Answer:
emf induces in both iron and copper is same
Explanation:
The induced emf in both the rings would be same.
The induced emf in any material is given by
[tex]\epsilon= -N\frac{d\phi}{dt}[/tex]
N= number of turns in the coil
dΦ= change in magnetic flux
dt= change in time
clearly, induced emf is independent of the material of the object,it only depends upon rate of change of flux.
therefore, emf induces in both iron and copper is same.
The induced emf and currents in the copper ring and the wooden ring are compared based on their conductivity.
Explanation:When a magnetic field changes, it induces an electric field and hence an emf (electromotive force) in a conducting loop. The magnitude of the induced emf depends on the rate of change of the magnetic field. In this case, both the copper ring and the wooden ring experience the same change in magnetic flux, which means they have the same induced electric fields. However, the copper ring, being a good conductor, has a much higher induced emf compared to the wooden ring, which has a lower conductivity.
Wooden Ring:
Conductivity: Wood is a poor conductor of electricity. Its conductivity is significantly lower compared to metals like copper.
Induced Emf and Currents: Due to its poor conductivity, a wooden ring will exhibit a much weaker induced emf and induce much smaller currents when subjected to a changing magnetic field.
In summary, the difference in conductivity between copper and wood directly influences the induced emf and currents in the respective rings. The high conductivity of copper facilitates efficient response to changes in magnetic fields, while the low conductivity of wood results in a much weaker and less efficient induction process.
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A 45.8-kg girl is standing on a 151-kg plank. Both originally at rest on a frozen lake that constitutes a frictionless, flat surface. The girl begins to walk along the plank at a constant velocity of 1.46î m/s relative to the plank.
(a) What is her velocity relative to the surface of ice?1 m/s(b) What is the velocity of the plank relative to the surface of ice?2 m/s
Answer:
a. 1.12m/s
b. 0.34m/s
Explanation:
Let Vgi = the velocity of the girl relative to the ice
Let Vgp = the velocity of the girl relative to the plank
Let Vpi = the velocity of the plank relative to the ice
Vgi = Vgp + Vpi
From the question, Vgp = 1.46/s
So, Vgi = 1.46 + Vpi
Conservation of Momentum (Relative to the ice), we have
Initial Momentum = Final Momentum
Because the girl and the plank are at rest initially, the initial Momentum = 0
So, 0 = Mg * Vgi + Mp * Vpi
Where Mg = Mass of the girl = 45.8kg
Mo = Mass of the Plank = 151kg
Make Vpi the subject of the formula, we then have
-Mg*Vgi = Mp*Vpi ---------- Divide through by Mp
Vpi = -Mg * Vgi/Mp
Vpi =( -45.8 * Vgi)/151
Vpi = -45.8Vgi/151
Remember that, we have (Vgi = 1.46 + Vpi)
We then substitute the expression of Vpi in the above equation
That is;
Vgi = 1.46 + (-45.8Vgi/151) ------- Open the bracket
Vgi = 1.46 - 45.8Vgi/151 ----------- Multiply through by 151
151 * Vgi = 151 * 1.46 - 45.8Vgj
151Vgi = 220.46 - 45.8Vgi --------- Collect like terms
151Vgi + 45.8Vgi = 220.46
196.8Vgi = 220.46 --------------- Divide through by 196.8
Vgi = 220.46/196.6
Vgi = 1.1213631739572736
Vgi = 1.12 m/s (Approximated)
So, the velocity of the girl relative to the ice is 1.12m/s
b. Velocity of the plank, relative to the ice
We can solve this using (Vgi = 1.46 + Vpi)
All we need to do is substitute 1.12 for Vgi in the equation
So, we have
Vgi = 1.46 + Vpi becomes
1.12 = 1.46 + Vpi -------- Collect like terms
1.12 - 1.46 = Vpi
-0.34 = Vpi
So, the Velocity of the plank is 0.34m/s to the left
Answer: (a) 2.095 m/s (b) 0.635 m/s
Explanation: The initial momentum is = 0. Also, the final momentum is = 0. At the initial momentum she is at rest hence it is = 0. Whereas, the final momentum is = 0 because it has to equal the initial momentum.
This helps us to understand that when final momentum is = 0 then the girl's direction would cancel the affect of the plank's momentum in the opposite direction.
