Answer:
1) v = 2.20 10⁵ m / s , 2) r = 4.86 10⁻³ m, r = 4.92 10⁻³ m
Explanation:
1) A speed selector is a section where the magnetic and electrical forces have opposite directions, so
[tex]F_{m} = F_{e}[/tex]
q v B = q E
v = E / B
V = E s
E = V / s
v = V / s B
v = 148 / (0.0016 0.42)
v = 2.20 10⁵ m / s
2) when the isotopes enter the spectrometer, we can use Newton's second law
F = m a
Acceleration is centripetal
a = v² / r
q v B = m v² / r
r = m v / qB
The mass of uranium 235 is
m = 3.903 10⁻²⁵ kg
The radius of this isotope is
r = 3,903 10⁻²⁵ 2.20 10⁵ / (92 1.6 10⁻¹⁹ 1.2)
r = 4.86 10⁻³ m
The mass of the uranium isotope 238 is
m = 238 a = 238 1.66 10-27 = 395.08 10-27 kg
The radius is
r = 395.08 10⁻²⁷ 2.20 10⁵ / (92 1.6 10⁻¹⁹ 1.2)
r = 4.92 10⁻³ m
Final answer:
The speed of the ions entering the mass spectrometer is 352.38 m/s. The difference in the diameters of the orbits of the 238U and 235U ions can be calculated using the formula for the radius of a charged particle's orbit in a magnetic field.
Explanation:
To find the speed of the ions entering the mass spectrometer, we can use the equation for the force on a charged particle in a magnetic field: F = qvB, where F is the force, q is the charge, v is the velocity, and B is the magnetic field. The force is equal to the electric force, qE, so we can set these two equations equal to each other and solve for v: qvB = qE. Since q is the charge of the ion and E is the electric field strength, we can rearrange this equation to solve for v, giving us v = E/B.
Plugging in the values given in the question, we find that the speed of the ions entering the mass spectrometer is 148 V / 0.42 T = 352.38 m/s.
To find the difference in the diameters of the orbits of the 238U and 235U ions, we can use the formula for the radius of a charged particle's orbit in a magnetic field: r = mv / (qB), where m is the mass of the ion, v is its velocity, q is its charge, and B is the magnetic field. Since the ions are singly ionized, their charges are +e, where e is the charge of an electron. Plugging in the values given in the question, we can calculate the radius of the orbits of the 238U and 235U ions, and then subtract these values to find the difference in their diameters.
A 70.0-cm length of a cylindrical silver wire with a radius of 0.175 mm is extended horizontally between two leads. The potential at the left end of the wire is 3.15 V, and the potential at the right end is zero. The resistivity of silver is 1.586 ✕ 10−8 Ω · m.
a. What are the magnitude and direction of the electric field in the wire?
b. What is the resistance of the wire?
c. What are the magnitude and direction of the current in the wire?
d. What is the current density in the wire?
Answer:
(a) the magnitude and direction of the electric field in the wire is 4.5 N/C towards the left end of the wire.
(b) the resistance of the wire is 0.1154 Ω
(c) the magnitude and direction of the current in the wire is 27.3 A towards the left end of the wire
(d) the current density in the wire is 4.053 x 10⁸ A/m³
Explanation:
Given;
Length of cylinder = 70cm = 0.7m
radius of cylinder = 1.75 x 10⁻⁴ m
potential V = 3.15 V
resistivity = 1.586 ✕ 10⁻⁸ Ω · m
Part (a) the magnitude and direction of the electric field in the wire
V = E x d
E = V/d
E = 3.15/0.7 = 4.5 N/C towards the left end of the wire.
Part (b) the resistance of the wire
[tex]R = \frac{\rho L}{A} \\\\R =\frac{\rho L}{\pi r^2} =R = \frac{1.586X10^{-8} X 0.7}{\pi (1.75X10^{-4})^2} = 0.1154 ohms[/tex]
R = 0.1154 Ω
Part (c) the magnitude and direction of the current in the wire
V = IR
I =V/R
I = 3.15/0.1154
I = 27.3 A towards the left end of the wire
Part (d) the current density in the wire
current density,J = current /volume
volume = πr²h = π x (1.75 × 10⁻⁴)² x 0.7 = 6.7357 x 10⁻⁸ m³
[tex]J = \frac{27.3}{6.7357 X10^{-8}} = 4.053 X10^8 \frac{A}{m^3}[/tex]
J = 4.053 x 10⁸ A/m³
Determine the effective spring constant of the suspension system of a car. Consider a load of 4 passengers, each with a mass of 70 kg. The car suspension system consists of four identical springs, each with a spring constant k. When all four people get into the car, the tires are depressed by about Δx = 2.0 cm.
Answer:
The value of spring constant for each spring of the suspension system of a car K = 343.35 [tex]\frac{N}{cm}[/tex]
Explanation:
Total force on the springs = weight of the four passengers
⇒ F = 4 × 70 × 9.81
⇒ F = 2746.8 N
In the suspension system of the car the four springs are connected in parallel. So Equivalent spring constant is given by,
⇒ [tex]K_{eq}[/tex] = 4 K -------- ( 1 )
Depression in the spring Δx = 2 cm
Now the force on the spring is given by
F = [tex]K_{eq}[/tex] × Δx
⇒ [tex]K_{eq}[/tex] = [tex]\frac{2746.8}{2}[/tex]
⇒ [tex]K_{eq}[/tex] = 1373.4 [tex]\frac{N}{cm}[/tex]
Now the spring constant for each spring = [tex]\frac{K_{eq}}{4}[/tex]
⇒ K = [tex]\frac{1373.4}{4}[/tex]
⇒ K = 343.35 [tex]\frac{N}{cm}[/tex]
This is the value of spring constant for each spring of the suspension system of a car.
The effective spring constant of the suspension system of a car. is mathematically given as
K = 343.35N/cm
What is the effective spring constant of the suspension system of a car.?
Question Parameter(s):
Consider a load of 4 passengers, each with a mass of 70 kg.
the tires are depressed by about dx = 2.0 cm.
Generally, the equation for the Force is mathematically given as
F = Keq × dx
Therefore
Keq= 2746.8/2
Keq= 1373.4N/cm
In conclusion, spring constant for each spring
K = 1373.4/4
K = 343.35N/cm
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A 2 kg package is released on a 53.1° incline, 4 m from a long spring with force constant k = 140 N/m that is attached at the bottom of the incline (Fig. 7-32). The coefficients of friction between the package and the incline are µs = 0.4 and µk= 0.2. The mass of the spring is negligible.A. What is the speed of the package just before it reaches the spring?
B. What is the maximum compression of the spring?
C. The package rebounds back up the incline. How close does it get to its initial position?
The problem can be addressed using energy conservation principles, factoring in gravitational potential, kinetic, and elastic potential energies, and work done by friction. Calculations will yield the speed of the package before hitting the spring, the maximum compression of the spring, and how close the package gets to its initial position on its return trip.
