To determine the amount of work done, we can use the work equation:
Work = Force × Displacement × cos(θ)
Work = amount of work done (in N-m or Joules),
Force is the applied force (in Newtons),
Displacement =distance moved in the direction of the force (in meters),
θ = angle between the force and the displacement.
Let's calculate the work done:
Force = 121 N
Displacement = 18 m
θ = 27.5 degrees
Work = 121 N × 18 m × cos(27.5 degrees)
Using a calculator, we can find the value of cos(27.5 degrees) ≈ 0.891.
Work = 121 N × 18 m × 0.891
Calculating the expression:
Work ≈ 2189.346 N-m
Rounding to 3 decimal places:
Work ≈ 2189.346 N-m
Therefore, the amount of work done by the 121 N force applied at an angle of 27.5 degrees to the horizontal to move the 55 kg object at a constant speed for a horizontal distance of 18 m is approximately 2189.346 N-m.
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A boutique beer brewery produces 2 types of beers, Dark-ale and
Light-ale daily with a total cost function given by T = 7 + ×
+ 5 where D is the quantity of Dark-ale beer (in
To minimize the total cost, the brewery should not produce any Dark-ale or Light-ale beer daily.
A boutique beer brewery produces two types of beers:
Dark-ale and Light-ale daily with a total cost function given by T = 7 + 5D + 6L where D is the quantity of Dark-ale beer and L is the quantity of Light-ale beer produced.
The brewery wants to determine the quantity of each type of beer to produce daily to minimize the total cost.
Let x be the quantity of Dark-ale beer and y be the quantity of Light-ale beer to produce daily, then the total cost function becomes:
T = 7 + 5xD + 6yTo minimize the total cost, we need to take the partial derivatives of T with respect to x and y and set them to zero.
Hence,dT/dx = 5d + 0 = 0 and
dT/dy = 0 + 6y
= 0
Solving for d and y respectively, we get:
d = 0y = 0
Thus, to minimize the total cost, the brewery should not produce any Dark-ale or Light-ale beer daily.
Note that this result is not practical and realistic.
Therefore, we need to find the second derivative of T with respect to x and y to verify whether the critical point (0,0) is a minimum or a maximum or a saddle point.
The second derivative test is as follows:
If d²T/dx² > 0 and dT/dx = 0, then the critical point is a minimum.
If d²T/dx² < 0 and dT/dx = 0, then the critical point is a maximum.
If d²T/dx² = 0, then the test is inconclusive and we need to try another method such as the first derivative test.To find the second derivative of T with respect to x, we differentiate dT/dx with respect to x as follows:
d²T/dx² = 5d²/dx² + 0
= 5(d²/dx²)
This shows that d²T/dx² > 0 for all values of d.
Hence, the critical point (0,0) is a minimum. Therefore, to minimize the total cost, the brewery should not produce any Dark-ale or Light-ale beer daily.
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A straight railway track is at a distance ‘d’
from you. A distant train approaches you
travelling at a speed u (< speed of sound)
and crosses you. How does the apparent
frequency (f) of the whi
A straight railway track is at a distance ‘d’ from you. A distant train approaches you traveling at a speed u (< speed of sound) and crosses you. How does the apparent frequency (f) of the which provided below When a train is moving with some speed towards a stationary observer
the observer hears the sound coming from the engine at a frequency which is greater than the actual frequency of the sound emitted by the engine. This phenomenon is called Doppler Effect. When the train is moving towards an observer, the frequency heard is greater than
the actual frequency and when the train is moving away from the observer, the frequency heard is lower than the actual frequency.
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What wavelength (in nanometers) is the peak intensity of the light coming from a star whose surface temperature is 8888 Kelvin? Calculate total energy radiated per unit area by a black body at this temperature
The peak wavelength of the light coming from a star whose surface temperature is 8888 K is given by Wien's law which states that λmaxT=2.898×10⁻³ mK.
Substituting λmax =3.28×10⁻⁷ m (1 nm = 10⁻⁹ m)
and T =8888 K,
λmaxT =2.898×10⁻³ mK
we get; λmax = (2.898 × 10⁻³)/(8888)
λmax = 3.27 × 10⁻⁷m
λmax = 327 nm.
So, the the peak wavelength of light coming from a star whose surface temperature is 8888 K is 327 nm.
The total energy radiated per unit area by a black body at this temperature is given by the Stefan-Boltzmann law which states that the total energy radiated per unit area of a black body per second is proportional to the fourth power of its absolute temperature.
The equation is given by; P = σAT⁴where P is the total energy radiated per unit area per second, σ is the Stefan-Boltzmann constant, A is the surface area of the body and T is the absolute temperature of the body.
Substituting σ = 5.67 × 10⁻⁸ Wm⁻²K⁻⁴ (Stefan-Boltzmann constant),
A=1 m² (unit surface area),
T=8888K,
we get; P = σAT⁴
P =5.67×10⁻⁸×1×(8888)⁴
P = 1.088×10⁸ Wm⁻²
Therefore, the conclusion is that the total energy radiated per unit area by a black body at this temperature is 1.088 × 10⁸ Wm⁻².
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Problem 1 (20\%). Find the Laplace transform a. L{t²e⁻⁹ᵗ} b. L{sin(4t)U(t−π/2 )}
The Laplace transform of [tex]t^2 e^{-9t}[/tex] is given by L{t²e^(-9t)} = 2! / (s + 9)³. The Laplace transform of sin(4t)U(t - π/2) is given by L{sin(4t)U(t - π/2)} = (4e^(-πs/2)) / (s² + 16).
a) The Laplace transform of [tex]t^2 e^{-9t}[/tex] is given by L{t²e^(-9t)} = 2! / (s + 9)³.
To find the Laplace transform, we use the properties and formulas of Laplace transforms. First, we apply the power rule to transform t² into (2!) / (s + 9)³, where s is the complex variable in the Laplace domain. Then, we use the exponential rule to transform e^(-9t) into 1 / (s + 9). Finally, we combine these two transformed terms to obtain the Laplace transform of t²e^(-9t).
b) The Laplace transform of sin(4t)U(t - π/2) is given by L{sin(4t)U(t - π/2)} = (4e^(-πs/2)) / (s² + 16).