Therefore,
Vg = Velocity of girl
Vp= Velocity of plank
(45.8)(Vg) + (151)(Vp) = 0
Girl's velocity relative to the plank is 1.46 m/s
Therefore, Vg - Vp = 1.46
Vg = 1.46 + Vp
(45.8)(1.46 + Vp) = 151Vp
66.868 + 45.8Vp = 151Vp
66.868 = 105.2Vp
Vp = 0.635 m/s
Vg = 1.46 + 0.635
Vg = 2.095
A small ball of mass 62 g is suspended from a string of length 72 cm and whirled in a circle lying in the horizontal plane. r v 9.8 m/s 2 72 cm 62 g 32◦ If the string makes an angle of 32◦ with the vertical, find the centripetal force experienced by the ball. The acceleration of gravity is ____.
Answer:
Centripetal force, [tex]F_x=0.379\ N[/tex]
Explanation:
It is given that,
Mass of the ball, m = 62 g = 0.062 kg
Length of the string, l = 72 cm = 0.72 m
Angle, [tex]\theta=32^{\circ}[/tex]
To find,
The centripetal force experienced by the ball.
Solution,
According to the attached free body diagram,
[tex]T\ cos\theta-mg=0[/tex]
[tex]T=\dfrac{mg}{cos\theta}[/tex]
[tex]T=\dfrac{0.062\times 9.8}{cos(32)}[/tex]
T = 0.716 N
The centripetal force will be experienced by the ball in horizontal direction. It is equal to :
[tex]F_x=T\ sin\theta[/tex]
[tex]F_x=0.716\ N\ sin(32)[/tex]
[tex]F_x=0.379\ N[/tex]
So, the centripetal force experienced by the ball is 0.379 N.
The centripetal force experienced by the small ball if mass 62 g and attached to the string is 0.379 N.
What is centripetal force?Centripetal force is the force which is required to keep rotate a body in a circular path. The direction of the centripetal force is inward of the circle towards the center of rotational path.
Given information-
The mass of the small bag is 62 grams.
The length of the string is 72 cm.
Suppose T is the tension in the string, m is the mass of ball and g is the gravitational force.
The horizontal force applied on the body is,
[tex]\sum F_x=T\cos \theta-mg=0\\T=\dfrac{mg}{\cos \theta}[/tex]
The centripetal force applied on the body is,
[tex]F_c=T\sin \theta\\T=\dfrac{F_c}{\sin \theta}[/tex]
Compare both the values of the T, we get,
[tex]\dfrac{F_c}{\sin\theta}=\dfrac{mg}{\cos \theta}\\{F_c}=\dfrac{mg\times{\sin\theta}}{\cos \theta}\\{F_c}={mg\times{\tan\theta}}[/tex]
Put the values as,
[tex]F_c=0.062\times9.8\times\tan(32)\\F_c=0.379\rm N[/tex]
Thus, the centripetal force experienced by the small ball if mass 62 g and attached to the string is 0.379 N.
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A flat, rectangular coil consisting of 60 turns measures 23.0 cm by 31.0 cm . It is in a uniform, 1.20-T, magnetic field, with the plane of the coil parallel to the field. In 0.210 s , it is rotated so that the plane of the coil is perpendicular to the field.Part A What is the change in the magnetic flux through the coil due to this rotation?Part B Find the magnitude of the average emf induced in the coil during this rotation.
The change in magnetic flux when the coil is rotated is equivalent to the initial flux through the coil. The magnitude of the average induced emf is found by dividing the change in magnetic flux by the time over which it occurs, multiplied by the number of turns in the coil, following Faraday's law.
The question is about applying Faraday's law of electromagnetic induction to calculate the change in magnetic flux and the induced electromotive force (emf).
Part A: Change in Magnetic Flux
The magnetic flux \\(\Phi\\) through a coil is given by \\(\Phi = B \cdot A \cdot \cos(\theta)\\), where \\(B\\) is the magnetic field strength, \\(A\\) is the area of the coil, and \\(\theta\\) is the angle between the field and the normal to the coil's plane. Initially, with the coil parallel to the field, \\(\theta = 0\\) degrees, so \\(\cos(\theta) = 1\\), and the initial flux is \\(\Phi_{initial} = B \cdot A\\). After rotation, \\(\theta = 90\\) degrees, so \\(\cos(\theta) = 0\\), and the final flux is \\(\Phi_{final} = 0\\). The change in flux is the final minus the initial. Therefore, the change in magnetic flux is equal to the initial flux \\(\Phi_{initial}\\).
