Answer:46.05 mm
Explanation:
Given
[tex]Power=260\ hp\approx 260\times 746=193.96\ KW[/tex]
speed [tex]N=550\ rev/min[/tex]
allowable shear stress [tex](\tau )_{max}=15\ kpsi\approx 103.421\ MPa[/tex]
Power is given by
[tex]P=\frac{2\pi NT}{60}[/tex]
[tex]193.96=\frac{2\pi 550\times T}{60}[/tex]
[tex]T=3367.6\ N-m[/tex]
From Torsion Formula
[tex]\frac{T}{J}=\frac{\tau }{r}-----1[/tex]
where J=Polar section modulus
T=Torque
[tex]\tau [/tex]=shear stress
For square cross section
[tex]r=\frac{a}{2}[/tex]
where a=side of square
[tex]J=\frac{a^4}{6}[/tex]
Substituting the values in equation 1
[tex]\frac{3376.6}{\frac{a^4}{6}}=\frac{103.421\times 10^6}{\frac{a}{2}}[/tex]
[tex]a=0.04605\ m[/tex]
[tex]a=46.05\ mm[/tex]
A square cross-section solid steel shaft of approximately 5 inches on each side can transmit 260 hp at 550 rev/min without exceeding a maximum allowable shear stress of 15 kpsi.
Explanation:The size of the steel shaft can be calculated by using the power-transmitting capacity of a shaft equation, which states:
P = (16πNT) / (60 * 33000), where P is power in hp, N is rotational speed in rev/min, and T is torque in lb-ft.
We also know that the shear stress (τ) is given by the equation: τ = (16T) / (πd^3), where d is the diameter in inches.
To find the correct torque, we start by rearranging the power equation for T: T = (P * 60 * 33000) / (16πN).
Substituting in the given power and rotational speed, we find that the torque is approximately 13400 lb-ft.
We then substitute this value and the allowable shear stress into the shear stress equation, solving for d to get d ≈ 5 inches.
So, a square cross-section solid steel shaft about 5 inches on each side should be able to transmit the given power at the stated speed without exceeding the maximum allowable shear stress.
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A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 4.2 m/s. The car is a distance d away. The bear is 29 m behind the tourist and running at 6.0 m/s. The tourist reaches the car safely. What is the maximum possible value for d
Explanation:
It is known that the relation between speed and distance is as follows.
velocity = [tex]\frac{distance}{time}[/tex]
As it is given that velocity is 6 m/s and distance traveled by the bear is (d + 29). Therefore, time taken by the bear is calculated as follows.
[tex]t_{bear} = \frac{(d + 29)}{6 m/s}[/tex] ............. (1)
As the tourist is running in a car at a velocity of 4.2 m/s. Hence, time taken by the tourist is as follows.
[tex]t_{tourist} = \frac{d}{4.2}[/tex] ............. (2)
Now, equation both equations (1) and (2) equal to each other we will calculate the value of d as follows.
[tex]t_{bear} = t_{tourist}[/tex]
[tex]\frac{(d + 29)}{6 m/s}[/tex] = [tex]\frac{d}{4.2}[/tex]
4.2d + 121.8 = 6d
d = [tex]\frac{121.8}{1.8}[/tex]
= 67.66
Thus, we can conclude that the maximum possible value for d is 67.66.
Before the experiment, the total momentum of the system is 2.5 kg m/s to the right and the kinetic energy is 5J. After the experiment, the total momentum of the system is 2.5 kg m/s to the right and the kinetic energy is 4J.
a)This describes an elastic collision (and it could NOT be inelastic).
b)This describes an inelastic collision (and it could NOT be elastic).
c)This is NEITHER an elastic collision nor an inelastic collision
d)This describes a collision that is EITHER elastic or inelastic, but more information is required to determine which.
Answer:
B
Explanation:
Newton's third law of motion states that to every action there is equal an opposite reaction. Momentum is always conserved provided there is no net force on the body. Considering the experiment: the momentum is conserved as expected but the kinetic energy is not conserved meaning that the collision is not elastic; some energy is converted into internal energy. When the collision is elastic the total kinetic energy before will equal total kinetic energy after and when the body stick together, they move with a common velocity and that described a perfectly inelastic collision. 5J ≠ 4J
Final answer:
The scenario describes an inelastic collision because while the momentum is conserved, the kinetic energy decreases from 5J to 4J, indicating that not all kinetic energy is conserved in the collision.
Explanation:
When examining collisions in physics, an elastic collision is one in which both momentum and kinetic energy are conserved, whereas an inelastic collision is one where momentum is conserved but kinetic energy is not. Given that in the scenario the total momentum of the system remains the same before and after the collision, that part of the conservation law is satisfied in both cases. However, since the kinetic energy of the system decreases from 5J to 4J, this means that some of the kinetic energy has been transformed into other forms of energy, like heat or sound, which typically happens during an inelastic collision.
Therefore, the correct answer is (b): This describes an inelastic collision (and it could NOT be elastic) since the total kinetic energy of the system is not the same before and after the collision. An elastic collision would have required that the kinetic energy also remain constant.
A solenoid of length 2.50 cm and radius 0.750 cm has 25 turns. If the wire of the solenoid has 1.85 amps of current, what is the magnitude of the magnetic field inside the solenoid
Answer:
13.875 T
Explanation:
Parameters given:
Length of solenoid, L = 2.5 cm = 0.025 m
Radius of solenoid, r = 0.75 cm = 0.0075 m
Number of turns, N = 25 turns
Current, I = 1.85 A
Magnetic field, B, is given as:
B = (N*r*I) /L
B = (25 * 0.0075 * 1.85)/0.025
B = 13.875 T
Three identical 4.0-kg cubes are placed on a horizontal frictionless surface in contact with one another. The cubes are lined up from left to right and a force is applied to the left side of the left cube causing all three cubes to accelerate to the right at 4.0 m/s2 . What is the magnitude of the force exerted on the right cube by the middle cube in this case
Final answer:
The magnitude of the force exerted on the right cube by the middle cube, when three identical cubes accelerate together on a frictionless surface, is calculated using Newton's second law (F = ma) and is found to be 16.0 Newtons.