To find the Laplace transform, we apply the Laplace transform properties and formulas. First, we transform sin(4t) using the standard Laplace transform table, which gives (4 / (s² + 16)). Then, we consider the unit step function U(t - π/2), which is 0 for t < π/2 and 1 for t ≥ π/2. Multiplying the transform of sin(4t) by e^(-πs/2) takes into account the time shift caused by the unit step function.
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An AISI 1018 steel has a yleld strength, 5y=295MPa Using the distortion-energy theory for the following given state of plane stress, determine the factor o safety. Write your final answer in two decimal places. σx = 82 Mpa, σy =32 Mpa, Txy =0
Hints: For distortion enerisy theory
a¹ = (n²ₓ - nₓnᵧ + n² ᵧ + 3n² ₓ ᵧ)¹/²
n = S/n ⁿ
the factor of safety is 11.8 (approx).
Given Data: AISI 1018 steel has a yield strength, 5y = 295 MPa, σx = 82 MPa, σy = 32 MPa, Txy = 0We need to calculate the factor of safety using the distortion-energy theory.
Formulae used: The formula used to find the factor of safety is as follows:
Factor of Safety (FoS) = Yield strength (5y)/ Maximum distortion energy
(a)The formula used to find the maximum distortion energy is as follows: Maximum distortion energy
(a) = [(nxx − nyy)² + 4nxy²]^(1/2) / 2
Here, nxx and nyy are normal stresses acting on the plane, and nxy is the shear stress acting on the plane.
Calculations:
Normal stress acting on the plane, nxx = σx = 82 MPa
Normal stress acting on the plane, nyy = σy = 32 MPa
Shear stress acting on the plane, nxy = Txy = 0
Maximum distortion energy (a) = [(nxx − nyy)² + 4nxy²]^(1/2) / 2= [(82 − 32)² + 4(0)²]^(1/2) / 2
= (50²)^(1/2) / 2= 50 / 2= 25 MPa
Factor of Safety (FoS) = Yield strength (5y)/ Maximum distortion energy (a)= 295 / 25= 11.8 (approx)
Therefore, the factor of safety is 11.8 (approx).
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Name: 19. If a wave has a peak amplitude of 17 cm, what is its RMS amplitude? NOTE: please calculate your answer in cm, not meters. 20. If a wave has an RMS amplitude of 240 mm, what is its peak ampli
The peak amplitude of the wave is approximately 339 mm.
19. If a wave has a peak amplitude of 17 cm, the RMS (Root Mean Square) amplitude can be calculated by dividing the peak amplitude by the square root of 2:
RMS amplitude = Peak amplitude / √2 = 17 cm / √2 ≈ 12 cm.
Therefore, the RMS amplitude of the wave is approximately 12 cm.
20. If a wave has an RMS amplitude of 240 mm, the peak amplitude can be calculated by multiplying the RMS amplitude by the square root of 2:
Peak amplitude = RMS amplitude * √2 = 240 mm * √2 ≈ 339 mm.
19. RMS (Root Mean Square) amplitude is a measure of the average amplitude of a wave. It is calculated by taking the square root of the average of the squares of the instantaneous amplitudes over a period of time.
In this case, if the wave has a peak amplitude of 17 cm, the RMS amplitude can be calculated by dividing the peak amplitude by the square root of 2 (√2). The factor of √2 is used because the RMS amplitude represents the equivalent steady or constant value of the wave.
20. The RMS (Root Mean Square) amplitude of a wave is a measure of the average amplitude over a period of time. It is often used to quantify the strength or intensity of a wave.
In this case, if the wave has an RMS amplitude of 240 mm, we can calculate the peak amplitude by multiplying the RMS amplitude by the square root of 2 (√2). The factor of √2 is used because the peak amplitude represents the maximum value reached by the wave.
By applying these calculations, we can determine the RMS and peak amplitudes of the given waves.
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Problem 3.26 Suppose the position of an object is given by 7 = (3.0t2 -6.0t³j)m. Where t in seconds.
Y Y Part A Determine its velocity as a function of time t Express your answer using two significa
The velocity of the object as a function of time `t` is given by `v= 6.0t² - 18.0t²j` where `t` is in seconds.
The position of an object is given by `x=7 = (3.0t²-6.0t³j)m`. Where `t` is in seconds.
The velocity of the object is the first derivative of its position with respect to time. So the velocity of the object `v` is given by: `[tex]v= dx/dt`[/tex]
Here, `x = 7 = (3.0t²-6.0t³j)m`
Taking the derivative with respect to time we have:
`v = dx/dt = d/dt(7 + (3.0t² - 6.0t³j))`
The derivative of 7 is zero. The derivative of `(3.0t² - 6.0t³j)` is `6.0t² - 18.0t²j`.
Therefore, the velocity of the object is `v = 6.0t² - 18.0t²j`.
To express the answer using two significant figures, we can round off to `6.0` and `-18.0`, giving the velocity of the object as `6.0t² - 18.0t²j`.
Therefore, the velocity of the object as a function of time `t` is given by `v= 6.0t² - 18.0t²j` where `t` is in seconds.
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When the three Milankovich cycles are in phase, their effects
are magnified, true or false?
"When the three Milankovitch cycles are in phase, their effects on Earth's climate are magnified" is true.
When the three Milankovitch cycles (eccentricity, axial tilt, and precession) are in phase, it means that their respective patterns align or coincide at certain points in time. This alignment can lead to a magnification of their combined effects on Earth's climate.
Each Milankovitch cycle individually affects the distribution and amount of solar radiation received by the Earth.
Eccentricity refers to changes in the Earth's orbit from more circular to more elliptical over long periods of time. Axial tilt refers to the tilt of the Earth's axis relative to its orbital plane, which results in varying amounts of sunlight reaching different latitudes during different seasons. Precession refers to the slow wobble or rotation of the Earth's axis, causing the orientation of Earth's axis to change over time.When these cycles align, their combined impact on the distribution of solar radiation can be more pronounced.