Part B: Magnitude of the Average Induced Emf
To find the magnitude of the average induced emf, Faraday's law is used, which states that the induced emf is proportional to the rate of change of magnetic flux. It is given by the equation \\(emf = -N \cdot \Delta\Phi / \Delta t\\), where \\(N\\) is the number of turns in the coil, \\(\Delta\Phi\\) is the change in flux, and \\(\Delta t\\) is the time over which the change occurs. Plugging in the values given, including the rate of change of flux from part A, we can calculate \\(emf\\).
. A satellite S travels in a circular orbit around the earth. A rocket is located at the apogee of its elliptical orbit for which = 0.58. (a) Determine the sudden change in speed that must occur at A so that the rocket can enter the satellite’s orbit while in free flight along the blue elliptical trajectory. (b) When it arrives at B, determine the sudden adjustment in speed that must be given to the rocket in order to maintain the circular orbit.
so it is not bad not good. pefectly balanced just as all things should be
The radius of the orbits and the rocket's mass, The changes would involve transitioning from the elliptical to the circular orbit, taking into account Earth's gravitational pull.
Explanation:This question delves into detail regarding the physics of celestial movements, in particular, satellites and their orbits. The problem pertains to the transition from an elliptical trajectory to a circular one.
The change in speed at point A would need to equal the difference between the speed in the elliptical orbit at the apogee and the speed necessary to maintain the desired circular orbit, factoring in the gravitational pull of Earth. The change in speed at point B would need to address any residual speed from the elliptical orbit to ensure the rocket's speed aligns with the necessary speed to maintain the circular orbit around the Earth.
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A 1400 kg car moving at 6.6 m/s is initially traveling north in the positive y direction.
After completing a 90° right-hand turn to the positive x direction in 4.9 s, the inattentive operator drives into a tree, which stops the car in 430 ms.
What is the magnitude of the impulse on the car :
(a) due to the turn and (b) due to the collision?
What is the magnitude of the average force that acts on the car (c) during the turn and (d) during the collision?
(e) What is the angle between the average force in (c) and the positive x direction?
Answer:
a) J = 13067 kg*m/s
b) J = 9240 kg*m/s
c) F = 2666.73 N
d) F = 21488.37 N
e) 135°
Explanation:
We know that the impulse could be calculated by:
J = ΔP
where ΔP is the change in the linear momentum:
ΔP = [tex]P_f -P_i[/tex]
Also:
P = MV
Where M is the mass and V is the velocity.
so:
J= [tex]MV_f-MV_i[/tex]
where [tex]V_f[/tex] is the final velocity and [tex]V_i[/tex] is the inicial velocity
a)The impluse from turn is:
[tex]J_x=MV_{fx}-MV_{ix}[/tex]
[tex]J_y=MV_{fy}-MV_{iy}[/tex]
On the turn, [tex]V_{ix}=0[/tex] and [tex]V_{fy}=0[/tex], the magnitude of the impulse on direction x and y are:
[tex]J_x=9240 kg*m/s[/tex]
[tex]J_y=-9240 kg*m/s[/tex]
So, using pythagoras theorem the magnitude of the impulse is:
J = [tex]\sqrt{9240^2+9240^2}[/tex]
J = 13067 kg*m/s
b) The impluse from the collision is:
[tex]J=MV_{f}-MV_{i}[/tex]
J = 0 - (1400)(6.6)
J = 9240 kg*m/s
c) Using the next equation:
FΔt = J
where F is the force, Δt is the time and J is the impulse.
Replacing J by the impulse due to the turn, Δt by 4.9s and solving for F we have that:
F = J / Δt
F = 13067 / 4.9 s
F = 2666.73 N
d) At the same way, replacing J by the impulse during the collision, Δt by 0.43s and solving for F we have that:
F = J / Δt
F = 9240 / 0.43
F = 21488.37 N
e) The force have the same direction than the impulse due to the turn, Then, if the impulse have a direction of -45°, the force have -45° or 135°
You are using a rotary encoder to determine the distance a robot has traveled. The rotary encoder is directly connected to one of the wheels that has a diameter of 10cm. Assume the encoder has 16 lines/revolution and that both the rising and falling edges (2x mode) are detected. What is the smallest detectable change of the rover traveling distance? If you want the rover to travel 2 meters before making a turn, how many encoder counts (rise and fall) are needed before sending a turn command? If you want to upgrade the encoder so that the smallest detectable distance change is ±0.20cm, what would be the required lines/revolution?
