave. The variables E_0and B_0 are the __________ of the electric and magnetic fields.
a. maximas
b. wavelenghts
c. amplitudes.
d. velocities
2. The variable omega is called the __________ of the wave.
a. wavenumber
b. wavelength
c. velocity
d. frequency
3. What is the mathematical expression for the electric field at the point x=0, y=0, z at time t?
4. For a given wave, what are the physical variables to which the wave responds?
Answer:
please the answer below
Explanation:
A general electromagnetic plane wave, traveling in the x direction, can be expressed in the form:
[tex]\vec{E}=E_0e^{-i(k\cdot x-\omega t)}\hat{j}\\\\\vec{B}=B_0e^{-i(k\cdot x-\omega t)}\hat{k}\\\\[/tex] (1)
1.
a. amplitudes
from (1) we can observe that E_0 and B_0 are the amplitudes.
2. frequency
3.
By replacing (1) we obtain:
[tex]\vec{E}=E_0e^{-i(k(0)-\omega t)}\hat{j}=E_0[cos\omega t+sin\omega t]\hat{j}[/tex]
4.
the wave respond to the followinf physical variables: amplitude, frequency, time and position, as we can see in (1).
hope this helps!!
A specimen of aluminum having a rectangular cross section 9.5 mm × 12.9 mm (0.3740 in. × 0.5079 in.) is pulled in tension with 35000 N (7868 lbf) force, producing only elastic deformation. The elastic modulus for aluminum is 69 GPa (or 10 × 106 psi). Calculate the resulting strain.
Answer:
The resultant strain in the aluminum specimen is [tex]4.14 \times {10^{ - 3}}[/tex]
Explanation:
Given that,
Dimension of specimen of aluminium, 9.5 mm × 12.9 mm
Area of cross section of aluminium specimen,
[tex]A=9.5\times 12.9=123.84\times 10^{-6}\ mm^2A=122.55\times 10^{-6}\ m^2[/tex]
Tension acting on object, T = 35000 N
The elastic modulus for aluminum is,[tex]E=69\ GPa=69\times 10^9\ Pa[/tex]
The stress acting on material is proportional to the strain. Its formula is given by :
[tex]\epsilon=\dfrac{\sigma}{E}[/tex]
[tex]\sigma[/tex] is the stress
[tex]\epsilon=\dfrac{F}{EA}\\\epsilon=\dfrac{35000}{69\times 10^9\times 122.5\times 10^{-6}}\\\epsilon=4.14\times 10^{-3}[/tex]
Thus, The resultant strain in the aluminum specimen is [tex]4.14 \times {10^{ - 3}}[/tex]
Answer:
Strain = 4.139 x 10^(-3)
Explanation:
We are given;
Dimension; 9.5 mm by 12.9 mm = 0.0095m by 0.0129m
Elastic Modulus; E = 69GPa = 69 x 10^(9)N/m²
Force = 35,000N
Now, Elastic modulus is given by;
E = σ/ε
Where
E is elastic modulus
σ is stress
ε is strain
Now, stress is given by the formula;
σ = F/A
Area = 0.0095m x 0.0129m = 0.00012255 m²
Thus, σ = 35000/0.00012255 = 285597715.218 N/m²
Now, we are looking for strain.
Let's make ε the subject;
E = σ/ε, Thus, ε = σ/E = 285597715.218/69 x 10^(9) = 0.00413909732 = 4.139 x 10^(-3)
To what temperature (in °C) must a cylindrical rod of one metal 10.083 mm in diameter and a plate of second metal having a circular hole 9.987 mm in diameter have to be heated for the rod to just fit into the hole? Assume that the initial temperature is 27°C and that the linear expansion coefficient values for metals one and two are 4.0 x 10-6 (°C)-1 and 16 x 10-6 (°C)-1, respectively.