For example, if the Earth's orbit is more elliptical (higher eccentricity), the variation in solar radiation received between aphelion (farthest distance from the Sun) and perihelion (closest distance to the Sun) will be greater. If the axial tilt is at its maximum during this time, the difference in sunlight reaching the Northern and Southern Hemispheres will also be maximized. Similarly, if the precession aligns with the other cycles, it can further amplify these effects.
This alignment of the Milankovitch cycles can lead to significant changes in climate patterns, such as more extreme seasons or variations in the distribution of heat across the globe. Scientists study these cycles and their alignment to better understand past and future climate changes on Earth.
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A string of length 2 m is fixed at both ends. The speed of waves on the string, is 30 m/s. What is the lowest frequency of vibration for the string in Hz? O a. 0.067 O b. 7.5 O c. 0.033 O d. 0.13 O e.
With a string of length 2 m that is fixed at both ends, and the speed of waves on the string is 30 m/s, then the lowest frequency of vibration for the string is 7.5 Hz. The correct option is b.
To find the lowest frequency of vibration for the string, we need to determine the fundamental frequency (also known as the first harmonic).
The fundamental frequency is given by the formula:
f = v / λ
Where:
f is the frequency of vibration,
v is the speed of waves on the string,
and λ is the wavelength of the wave.
In this case, the string length is given as 2m. For the first harmonic, the wavelength will be twice the length of the string (λ = 2L), since the wave must complete one full cycle along the length of the string.
λ = 2 * 2m = 4m
v = 30 m/s
Substituting these values into the formula:
f = v / λ
f = 30 m/s / 4 m
f = 7.5 Hz
Therefore, the lowest frequency of vibration for the string is 7.5 Hertz. The correct answer is option b. 7.5 Hz.
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1. (15 pts) Solar System Systematics. The gravitational force between two objects of mass m, m,m, which are separated by a distance r is equal to F = G- The dimensions of 2 and m₂ force are [M][L]/[
The dimensions of the gravitational force F are [M][L]/[T]^2, as expected.
Given:
F = gravitational force
G = gravitational constant
m₁, m₂ = masses of the objects
r = distance between the objects
The dimensions of the gravitational force can be expressed as [M][L]/[T]^2, where [M] represents mass, [L] represents length, and [T] represents time.
Let's analyze the dimensions of each term in the equation F = G(m₁m₂)/r²:
G: The gravitational constant has dimensions [M]^-1[L]^3[T]^-2.
m₁m₂: The product of the masses has dimensions [M]².
r²: The square of the distance has dimensions [L]^2.
Now, let's calculate the dimensions of the entire equation:
F = G(m₁m₂)/r² = [M]^-1[L]^3[T]^-2 * [M]² / [L]^2
Simplifying, we get:
F = [M]^-1[L]^[3-2+2][T]^-2 = [M]^[0][L]^[3][T]^-2
Thus, the dimensions of the gravitational force F are [M][L]/[T]^2, as expected.
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simple answer
Q3: A proton moves at 3X105 m/s in positive X-axis direction through a magnetic field of 4.5 T which is in the negative Y-axis direction. Calculate the magnitude of the force exerted on the proton. Th
The magnitude of the force exerted on the proton can be calculated using the formula for the magnetic force experienced by a charged particle in a magnetic field. The calculated force is 1.35 × 10^(-13) N.
The magnetic force experienced by a charged particle moving through a magnetic field is given by the formula F = qvBsinθ, where F is the force, q is the charge of the particle, v is its velocity, B is the magnetic field strength, and θ is the angle between the velocity vector and the magnetic field vector.
In this case, the proton has a positive charge of 1.6 × 10^(-19) C, a velocity of 3 × 10^5 m/s in the positive X-axis direction, and the magnetic field has a strength of 4.5 T in the negative Y-axis direction.
Since the proton is moving parallel to the X-axis and the magnetic field is perpendicular to the Y-axis, the angle between the velocity and the magnetic field is 90 degrees. Therefore, sinθ = 1.
Substituting the given values into the formula, we have F = (1.6 × 10^(-19) C)(3 × 10^5 m/s)(4.5 T)(1) = 1.35 × 10^(-13) N.
Hence, the magnitude of the force exerted on the proton is 1.35 × 10^(-13) N.
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3. On Mars, the acceleration of gravity is 12 meters/second². A rock is thrown vertically from a height of 80 feet and an initial speed of 64 feet/second. a) Write the quadratic function in standard
The quadratic function in standard form is:h(t) = -6t² + 19.5072t + 24.384 meters.
The acceleration of gravity on Mars is 12 meters/second²A rock is thrown vertically from a height of 80 feet with an initial speed of 64 feet/second. The given values are in two different units, we should convert them into the same unit.1 feet = 0.3048 meterTherefore,80 feet = 80 × 0.3048 = 24.384 meters64 feet/second = 64 × 0.3048 = 19.5072 meters/second
The quadratic function for the given problem can be found using the formula:
h = -1/2gt² + v₀t + h₀
whereh₀ = initial height of rock = 24.384 mv₀ = initial velocity of rock = 19.5072 m/st = time after which the rock hits the groundg = acceleration due to gravity = 12 m/s²
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For the circuit given below, where V-9 V, what resistor connected across terminals ab will absorb maximum power from the circuit? What is that power? R= ps 3kQ kQ W 1kQ 10 k wwwwww 120 40 k ob B
To determine resistor that will absorb maximum power from circuit, we need to find value that matches load resistance with internal resistance.Maximum power absorbed by resistor is 27 mW.
The power absorbed by a resistor can be calculated using the formula P = V^2 / R, where P is the power, V is the voltage across the resistor, and R is the resistance.
Since the voltage across the resistor is given as 9 V and the resistance is 3 kΩ, we can substitute these values into the formula: P = (9 V)^2 / (3 kΩ) = 81 V^2 / 3 kΩ = 27 W / kΩ = 27 mW.
Therefore, the maximum power absorbed by the resistor connected across terminals ab is 27 mW.