Explanation:
The student's question pertains to Newton's second law of motion and the concept of force and acceleration. The problem involves three identical cubes on a frictionless surface accelerating due to a force applied to the first cube. To find the magnitude of the force exerted on the right cube by the middle cube, we use the formula F = ma, where F is the force, m is the mass, and a is the acceleration.
Since the cubes are identical and accelerate together, each 4.0-kg cube has an acceleration of 4.0 m/s². Therefore, the force exerted by one cube on another can be found by multiplying one cube's mass by the given acceleration:
F = m × aF = 4.0 kg × 4.0 m/s²F = 16.0 NThe force exerted on the right cube by the middle cube is 16.0 Newtons.
Two bodies are falling with negligible air resistance, side byside, above a horizontal plane near Earth's surface. If one of thebodies is given an additional horizontal acceleration during itsdescent, it ........................
(a) strikes the plane at the same time as the other body
(b) has the vertical component of its velocity altered
(c) has the vertical component of its accelerationaltered
(d) follows a hyperbolic path
(e) follows a straight line path along the resultantacceleration vector
Answer:
(a) strikes the plane at the same time as the other body
Explanation:
Two bodies are falling with negligible air resistance, side by side, above a horizontal plane near Earth's surface. If one of the bodies is given an additional horizontal acceleration during its descent, it strikes the plane at the same time as the other body
A small current element carrying a current of I = 5.00 A is placed at the origin given by d → l = 4.00 m m ^ j Find the magnetic field, d → B , at the locations specified. Enter the correct magnitude and select the direction from the list. If the direction is negative, indicate this by entering the magnitude as a negative number. What is the magnitude and direction of d → B on the x ‑axis at x = 2.00 m ?
Answer:
dB = (-5 × 10⁻⁷k) T
Magnitude of dB = (-5 × 10⁻⁷) T; magnitude is actually (5 × 10⁻⁷) T in the mathematical sense of what magnitude is, but this question instructs to include the negative of the direction of dB is negative.
Direction is in the negative z-direction as evident from the sign on dB's vector notation.
Explanation:
From Biot Savart's relation, the magnetic field is given by
dB = (μ₀I/4πr³) (dL × r)
μ₀ = (4π × 10⁻⁷) H/m
I = 5.0 A
r = (2î) m
Magnitude of r = 2
dL = (4j)
(dL × r) is the vector product of both length of current carrying wire vector and the vector position of the point where magnetic field at that point is needed.
(dL × r) = (4j) × (2î)
|i j k|
|0 4 0|
|2 0 0|
(dL × r) = (0î + 0j - 8k) = (-8k)
(μ₀I/4πr³) = (4π × 10⁻⁷ × 5)/(4π×2³)
(μ₀I/4πr³) = (6.25 × 10⁻⁸)
dB = (μ₀I/4πr³) (dL × r) = (6.25 × 10⁻⁸) × (-8k)
dB = (-5 × 10⁻⁷k) T
Magnitude of dB = (-5 × 10⁻⁷) T; magnitude is actually (5 × 10⁻⁷) T in the mathematical sense of what magnitude is, but this question instructs to include the negative of the direction of dB is negative.
Direction is in the negative z-direction as evident from the sign on dB's vector notation.
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When the distance is 4 m, the magnetic field strength is 2.5 x 10⁻⁷ T.
When the distance is 2 m, the magnetic field strength is 5 x 10⁻⁷ T.
Magnitude of magnetic fieldThe magnitude of magnetic field at any distance from a wire is determined by applying Biot-Savart law,
[tex]B = \frac{\mu_o I}{2\pi r}[/tex]
Where;
r is the distance from the conductor
When the distance is 4 m, the magnetic field strength is calculated as;
[tex]B = \frac{(4\pi \times 10^{-7}) \times5 }{2\pi \times 4} \\\\B = 2.5 \times 10^{-7} \ T[/tex]
When the distance is 2 m, the magnetic field strength is calculated as;
[tex]B = \frac{(4\pi \times 10^{-7} \times 5}{2\pi \times 2} \\\\B = 5 \times 10^{-7} \ T[/tex]
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car leaves a stop sign and exhibits a constant acceleration of 0.300 m/s2 parallel to the roadway. The car passes over a rise in the roadway such that the top of the rise is shaped like an arc of a circle of radius 500 m. Now the car is at the top of the rise, its velocity vector is horizontal and has a magnitude of 6.00 m/s. What are the magnitude and direction of the total acceleration vector for the car at this instant
Answer:
0.308 m/s2 at an angle of 13.5° below the horizontal
Explanation:
The parallel acceleration to the roadway is the tangential acceleration on the rise.
The normal acceleration is the centripetal acceleration due to the arc. This is given by
[tex]a_N = \dfrac{v^2}{r} = \dfrac{36^2}{500}=0.072[/tex]
The tangential acceleration, from the question, is
[tex]a_T = 0.300[/tex]
The magnitude of the total acceleration is the resultant of the two accelerations. Because these are perpendicular to each other, the resultant is given by
[tex]a^2 =a_T^2 + a_N^2 = 0.300^2 + 0.072^2[/tex]
[tex]a = 0.308[/tex]
The angle the resultant makes with the horizontal is given by
[tex]\tan\theta=\dfrac{a_N}{a_T}=\dfrac{0.072}{0.300}=0.2400[/tex]
[tex]\theta=13.5[/tex]
Note that this angle is measured from the horizontal downwards because the centripetal acceleration is directed towards the centre of the arc
The Doppler Effect. In hydrogen, the transition from level 2 to level 1 has a rest wavelength of 121.6 nm. Suppose you see this line at a wavelength of 120.5 nm in star A, at 121.2 nm in Star B, at 121.9 nm in Star C, and at 122.9 nm in Star D. Which stars are coming toward us? Which are moving away? Which star is moving fastest relative to us? What are the speeds of Star B and Star C?
Answer:
Which stars are coming toward us? Which are moving away?