The smallest detectable distance with the current setup is 0.98125cm. To make the robot travel 2m, the encoder needs to count roughly 204 times. If the smallest detectable change should be ±0.20cm, the encoder requires 80 lines per revolution.
Explanation:The smallest detectable change in travel distance for the robot can be calculated based on the circumference of the wheel and the resolution of the rotary encoder. Given the diameter of the wheel is 10cm, the circumference can be calculated using the formula 2πr (r= radius of wheel), which results in 31.4cm. With 16 lines per revolution, this yields a distance per count (rise and fall) of 31.4cm/32 = 0.98125cm.
For the robot to travel 2m before making a turn, we will need to calculate the number of encoder counts. 2m equals 200cm, and if each count is representing a distance of 0.98125cm, then the number of encoder counts would be 200/0.98125, roughly equal to 204 counts.
Upgrading the EncoderIf the goal is to reduce the smallest detectable distance to ±0.20cm, then we need to increase the lines per revolution of the encoder. If one line equates to 0.98125cm right now, that means we need roughly 5 times more lines on the encoder to achieve a resolution of 0.20cm. Therefore, the required lines per revolution would be 5*16 = 80 lines.
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The heating element of a water heater in an apartment building has a maximum power output of 30 kW. Four residents of the building take showers at the same time, and each receives heated water at a volume flow rate of 1.49 10-4 m3/s. If the water going into the heater has a temperature of 10° C, what is the maximum possible temperature of the hot water that each showering resident receives
Answer:
22.02°C
Explanation:
P = Power output = 30 kW
Volume flow rate of water = [tex]1.49\times 10^{-4}\ m^3/s[/tex]
S = Specific heat of water = 4.186 kJ/mol°C
[tex]T_i[/tex] = Initial temperature = 10°C
[tex]T_f[/tex] = Final temperature
Total volume of water is
[tex]v=4\times 1.49\times 10^{-4}\\\Rightarrow v=5.96\times 10^{-4}\ m^3/s[/tex]
Mass flow rate is
[tex]m=\rho\times v\\\Rightarrow m=1000\times 5.96\times 10^{-4}\\\Rightarrow m=0.596\ kg/s[/tex]
Heat is given by
[tex]\frac{q}{t}=\frac{m}{t}S\Delta T\\\Rightarrow 30=0.596\times 4.186(T_f-10)\\\Rightarrow T_f=\frac{30}{0.596\times 4.186}+10\\\Rightarrow T_f=22.02\ ^{\circ}C[/tex]
The maximum possible temperature of the hot water that each showering resident receives is 22.02°C
A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .
(a) What is the orbital period of the spacecraft?
(b) What is the spacecraft’s kinetic energy?
The orbital period of a spacecraft can be calculated using Kepler's Third Law and the kinetic energy can be determined using the equation K = GMm/(2r).
Explanation:The orbital period of the spacecraft can be found using Kepler's Third Law, which states that the square of the period of a body moving in a circular orbit is proportional to the cube of the average distance from the center of the force. If we let RE and m denote Earth's radius (6.371 x 10^6 m) and mass (5.972 x 10^24 kg), and G define the gravitational constant (6.67 x 10^-11 N m^2/kg^2), the period T (in seconds) can be calculated as T = 2π √[(4π^2(R+2h)^3)/Gm], or T = 2π √[(4π^2(3RE)^3)/(GmE)].
The kinetic energy K of an object moving in a circular orbit is given by K = GMm/(2r), where M is the mass of the central body (in this case Earth), m is the mass of the object (the spacecraft), and r is the distance of the object from the center of the force (in this case the Earth's center). In this instance, K = GMm/(2R), with R being the radius of Earth, should suffice.
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The Tevatron acceleator at the Fermi National Accelerator Laboratory (Fermilab) outside Chicago boosts protons to 1 TeV (1000 GeV) in five stages (the numbers given in parentheses represent the total kinetic energy at the end of each stage): Cockcroft-Walton (750 keV), Linac (400 MeV), Booster (8 GeV), Main ring or injector (150 Gev) and finally the Tevatron itself (1 TeV). What is the speed of the proton at the end of each stage?
Answer:
a) [tex] v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s[/tex]
b) [tex] v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s[/tex]
c) [tex] v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s[/tex]
d) [tex] v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s[/tex]
e) [tex] v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s[/tex]
Explanation:
At that energies, the speed of proton is in the relativistic theory field, so we need to use the relativistic kinetic energy equation.