The student has presented a physics problem involving dynamics, energy conservation, and spring compression. This type of problem can be solved using the principles of mechanics, specifically the conservation of mechanical energy and motion on an incline with friction.
To find the speed of the package just before it reaches the spring, we would apply conservation of energy, taking into account the work done by friction. Initially, the package has potential energy due to its height on the incline, and this potential energy is converted into kinetic energy and work done against friction as the package slides down.
To determine the maximum compression of the spring, we would again use conservation of energy, where now the kinetic energy of the package is converted into elastic potential energy stored in the spring at maximum compression.
The package's return journey up the incline involves energy transformation from the spring's potential energy back to the package's kinetic energy and finally to gravitational potential energy. The effects of kinetic friction will again play a role in how far the package travels back up the incline.
- (A) Speed before hitting the spring: [tex]\( 7.3 \, \text{m/s} \)[/tex]
- (B) Maximum compression of the spring: [tex]\( 0.855 \, \text{m} \)[/tex]
- (C) Distance from initial position after rebound: [tex]\( 1.2 \, \text{m} \)[/tex]
To solve this problem, we'll address each part step-by-step.
Part A: Speed of the Package Just Before It Reaches the Spring
First, we need to find the acceleration of the package as it slides down the incline, considering friction.
Forces Acting on the Package:
1. Gravitational force parallel to the incline: [tex]\( F_g = mg \sin(\theta) \)[/tex]
2. Frictional force opposing the motion: [tex]\( F_f = \mu_k mg \cos(\theta) \)[/tex]
Calculating the Net Force:
[tex]\[ F_{\text{net}} = mg \sin(\theta) - \mu_k mg \cos(\theta) \][/tex]
[tex]\[ F_{\text{net}} = mg (\sin(\theta) - \mu_k \cos(\theta)) \][/tex]
Net Force Calculation:
[tex]\[ F_{\text{net}} = 2 \times 9.8 \times (\sin(53.1^\circ) - 0.2 \cos(53.1^\circ)) \]\[ \sin(53.1^\circ) \approx 0.8 \]\[ \cos(53.1^\circ) \approx 0.6 \]\[ F_{\text{net}} = 2 \times 9.8 \times (0.8 - 0.2 \times 0.6) \]\[ F_{\text{net}} = 2 \times 9.8 \times (0.8 - 0.12) \]\[ F_{\text{net}} = 2 \times 9.8 \times 0.68 \]\[ F_{\text{net}} = 13.328 \, \text{N} \][/tex]
Acceleration:
[tex]\[ a = \frac{F_{\text{net}}}{m} = \frac{13.328}{2} = 6.664 \, \text{m/s}^2 \][/tex]
Using Kinematics to Find Final Speed:
The package starts from rest and travels 4 meters.
[tex]\[ v^2 = u^2 + 2as \]\[ v^2 = 0 + 2 \times 6.664 \times 4 \]\[ v^2 = 53.312 \]\[ v = \sqrt{53.312} \]\[ v \approx 7.3 \, \text{m/s} \][/tex]
Part B: Maximum Compression of the Spring
When the package hits the spring, it compresses the spring until all kinetic energy is converted to potential energy in the spring and work done against friction.
Initial Kinetic Energy:
[tex]\[ KE_{\text{initial}} = \frac{1}{2} m v^2 = \frac{1}{2} \times 2 \times (7.3)^2 \]\[ KE_{\text{initial}} = \frac{1}{2} \times 2 \times 53.29 \]\[ KE_{\text{initial}} = 53.29 \, \text{J} \][/tex]
Work Done Against Friction During Compression:
Let's denote the compression of the spring as [tex]\( x \)[/tex].
[tex]\[ F_f = \mu_k m g \cos(\theta) = 0.2 \times 2 \times 9.8 \times 0.6 = 2.352 \, \text{N} \][/tex]
The work done by friction:
[tex]\[ W_f = F_f \cdot d = 2.352 \cdot x \][/tex]
Spring Potential Energy:
[tex]\[ PE_{\text{spring}} = \frac{1}{2} k x^2 = \frac{1}{2} \times 140 \times x^2 = 70x^2 \][/tex]
Energy Balance:
[tex]\[ KE_{\text{initial}} = PE_{\text{spring}} + W_f \]\[ 53.29 = 70x^2 + 2.352x \][/tex]
Solving this quadratic equation:
[tex]\[ 70x^2 + 2.352x - 53.29 = 0 \][/tex]
Using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
[tex]\[ x = \frac{-2.352 \pm \sqrt{2.352^2 - 4 \times 70 \times (-53.29)}}{2 \times 70} \]\[ x = \frac{-2.352 \pm \sqrt{5.533 + 14921.2}}{140} \]\[ x = \frac{-2.352 \pm \sqrt{14926.733}}{140} \]\[ x = \frac{-2.352 \pm 122.1}{140} \][/tex]
Taking the positive root:
[tex]\[ x = \frac{119.748}{140} \]\[ x \approx 0.855 \, \text{m} \][/tex]
Part C: Distance Package Rebounds Back Up the Incline
Potential Energy in Spring:
[tex]\[ PE_{\text{spring}} = 70x^2 = 70 \times (0.855)^2 = 70 \times 0.731 \approx 51.17 \, \text{J} \][/tex]
Work Done Against Friction During Rebound:
[tex]\[ W_f = F_f \cdot d = 2.352 \times 0.855 = 2.01 \, \text{J} \][/tex]
Total energy after rebound:
[tex]\[ PE_{\text{spring}} - W_f = 51.17 - 2.01 = 49.16 \, \text{J} \][/tex]
Kinetic Energy at Rebound:
[tex]\[ KE_{\text{rebound}} = 49.16 \, \text{J} \][/tex]
Using Kinematics:
[tex]\[ KE_{\text{rebound}} = \frac{1}{2} m v^2 \]\[ 49.16 = \frac{1}{2} \times 2 \times v^2 \]\[ 49.16 = v^2 \]\[ v = \sqrt{49.16} \approx 7.01 \, \text{m/s} \][/tex]
The package travels back up the incline with this velocity until it stops due to friction and gravity.
Distance Traveled Up the Incline:
Using energy conservation:
[tex]\[ KE_{\text{rebound}} = mgh + W_f \]\[ 49.16 = 2 \times 9.8 \times h + 2.352 \times h \]\[ 49.16 = 19.6h + 2.352h \]\[ 49.16 = 21.952h \]\[ h = \frac{49.16}{21.952} \approx 2.24 \, \text{m} \][/tex]
The distance along the incline:
[tex]\[ d = \frac{h}{\sin(\theta)} \approx \frac{2.24}{0.8} \approx 2.8 \, \text{m} \][/tex]
Distance from Initial Position:
[tex]\[ \text{Distance from initial position} = 4 - 2.8 = 1.2 \, \text{m} \][/tex]
The complete question is:
A 2 kg package is released on a 53.1° incline, 4 m from a long spring with force constant k = 140 N/m that is attached at the bottom of the incline (Fig. 7-32). The coefficients of friction between the package and the incline are µs = 0.4 and µk= 0.2. The mass of the spring is negligible.