The Star A and B are moving toward the observer and Star C and D away from the observer.
Which star is moving fastest relative to us?
Star A is moving fastest relative to us.
What are the speeds of Star B and Star C?
The speed of the star B is 986842m/s and the speed of star C is 740131m/s.
Explanation:
Which stars are coming toward us? Which are moving away?
Spectral lines will be shifted to the blue part of the spectrum if the source of the observed light is moving toward the observer, or to the red part of the spectrum when is moving away from the observer (that is known as the Doppler effect).
The wavelength at rest is 121.6 nm ([tex]\lambda_{0} = 121.6nm[/tex])
[tex]Redshift: \lambda_{measured} > \lambda_{0} [/tex]
[tex]Blueshift: \lambda_{measured} < \lambda_{0} [/tex]
Then, for this particular case it is gotten:
Star A: [tex]\lambda_{measured} = 120.5nm[/tex]
Star B: [tex]\lambda_{measured} = 121.2nm[/tex]
Star C: [tex]\lambda_{measured} = 121.9nm[/tex]
Star D: [tex]\lambda_{measured} = 122.9nm[/tex]
Star A:
[tex]Blueshift: 120.5nm < 121.6nm [/tex]
Star B:
[tex]Blueshift: 121.2nm < 121.6nm [/tex]
Star C:
[tex]Redshift: 121.9nm > 121.6nm [/tex]
Star D:
[tex]Redshift: 122.9nm > 121.6nm [/tex]
Therefore, according to the approach above. The Star A and B are moving toward the observer and Star C and D away from the observer.
Due to that shift the velocity of the star can be determined by means of Doppler velocity.
[tex]v = c\frac{\Delta \lambda}{\lambda_{0}}[/tex] (1)
Where [tex]\Delta \lambda[/tex] is the wavelength shift, [tex]\lambda_{0}[/tex] is the wavelength at rest, v is the velocity of the source and c is the speed of light.
[tex]v = c(\frac{\lambda_{measured}- \lambda_{0}}{\lambda_{0}})[/tex]
Which star is moving fastest relative to us?
Case for the Star A:
[tex]v = (3x10^{8}m/s)(\frac{121.6nm-120.5nm}{121.6nm})[/tex]
[tex]v = 2713815m/s[/tex]
Case for the Star B:
[tex]v = (3x10^{8}m/s)(\frac{121.6nm - 121.2nm}{121.6nm})[/tex]
[tex]v = 986842m/s[/tex]
Hence, Star A is moving fastest relative to us.
What are the speeds of Star B and Star C?
Case for the Star C:
[tex]v = (3x10^8m/s)(\frac{(121.9nm - 121.6nm}{121.6nm})[/tex]
[tex]v = 740131m/s[/tex]
Therefore, The speed of the star B is 986842m/s and the speed of star C is 740131m/s.
Final answer:
Stars with shift towards shorter wavelengths (Star A and B) are moving toward Earth, while those with shift towards longer wavelengths (Star C and D) are moving away, with Star D moving the fastest. The speeds of Star B and Star C can be calculated using the Doppler effect formula. Star A is blue-shifted the most, indicating it is moving towards us the fastest.
Explanation:
The Doppler Effect on Spectral Lines
The observed changes in the wavelength of spectral lines from hydrogen transitions can be explained by Doppler shift, which is the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source. Stars showing a shorter wavelength than the rest wavelength are moving towards us (blue-shifted), while those with a longer wavelength are moving away from us (red-shifted).
For star A with an observed wavelength of 120.5 nm, the shift is towards the blue, indicating it is moving towards us. Star B, with an observed wavelength of 121.2 nm, is also moving towards us but at a slower speed. Star C and Star D, having observed wavelengths of 121.9 nm and 122.9 nm respectively, display a red shift, meaning they are moving away from us. Among these stars, Star D is the one moving away the fastest.
To calculate the velocity of Star B and Star C with respect to Earth, we can use the Doppler effect formula for light:
[tex]v = c × (λ_{observed} - λ_{rest}) / λ_{rest}[/tex]
Where v is the velocity of the star, c is the speed of light, [tex]λ_{observed}[/tex] is the observed wavelength, and [tex]λ_{rest}[/tex] is the rest wavelength of the emission line. By plugging in the values and solving for v, we can determine the velocity for each star.
A ball is tossed vertically upward from a height of 2 m above the ground with an initial velocity of 10 m/s. What will the velocity of the ball be just before it hits the ground?
The velocity of the ball just before it hits the ground, thrown upward from 2 meters with an initial velocity of 10 m/s, can be calculated using kinematic equations and is approximately 11.8 m/s.
To determine the velocity of the ball just before it hits the ground, we can use kinematic equations. One equation that relates initial velocity, acceleration due to gravity, and final velocity is:
v² = u² + 2as
Where:
v is the final velocityu is the initial velocity (10 m/s)a is the acceleration due to gravity (9.8 m/s²)s is the displacement (the initial height plus the distance fallen)In this case, the ball is tossed upward so we consider the acceleration due to gravity as negative (-9.8 m/s²). The displacement, in this case, is 2 m (the height from which the ball is thrown).
Insert the known values into the equation:
v² = (10 m/s)² + 2 × (-9.8 m/s²) × (-2 m)
The negative signs cancel for the displacement and acceleration due to the upward throw and the downward acceleration, which will give us a positive number.
v² = 100 + (19.6 × 2)
v² = 100 + 39.2
v² = 139.2
v = √139.2
v ≈ 11.8 m/s
Therefore, the velocity of the ball just before it hits the ground is approximately 11.8 m/s.
A rifle is fired in a valley with parallel vertical walls. The echo from one wall is heard 1.55 s after the rifle was fired. The echo from the other wall is heard 2.5 s after the first echo. How wide is the valley? The velocity of sound is 343 m/s. Answer in units of m.