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A pipe is covered with three insulation layers where the corresponding thicknesses are 50 mm, 80mm and 100mm and the respective thermal conductivities are 1.15 W/m•C, 1.45 W/m°C and 2.8 W/mC. The inner side of the pipe which has a diameter of 30 cm is exposed to a hot gas at 1200 °C with convection coefficient of 50 W/m²°C and the temperature of the inner side of the pipe surface is 900 °C. The air outside the pipe is at 25°C with a convection coefficient of 20 W/m²°C. a. Draw a schematic diagram which represents the heat transfer process [1 mark] b. Calculate the Heat transfer rate [3 mark] c. The overall heat transfer coefficient "U" of the system based on the inner pipe [3 mark] d. Temperature at each of the layers and at the outermost surface of the pipe. [3 mark]
The temperature at each layer and at the outermost surface of the pipe is 903.543°C
Calculate the heat transfer rate with the help of formula;
[tex]Q = h1 . A . (Ts1 − T∞1 )[/tex]
= h2 . A . (Ts2 − Ts1)
= h3 . A . (Ts3 − Ts2) ... (1)
Where; h1 = 50 W/m²°C,
h2 = U2 = 4.59 W/m²°C,
h3 = U3 = 1.24 W/m²°C and
A = π DL,
Here, the diameter of the pipe (D) is 30cm or 0.3 m.
The length (L) of the pipe can be assumed as 1m.
Therefore,
A = π DL
= 3.14 x 0.3 x 1
= 0.942 m²
Substituting the respective values in equation
(1);Q = 50 x 0.942 x (900 - 1200)
= 70,650 W
= 70.65 kW
Therefore, the heat transfer rate is 70.65 kW.C.
Calculation of overall heat transfer coefficient:
Calculate the overall heat transfer coefficient (U) based on the inner pipe with the help of formula:
1/U = 1/h1 + t1/k1 ln(r2/r1) + t2/k2 ln(r3/r2) + t3/k3 ln(ro/r3) ... (2)
Where; t1 = 50mm,
k1 = 1.15 W/m°C,
t2 = 80mm,
k2 = 1.45 W/m°C,
t3 = 100mm,
k3 = 2.8 W/m°C,
r1 = (0.3/2) + 0.05 = 0.2m,
r2 = (0.3/2) + 0.05 + 0.08 = 0.33m,
r3 = (0.3/2) + 0.05 + 0.08 + 0.1 = 0.43m,
ro = (0.3/2) + 0.05 + 0.08 + 0.1 + 0.05 = 0.48m
Substituting the respective values in equation (2);
1/U = 1/50 + 0.05/1.15 ln(0.33/0.2) + 0.08/1.45
ln(0.43/0.33) + 0.1/2.8 ln(0.48/0.43)1/U = 0.02
Therefore,
U = 50 W/m²°C.D.
Calculation of temperature at each layer and at the outermost surface of the pipe:
Calculate the temperature at each layer and at the outermost surface of the pipe using the formula;
Ts - T∞ = Q / h . A ...(3)
Where; h1 = 50 W/m²°C,
h2 = 4.59 W/m²°C and
h3 = 1.24 W/m²°C.
Calculation of Temperature at each layer;
For layer 1,
Ts1 - T∞1 = Q / h1 . A
= 70.65 / (50 x 0.942)
= 1.49°C
Due to symmetry, temperature at the outer surface of layer 1 will be equal to that of layer 2,
i.e.,Ts2 - Ts1 = Ts1 - T∞1 = 1.49°C
Therefore, Ts2 = Ts1 + 1.49 = 901.49°C
Due to symmetry, temperature at the outer surface of layer 2 will be equal to that of layer 3, i.e.,
Ts3 - Ts2 = Ts2 - Ts1
= 1.49°C
Therefore, Ts3 = Ts2 + 1.49
= 902.98°C
For outermost surface of the pipe,
Ts4 - Ts3 = Ts3 - T∞2
= (70.65 / 20 x π DL)
= 0.563°C
Therefore, Ts4 = Ts3 + 0.563
= 903.543°C
Therefore, the temperature at each layer and at the outermost surface of the pipe is as follows;
Ts1 = 901.49°C
Ts2 = 902.98°C
Ts3 = 903.543°C
Ts4 = 903.543°C
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Detimine the minimum plate thickness requied for plane strain conditions to prevail at the crack tio for the following steels and calculate the plastic zone size, use v=0.3, (a) Stee A, Kc = 100MPavm and yield strength =660MPa. (b) Sizal 3,hic = 180MPav m and yield strength =350MPa.
For plane strain conditions to prevail, the thickness of the plate can be determined using the given parameters for steel A and Sizal 3. (a) Steel A The minimum plate thickness can be calculated using the expression given below:
[tex]$$b=\frac{1.12(K_c/\sigma_{y})^2}{1-\nu^2}$$[/tex]
where b is the minimum thickness, Kc is the fracture toughness, [tex]σy[/tex] is the yield strength, and ν is the Poisson's ratio. For steel A,[tex]Kc = 100 MPa√m[/tex]and yield strength = [tex]660 MPa[/tex], therefore:
[tex]$$b=\frac{1.12(100/660)^2}{1-0.3^2}= 8.28 \space mm$$[/tex]
The plastic zone size can be calculated as:
[tex]$$r=\frac{K_c^2}{\sigma_y^2}=\frac{100^2}{660^2}=0.0236\space m=23.6\space mm$$[/tex] Therefore, the minimum thickness of the plate for plane strain conditions to prevail at the crack tip is 8.28 mm and the plastic zone size is 23.6 mm for steel A.
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need answers in details like a 10 mark ques ans
3. Calculate the de-Broglie wavelength of electron whose energy is 15 eV. 4. An electron confined to move between two rigid walls separated by10-9m. Find the first three allowed energy states of the e
The first three allowed energy states of an electron confined to move between two rigid walls separated by 10^-9 m are 4.89 x 10^-19 J, 1.96 x 10^-18 J, and 4.41 x 10^-18 J, respectively.
Question 3: Calculate the de-Broglie wavelength of an electron whose energy is 15 eV. The energy of an electron can be represented in terms of wavelength according to de-Broglie's principle.
We can use the following formula to calculate the wavelength of an electron with an energy of 15 eV:[tex]λ = h/p[/tex], where h is Planck's constant (6.626 x 10^-34 J.s) and p is the momentum of the electron.
[tex]p = sqrt(2*m*E)[/tex], where m is the mass of the electron and E is the energy of the electron. The mass of an electron is 9.109 x 10^-31 kg.