A. What is the speed of the package just before it reaches the spring?
B. What is the maximum compression of the spring?
C. The package rebounds back up the incline. How close does it get to its initial position?
A large block is being pushed against a smaller block such that the smaller block remains elevated while being pushed. The mass of the smaller block is m = 0.45 kg. It is found through repeated experimentation that the blocks need to have a minimum acceleration of a = 13 m / s 2 in order for the smaller block to remain elevated and not slide down. What is the coefficient of static friction between the two blocks?
Explanation:
According to the free body diagram a block of mass m will have expression for force as follows.
N = ma
and, [tex]f_{c} - mg[/tex] = 0
[tex]\mu_{s}N - mg[/tex] = 0
[tex]\mu_{s} = \frac{mg}{N}[/tex] = [tex]\frac{mg}{ma}[/tex]
= [tex]\frac{g}{a}[/tex]
= [tex]\frac{9.8}{13}[/tex]
= 0.75
Therefore, we can conclude that the value of coefficient of static friction between the two blocks is 0.75.
The coefficient of static friction between the large block and smaller block is equal to 0.754.
Given the following data:
Mass of smaller block (m) = 0.45 kg.Acceleration (a) = [tex]13 \;m/s^2[/tex]Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]
To determine the coefficient of static friction between the large block and smaller block:
A force of static friction can be defined as the frictional force that resists the relative motion of two (2) surfaces.
Hence, a force of static friction is a frictional force that keeps an object at rest or stationary rather than being in relative motion.
Mathematically, the force of static friction is given by the formula;
[tex]Fs = uFn[/tex]
Where;
Fs represents the force of static friction.μ represents the coefficient of friction.Fn represents the normal force.For these block systems, the forces acting on them is given by:
[tex]uma - mg = 0\\\\uma = mg\\\\ua =g\\\\u=\frac{g}{a}[/tex]
Substituting the parameters into the formula, we have;
[tex]u=\frac{9.8}{13}[/tex]
u = 0.754
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You give a book a brief push to make it slide up a rough inclined surface. It comes to a stop and slides back down to the starting point.
Does it take the same amount of time to go up as it does to come down?
Answer:No, it will take a longer time to slide up than to slide down.
Explanation: An inclined plane is a plane that slides at an angle to the ground,this angle can be 45 degrees,60degrees etc. WHEN AN OBJECT IS SLIDING UP A ROUGH INCLINED SURFACE, IT WILL HAVE TO OVERCOME THE FRICTION OR RESISTANCE OF THE ROUGH SURFACE AND THE OPPOSITION OF THE FORCE IF GRAVITY, WHICH WILL CAUSE IT TO TAKE A LONGER TIME TO GO UP WHILE WHEN SLIDING DOWN IT WILL ONLY HAVE TO OVERCOME THE FRICTION OF THE ROUGH SURFACE AS THE FORCE OF GRAVITY ACTS TO PULL IT DOWN THE SLOPE.
Assuming the passive sign convention and an operating frequency of 314 rad/s, calculate the phasor voltage V which appears across each of the following when driven by the phasor current I = 1020° mA: (a) a 2 a resistor; (b) a 1 F capacitor; (c) a 1 Hinductor; (d) a 2 resistor in series with a 1 F capacitor; (e) a 2. resistor in series with a 1 H inductor. (f) Calculate the instantaneous value of each voltage determined in part (e) at t = 0.
Answer:
note:
solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word because to different version of MS Office please find the attachment
A wire of length L is wound into a square coil with 167 turns and used in a generator that operates at 60.0 Hz and 120 V rms value in a 0.041-T magnetic field. What is the length L of the wire used to construct the coil
Answer:
[tex]171.43m[/tex]
Explanation:
First, define [tex]emf[/tex]( electromotive force )-is the unit electric charge imparted by an energy source. In this case the generator.
The peak emf is:
[tex]E_p_e_a_k=\sqrt2(E_m_a_x)[/tex]=[tex]\sqrt(2\times 120V)=170V[/tex]
Substituting [tex]w=2\pi f[/tex] and the value for peaf [tex]E[/tex] gives:
Total length=[tex]4\sqrt {\frac{NE_p_e_a_k}{Bw}[/tex]=[tex]4\sqrt{\frac{167\times 170}{0.041T\times 2pi \times 60.0Hz}[/tex]
=[tex]171.43m[/tex]
Hence, wire's length is 171.43m
While traveling on a horizontal road at speed vi, a driver sees a large rabbit ahead and slams on the brakes. The wheels lock and the car begins to slide against the road. The car collides with the rabbit at a final speed of vf, after T seconds of braking. What is the coefficient of kinetic friction of the tires against the road?
Answer:
μk = (Vf - Vc)/(T×g)
Explanation:
Given
Vi = initial velocity of the car
Vf = final velocity of the car
T = Time of application of brakes
g = acceleration due to gravity (known constant)
Let the mass of the car be Mc
Assuming only kinetic frictional force acts on the car as the driver applies the brakes,
The n from Newtown's second law of motion.
Fk = Mc×a
Fk = μk×Mc×g
a = (Vf - Vc)/T
Equating both preceding equation.
μk×Mc×g = Mc × (Vf - Vc)/T
Mc cancels out.
μk = (Vf - Vc)/(T×g)
Final answer:
To find the coefficient of kinetic friction of the tires against the road, you need the initial and final speeds of the car, the time it took to decelerate, and the acceleration due to gravity. Without specific values for these variables, the problem cannot be solved directly, but the formula given provides a method to calculate the coefficient if such values are known.
Explanation:
The question asks for the coefficient of kinetic friction between the tires and the road when a car, initially traveling at speed vi, decelerates to a final speed vf over time T seconds due to the driver slamming on the brakes upon seeing a rabbit.
The formula to calculate the coefficient of kinetic friction (μ_k) can be derived from Newton's second law of motion and the equation of motion that relates initial velocity, final velocity, acceleration, and time. The formula for the coefficient of kinetic friction is μ_k = (vi - vf) / (g × T), where g is the acceleration due to gravity (9.81 m/s2).
To solve this problem, you need the initial and final velocities of the car (vi and vf), the deceleration time (T), and knowledge that the acceleration due to gravity (g) is approximately 9.81 m/s2. However, actual calculations cannot be performed without specific values for vi, vf, and T.
This equation illustrates that the coefficient of kinetic friction is directly related to the deceleration rate of the car on the road surface.
ball of mass 0.4 kg is attached to the end of a light stringand whirled in a vertical circle of radius R = 2.9 m abouta fixed point. Find the magnitude of the tension when themass is at the top if its speed at the top is 8.5 m/s.(
Answer:
[tex]6.046N[/tex]
Explanation:
The net force exerted on the mass is the sum of tension force and the external force of gravity.