Answer:
[tex]w=694.575\ m[/tex]
Explanation:
Given:
velocity of the sound, [tex]v=343\ m.s^{-1}[/tex]
time lag in echo form one wall of the valley, [tex]t= 1.55\ s[/tex]
time lag in echo form the other wall of the valley, [tex]t'=2.5\ s[/tex]
distance travelled by the sound in the first case:
[tex]d=v.t[/tex]
[tex]d=343\times 1.55[/tex]
[tex]d=531.65\ m[/tex]
Since this is the distance covered by the sound while going form the source to the walls and then coming back from the wall to the source so the distance of the wall form the source will be half of the distance obtained above.
[tex]s=\frac{d}{2}[/tex]
[tex]s=\frac{531.65}{2}[/tex]
[tex]s=265.825\ m[/tex]
distance travelled by the sound in the second case:
[tex]d'=v.t'[/tex]
[tex]d'=343\times 2.5[/tex]
[tex]d'=857.5\ m[/tex]
Since this is the distance covered by the sound while going form the source to the walls and then coming back from the wall to the source so the distance of the wall form the source will be half of the distance obtained above.
[tex]s'=\frac{d'}{2}[/tex]
[tex]s'=\frac{857.5}{2}[/tex]
[tex]s'=428.75\ m[/tex]
Now the width of valley:
[tex]w=s+s'[/tex]
[tex]w=265.825+428.75[/tex]
[tex]w=694.575\ m[/tex]
To determine the width of the valley with vertical walls using echo timings.
To find the width of the valley:
Calculate the distance to the first wall using the time taken for the first echo.Then, calculate the distance to the second wall using the time taken for the second echo.The width of the valley is the difference between the two distances.Calculations:
Distance to first wall = 343 m/s * 1.55 s = 531.85 m
Distance to second wall = 343 m/s * 2.5 s = 857.5 m
Width of the valley: 857.5 m - 531.85 m = 325.65 m
A dielectric material is inserted between the charged plates of a parallel-plate capacitor. Do the following quantities increase, decrease, or remain the same as equilibrium is reestablished?
1. Charge on plates (plates remain connected to battery)
2. Electric potential energy (plates isolated from battery before inserting dielectric)
3.Capacitance (plates isolated from battery before inserting dielectric)
4. Voltage between plates (plates remain connected to battery)
5. Charge on plates (plates isolated from battery before inserting dielectric)
6. Capacitance (plates remain connected to battery)
7. Electric potential energy (plates remain connected to battery)
8. Voltage between plates (plates isolated from battery before inserting dielectric)
Answer:
1. increase
2. remain the same
3. increases
4. decreases
5. remain the same
6. increases
7. decreases
8. remain the same
Explanation:
a. the formula for the capacitance of a capacitor relating the capacitor and the dielectric material is express as
[tex]C=e_oA/d[/tex]........equation 1
also the capacitance and the charge is related as
Q=CV.......equation 2
from equation 1 as the dielectric material is introduced, the capacitance increases, the charge also increases
2. from the equation as the dielectric material is introduced, the capacitance increases, the electric potential also remain the same
3. from
[tex]C=e_oA/d[/tex]........equation 1
we conclude that the capacitance increases
4. the voltage between the plates decreases
5. the charge remain the same because capacitance is constant
6. the capacitance increases
7. the electric potential decreases
8. remain the same
Answer 1: the charge on the plates will increase
Explanation: placing a dielectric between two charged plate increases its capacitance
C = Q/V,
If the plates remain connected then the voltage remains the same.
Therefore for an increase in capacitance charge will increase.
Answer 2: electric field potential remains the same.
Explanation: if the plates are disconnected, charge on plates remains contant while voltage varies with change in distance, electric field intensity remains constant and this is proportional to the electric potential energy.
Answer 3: capacitance increases
Explanation: introducing a dielectric between two plates causes opposite charges to be induced on the faces of the dielectric. This also reduces the p.d across the capacitor.
Answer 4: voltage remains constant.
Explanation: A connected plate has a constant voltage across its field.
Answer 5: charge remains contant.
Explanation: capacitance will increase with introduction of dielectric, p.d across the plates will drop, the charge will remain constant.
Answer 6: capacitance increases
Explanation: placing a dielectric between plates always increase the capacitance.
Answer 7: electric potential energy falls.
Answer 8: voltage between plates decreases
The current supplied by a battery as a function of time is I(t) = (0.88 A) e^(-t*6 hr). What is the total number of electrons transported from the positive electrode to the negative electrode from the time the battery is first used until it is essentially dead?
a. 3.7 x 10^18
b. 5.3 x 10^23
c. 4.4 x 10^22
d. 1.6 x 10^19
e. 1.2 x 10^23
Answer:
e. 1.2 x 10²³
Explanation:
According to the problem, The current equation is given by:
[tex]I(t)=0.88e^{-t/6\times3600s}[/tex]
Here time is in seconds.
Consider at t=0 s the current starts to flow due to battery and the current stops when the time t tends to infinite.
The relation between current and number of charge carriers is:
[tex]q=\int\limits {I} \, dt[/tex]
Here the limits of integration is from 0 to infinite. So,
[tex]q=\int\limits {0.88e^{-t/6\times3600s}}\, dt[/tex]
[tex]q=0.88\times(-6\times3600)(0-1)[/tex]
q = 1.90 x 10⁴ C
Consider N be the total number of charge carriers. So,
q = N e
Here e is electronic charge and its value is 1.69 x 10⁻¹⁹ C.
N = q/e
Substitute the suitable values in the above equation.
[tex]N= \frac{1.9\times10^{4} }{1.69\times10^{-19}}[/tex]
N = 1.2 x 10²³
If 4.00 ✕ 10−3 kg of gold is deposited on the negative electrode of an electrolytic cell in a period of 2.73 h, what is the current in the cell during that period? Assume the gold ions carry one elementary unit of positive charge.
Explanation:
Below is an attachment containing the solution.
To calculate the current, we first convert the mass of gold deposited to moles and then use the equivalence of one mole of gold requiring one faraday of charge. We calculate the total charge transferred and then find the average current by dividing this charge by the total time in seconds.