[tex] KE=mc^{2}(\gamma -1) = mc^{2}(\frac{1}{\sqrt{1-\beta^{2}}} -1)[/tex] (1)
Here β = v/c, when v is the speed of the particle and c is the speed of light in vacuum.
Let's solve (1) for β.
[tex] \beta = \sqrt{1-\frac{1}{\left (\frac{KE}{mc^{2}}+1 \right )^{2}}} [/tex]
We can write the mass of a proton in MeV/c².
[tex] m_{p}=938.28 MeV/c^{2} [/tex]
Now we can calculate the speed in each stage.
a) Cockcroft-Walton (750 keV)
[tex] \beta = \sqrt{1-\frac{1}{\left (\frac{0.75 MeV}{938.28 MeV}+1 \right )^{2}}} [/tex]
[tex] \beta = 0.04 [/tex]
[tex] v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s[/tex]
b) Linac (400 MeV)
[tex] \beta = \sqrt{1-\frac{1}{\left (\frac{400 MeV}{938.28 MeV}+1 \right )^{2}}} [/tex]
[tex] \beta = 0.71 [/tex]
[tex] v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s[/tex]
c) Booster (8 GeV)
[tex] \beta = \sqrt{1-\frac{1}{\left (\frac{8000 MeV}{938.28 MeV}+1 \right )^{2}}} [/tex]
[tex] \beta = 0.994 [/tex]
[tex] v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s[/tex]
d) Main ring or injector (150 Gev)
[tex] \beta = \sqrt{1-\frac{1}{\left (\frac{150000 MeV}{938.28 MeV}+1 \right )^{2}}} [/tex]
[tex] \beta = 0.999 [/tex]
[tex] v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s[/tex]
e) Tevatron (1 TeV)
[tex] \beta = \sqrt{1-\frac{1}{\left (\frac{1000000 MeV}{938.28 MeV}+1 \right )^{2}}} [/tex]
[tex] \beta = 0.9999 [/tex]
[tex] v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s[/tex]
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A square plate has mass 0.600 kg and sides of length 0.150 m. It is free to rotate without friction around an axis through its center and perpendicular to the plane of the plate. How much work must you do on the plate to change its angular speed from 0 to 40.0 rad/s?
Answer:
W = 1.8 J
Explanation:
given,
mass of square plate = 0.6 Kg
length of side of square plate = 0.15 m
speed is change from 0 to 40.0 rad/s
work done = ?
moment of inertia through the plane perpendicular to plane
[tex]I = \dfrac{Mr^2}{6}[/tex]
r will be equal to 0.15 m
[tex]I = \dfrac{0.6 \times 0.15^2}{6}[/tex]
[tex]I =0.00225\ kgm^2[/tex]
work done is equal to change in kinetic energy
[tex]W = \dfrac{1}{2}I(\omega_f^2-\omega_i^2)[/tex]
[tex]W = \dfrac{1}{2}\times 0.00225(40^2-0^2)[/tex]
W = 1.8 J
work done when speed changes from 0 to 40 rad/s is equal to W = 1.8 J
In order to rendezvous with an asteroid passing close to the earth, a spacecraft must be moving at 8.50×103m/s relative to the earth at a distance 2.50×108m from the center of the earth. At what speed must this spacecraft be launched from the earth's surface? Neglect air resistance and the gravitational pull of the moon.
Answer:
v₀ = 13.9 10³ m / s
Explanation:
Let's analyze this exercise we can use the basic kinematics relationships to love the initial velocity and the acceleration we can look for from Newton's second law where force is gravitational attraction.
F = m a
G m M / x² = m dv / dt = m dv/dx dx/dt
G M / x² = dv/dx v
GM dx / x² = v dv
We integrate
v² / 2 = GM (-1 / x)
We evaluate between the lower limits where x = Re = 6.37 10⁶m and the velocity v = vo and the upper limit x = 2.50 10⁸m with a velocity of v = 8.50 10³ m/s
½ ((8.5 10³)² - v₀²) = GM (-1 /(2.50 10⁸) + 1 / (6.37 10⁶))
72.25 10⁶ - v₀² = 2 G M (+0.4 10⁻⁸ - 1.57 10⁻⁷)
72.25 10⁶ - v₀² = 2 6.63 10⁻¹¹ 5.98 10²⁴ (-15.3 10⁻⁸)
72.25 10⁶ - v₀² = -1.213 10⁸
v₀² = 72.25 10⁶ + 1,213 10⁸
v₀² = 193.6 10⁶
v₀ = 13.9 10³ m / s
A large uniform chain is hanging from the ceiling, supporting a block of mass 46 kg. The mass of the chain itself is 19 kg, and the length of the chain is 1.9 m. The acceleration of gravity is 9.81 m/s 2 . Find the tension in the chain at the point where the chain is supporting the block. Answer in units of N.