Answer:
834°C
Explanation:
By setting the equation of final diameter of one metal and another metal equal to one another in order to determine final temperature, the equation for final diameter is given by,
[tex]d_{f}[/tex] =[tex]d_{o[/tex](1 + α([tex]t_{f}[/tex]-[tex]t_{o[/tex]))
[tex]d_{f1[/tex]=[tex]d_{f2[/tex]
9.987(1 + 16 x [tex]10^{-6}[/tex]([tex]t_{f}[/tex]-27)) = 10.083( 1 + (4 x [tex]10^{-6}[/tex])([tex]t_{f}[/tex]-27))
9.987(1 + 16 x [tex]10^{-6}[/tex][tex]t_{f}[/tex] - 432 x [tex]10^{-6}[/tex]) = 10.083( 1 + 4 x [tex]10^{-6}[/tex][tex]t_{f}[/tex] - 108 x [tex]10^{-6}[/tex])
9.987 + 1.59 x [tex]10^{-4[/tex][tex]t_{f}[/tex] - 4.31 x [tex]10^{-3[/tex] = 10.083 + 4.033 x [tex]10^{-5[/tex] [tex]t_{f}[/tex] - 1.088 x [tex]10^{-3[/tex]
1.59 x [tex]10^{-4[/tex][tex]t_{f}[/tex] - 4.033 x [tex]10^{-5[/tex] [tex]t_{f}[/tex] = 10.083-1.088 x [tex]10^{-3[/tex]- 9.987+4.31 x [tex]10^{-3[/tex]
1.187 x [tex]10^{-4[/tex][tex]t_{f}[/tex]= 0.099
[tex]t_{f}[/tex] = 0.099/ 1.187 x [tex]10^{-4[/tex]
[tex]t_{f}[/tex] = 834°C
Find the force necessary to start the crate moving, given that the mass of the crate is 32 kg and the coefficient of static friction between the crate and the floor is 0.57. Express your answer using two significant figures.
To move a 32 kg crate with a coefficient of static friction of 0.57 with the floor, the force necessary to start the crate moving is 180 N after rounding to two significant figures.
Explanation:To find the force necessary to start the crate moving, we need to use the concept of static friction. The force of static friction is given by the equation fs=μsN, where fs is the static friction, μs is the coefficient of static friction and N is the normal force. The normal force is equal to the product of the mass and gravitational acceleration (N = mg), where m is the mass and g is the acceleration due to gravity. Given that the mass of the crate is 32 kg and the coefficient of static friction is 0.57, we first calculate N = (32kg)(9.8m/s²) = 313.6 N.
Then, by substituting these values into the equation, we have fs = (0.57)(313.6 N) =178.85 N. Therefore, the force necessary to start the crate moving is 178.85 N, but we express the answer using two significant figures, so it is 180 N.
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In the formula 4h²o how many total hydrogen atoms are there
In the United States, the average person eats about 3000 food Calories per day. What is the average power of this energy intake? 14.5 watts 14.5 watts 145 watts 145 watts 1450 watts 1450 watts 14 , 500 watts
Answer:
Explanation:
3000 Calorie = 3000 x 1000 calorie
1 calorie = 4.2 Joule
3000 x 1000 calorie = 4.2 x 3000 x 1000 J
= 12.6 x 10⁶ J
It is taken in one day
time period of one day = 24 x 60 x 60 second
= 86400 s
energy given by this intake in 86400s is 12.6 x 10⁶ J
power = energy / time
= 12.6 x 10⁶ / 86400 J/s
= 145.8 J/s or W
= 145 W.
Final answer:
The average power of energy intake for a person consuming 1,300 to 3,000 kcals per day translates to approximately 63 to 145 watts. This calculation is done by converting calories to joules and then dividing by the total number of seconds in a day.
Explanation:
In order to calculate the average power of the energy intake from food, we can convert the amount of energy consumed per day into watts. For a range of 1,300 to 3,000 kcal per day, we need to convert these values to joules, since 1 kcal = 4.184 kJ. Therefore, 1,300 kcal converts to 1,300 x 4.184 kJ = 5,439.2 kJ and 3,000 kcal to 3,000 x 4.184 kJ = 12,552 kJ.
Next, we divide each amount by the number of seconds in one day to find the average power in watts. There are 86,400 seconds in a day. Thus, the power for 1,300 kcal/day is 5,439.2 kJ / 86,400 s = approximately 63 watts, and for 3,000 kcal/day is 12,552 kJ / 86,400 s = approximately 145 watts.
The power range for 1,300 to 3,000 kcal/day is thus approximately 63 watts to 145 watts.