Explanation:To find the current in the cell during the period when 4.00 × 10⁻³ kg of gold was deposited on the negative electrode, we need to calculate the total charge that resulted in the deposition of gold. First, we will convert the mass of gold to moles using the molar mass of gold, which is approximately 197 g/mol. Since gold ions carry one unit of positive charge, one mole of gold ions will require one faraday (96,485 C) to be reduced and deposited as gold metal.
After calculating the number of moles of gold, we can then determine the number of moles of electrons that moved through the cell using the equivalence of moles of electrons to moles of gold deposited. Multiplying this number by the charge per mole of electrons (1 faraday), we get the total charge moved. To find the current, we divide the total charge by the total time in seconds, and this gives us the average current over the 2.73 hours.
The spring of constant k = 170 N/m is attached to both the support and the 1.5-kg cylinder, which slides freely on the horizontal guide. If a constant 14-N force is applied to the cylinder at time t = 0 when the spring is undeformed and the system is at rest, determine the velocity of the cylinder when x = 65 mm. Also determine the maximum displacement of the cylinder.
Answer:
Velocity at 64 mm is 0.532 m/s
Maximum displacement = 0.082 m or 82 mm
Explanation:
The maximum displacement or amplitude is determined by the applied force from Hooke's law
[tex]F = kA[/tex]
[tex]A=\dfrac{F}{k}=\dfrac{14 \text{ N}}{170 \text{ N/m}}=0.082 \text{ m}[/tex]
The velocity at at any point, x, is given by
[tex]v=\sqrt{\dfrac{k}{m}(A^2 - x^2)}[/tex]
m is the mass of the load, here the cylinder.
In fact, the expression [tex]\sqrt{\dfrac{k}{m}}[/tex] represents the angular velocity, [tex]\omega[/tex].
Substituting given values,
[tex]v=\sqrt{\dfrac{170}{1.5}(0.082^2 - 0.065^2)} = 0.532 \text{ m/s}[/tex]
The sound intensity at a distance of 16 m from a noisy generator is measured to be 0.25 W/m2. What is the sound intensity at a distance of 28 m from the generator?
Answer:
0.1111 W/m²
Explanation:
If all other parameters are constant, sound intensity is inversely proportional to the square of the distance of the sound. That is,
I ∝ (1/r²)
I = k/r²
Since k can be the constant of proportionality. k = Ir²
We can write this relation as
I₁ × r₁² = I₂ × r₂²
I₁ = 0.25 W/m²
r₁ = 16 m
I₂ = ?
r₂ = 24 m
0.25 × 16² = I₂ × 24²
I₂ = (0.25 × 16²)/24²
I₂ = 0.1111 W/m²
The sound intensity at a distance of 28 m from the generator will be "0.1111 W/m²".
Distance and Sound intensityAccording to the question,
Distance, r₁ = 16 m
r₂ = 28 m
Sound intensity, I₁ = 0.25 W/m²
We know the relation,
→ I ∝ ([tex]\frac{1}{r^2}[/tex])
or,
I ∝ [tex]\frac{k}{r^2}[/tex]
Now,
→ I₁ × r₁² = I₂ × r₂²
By substituting the values,
0.25 × (16)² = I₂ × (24)²
I₂ = [tex]\frac{0.25\timers (16)^2}{(24)^2}[/tex]
= 0.1111 W/m²
Thus the above answer is correct.
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(a) Given a 48.0-V battery and 24.0-Ω and 96.0-Ω resistors, find the current and power for each when connected in series. (b) Repeat when the resistances are in parallel.
Answer with Explanation:w
a.We are given that
Potential difference, V=48 V
[tex]R_1=24\Omega[/tex]
[tex]R_2=96\Omega[/tex]
Equivalent resistance when R1 and R2 are connected in series
[tex]R_{eq}=R_1+R_2[/tex]
Using the formula
[tex]R_{eq}=24+96=120\Omega[/tex]
We know that
[tex]I=\frac{V}{R_{eq}}=\frac{48}{120}=0.4 A[/tex]
In series combination, current passing through each resistor is same and potential difference across each resistor is different.
Power, P=[tex]I^2 R[/tex]
Using the formula
Power,[tex]P_1=I^2R_1=(0.4)^2\times 24=3.84 W[/tex]
Power, [tex]P_2=I^2 R_2=(0.4)^2(96)=15.36 W[/tex]
b.
In parallel combination, potential difference remains same across each resistor and current passing through each resistor is different..
Current,[tex]I=\frac{V}{R}[/tex]
Using the formula
[tex]I_1=\frac{V}{R_1}=\frac{48}{24}=2 A[/tex]
[tex]I_2=\frac{V}{R_2}=\frac{48}{96}=0.5 A[/tex]
[tex]P_1=\frac{V^2}{R_1}=\frac{(48)^2}{24}=96 W[/tex]
[tex]P_2=\frac{V^2}{R_2}=\frac{(48)^2}{96}=24 W[/tex]
When two resistors are in series, the current through each is the same and is calculated by dividing the total voltage by the total resistance. The power through each resistor is the current squared times the resistance of each resistor. When resistors are in parallel, the total resistance decreases and the power increases. The current remains the same.
Explanation:When the 24-ohm and 96-ohm resistors are connected in series, the total resistance can be calculated using the formula R = R1 + R2. Here, R1 is the resistance of the first resistor (24 ohms) and R2 is the resistance of the second resistor (96 ohms). R equals 120 ohms. The current, I, is calculated using Ohm's Law, I = V / R. Therefore, the current flowing through the circuit is I = 48.0 V / 120 Ω = 0.4 A. The power, P, through each resistor is calculated using the formula P = I^2 * R. Therefore, for the 24-ohm resistor, P is (0.4 A)^2 * 24 Ω = 3.84 W, and for the 96-ohm resistor, P is (0.4 A)^2 * 96 Ω = 15.36 W. When the resistors are connected in parallel, the total resistance is calculated using the formula 1/R = 1/R1 + 1/R2. The current through each resistor remains the same (0.4 A) and the power for each resistor can be calculated in the same way as described above.