Therefore, p = sqrt(2*9.109 x 10^-31 kg * 15 eV * 1.602 x 10^-19 J/eV)
= 4.79 x 10^-23 kg.m/s.
Substituting the value of p into the formula for wavelength, we get:
λ = h/p = 6.626 x 10^-34 J.s / 4.79 x 10^-23 kg.m/s = 1.39 x 10^-10 m.
Therefore, the de-Broglie wavelength of an electron whose energy is 15 eV is 1.39 x 10^-10 m.
Question 4: An electron is confined to move between two rigid walls separated by 10^-9 m. Find the first three allowed energy states of the electron.
The allowed energy states of an electron in a one-dimensional box of length L are given by the following equation:
E = (n^2 * h^2)/(8*m*L^2), where n is the quantum number (1, 2, 3, ...), h is Planck's constant (6.626 x 10^-34 J.s), m is the mass of the electron (9.109 x 10^-31 kg), and L is the length of the box (10^-9 m).
To find the first three allowed energy states, we need to substitute n = 1, 2, and 3 into the equation and solve for E.
For n = 1, E = (1^2 * 6.626 x 10^-34 J.s)^2 / (8 * 9.109 x 10^-31 kg * (10^-9 m)^2)
= 4.89 x 10^-19 J.
For n = 2,
E = (2^2 * 6.626 x 10^-34 J.s)^2 / (8 * 9.109 x 10^-31 kg * (10^-9 m)^2)
= 1.96 x 10^-18 J.
For n = 3,
E = (3^2 * 6.626 x 10^-34 J.s)^2 / (8 * 9.109 x 10^-31 kg * (10^-9 m)^2)
= 4.41 x 10^-18 J.
Therefore, the first three allowed energy states of an electron confined to move between two rigid walls separated by 10^-9 m are 4.89 x 10^-19 J, 1.96 x 10^-18 J, and 4.41 x 10^-18 J, respectively.
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this answer is incorrect
A 23.0-V battery is connected to a 3.80-µF capacitor. How much energy is stored in the capacitor? 1005 1 X Did you accidentally divide or take the inverse in your calculation?) Need Help? Read It
When a battery is connected to a capacitor, it charges the capacitor by transferring energy. The energy stored in a capacitor can be calculated using the formula: E = 0.5 * C * [tex]V^2[/tex], where E represents the energy stored, C is the capacitance, and V is the voltage.
In this case, the capacitance is given as 3.80 µF and the voltage of the battery is 23.0 V. By substituting these values into the formula, we can calculate the energy stored in the capacitor.
Energy (E) = 0.5 * 3.80 µF * [tex](23.0 V)^2[/tex]
After performing the necessary calculations, we can determine the energy stored in the capacitor.
The energy stored in the capacitor connected to a 23.0-V battery and having a capacitance of 3.80 µF is determined to be the value calculated using the formula mentioned above.
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Two deuterons undergo nuclear fusion to form a Helium nucleus.
Calculate the Energy
released in this process (binding energy per nucleon for
deuteron = 1.1 MeV and for helium
= 7.0 MeV)
(ii) Two deuterons undergo nuclear fusion to form a Helium nucleus. Calculate the Energy released in this process (binding energy per nucleon for deuteron = 1.1 MeV and for helium = 7.0 MeV)
The energy released during the fusion of two deuterons into a helium nucleus is 25.8 MeV.
(i)Binding energy per nucleon is the amount of energy needed to separate the nucleus of an atom into individual protons and neutrons. The binding energy per nucleon for deuteron is 1.1 MeV and that for helium is 7.0 MeV. Therefore, to find the energy released during the fusion of two deuterons into a helium nucleus, we need to calculate the total binding energy of the initial two deuterons and compare it with the binding energy of the final helium nucleus.
The binding energy of a deuteron with one proton and one neutron is given by:
Binding energy of a deuteron = (1 nucleon) × (1.1 MeV/nucleon)
= 1.1 MeV
Therefore, the total binding energy of two deuterons is:
Total binding energy of two deuterons = 2 × 1.1 MeV
= 2.2 MeV
The binding energy of a helium nucleus with two protons and two neutrons is given by:
Binding energy of a helium nucleus = (4 nucleons) × (7.0 MeV/nucleon)
= 28 MeV
The difference in the binding energies of the initial two deuterons and the final helium nucleus is the energy released during the fusion process:
Energy released during fusion = binding energy of initial deuterons - binding energy of final helium nucleus
= 2.2 MeV - 28 MeV
= -25.8 MeV
Therefore, the energy released during the fusion of two deuterons into a helium nucleus is 25.8 MeV. Note that the energy released is negative, which means that energy is required to break apart a helium nucleus into its individual protons and neutrons.
(ii) The energy released during the fusion of two deuterons into a helium nucleus is 25.8 MeV.
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1. Given the known decay constant λ of a radioactive nucleus calculate (a) the probability of decay of the nucleus during time to (from t = 0) to =): (b) the mean lifetime of the nucleus. Answer :a)1
Given the known decay constant λ of a radioactive nucleus, we can calculate (a) the probability of decay of the nucleus during time t0 (from t = 0) to t: (b) the mean lifetime of the nucleus.
(a) The probability of decay of the nucleus is given by:Where N(t) is the number of radioactive nuclei at time t and N(0) is the number of radioactive nuclei at time t = 0.
Therefore,The probability of decay of the nucleus during time t is given by P(t) = 1 - P0(t). b) The mean lifetime of the nucleus is defined as the average time it takes for a radioactive nucleus to decay. It is denoted by τ and is given by:τ = 1/λWe can also express the mean lifetime as:T1/2 = τ ln(2) where T1/2 is the half-life of the radioactive nucleus.
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A block of wood 0. 20m thick is floating in seawater. The specific gravity of wood is 0.65 while that of seawater is 1. 03. Find the minimum area of a block which will support a man weighing 80kg.
The minimum area of the block that will support the man weighing 80 kg is 343.28 m².
Given, The thickness of the block of wood = 0.20mSpecific gravity of wood = 0.65Specific gravity of seawater
= 1.03Weight of the man
= 80kgWe need to find the minimum area of a block which will support the man.