Answer:
T = 451.26 N
Explanation:
It is given that,
The mass of block, m = 46 kg
Mass of the chain, m' = 19 kg
Length of the chain, l = 1.9 m
Let T is the the tension in the chain at the point where the chain is supporting the block. It is clearly equal to the product of mass and acceleration.
[tex]T=mg[/tex]
[tex]T=46\ kg\times 9.81\ m/s^2[/tex]
T = 451.26 N
So, the tension in the chain at the point where the chain is supporting the block is 451.26 N. Hence, this is the required solution.
calculate the speed of a car that covers a distance of 108km in 30 mins
The speed of the car is 60 m/s
Explanation:
The speed of an object is a scalar quantity telling "how fast" the object is moving, regardless of its direction. It can be calculated as follows:
[tex]speed=\frac{d}{t}[/tex]
where
d is the distance covered
t is the time taken
For the car in this problem, we have
[tex]d=108 km =1.08\cdot 10^5 m[/tex] is the distance covered
[tex]t=30 min \cdot 60 =1800 s[/tex] is the time taken
Subsituting into the equation, we find the speed:
[tex]speed=\frac{1.08\cdot 10^5}{1800}=60 m/s[/tex]
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The brass pendulum of a particular grandfather clock has a mass of 2 kilograms and completes its swing once every 2 seconds.
If you were to replace the brass pendulum with an identically shaped gold pendulum having a mass of 4 kilograms, it would complete its swing once every
2 seconds.
4 seconds.
8 seconds.
16 seconds.
To solve this problem it is necessary to apply the concept of period in a pendulum. Its definition is framed in the equation
[tex]T= 2\pi \sqrt{\frac{l}{g}}[/tex]
Where,
l = Length
g = Gravity
As you can see, the Period is dependent on both the gravitational acceleration and the length of the string. The period does not depend on mass. Therefore the period must remain the same at 2 seconds.
A scientist makes a device to catch baseballs. A long bar of total mass 2.2kg and length 1.2m is fixed at its center. It catches a ball of mass 0.15kg moving at v=16m/s in a box at one of its ends. Ignore the mass of the catcher box for this problem. What is the angular velocity of the system after the ball is caught?
Answer:
ωf = 4.53 rad/s
Explanation:
By conservation of the angular momentum:
Ib*ωb = (Ib + Ic)*ωf
Where
Ib is the inertia of the ball
ωb is the initial angular velocity of the ball
Ic is the inertia of the catcher
ωf is the final angular velocity of the system
We need to calculate first Ib, Ic, ωb:
[tex]Ib = mb*(L/2)^2=0.15*(1.2/2)^2=0.054 kg.m^2[/tex]
[tex]Ic = mc/12*L^2=2.2/12*1.2^2=0.264 kg.m^2[/tex]
ωb = Vb / (L/2) = 16 / (1.2/2) = 26.67 m/s
Now, ωf will be:
[tex]\omega f = \frac{Ib*\omega b}{Ib + Ic} = 4.53rad/s[/tex]
Find a recurrence relation for the number of ways to go n miles by foot walking at 2 miles per hour or jogging at 4 miles per hour or running at 8 miles per hour; at the end of each hour a choice is made of how to go the next hour. (b) How many ways are there to go 12 miles?
Answer:
(a) [tex]x_{n} = x_{n - 2} + x_{n - 4} + x_{n - 8}[/tex]
(b) 18 ways
Solution:
As per the question:
We need to find the no.of ways to go 'n' miles by foot:
From the question it is clear that, at each hours even miles can be traveled by a person, If the distance to be covered is odd, then there in no way 3 miles can be covered.