[tex]F_n_e_t=F_g+F_t[/tex]
[tex]F_t[/tex] is the tension force.[tex]F_g=9.8N/kg[/tex] is the force of gravity.
[tex]F_n_e_t=ma_c=mv^2/r\\[/tex]
where [tex]r[/tex] is the rope's radius from the fixed point.
From the net force equation above:
[tex]F_t=F_n_e_t-F_g\\=mv^2/r-mg\\=0.4\times(8.5^2/2.9)-0.4\times9.8\\=6.046N[/tex]
Hence the tension force is 6.046N
A 4.0 cm × 4.2 cm rectangle lies in the xy-plane. You may want to review (Pages 664 - 668) . Part A What is the electric flux through the rectangle if E⃗ =(150ı^−200k^)N/C? Φe = N⋅m2/C Previous AnswersRequest Answer Incorrect; Try Again; 5 attempts remaining Part BPart complete What is the electric flux through the rectangle if E⃗ =(150ı^−200ȷ^)N/C? Φe = 0 N⋅m2/C
Answer:
Explanation:
Area, A = 4 cm x 4.2 cm = 16.8 cm²
A).
[tex]\overrightarrow{E}=150\widehat{i}-200\widehat{k}[/tex]
Area is in x y plane so
[tex]\overrightarrow{A}=16.8\times 10^{-4}\widehat{k}[/tex]
Electric flux,
[tex]\phi =\overrightarrow{E}.\overrightarrow{A}[/tex]
[tex]\phi =\left ( 150\widehat{i}-200\widehat{k} \right ).\left (16.8\times 10^{-4}\widehat{k} \right )[/tex]
Ф = 0.336 Nm²/C
B).
[tex]\overrightarrow{E}=150\widehat{i}-200\widehat{j}[/tex]
[tex]\phi =\overrightarrow{E}.\overrightarrow{A}[/tex]
[tex]\phi =\left ( 150\widehat{i}-200\widehat{j} \right ).\left (16.8\times 10^{-4}\widehat{k} \right )[/tex]
Ф = 0 Nm²/C
You need to push a heavy box across a rough floor, and you want to minimize the average force applied to the box during the time the box is being pushed. Which method of pushing results in the minimum average force being applied to the box?
A. Keep pushing the box forward at a steady speed.
B. Push the box forward a short distance, rest, then repeat until finished,
C. Push the box so that it accelerates forward at a constant rate.
Answer:A. Keep pushing the box forward at a steady speed.
Explanation: Frictional force is a resistant force which oppose the Movement of an object, frictional force can emanate from a rough surface.
When an object that is moving with a consistent force is opposed by the roughness of the surface through which it is moving,it will cause the object to continue to move with a reduced speed as it goes along.
WHEN YOU APPLY A CONSTANT FORCE ON A MOVING OBJECT THAT IS OPPOSED BY A ROUGH SURFACE IT WILL RESULT IN AN AVERAGE MINIMAL FORCE BEING APPLIED TO THE OBJECT.
The most soaring vocal melody is in Johann Sebastian Bach's Mass in B minor. In one section, the basses, tenors, altos, and sopranos carry the melody from a low D to a high A. In concert pitch, these notes are now assigned frequencies of 146.8 Hz and 880.0 Hz. (Use 343 m/s as the speed of sound, and 1.20 kg/m3 as the density of air.)
Find the wavelength of the initial note.
Answer:
2.33651226158 m
Explanation:
From the question the required data is as follows
f = Frequency of the initial note = 146.8 Hz
v = Velocity of sound in air = 343 m/s
The wavelength of a wave is given by
[tex]\lambda=\dfrac{v}{f}[/tex]
[tex]\Rightarrow \lambda=\dfrac{343}{146.8}[/tex]
[tex]\Rightarrow \lambda=2.33651226158\ m[/tex]
The wavelength of the initial note is 2.33651226158 m
You have a 78.7 mF capacitor initially charged to a potential difference of 11.5 V. You discharge the capacitor through a 3.03 Ω resistor. What is the time constant?
Answer:
[tex]\tau=0.23\;second[/tex]
Explanation:
Given,
[tex]C=78.7\;mF\\V=11.5\;V\\R=3.03\;\Omega\\[/tex]
Time constant
[tex]\tau=RC\\\tau=78.7\times10^{-3}\times3.03\\\tau=238.461\times10^{-3}\;second\\\tau=0.23\;second[/tex]
If the moment acting on the cross section is M=630N⋅m, determine the maximum bending stress in the beam. Express your answer to three significant figures and include appropriate units.
Answer:
2.17 Mpa
Explanation:
The location of neutral axis from the top will be
[tex]\bar y=\frac {(240\times 25)\times \frac {25}{2}+2\times (20\times 150)\times (25+(\frac {150}{2}))}{(240\times 25)+2\times (20\times 150)}=56.25 mm[/tex]
Moment of inertia from neutral axis will be given by [tex]\frac {bd^{3}}{12}+ ay^{2}[/tex]
Therefore, moment of inertia will be
[tex]\frac {240\times 25^{3}}{12}+(240\times 25)\times (56.25-25/2)^{2}+2\times [\frac {20\times 150^{3}}{12}+(20\times 150)\times ((25+150/2)-56.25)^{2}]=34.5313\times 10^{6} mm^{4}}[/tex]
Bending stress at top= [tex]\frac {630\times 10^{3}\times (175-56.25)}{34.5313\times 10^{6}}=2.1665127\approx 2.17 Mpa[/tex]
Bending stress at bottom=[tex]\frac {630\times 10^{3}\times 56.25}{34.5313\times 10^{6}}=1.026242858\approx 1.03[/tex] Mpa
Comparing the two stresses, the maximum stress occurs at the bottom and is 2.17 Mpa
Without the shape and dimensions of the beam's cross-section, we cannot accurately calculate the maximum bending stress from the given moment of 630 N·m.
Explanation:To determine the maximum bending stress in the beam with a moment acting on the cross section (M) of 630 N·m, we need to use the formula for bending stress, which is σ = M·c/I, where σ is the stress, M is the moment, c is the distance from the neutral axis to the outer fiber, and I is the moment of inertia of the beam section. Unfortunately, we do not have the values for c and I in the question provided. For a circular cross section, c would be the radius, and I can be calculated with π·r⁴/4 where r is the radius. However, without additional information such as the shape and dimensions of the cross-sectional area of the beam, we cannot proceed further.
Since the given moment is 630 N⋅m, we need to find the values of c and I for the specific beam. Once we have these values, we can substitute them into the formula to calculate the maximum bending stress in the beam.
Without further information about the cross section of the beam, we cannot determine the exact values of c and I, and therefore, we cannot calculate the maximum bending stress.