To begin with the solution, we can first find the volume of the block of wood. Volume of the block
= thickness x area
= 0.20m x A
(where A is the area of the block)
Now, we know that the block is floating in seawater. This means that the weight of the block of wood is equal to the weight of the water displaced by it.
We can use Archimedes' principle to find the weight of the water displaced.
Wood's weight = Volume of water displaced x specific gravity of seawater
= Volume of water displaced x 1.03
Also, we know that the weight of the man should be supported by the block. This means that the weight of the block of wood + the weight of the water displaced should be greater than or equal to the weight of the man. Wood's weight + Water's weight >
= Man's weight0.65 x (0.20 x A) + 1.03 x (0.20 x A) >
= 80We can solve this equation for A to find the minimum area of the block.0.13A + 0.206A >= 80 / 1.68A >
= 343.28 m²
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calculate approximate ratio of the binding energy of o2 to the rest
energy of o2
Checkpoint 16 Calculate the approximate ratio of the binding energy of O2 (about 5 eV) to the rest energy of O2. Most oxygen nuclei contain 8 protons and 8 neutrons, and the rest energy of a proton or
The approximate ratio of the binding energy of O₂ to the rest energy of O₂ is (8 x [tex]10^{-19}[/tex] MeV) / (2 x 8 x 940 MeV), which simplifies to 5 x [tex]10^{-23}[/tex].
The approximate ratio of the binding energy of O₂ to the rest energy of O₂ can be calculated, considering that most oxygen nuclei contain 8 protons and 8 neutrons. The rest energy of a proton or neutron is about 940 MeV. However, due to the small magnitude of the binding energy compared to the rest energy, it would be challenging to detect the difference in mass between a mole of molecular oxygen (O₂) and two moles of atomic oxygen using a laboratory scale.
The binding energy of a nucleus represents the energy required to separate its constituent nucleons (protons and neutrons). In this case, we consider the binding energy of O₂, which is approximately 5 eV (electron volts).
The rest energy of a proton or neutron is approximately 940 MeV (mega-electron volts), which is significantly larger than the binding energy of O₂. To calculate the ratio, we convert the binding energy to MeV by multiplying it by the conversion factor (1 eV = 1.6 x [tex]10^{-19}[/tex] J = 1.6 x [tex]10^{-19}[/tex] * 6.242 x [tex]10^{18}[/tex] MeV), resulting in a binding energy of approximately 8 x [tex]10^{-19}[/tex] MeV.
The approximate ratio of the binding energy of O₂ to the rest energy of O₂ is (8 x [tex]10^{-19}[/tex] MeV) / (2 x 8 x 940 MeV), which simplifies to 5 x [tex]10^{-23}[/tex].
Due to the extremely small magnitude of this ratio, it would be exceedingly difficult to detect the difference in mass between a mole of molecular oxygen (O₂) and two moles of atomic oxygen using a laboratory scale. The difference is too minuscule to be measured with the precision of typical laboratory instruments.
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The complete question is: <Calculate the approximate ratio of the binding energy of O2 (about 5 eV) to the rest energy of O2. Most oxygen nuclei contain 8 protons and 8 neutrons, and the rest energy of a proton or neutron is about 940 MeV. Do you think you could use a laboratory scale to detect the difference in mass between a mole of molecular oxygen (O₂) and two moles of atomic oxygen?>
2. Determine the moment of inertia of the shaded area about the x axis using the method of integration. Show the elemental area on your FBD.
ya 20
The moment of Inertia in the shaded area will be 4.266 [tex]ln^{4}[/tex]
Consider the equation for the curve
[tex]y^{2} =x[/tex]
[tex]y= x^{1/2}[/tex]
Draw the free-body diagram showing the differential elements of the shaded area mentioned in the figure below.
Consider a rectangular different element with respect to the x-axis with a thickness dx and interacts with the boundary at (x, y)
Express the area of the differential, element parallel to the y-axis.
dA=Ydx
Calculate the moment of Inertia of different elements about the X -axis.
Calculate the movement of inertia of the different elements about X -the axis
dIx= dIx+dA[tex]y^{2}[/tex]
The moment of Inertia of the element about the centroidal axis.
dIx=1/12(dx)[tex]y^{3}[/tex]
From the moment of inertia of different elements about x-axis,
dIx=1/12(dx)y³+[tex]{\frac{y}{2}}^{2}[/tex]
=1/3[tex]y^{3}[/tex]dx
=1/3[tex]x^{3/2}[/tex]dx
From the moment of Inertia for the shaded area about X -a xis.
Ix=∫dIx
= ₄⁰∫ 1/3[tex]x^{3/2}[/tex].dx=2/15.x [tex]x^{5/2}[/tex]₄⁰∫
=4.266[tex]In^{4}[/tex]
Therefore, the moment of Inertia will be =4.266[tex]In^{4}[/tex].
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Using your knowledge of X-ray line broadening, explain how it is possible to determine of line broadening is attributable to strain or crystallite size or both. (10) b) In backscattered electron diffraction, explain the important features of Kikuchi bands (e.g. what they represent, what their intersections represent, what is represented by their widths and the angles between them). (5) c) Describe why EBSD can give very good orientation information for nickel electrodeposited on rolled brass plate.
EBSD can provide excellent orientation information for nickel electrodeposited on rolled brass plates for several reasons. Nickel electrodeposition on brass plates generates a columnar grain structure that is textured and anisotropic, allowing EBSD to measure the crystal orientation distribution.