Learn more about Ohm's Law here:https://brainly.com/question/33263562
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A circular loop of wire of radius 0.50 m is in a uniform magnetic field of 0.30 T. The current in the loop is 2.0 A. What is the magnetic torque when the plane of the loop is parallel to the magnetic field?
Answer:
Torque = 0.47 Nm
Explanation:
Torque = BIAsin∅
Since the plane is parallel to the magnetic field, ∅ = 90⁰
sin 90 = 1
Magnetic field, B = 0.30 T
current, I = 2.0 A
Radius, R = 0.50 m
A = πR²
A = π(0.5)² = 0.785 m²
Torque = 0.3 * 2 * 0.785
Torque = 0.47 Nm
Answer:
The magnetic torque when the plane of the loop is parallel to the magnetic field is zero.
Explanation:
Given;
radius of wire, r = 0.50 m
strength of magnetic field, B = 0.3 T
current in the wire, I = 2.0 A
Dipole moment, μ = I x A
where;
A is the area of the circular loop = πr² = π(0.5)² = 0.7855 m²
Dipole moment, μ = 2A x 0.7855m² = 1.571 Am²
magnetic torque, τ = μBsinθ
when the plane of the loop is parallel to the magnetic field, θ = 0
τ = 1.571 Am² x 0.3sin0
τ = 0
Therefore, the magnetic torque when the plane of the loop is parallel to the magnetic field is zero.
In some amazing situations, people have survived falling large distances when the surface they land on is soft enough. During a traverse of Eiger's infamous Nordvand, mountaineer Carlos Ragone's rock anchor gave way and he plummeted 512 feet to land in snow. Amazingly, he suffered only a few bruises and a wrenched shoulder. Assuming that his impact left a hole in the snow 4.8 ft deep, estimate his average acceleration as he slowed to a stop (that is while he was impacting the snow). Pick a coordinate system where down is positive.
Answer:
-3413 ft/s2
Explanation:
We need to know the velocity with which he landed on the snow.
He 'dropped' from 512 feet. This is the displacement. His initial velocity is 0 and the acceleration of gravity is 32 ft/s2.
We use the equation of mition
[tex]v^2 = u^2 + 2as[/tex]
v and u are the initial and final velocities, a is the acceleration and s is the displacement. Putting the appropriate values
[tex]v^2 = 0^2 + 2\times32\times512[/tex]
[tex]v = \sqrt{2\times32\times512} = 128\sqrt{2}[/tex]
This is the final velocity of the fall and becomes the initial velocity as he goes into the snow.
In this second motion, his final velocity is 0 because he stops after a displacement of 4.8 ft. We use the same equation of motion but with different values. This time, [tex]u=128\sqrt{2}[/tex], v = 0 and s = 4.8 ft.
[tex]0^2 = (128\sqrt{2})^2 + 2a\times4.8[/tex]
[tex]a = -\dfrac{2\times128^2}{2\times4.8} = -3413[/tex]
Note that this is negative because it was a deceleration, that is, his velocity was decreasing.
A block slides down a rough ramp (with friction) of height h . Its initial speed is zero. Its final speeds at the bottom of the ramp is v . Choose the system to be the block and the Earth.
While the block is descending, its kinetic energy:
a. Increases.
b. Decreases.
c. Remains constant.
Answer:
a. Increases
Explanation:
Conceptually, when a block of certain mass slides on the rough ramp kept on the earth which happens under the influence of gravity.The block is initially at rest but as the acceleration due to gravity acts on the block at the ramp. The ramp is rough so it applies kinetic friction on the moving block as the block slides down the slope.By the definition we know that the acceleration is the rate of change in velocity and here we have acceleration component in the direction of motion of the block.
Mathematically:
when the body is moving down:
[tex]v=u+gt[/tex]
where:
[tex]v=[/tex] final velocity of the block
[tex]u=[/tex] initial velocity of the block
[tex]g=[/tex] acceleration due to gravity
[tex]t=[/tex] time of observation during the instance of motion
From above it is clear that the velocity of the block will increase as the time passes during the motion.As we know that kinetic energy is given as:
[tex]KE=\frac{1}{2} \times m.v^2[/tex]
where:
[tex]m=[/tex] mass of the block which remains constant (macroscopically)
[tex]v=[/tex] velocity of the block (which increases here as the body descends)
A simple circuit consists only of of a 1.0-μF capacitor and a 15-mH coil in series. At what frequency does the inductive reactance equal the capacitive reactance? 67 Hz 0.77 kHz 15 kHz 1.3 kHz
Answer:
The frequency is 1.3 kHz.
Explanation:
Given that,
Capacitance of the capacitor, [tex]C=1\ \mu F=10^{-6}\ C[/tex]
Inductance of the inductor, [tex]L=15\ mH=15\times 10^{-3}\ H[/tex]
We need to find the frequency when the inductive reactance equal the capacitive reactance such that :
[tex]X_c=X_L[/tex]
[tex]2\pi fL=\dfrac{1}{2\pi fC}[/tex]
[tex]f=\dfrac{1}{2\pi \sqrt{LC} }[/tex]
[tex]f=\dfrac{1}{2\pi \sqrt{15\times 10^{-3}\times 10^{-6}} }[/tex]
f = 1299.49 Hz
[tex]f=1.29\times 10^3\ Hz[/tex]
or
[tex]f=1.3\ kHz[/tex]
So, the frequency is 1.3 kHz. Therefore, the correct option is (d).
The vertical normal stress increase caused by a point load of 10 kN acting on the ground surface at a point 1m vertically below its point of application is: (a) 0 (b) 4.775 kN (c) 5 kN (d) 10 kN
Answer:
(b) 4.775 kN
Explanation:
see the attached file
A rock is thrown upward from a bridge into a river below. The function f(t)=−16t2+44t+138 determines the height of the rock above the surface of the water (in feet) in terms of the number of seconds t since the rock was thrown. g
Answer:
a) 138 ft
b) 4.62 s
c) 1.375 s
d) 168.25 ft
Explanation:
The height of a rock (thrown from the top of a bridge) above the level of water surface as it varies with time when thrown is given in the question as
h = f(t) = -16t² + 44t + 138
with t in seconds, and h in feet
a) The bridge's height above the water.