Thus defining recurrence as:
[tex]x_{o} = 1[/tex]
[tex]x_{n} = 0[/tex]
where
n: negative or odd
Then
[tex]x_{n} = x_{n - 2} + x_{n - 4} + x_{n - 8}[/tex]
(b) No. of ways to travel 12 miles:
[tex]x_{o} = 1[/tex]
[tex]x_{2} = 1 + 0 = 1[/tex]
[tex]x_{4} = x_{2} + x_{o} + 0 = 1 + 1 = 2[/tex]
[tex]x_{6} = x_{4} + x_{2} + 0 = 2 + 1 + 0 = 3[/tex]
[tex]x_{8} = x_{6} + x_{4} + 0 = 3 + 2 = 6[/tex]
[tex]x_{10} = x_{8} + x_{6} + x_{2} = 6 + 3 + 1 = 10[/tex]
[tex]x_{12} = x_{10} + x_{8} + x_{4} = 10 + 6 + 2 = 18[/tex]
Thus clearly there are 18 ways to travel 12 miles
The recurrence relation for the number of ways to go n miles by foot at different speeds is An = An-2 + An-4 + An-8. To find the number of ways to go 12 miles, we can compute the values of An using the recurrence relation.
Explanation:To find a recurrence relation for the number of ways to go n miles by foot walking at 2 miles per hour or jogging at 4 miles per hour or running at 8 miles per hour, we need to consider the choices made at each hour. Let's denote the number of ways to go n miles as An. At each hour, there are three choices: walk, jog, or run. If we choose to walk, we have An-2 ways to go n miles. If we choose to jog, we have An-4 ways. If we choose to run, we have An-8 ways. Therefore, the recurrence relation is:
An = An-2 + An-4 + An-8
To find the number of ways to go 12 miles, we need to find A12. Starting from the base cases, A0 = 1, A2 = 1, A4 = 2, and A8 = 4. Using the recurrence relation, we can compute the values of A12 as follows:
A12 = A10 + A8 + A4= A8 + A6 + A2 + A6 + A4 + A0 + A4 + A2 + A0 + A2 + A0= 4 + 1 + 1 + 1 + 2 + 1 + 2 + 1 + 1 + 1 + 1= 16Learn more about Recurrence relation here:https://brainly.com/question/36494057
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A coil has N_a turns enclosing an area of A. In a physics laboratory experiment, the coilis rotated during the time interval Deltat from a position in which theplane of each turn is perpendicular to Earth's magnetic field toone in which the plane of each turn is parallel to the field. Themagnitude of Earth's magnetic field at the lab location isB.
a.)What is the total magnitude Phi_initial of the magneticflux through the coil before it is rotated?
b.)What is the magnitude Phi_final of the total magneticflux through the coil after it is rotated?
c.)What is the magnitude of the average emf induced in the coil?
Answer:
Answered
Explanation:
a) The magnitude of the magnetic flux through the coil before it is rotated is,
[tex]\phi_{initial}= BAsin\theta[/tex]
=BAsin90° = BA
therefore, the magnitude of the magnetic flux= BA
b) The magnitude of the magnetic flux through the coil after it is rotated is, Ф_final = BAsinθ
= BA sin0= 0
therefore, the magnitude of magnetic flux through the coil after it is rotated is, zero.
c) The magnitude of the average emf induced in the coil is,
[tex]\epsilon = N\frac{d\phi}{dt}[/tex]
[tex]\epsilon = N\frac{NBA}{\Delta t}[/tex]
because Ф = BA
On the other side of the gorge, at the highest point of his swing, the vine makes an angle of \theta=40^\circθ=40 ∘ from the vertical. At this point, Tarzan's speed is instantaneously zero (as he transitions from swinging up to swinging back again). What is the tension in the vine at the highest point of his swing?
Answer:
[tex]T = m*g*cos\theta[/tex]
Explanation:
Since tarzan moves in a circular trajectory we can sum all forces on the centripetal-axis:
[tex]T - m*g*cos\theta = m*a_c[/tex]
[tex]T - m*g*cos\theta = m*V^2/R[/tex] Since the speed is zero:
[tex]T - m*g*cos\theta = 0[/tex]
[tex]T = m*g*cos\theta[/tex] Having Tarzan's mass we could calculate the module of the tension in the vine.
The tension in Tarzan's vine can be determined at the highest point of his swing, when his velocity is zero, using Newton's laws of motion. At this point, the tension's vertical component equals his weight, while the horizontal component, providing centripetal force, is zero. The equation T=mg/cosθ can be used to calculate the tension.