The X-ray line broadening is an essential tool used in determining whether the line broadening is due to crystallite size or strain, or both. The peak width is directly proportional to the variance of the crystal planes' interplanar distance. The diffraction peak is widened due to small crystallite sizes, while the X-ray line is broadened by micro-strain in the crystal lattice. line broadening due to strain is distinguished from that due to crystallite size by comparing the X-ray line profile to that of the standard broadening line profile. To extract the information needed to calculate the size or strain of the diffraction peaks, two different methods of X-ray line broadening analysis are used. The Williamson-Hall (W-H) approach and the Warren-Averbach (W-A) technique are the two most common ones. The W-H approach separates the size and strain contributions to the broadening, while the W-A method is a Fourier transform of the X-ray peak broadening. b) Backscattered electron diffraction (EBSD) is a powerful technique for determining crystalline structure and texture in materials. Kikuchi bands are bright, curved lines observed in the EBSD pattern, which reflects the diffracted electron paths in the crystal lattice. These bands represent the intersection between the diffracted electrons and the reciprocal lattice points. As the diffraction angle changes, the Kikuchi bands’ spacing widens and shrinks, representing the change in the diffraction path. When the Kikuchi bands intersect, it represents a reciprocal lattice point. Furthermore, the angle between the Kikuchi lines is directly proportional to the angle between the diffracting planes. The width of the Kikuchi bands is related to the direction and density of the lattice points, with the bands' width decreasing as the lattice spacing increases. c) EBSD can provide excellent orientation information for nickel electrodeposited on rolled brass plates for several reasons. Nickel electrodeposition on brass plates generates a columnar grain structure that is textured and anisotropic, allowing EBSD to measure the crystal orientation distribution. The electrodeposition process used for nickel produces large grain sizes, and EBSD can detect these large grains by the high scattering power of nickel.
The high scattering power of nickel combined with the large grain size makes EBSD an efficient method of measuring crystal orientation distributions in nickel electrodeposits on brass plates. EBSD is also able to distinguish between large grains and single crystals, allowing for greater orientation accuracy. In conclusion, EBSD is a highly valuable tool for measuring crystal orientation distributions in nickel electrodeposits on brass plates due to its ability to detect large grains and distinguish between large grains and single crystals.
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Given a rod with two varying sections as shown below: Triangular distributed load with intensity w=2&N/m C /le → → → → B fincm Acm. w with E = 70Gpa; Asc = 100mm?; Agc = 50mm%; w = 2 KN/m trin
A rod with two varying sections and a triangular distributed load with intensity w=2 N/m is given below:Triangular distributed load with intensity w = 2 N/m has been applied on the rod as shown in the figure below. Here, E = 70 GPa, Asc = 100 mm², Agc = 50 mm² and triangular load with w = 2 kN/m.A triangular distributed load may be considered as a superposition of two rectangular distributed loads, one in the positive y direction and one in the negative y direction.
The midpoint of these loads corresponds to the location of the vertex of the triangular load.In this question, the section BC and the section CD have different cross-sectional areas. Due to this, we cannot consider this rod as a uniform rod. We will need to calculate the bending moments for both sections separately.For section BC:Calculation of the vertical reaction force at point B,Vb = 8.33 kN Calculation of the shear force at section C-Splitting the triangle and applying the load component on the section A-C Shear force at section C,VC = 2 kNFor bending moment at section C,BM_C = 2 * (5/2) - 2 * (5/3) = 1.67 kNm For bending moment at section B,BM_B = (8.33 * 2) - (2 * 5) - (1.67) = 8.99 kNm.
For section CD:Calculation of the vertical reaction force at point C,VC = 2.67 kN Calculation of the shear force at section D-Splitting the triangle and applying the load component on the section A-D Shear force at section D,VD = 1.33 kNFor bending moment at section D,BM_D = 1.33 * (5/3) = 2.22 kNm For bending moment at section C,BM_C = (2.67 * 2) - (2 * 5) - (2.22) = -2.78 kNm Therefore, the bending moment for section BC and section CD are 8.99 kNm and -2.78 kNm, respectively.
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There are already several questions asking for this
question but the answers are neither complete nor correct. Please
read the question and help me with all the questions! Do not forget
= < 1. A uniform surface current flowing in the xy plane, described by surface current K = Kî generates a magnetic field MoK -î for z> 0 2 В. MOK -î for z < 0 2 a) Is it possible to find a magneti
The question is asking whether it is possible to find a magnetic vector potential for a given uniform surface current flowing in the xy plane and generating a magnetic field for different regions of space.
To determine whether it is possible to find a magnetic vector potential for the given scenario, we need to consider the conditions that must be satisfied. In general, a magnetic vector potential A can be found if the magnetic field B satisfies the condition ∇ × A = B. This is known as the magnetic vector potential equation.
In the given situation, the magnetic field is different for the regions above and below the xy plane. For z > 0, the magnetic field is described as B = MoK -î, and for z < 0, it is described as B = -MoK -î. To find the magnetic vector potential, we need to determine if there exists a vector potential A that satisfies the equation ∇ × A = B in each region.
By calculating the curl of A, we can check if it matches the given magnetic field expressions. If the curl of A matches the magnetic field expressions for both regions, then it is possible to find a magnetic vector potential for the given scenario. However, if the curl of A does not match the magnetic field expressions, then it is not possible to find a magnetic vector potential that satisfies the conditions.
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A Question 59 (3 points) Retake question What is the power of a lens that has a focal length of 175 cm? NOTE that the centimeter units here. Give your answer to the nearest hundredth (0.01) diopter. D
The power of the lens that has a focal length of 175 cm is 0.57 D.
The formula for power of a lens is given by
P = 1/f
where, f is the focal length of the lens
We are given that the focal length of the lens is 175 cm.
Thus, the power of the lens is
P = 1/f
= 1/175 cm
= 0.0057 cm⁻¹
Since we need the answer in diopters, we need to multiply the above answer by 100.
We get
P = 0.57 D
The power of the lens can be calculated by using the formula
P = 1/f
where f is the focal length of the lens.
Given that the focal length of the lens is 175 cm, we can calculate the power of the lens.
Therefore, the power of the lens is
P = 1/f
= 1/175 cm
= 0.0057 cm⁻¹.
To get the answer in diopters, we need to multiply the answer by 100.
Hence, the power of the lens is P = 0.57 D.
Therefore, the power of the lens that has a focal length of 175 cm is 0.57 D.
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Estimate the length and width that would be needed for a gravitational settler to remove 1 µm diameter particles for a gas stream with a flow rate of 3000 m3/min. The height of the settler is 2 m, and the width is 10 m. Are these dimensions reasonable? Would it help to make the settler shorter? Why or why not?
The dimensions of the gravitational settler mentioned are not suitable. To estimate the required length and width, more information is needed. However, simply making the settler shorter would not be helpful.