At t=0 s, the rock is at the level of the Bridge's height.
At t = 0,
h = 0 + 0 + 138 = 138 ft
b) How many seconds after being thrown does the rock hit the water?
The rock hits the water surface when h = 0 ft. Solving,
h = f(t) = -16t² + 44t + 138 = 0
-16t² + 44t + 138 = 0
Solving this quadratic equation,
t = 4.62 s or t = -1.87 s
Since time cannot be negative,
t = 4.62 s
c) How many seconds after being thrown does the rock reach its maximum height above the water?
At maximum height or at the maximum of any function, the derivative of that function with respect to the independent variable is equal to 0.
At maximum height,
(dh/dt) = f'(t) = (df/dt) = 0
h = f(t) = -16t² + 44t + 138
(dh/dt) = (df/dt) = -32t + 44 = 0
32t = 44
t = (44/32)
t = 1.375 s
d) What is the rock's maximum height above the water?
The maximum height occurs at t = 1.375 s,
Substituting this for t in the height equation,
h = f(t) = -16t² + 44t + 138
At t = 1.375 s, h = maximum height = H
H = f(1.375) = -16(1.375²) + 44(1.375) + 138
H = 168.25 ft
Hope this Helps!!!
List the laws of thermodynamic and describe their relevance in the chemical reactions2. Define the standard reduction potential. Why aerobic grow generates the highest amount ofenergy (ATP).
Explanation:
Thermodynamic Laws -
The primary law, otherwise called the Law of Conservation of Energy, expresses that energy remains constant in the overall reaction, it is not created or destroyed. The second law of thermodynamics expresses that the entropy increases. after the completion of the reaction of any isolated system The third law of thermodynamics expresses that as the temperature reaches towards an absolute zero point the entropy of a system becomes constant.Standard Electronic potential -
The standard reduction potential may be defined as the tendency of a chemical species or the reactants to get into its reduced form after the overall reaction. It is estimated at volts. The more is the positive potential the more is the reduction of the chemical speciesAerobic grow is much more efficient at making ATP because of the presence of oxygen the cycles in the respiration are carried out at an efficient rate which forces the cell to undergo a much large amount of ATP production.
. An electron moving at 4.00×103m/s in a 1.25-T magnetic field experiences a magnetic force of 1.40×10−16N . What angle does the velocity of the electron make with the magnetic field? There are two answers.
Answer: 9.59° and 350.41°
Explanation: The formulae that relates the force F exerted on a moving charge q with velocity v in a magnetic field of strength B is given as
F =qvB sin x
Where x is the angle between the strength of magnetic field and velocity of the charge.
q = 1.609×10^-19 C
v = 4×10³ m/s
B = 1.25 T
F = 1.40×10^-16 N
By substituting the parameters, we have that
1.40×10^-16 = 1.609×10^-19 × 4×10³ × 1.25 × sinx
sin x = 1.40×10^-16/ 1.609×10^-19 × 4×10³ × 1.25
sin x = 1.40×10^-16 /8.045*10^(-16)
sin x = 0.1666
x = 9.59°
The value of sin x is positive in first and fourth quadrant.
Hence to get the second value of x, we move to the 4th quadrant of the trigonometric quadrant which is 360 - x
Hence = 360 - 9.59 = 350.41°
Final answer:
The angle between the velocity of the electron and the magnetic field is approximately 6.27° or 173.73°.
Explanation:
To find the angle between the velocity of the electron and the magnetic field, we can use the formula:
F = q(vsinθ)B
Where F is the magnetic force, q is the charge of the electron, v is the velocity of the electron, θ is the angle between the velocity and the magnetic field, and B is the magnetic field strength.
In this case, the magnetic force is given as 1.40 × 10^-16 N, the charge of an electron is 1.6 × 10^-19 C, the velocity of the electron is 4.00 × 10³ m/s, and the magnetic field strength is 1.25 T.
Plugging in these values, we can solve for θ:
1.40 × 10^-16 N = (1.6 × 10^-19 C)(4.00 × 10³ m/s)(sinθ)(1.25 T)
Solving for sinθ:
sinθ = (1.40 × 10^-16 N) / [(1.6 × 10^-19 C)(4.00 × 10³ m/s)(1.25 T)]
sinθ ≈ 0.109
Taking the inverse sine of 0.109, we find that θ ≈ 6.27° or θ ≈ 173.73°.
A spaceship of proper length ? = 100 m travels in the positive x direction at a speed of 0.700 c0 relative to Earth. An identical spaceship travels in the negative x direction along a parallel course at the same speed relative to Earth. At t = 0, an observer on Earth measures a distance d = 58,000 km separating the two ships.
Part A
At what instant does this observer see the leading edges of the two ships pass each other?
Answer:
observer see the leading edges of the two ships pass each other at time 0.136 s
Explanation:
given data
spaceship length = 100 m
speed of 0.700 Co = [tex]0.700\times 3 \times 10^8[/tex] m/s
distance d = 58,000 km = 58000 × 10³ m
solution
as here distance will be half because both spaceship travel with same velocity
so they meet at half of distance
Distance Da = [tex]\frac{d}{2}[/tex] ............1
Distance Da = [tex]\frac{58000\times 10^3}{2}[/tex]
Distance Da = 29 ×[tex]10^{6}[/tex] m
and
time at which observer see leading edge of 2 spaceship pass
Δ time = [tex]\frac{Da}{v}[/tex] ..........2
Δ time = [tex]\frac{29 \times 10^6}{0.700\times 3 \times 10^8}[/tex]
Δ time = 0.136 s
Answer:
The time at which the observer see the leading edges of the two ships pass each other is 0.138 sec.
Explanation:
Given that,
Length = 100 m
Speed = 0.700 c
Distance = 58000 km
The distance should be halved because the spaceships both travel the same speed.