Two large metal plates are separated by 2.67 cm. One plate has a negative charge, and the other plate has a positive charge. The negative plate is heated until an electron barely comes loose from its surface. The electron then accelerates between the plates and strikes the positive plate moving at a speed of 1.32x107 m/s. What is the voltage between the plates
Answer:
voltage between the plates is 4.952 × [tex]10^{-26}[/tex] V
Explanation:
given data
plate separated distance = 2.67 cm
electron speed = 1.32 × [tex]10^{7}[/tex] m/s
solution
we will get here first force that is express as
force in parallel plate F = [tex]\frac{eV}{d}[/tex] ..............1
and force by Newton second law F = ma .............2
equate equation 1 and 2
ma = [tex]\frac{eV}{d}[/tex] .................3
and here we know as kinematic equation
v²- u² = 2 × a × s ...........4
so for initial speed acceleration will be
a = [tex]\frac{v^2-u^2}{2\times s}[/tex]
a = [tex]\frac{(1.32 \times 10^7)^2}{2\times 2.67 \times 10^{-2}}[/tex]
a = 3.262 × [tex]10^{-13}[/tex] m/s²
now we put a in equation 3 and we get v
ma = [tex]\frac{eV}{d}[/tex]
9.1093 × [tex]10^{-31}[/tex] × 3.262 × [tex]10^{-13}[/tex] = [tex]\frac{1.602 \times 10^{-19} V}{2.67 \times 10^{-2}}[/tex]
solve it we get
v = 4.952 × [tex]10^{-26}[/tex] V
Nitrogen at an initial state of 300 K, 150 kPa, and 0.2 m3is compressed slowly in an isothermal process to a final pressure of 800 kPa. Determine the work done during this process.
Answer:
[tex]W=-251096.465\ J[/tex] negativesign denotes thatthe work is consumed by the system.
Explanation:
Given:
Isothermal process.
initial temperature, [tex]T_1=300\ K[/tex]
initial pressure, [tex]P_1=150kPa[/tex]
initial volume, [tex]V_1=0.2\ m^3[/tex]
final pressure, [tex]P_2=800\ kPa[/tex]
The work done during an isothermal process is given by:
[tex]W=P_1.V_1\times ln(\frac{P_1}{P_2} )[/tex]
[tex]W=150\times 1000\times \ln\frac{150}{800}[/tex]
[tex]W=-251096.465\ J[/tex] negativesign denotes thatthe work is consumed by the system.
A piston–cylinder device initially contains 2 L of air at 100 kPa and 25°C. Air is now compressed to a final state of 600 kPa and 150°C. The useful work input is 1.2 kJ. Assuming the surroundings are at 100 kPa and 25°C, determine (a) the exergy of the air at the initial and the final states, (b) the minimum work that must be supplied to accomplish this compression process, and (c) the second-law efficiency of this process.
Answer:
a. The energy of the air at the initial and the final states is 0kJ and 0.171kJ respectively
b. 0.171kJ
c. 0.143
Explanation:
a.
Because there are same conditions of the state of air at the surroundings and at the Initial stage, the energy of air at the Initial stage is 0kJ.
Calculating energy at the final state;
We start by calculating the specific volume of air in the environment and at the final state.
U2 = At the final state, it is given by
RT2/P2
U1= At the Initial state, it is given by
RT1/P1
Where R = The gas constant of air is 0.287 kPa.m3/kg
T2 = 150 + 273 = 423K
T1 = 25 + 273 = 298K
P2 = 600KPa
P1 = 100KPa
U2 = 0.287 * 423/600
U2 = 0.202335m³/kg
U1 = 0.287 * 298/100
U1 = 0.85526m³/kg
Then we Calculate the mass of air using ideal gas relation
PV = mRT
m = P1V/RT1 where V = 2*10^-3kg
m = 100 * 2 * 10^-3/(0.287 * 298)
m = 0.00234kg
Then we calculate the entropy difference, ∆s. Which is given by
cp2 * ln(T2/T1) - R * ln(P2/P1)
Where cp2 = cycle constant pressure = 1.005
∆s = 1.005 * ln (423/298) - 0.287 * ln(600/100)
∆s = -0.1622kJ/kg
Energy at the final state =
m(E2 - E1 + Po(U2 - U1) -T0 * ∆s)
E2 and E1 are gotten from energy table as 302.88 and 212.64 respectively
Energy at the final state
= 0.00234(302.88 - 212.64 + 100(0.202335 - 0.85526) - 298 * -0.1622)
Energy at the final state = 0.171kJ
b.
Minimum Work = ∆Energy
Minimum Work = Energy at the final state - Energy at the initial state
Minimum Work = 0.171 - 0
Minimum Work done = 0.171kJ
c. The second-law efficiency of this process is calculated by ratio of meaningful and useful work
= 0.171/1.2
= 0.143
Exergy measures the maximum work a system can produce. To calculate its change and consequently the minimum work supplied and second-law efficiency, additional data like specific heats are required.
Explanation:This involves thermodynamics, a branch of physics that deals with energy transfer. Specifically, this question is about the concept of exergy, a measure of the maximum amount of work a system can produce with respect to its environment.
(a) The exergy (or available energy) of a system in a given state is the maximum theoretical work that can be obtained as the system communicates with an equilibrium state. In this case, the initial and final states of the system are given, but we need more data such as the specific heats, to compute the initial and final exergies.
(b) The minimum work that must be supplied is equivalent to the change in exergy from the initial to the final state, but again, it cannot be determined without knowing the specific heat values of air
(c) The second-law efficiency is defined as the ratio of the actual work to the work done in a reversible process. Here, it is the ratio of the useful work input (1.2 kJ) to the minimum work needed for the compression process. To find the exact efficiency, we need to compute the minimum required work, which would require the specific heat values.
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Under what circumstances are two circuits considered equivalent? A. their input values are the same B. the output of one is the inverse of the output of the other C. their output values are the same for all possible input combinations D. their output values are always 1 E. the input of one matches the output of the other
Two circuits are considered equivalent when their output values are the same for all possible input combinations.
A circuit equivalent to another is one that meets the same conditions, (eg same current), under a different configuration.
The equivalent circuit made in this way is not the same as the original one, but if the total voltages, the total currents and the total resistance of the circuit will be equal, which will be the equivalent.The concept of equivalent circuit is used when saying, for example, that a real generator is equivalent to an ideal one with its internal resistance in series.Therefore, we can conclude that two circuits are considered equivalent when their output values are the same for all possible input combinations.
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Final answer:
Two circuits are considered equivalent if they produce the same output for all possible input combinations, as can be demonstrated using a truth table in digital logic. The correct option is C.
Explanation:
Two circuits are considered equivalent when their output values are the same for all possible input combinations. This means that regardless of the input values, both circuits will produce the same output across all scenarios. For instance, in digital logic, two circuits using different logic gates such as AND, OR, NAND, or XOR would be equivalent if they yield the same output for each combination of inputs, as represented in a truth table. An example of this is the use of De Morgan's laws where an AND gate followed by a NOT gate can be equivalent to a NAND gate, producing the same outputs. Understanding these principles is essential for fields like electronics, computer engineering, and programming. Hence, Option C is correct.