To estimate the dimensions of a gravitational settler, we need to consider the settling velocity of the particles and the residence time required for effective separation. The settling velocity of a particle depends on its size and density as well as the properties of the gas stream. Smaller particles have lower settling velocities, making their separation more challenging.
The height of the settler is typically designed to provide sufficient residence time for particles to settle. A shorter settler height would reduce the residence time and might not allow adequate separation. In this case, a height of 2 m seems reasonable but could be adjusted based on the settling velocity of the 1 µm diameter particles.
The width of the settler is not the primary dimension that determines particle separation. Instead, it determines the cross-sectional area available for gas flow, which should be large enough to accommodate the desired flow rate without excessive pressure drop. However, the length of the settler plays a crucial role in particle separation.
To estimate the required length, we need to calculate the settling distance needed for the particles to reach the bottom of the settler. This settling distance depends on the settling velocity of the particles and the desired efficiency of particle removal. Without knowing the settling velocity or the desired efficiency, it is not possible to provide a specific length.
In summary, the dimensions of the gravitational settler mentioned are insufficient for effectively removing 1 µm diameter particles from a gas stream with a flow rate of 3000 m3/min. Simply making the settler shorter would not be helpful because it would reduce the residence time and potentially compromise the separation efficiency. The length and width of the settler should be determined based on the settling velocity of the particles, desired efficiency, and other design considerations to ensure effective particle removal.
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what is the greatest distance you can be from base camp at the end of the third displacement regardless of direction
To determine the greatest distance you can be from the base camp at the end of the third displacement, regardless of direction, we need more specific information about the magnitudes and directions of the displacements.
Displacement is a vector quantity that has both magnitude and direction. The distance covered during multiple displacements depends on the individual magnitudes and directions of each displacement. Without specific values, it is not possible to determine the exact greatest distance from the base camp.
If you provide the magnitudes and directions of the three displacements, I can help you calculate the total distance and determine the maximum possible distance from the base camp at the end of the third displacement.
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"A prototype oil pipeline with a diameter of 2 m provides at a
flow rate of Qoil = 2 m3/s. The performance of the pipeline is
tested in a physical model using water and a 10 cm diameter pipe.
These tests show a pressure loss per meter length of pipe of ∆p/L = 2000 Pa/m. Use standard fluid properties for water; for the oil, use rhooil = 860 kg/m3 and μoil = 0.005 kg/m s.Complete the following:a) Based upon the relevant parameters listed above, use dimensional analysis to form appropriate dimensionless group(s) for the pressure loss along the pipeline.[9 MARKS]b) Considering the provided pressure loss per meter length for the pipe in the physical model, calculate the pressure loss per meter length for the prototype pipeline.
The friction factor is 0.0193 for turbulent flow. Using the friction factor, the pressure loss per meter length for the prototype pipeline was determined to be approximately 76 Pa/m.
a) Based on the relevant parameters listed, dimensionless group(s) for the pressure loss along the pipeline can be represented by dimensionless groups Froude Number and Reynolds Number. The Froude number is Fr = V / (gL) and the Reynolds number is Re = ρVD / μ. Where, V is the fluid velocity, D is the pipe diameter, μ is the fluid viscosity, ρ is the fluid density, g is the acceleration due to gravity and L is the length of the pipeline. The dimensionless group of pressure loss per unit length of the pipe is expressed as ∆p / ρV2D. Therefore, applying the Buckingham Pi theorem we can write the pressure loss of the pipeline as ∆p / ρV2D = f (Re, Fr) where f is an unknown function of Reynolds number and Froude number.
b) The prototype oil pipeline has a diameter of 2m and it provides a flow rate of Q oil = 2m3/s. The pressure loss per meter length of pipe of the physical model is given as ∆p/L = 2000 Pa/m. Using the Reynolds number as the dimensionless group to analyze the pressure drop, we have Re = ρVD / μ = (860 kg/m³ x 2 m³/s x 2 m) / 0.005 kg/ms ≈ 688,000. Therefore, the flow through the physical model of the pipe is turbulent flow. Now using the known value of Reynolds number, we can determine the friction factor. The friction factor is given by f = (0.3164 / Re^(1/4)). Therefore, we have f = (0.3164 / 688000^(1/4)) ≈ 0.0193. Using the dimensionless group for pressure loss per unit length of the pipe, we can determine the pressure loss per meter length for the prototype pipeline. ∆p / ρV²D = f(Re) = 0.0193 (688,000)^(-1/4). Since the prototype oil pipeline has a diameter of 2m, the velocity can be calculated as V = Q / A = (2m³/s) / (π(2m/2)²) ≈ 0.319 m/s. Thus, the pressure loss per meter length for the prototype pipeline is approximately, ∆p/L = (∆p / ρV²D) x ρV²D / 2 = (0.0193(688,000)^(-1/4)) x 860 kg/m³ x (0.319 m/s)² x (2m/2) ≈ 76 Pa/m.
The dimensionless groups for the pressure loss along the pipeline were calculated as Froude number and Reynolds number using Buckingham Pi theorem. The friction factor is 0.0193 for turbulent flow. Using the friction factor, the pressure loss per meter length for the prototype pipeline was determined to be approximately 76 Pa/m.
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solve alll all questions i need your help .please i will give you a
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If B = 1.5 T, The bar length is 60 cm, Which point a or b, should be the positive terminal of the battery? If Vmax= 175 V what is the greatest mass m that can be measured? Stan a b Battery 5.00 Ω www
The greatest mass m that can be measured is determined as 3.21 kg.
What is the greatest mass that can be measured?The value of the greatest mass that can be measured is calculated from the greatest electric force that can be measured as follows;
The maximum current flowing in the circuit is calculated as;
Imax = Vmax / R
Imax = 175 V / 5Ω
Imax = 35 A
The maximum electric force in the circuit is calculated as follows;
F = BIL
where;
B is the magnetic fieldI is the maximum current flowing in the wireL is the length of the wire = 60 cm = 0.6 mF = 1.5 T x 35 A x 0.6 m
F = 31.5 N
The greatest mass m that can be measured is calculated as follows;
F = mg
m = F /g
m = (31.5 N ) / ( 9.8 m/s²)
m = 3.21 kg
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