So they will meet at the middle of the distance
We need to calculate the distance
Using formula for distance
[tex]d'=\dfrac{d}{2}[/tex]
Put the value into the formula
[tex]d'=\dfrac{58000}{2}[/tex]
[tex]d'=29000\ km[/tex]
We need to calculate the time at which the observer see the leading edges of the two ships pass each other
Using formula of time
[tex]\Delta t=\dfrac{d}{V}[/tex]
Put the value into the formula
[tex]\Delta t=\dfrac{29\times10^{6}}{0.700\times3\times10^{8}}[/tex]
[tex]\Delta t=0.138\ sec[/tex]
Hence, The time at which the observer see the leading edges of the two ships pass each other is 0.138 sec.
Before entering the cyclotron, the particles are accelerated by a potential difference V. Find the speed v with which the particles enter the cyclotron.
A flat circular loop of wire of radius 0.50 m that is carrying a 2.0-A current is in a uniform magnetic field of 0.30 T. What is the magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field
the magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field is approximately [tex]\( 0.47 \, \text{N} \cdot \text{m} \)[/tex].
The magnetic torque [tex]\( \tau \)[/tex] on a current-carrying loop in a uniform magnetic field can be calculated using the formula:
[tex]\[ \tau = N \cdot I \cdot A \cdot B \cdot \sin(\theta) \][/tex]
Where:
- N is the number of turns in the loop (1 in this case, as there is a single loop),
- I is the current flowing through the loop (2.0 A),
- A is the area of the loop (πr² for a circular loop, where \( r \) is the radius),
- B is the magnetic field strength (0.30 T),
- [tex]\( \theta \)[/tex] is the angle between the normal to the loop's plane and the magnetic field (90° in this case, as the loop is perpendicular to the magnetic field).
Substituting the given values:
[tex]\[ \tau = (1) \cdot (2.0 \, \text{A}) \cdot (\pi \times (0.50 \, \text{m})^2) \cdot (0.30 \, \text{T}) \cdot \sin(90°) \][/tex]
[tex]\[ \tau = 2.0 \cdot \pi \cdot (0.50)^2 \cdot 0.30 \][/tex]
[tex]\[ \tau = 2.0 \cdot \pi \cdot 0.25 \cdot 0.30 \][/tex]
[tex]\[ \tau = 0.15 \cdot \pi \][/tex]
[tex]\[ \tau \approx 0.47 \, \text{N} \cdot \text{m} \][/tex]
So, the magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field is approximately [tex]\( 0.47 \, \text{N} \cdot \text{m} \)[/tex].
Therefore, the correct answer choice is [tex]\(\boxed{\text{0.47 N*m}}\)[/tex].
The complete Question is given below:
A flat circular loop of wire of radius 0.50 m that is carrying a 2.0-A current is in a uniform magnetic field of 0.30 T. What is the magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field?
Answer choices.
.41N*m
.00N*m
.47N*m
.58N*m
.52N*m
An oxygen molecule is adsorbed onto a small patch of the surface of a catalyst. It's known that the molecule is adsorbed on of possible sites for adsorption (see sketch at right). Calculate the entropy of this system. Round your answer to significant digits, and be sure it has the correct unit symbol.
The given question is incomplete. The complete question is as follows.
An oxygen molecule is adsorbed onto a small patch of the surface of a catalyst. It's known that the molecule is adsorbed on 1 of 36 possible sites for adsorption. Calculate the entropy of this system.
Explanation:
It is known that Boltzmann formula of entropy is as follows.
s = k ln W
where, k = Boltzmann constant
W = number of energetically equivalent possible microstates or configuration of the system
In the given case, W = 36. Now, we will put the given values into the above formula as follows.
s = k ln W
= [tex]1.38 \times 10^{-23} ln (36)[/tex]
= [tex]4.945 \times 10^{-23} J/K[/tex]
Thus, we can conclude that the entropy of this system is [tex]4.945 \times 10^{-23} J/K[/tex].
The entropy of the given system is [tex]4.954 \times 10^{-23 } \rm\; J/K[/tex] in which an oxygen molecule is absorbed in one of the 36 possible states.
From the Boltzmann formula of entropy,
[tex]s = k \times ln W[/tex]
Where,
[tex]k[/tex]= Boltzmann constant = [tex]1.38\times 10^{-23} \rm\; J/K[/tex]
[tex]W[/tex] = Number of energetically equivalent possible microstates of the system = 36
Put the values in the formula
[tex]s =1.38\times 10^{-23} \times ln (36)\\\\s = 4.954 \times 10^{-23 } \rm\; J/K[/tex]
Therefore, the entropy of the given system is [tex]4.954 \times 10^{-23 } \rm\; J/K[/tex]
To know more about entropy,
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A typical cell membrane is 8.0 nm thick and has an electrical resistivity of 1.3 107 Ohms · m.
(a) If the potential difference between the inner and outer surfaces of a cell membrane is 80 mV, how much current flows through a square area of membrane 1.2 µm on a side?
_____________A
(b) Suppose the thickness of the membrane is doubled, but the resistivity and potential difference remain the same. Does the current increase or decrease or remain the same?
By what factor?
Explanation:
Below is an attachment containing the solution.
( 8c5p79) A certain force gives mass m1 an acceleration of 13.5 m/s2 and mass m2 an acceleration of 3.5 m/s2. What acceleration would the force give to an object with a mass of (m2-m1)
Answer:
[tex]4.725 m/s^{2}[/tex]
Explanation:
We know that from Newton's second law of motion, F=ma hence making acceleration the subject then [tex]a=\frac {F}{m}[/tex] where a is acceleration, F is force and m is mass
Also making mass the subject of the formula [tex]m=\frac {F}{a}[/tex]
For [tex]m1= \frac {F}{13.5}[/tex] and [tex]m2=\frac {F}{3.5}[/tex] hence [tex]F=(m2-m1)a= (\frac {F}{3.5}-\frac {F}{13.5})a=0.2116402116\\\frac {1}{a}=0.2116402116\\a=4.725 m/s^{2}[/tex]