A liquid in a test tube has a curved surface such that the edges touching the glass are higher than the surface at the center. This must mean that the cohesive forces are less than the adhesive forces. 1. False 2. True
Answer:
True
Explanation:
Cohesive forces are forces that exist between the molecules of a substance of the same material while adhesive forces are forces that exists between the molecules of the substances of different materials.
This is responsible for the nature of menisci formed by different liquids when they are filled into containers or glass tubes.
If the cohesive forces of the liquid molecules are stronger than the adhesive forces between the liquid molecules and the glass material, a convex meniscus will be formed. This means that the edges of the liquid touching the glasses are lower than the surface at the centre. The meniscus formed by mercury in a glass tube is an example of this.
However, If the cohesive forces of the liquid molecules are weaker than the adhesive forces between the liquid molecules and the glass material, a concave meniscus will be formed. This means that the edges of the liquid touching the glasses are higher than the surface at the centre. The meniscus formed by water in a glass tube is an example of this.
Please view the attached diagram:
Please note: I got the diagram online, it was not drawn by me. I Just needed to quickly get something to illustrate my explanations. Thanks.
) A stone initially moving at 8.0 m/s on a level surface comes to rest due to friction after it travels 11 m. What is the coefficient of kinetic friction between the stone and the surface
Answer:
-0.3
Explanation:
F' = μmg ........... Equation 1
Where F' = Frictional force, μ = coefficient of kinetic friction, m = mass of the stone, g = acceleration due to gravity.
But,
F' = ma ............ Equation 2
Where a = acceleration of the stone.
Substitute equation 2 into equation 1
ma = μmg
dividing both side of the equation by m
a = μg
make μ the subject of the equation
μ = a/g............... Equation 3
From the equation of motion,
v² = u²+2as................. Equation 4
Where v and u are the final and the initial velocity respectively, s = distance.
Given: v = 0 m/s (to rest), u = 8.0 m/s, s = 11 m.
Substitute into equation 4
0² = 8² + 2×11×a
22a = -64
a = -64/22
a = -32/11 m/s² = -2.91 m/s²
substitute the values of a and g into equation 3
μ = -2.91/9.8
μ = -0.297
μ ≈ -0.3
Final answer:
The coefficient of kinetic friction is found using the work-energy principle by equating the initial kinetic energy of the stone to the work done by friction. Given that the stone travels 11 meters and comes to rest, the coefficient of kinetic friction is calculated as approximately 0.296.
Explanation:
To find the coefficient of kinetic friction between the stone and the surface, we need to use the work-energy principle, which states that the work done by all the forces acting on an object is equal to the change in its kinetic energy. Since the stone comes to rest, all of its initial kinetic energy has been converted into work done against friction.
First, let's calculate the initial kinetic energy (KE) of the stone:
KE = (1/2)mv²KE = (1/2)(8.0 m/s)²KE = 32 J (joules)This energy is equal to the work done by friction (Wf):
Wf = Frictional force (f) x Distance (d)
Since the frictional force is equal to the kinetic friction coefficient (µk) multiplied by the normal force (N), and the normal force is equal to the weight of the stone (mg, where g is the acceleration due to gravity), we can express the work done by friction as:
Wf = µkmgd
Setting the work done by friction equal to the initial kinetic energy gives us:
32 J = µkmg(11 m)
Solving for the coefficient of kinetic friction:
µk = 32 J / (mg x 11 m)
Now, assuming the acceleration due to gravity (g) is 9.8 m/s²:
µk = 32 J / (m x 9.8 m/s² x 11 m)
Since the mass (m) cancels out, we don't need to know it:
µk = 32 J / (9.8 m/s² x 11 m)
µk = 0.296
Therefore, the coefficient of kinetic friction between the stone and the level surface is approximately 0.296.
A very thin 19.0 cm copper bar is aligned horizontally along the east-west direction. If it moves horizontally from south to north at velocity = 11.0 m/s in a vertically upward magnetic field and B = 1.18 T , what potential difference is induced across its ends ? which end (east or west) is at a higher potential ? a) East b) West
Answer:
2.47 V,East
Explanation:
We are given that
l=19 cm=[tex]19\times 10^{-2} m[/tex]
[tex] 1 cm=10^{-2} m[/tex]
[tex]v=11 m/s[/tex]
B=1.18 T
We have to find the potential difference induced across its ends.
[tex]E=Bvl[/tex]
Using the formula
[tex]E=1.18\times 11\times 19\times 10^{-2}[/tex]
[tex]E=2.47 V[/tex]
Hence, the potential difference induces across its ends=2.47 V
The positive charge will move towards east direction and the negative charge will move towards west direction because the direction of force will be east.Therefore, the potential at east end will be high.
The induced potential difference in the copper bar is 2.47 V, with the east end being at a higher potential.
To determine the potential difference induced across the ends of a copper bar moving through a magnetic field, we use the formula:
V = B * l * v
where:
B is the magnetic field strength (1.18 T).l is the length of the bar (0.19 m).v is the velocity (11.0 m/s).Substituting the given values:
V = 1.18 T * 0.19 m * 11.0 m/s = 2.47 V
The potential difference across the ends of the bar is 2.47 V.
To determine which end is at a higher potential, we apply the right-hand rule. Pointing the thumb of your right hand in the direction of the velocity (north), and your fingers in the direction of the magnetic field (upwards), the palm points towards the force acting on positive charges (from west to east).
Therefore, the east end is at a higher potential.
The correct answer is: a) East
X rays of 25 keV are received with an X - ray intensifying screen that produces light photons at 425 nm. If the conversion effi ciency of intensifying screen is 20%, calculate how many light photons will be generated by an X - ray photon.
Answer:
Photons Generated= ~1715 photons
Explanation:
The detailed explanation of answer is given in attached file.
Mt. Asama, Japan, is an active volcano. In 2009, an eruption threw solid volcanic rocks that landed 1 km horizontally from the crater. If the volcanic rocks were launched at an angle of 40° with respect to the horizontal and landed 900 m below the crater, (a) what would be their initial velocity and (b) what is their time of flight?
Answer:
a) 69.3 m/s
b) 18.84 s
Explanation:
Let the initial velocity = u
The vertical and horizontal components of the velocity is given by uᵧ and uₓ respectively
uᵧ = u sin 40° = 0.6428 u
uₓ = u cos 40° = 0.766 u
We're given that the horizontal distance travelled by the projectile rock (Range) = 1 km = 1000 m
The range of a projectile motion is given as
R = uₓt
where t = total time of flight
1000 = 0.766 ut
ut = 1305.5
The vertical distance travelled by the projectile rocks,
y = uᵧ t - (1/2)gt²
y = - 900 m (900 m below the crater's level)
-900 = 0.6428 ut - 4.9t²
Recall, ut = 1305.5
-900 = 0.6428(1305.5) - 4.9 t²
4.9t² = 839.1754 + 900
4.9t² = 1739.1754
t = 18.84 s
Recall again, ut = 1305.5
u = 1305.5/18.84 = 69.3 m/s
A small ranger vehicle has a soft, ragtop roof. When the car is at rest, the roof is flat. When the car is cruising at highway speeds with its windows rolled up, does the roof bow upwards, remain flat, or bow downwards
Answer:
roof bow upwards
Explanation:
The top of the roof of the small ranger vehicle will bow upwards. This is as a result of gas pressure on the soft ragtop roof.
As air begins to fill the vehicle, pressure resonates in all direction proportionally. The pressure of the air will be greater than that which the roof can withstand and this forces the roof sky up. It is a common scene when we see roof of ragtop vehicles bowing upwards into the sky.Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3, 550 J/kg·K, 48 W/m·K), which is initially at a uniform temperature of 170°C and is to be heated to a minimum temperature of 550°C. Heating is effected in a gas-fired furnace, where products of combustion at [infinity] 800°C maintain a convection coefficient of 250 W/m2·K on both surfaces of the plate. How long should the plate be left in the furnace?
Answer:
The answer to the question is;
The plate be left in the furnace for 905.69 seconds.
Explanation:
To solve the question, we have to check the Bi number as follows
Bi = [tex]\frac{hL}{k} = \frac{250\frac{W}{m^{2} K} *0.05 m}{48\frac{W}{mK} } = 0.2604[/tex]
As the Bi number is > 0.1 we have to account for the variation of temperature with location in the mass.
We perform nonlumped analysis
The relation for heat transfer given by
Y = [tex]\frac{T_f-T_{inf}}{T_i- T_{inf}}[/tex]
=[tex]\frac{550-800}{170- 800}[/tex] = 0.3968 = C₁ exp (ζ₁² F₀)
where
C₁ and ζ₁ are coefficients of a series solution
We therefore look for the values of C₁ and ζ₁ from Bi tables to be
ζ₁ = 0.4801 +(0.26-0.25) (0.5218-0.4801)/(0.3-0.25) ≈ 0.4884 and
C₁ = 0.4801 +(0.26-0.25) (1.0450 - 1.0382)/(0.3-0.25) ≈ 1.03956 and
This gives the relation
0.3968 = 1.03956 exp (ζ₁² F₀)
or ζ₁² [tex](\frac{\alpha t}{L^2})[/tex]
where
α = Thermal diffusivity of solid = k/(ρ·c[tex]_p[/tex]) = [tex]\frac{48}{7830*550}[/tex] = 1.1146×10⁻⁵
c[tex]_p[/tex] = Specific heat capacity of solid at constant pressure = 550 J/kg·K
ρ = Density of the solid = 7830 kg/m³
=㏑[tex](\frac{0.3968 }{1.03956 })[/tex] = -0.9631 from where we have
t = [tex]\frac{0.9631 *0.05^{2} }{0.4884^2*1.11*10^{-5}}[/tex] = 905 seconds.
An initially uncharged 4.07 μF capacitor and a 7.71 k Ω resistor are connected in series to a 1.50 V battery that has negligible internal resistance. What is the initial current in the circuit, expressed in milliamperes?
Answer:
I₀ = 0.2 mA
Explanation:
Just after being connected, as the voltage between plates of a capacitor can't change instantanously, the initial voltage through the capacitor must be zero, so it presents like a short to the battery.So, in these circumstances, as the battery only "sees"a resistor, the initial current can be found applying Ohm's law to the resistor, as follows:[tex]V = I_{0}*R \\\\ I_{0} = \frac{V}{R} = \frac{1.50V}{7.71e3\Omega} = 0.2 mA[/tex]
The initial current (that will be diminishing as the capacitor charges), is 0.2 mA.0.19mA
Explanation:Given;
Capacitor of capacitance = 4.07 μF
Resistor of resistance = 7.71 kΩ = 7.71 x 1000Ω = 7710Ω
Voltage = 1.50V
Since the capacitor is initially uncharged, it behaves like a short circuit. Therefore, the only element drawing current at that instant is the resistor. This means that the initial current in the circuit is the one due to (flowing through) the resistor.
And since there is negligible internal resistance, the emf of the battery is equal to the voltage supplied by the battery and is used to supply current to the resistor. Therefore, according to Ohm's law;
V = I x R ---------------(i)
Where;
V = voltage supplied or the emf
I = current through the resistor
R = resistance of the resistor
Substitute the values of V and R into equation (i) as follows;
1.50 = I x 7710
I = [tex]\frac{1.5}{7710}[/tex]
I = 0.00019A
Multiply the result by 1000 to convert it to milliamperes as follows;
0.00019 x 1000 mA = 0.19mA
Therefore, the initial current in the circuit, expressed in milliamperes is 0.19
"It is not correct to say that a body contains a certain amount of heat, yet a body can transfer heat to another body. How can a body give away something it does not have in the first place
Answer:
Because heat is a path function or the energy in transit.
Explanation:
It is not correct to say that a body contains a certain amount of heat because the heat is a path function and not a property of the system. It is the energy in transit which can be encountered only when it crosses the system boundary.Heat is the energy in transit of a matter which flows by the virtue of temperature difference. The heat energy in a body is stored in the form of kinetic energy of the molecules which gets converted into heat that we know as the responsible factor for the rise in temperature usually.Heat is not a substance a body possesses; it's energy transfer. Describing heat as something a body "contains" is inaccurate. Heat is energy in transit during temperature differences.
The statement that a body contains a certain amount of heat is inaccurate in the context of thermodynamics. Heat is not a substance that a body can possess like a tangible quantity. Instead, heat is a form of energy transfer between systems due to a temperature difference. When we say a body transfers heat, it implies an exchange of thermal energy between the body and its surroundings.
A body does not "contain" heat in the way it contains mass or volume. Rather, it possesses internal energy, and heat is the energy in transit. This distinction is crucial in understanding the principles of thermodynamics. When a body transfers heat, it signifies a change in its internal energy, which could result from molecular motion or other energy interactions.
Therefore, the expression "giving away something it does not have" is a misconception. The body has internal energy, and during a heat transfer, this internal energy changes, affecting the body's temperature. It is crucial to frame discussions about heat in terms of energy transfer rather than possession, aligning with the principles of thermodynamics.
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The question probable may be:
It is not correct to say that a body contains a certain amount of heat, yet a body can transfer heat to another body. How can a body give away something it does not have in the first place?
A hot water stream at 80 oC enters a mixing chamber with mass flow rate of 3.6 kg/s and mixed with cold water at 20 oC. If the mixture temperature is 42 oC, determine the mass flow rate of cold stream in kg/min. Assume all the streams being at 250 kPa
Explanation:
The mixing chamber will be well insulated when steady operating conditions exist such that there will be negligible heat loss to the surroundings. Therefore, changes in the kinetic and potential energies of the fluid streams will be negligible and there are constant fluid properties with no work interactions.
[tex]T < T_{sat}[/tex] at 250 kPa = [tex]127.41^{o}C[/tex]
[tex]h_{1}[/tex] approx equal to [tex]h_{f}[/tex] at [tex]80^{o}C[/tex]
= 335.02 kJ/kg
[tex]h_{2}[/tex] ≈ [tex]h_{f}[/tex] at [tex]20^{o}C[/tex]
= 83.915 kJ/kg
and, [tex]h_{3}[/tex] ≈ [tex]h_{f}[/tex] at [tex]42^{o}C[/tex] = 175.90 kJ/kg
Therefore, mass balance will be calculated as follows.
[tex]m^{o}_{in} - m^{o}_{out} = \Delta m^{o}_{system} \rightarrow m^{o}_{1} + m^{o}_{2} = m^{o}_{3}[/tex]
And, energy balance will be given as follows.
[tex]E^{o}_{in} - E^{o}_{out} = \Delta E^{o}_{system}[/tex]
As we are stating steady conditions,
[tex]\Delta m^{o}_{system}[/tex] and [tex]\Delta E^{o}_{system}[/tex] cancel out to zero.
So, [tex]E^{o}_{in} = E^{o}_{out}[/tex]
[tex]m^{o}_{1}(h_{1}) + m^{o}_{2}(h_{2}) = m^{o}_{3}(h_{3})[/tex]
On combining the relations, we solve for [tex]m^{o}_{2}[/tex] as follows.
[tex]m^{o}_{1}(h_{1}) + m^{o}_{2}(h_{2}) = (m^{o}_{1} + m^{o}_{2})(h_{3})[/tex]
[tex]m^{o}_{2} = (\frac{(h_{1} - h_{3})}{(h_{3} - h_{2})}) \times m^{o}_{1}[/tex]
= [tex]\frac{(335.02 - 175.90)}{(175.90 - 83.915)} \times 0.5[/tex]
[tex]m^{o}_{2}[/tex] = 0.865 kg/s
= 51.9 kg/min (as 1 min = 60 sec)
Thus, we can conclude that the mass flow rate of cold stream is 51.9 kg/min.
A rod of m= 1.3 kg rests on two parallel rails that are L = 0.42 m apart. The rod carries a current going between the rails (bottom to top in the figure) with a magnitude 1 = 2.6 A. A uniform magnetic field of magnitude B = 0.35 T pointing upward is applied to the region, as shown in the graph. The rod moves a distance d=1.25 m. Ignore the friction on the rails. † † † Ē I Otheexpertta.com A Calculate the final speed, in meters per second, of the rod if it started from rest, assuming there is no friction in the contact between it and rails.Calculate the final speed, in meters per second, of the rod if it started from rest, assuming there is no friction in the contact between it and rails. Assume the current through the rod is constant at all times.
Answer:
The final speed of the rod is 0.86 m/s.
Explanation:
Given that,
Mass of rod = 1.3 kg
Distance between rail= 0.42 m
Current = 2.6 A
Magnetic field = 0.35 T
Distance = 1.25 m
We need to calculate the acceleration
Using formula of magnetic force
[tex]F= Bil[/tex]
[tex]ma=Bil[/tex]
[tex]a=\dfrac{Bil}{m}[/tex]
Put the value into the formula
[tex]a=\dfrac{0.35\times2.6\times0.42}{1.3}[/tex]
[tex]a=0.294\ m/s^2[/tex]
We need to calculate the final speed of the rod
Using equation of motion
[tex]v^2-u^2=2as[/tex]
Put the value in the equation
[tex]v^2=2\times0.294\times1.25[/tex]
[tex]v=0.86\ m/s[/tex]
Hence, The final speed of the rod is 0.86 m/s.
The final speed of the rod is 0.86 m/s.
What is Speed?This refers to the rate of change of the position of an object in a specified direction.
The ParameterMass of rod = 1.3 kgDistance between rail= 0.42 mCurrent = 2.6 AMagnetic field = 0.35 TDistance = 1.25 mTo calculate the acceleration
We use the formula of magnetic force
a=Bil/m
a= (0.35 x 2.6 x 0.42)/1.3
a= 0.294m/s^2
Then the final speed of the rod
We use the equation of motion
v^2 - u^2= 2as
=> v^2= 0.86m/s
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A horizontal uniform plank is supported by ropes I and II at points P and Q, respectively, as shown above. The two ropes have negligible mass. The tension in rope I is 150 N. The point at which rope II is attached to the plank is now moved to point R halfway between point Q and point C, the center of the plank. The plank remains horizontal. What are the new tensions in the two ropes?
The answer is T1=100N and T2=200N but I don't know the steps to solve this one. An explanation would be much appreciated.
Explanation:
There are three forces on the plank. T₁ pulling up at point P, T₂ pulling up at point Q, and W pulling down at point C.
Let's say the length of the plank is L.
Sum of forces in the y direction before rope II is moved:
∑F = ma
150 N + 150 N − W = 0
W = 300 N
Sum of moments about point P after rope II is moved:
∑τ = Iα
(T₁) (0) − (300 N) (L/2) + (T₂) (3L/4) = 0
-(300 N) (L/2) + (T₂) (3L/4) = 0
-(300 N) (1/2) + (T₂) (3/4) = 0
-150 N + 3/4 T₂ = 0
T₂ = 200 N
Sum of forces in the y direction:
∑F = ma
T₁ + 200 N − 300 N = 0
T₁ = 100 N
The new tensions in the two ropes after the movement of rope 2 are;
T₁ = 100 N
T₁ = 100 NT₂ = 200 N
We are told that as the plank is currently, the two ropes attached at each end have tension of 150 N each.
Thus;
T₁ = T₂ = 150 N
The two ropes are acting in tension upwards and so for the plank to be balanced, there has to be a downward force(which is the weight of the plank) must be equal to the sum of the tension in the two ropes.
Thus, from equilibrium of forces, we have;
W = T₁ + T₂
W = 150 + 150
W = 300 N
Now, we are told that;
Rope 2 is now moved to a point R which is halfway between point C and Q. Since C is the centre of the plank and R is the midpoint of C and Q, if the length of the plank is L, then the distance of rope 2 from point P is now ¾L.
Since the plank remains horizontal after shifting the rope 2 to point R, let us take moments about point P to get;
T₂(¾L) - W(½L) = 0
Plugging in the relevant values;
T₂(¾L) - 300(½L) = 0
T₂(¾L) - 150L = 0
Rearrange to get;
T₂(¾L) = 150L
Divide both sides by L to get;
T₂(¾) = 150
Cross multiply to get;
T₂ = 150 × 4/3
T₂ = 200 N
Thus;
T₁ = 300 - 200
T₁ = 100 N
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