Explanation:The subject of your question involves physics, particularly mechanics. The tension in the vine at the highest point of Tarzan's swing, when his velocity is zero and the vine makes an angle of 40 degrees with the vertical, can be found using Newton's second law of motion, which can be written as ΣF=ma, where F represents force, m is mass, and a is acceleration.
At the highest point of the swing, the only forces acting on Tarzan are the tension in the vine (T) and his weight (mg), where m is his mass and g is the acceleration due to gravity. These two forces result in a net force that provides the centripetal acceleration necessary for Tarzan to change direction and swing back down. We have:
The vertical component of tension (Tcosθ) equals mg.The horizontal component of tension (Tsinθ) provides the centripetal force, which is zero at the highest point.So we can get the tension in vine (T) from the equation Tcosθ=mg. Rearranging, we have T=mg/cosθ.
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A woman drops a vibrating tuning fork, which is vibrating at 513 Hertz, from a tall building. Through what distance (in m) has the tuning fork fallen when the frequency detected at the starting point is 489 Hertz? (Assume the speed of sound in air is 343 m/s.)
Answer:
h = 15.34 m
Explanation:
given,
tuning fork vibration = 513 Hz
speed of sound = 343 m/s
frequency after deflection = 489 Hz
the source (the fork) moves away from the observer, its speed increases and hence the apparent frequency decreases
[tex]f_{apparent} = \dfrac{v}{v+u}f_0[/tex]
[tex]489 = \dfrac{343}{343+u}\times 513[/tex]
[tex]0.953 = \dfrac{343}{343+u}[/tex]
[tex]343+u = \dfrac{343}{0.953}[/tex]
[tex]343+u = 359.92[/tex]
u = 16.92 m/s
height of the building
v² = u² + 2 g s
16.92² = 2 x 9.8 x h
h = 14.61 m
time taken by sound to reach observer
[tex]t = \dfrac{14.61}{343}[/tex]
[tex]t =0.0426\ s[/tex]
in this time tuning fork has fallen one more now,
[tex]h' = u t + \dfrac{1}{2}gt^2[/tex]
[tex]h' = 16.92\times 0.0426 + \dfrac{1}{2}\times 9.8 \times 0.0426^2[/tex]
h' = 0.7296 m = 0.73 m
total distance
h = 14.61 + 0.73
h = 15.34 m
Final answer:
The tuning fork has fallen through a distance of 2,933.77 m.
Explanation:
To calculate the distance fallen by the vibrating tuning fork, we can use the equation:
f' = f(v/(v + vo))
where f' is the observed frequency, f is the original frequency of the tuning fork, v is the speed of sound, and vo is the velocity of the tuning fork as it falls.
Plugging in the given values: f = 513 Hz, v = 343 m/s, and f' = 489 Hz, we can rearrange the equation to solve for vo:
vo = v(f/f' - 1)
Substituting the values and solving for vo:
vo = 343((513/489) - 1) = 12.3 m/s
Since the equation for distance fallen is:
d = vo*t + (1/2)*g*t^2
and we are assuming the tuning fork is dropped from rest, we can simplify the equation to:
d = (1/2)*g*t^2
where g is the acceleration due to gravity and t is the time the tuning fork has been falling. Since we are solving for distance, we can rearrange the equation to solve for t:
t = sqrt(2*d/g)
Substituting the values and solving for t:
t = sqrt(2*d/9.8) = sqrt(2*d/10) = sqrt(d/5)
Now, we can substitute the value of vo and the known value of f' = 489 Hz into the equation:
vo*t = sqrt(d/5)*12.3 = d
Plugging in the known value of f' = 489 Hz and solving for d:
sqrt(d/5)*12.3 = d/489
Squaring both sides of the equation and solving for d:
d = (12.3^2 * 489^2)/(489^2 - 12.3^2*5) = 2933.77 m
Orangutans can move by brachiation, swinging like a pendulum beneath successive handholds. If an orangutan has arms that are 0.90 m long and repeatedly swings to a 20° angle, taking one swing immediately after another, estimate how fast it is moving in m/s.
To estimate the speed of the swinging orangutan, we can use the principles of rotational motion and trigonometry. By considering the angle of swing and the length of the orangutan's arm, we can calculate the horizontal distance traveled in one swing. However, without the time taken for one swing, we cannot provide an exact value for the speed.
Explanation:To estimate the speed at which the orangutan is moving while swinging from branch to branch, we can use the principles of rotational motion and trigonometry. When the orangutan swings to a 20° angle, it forms a right triangle with the vertical height being 0.90 m and the hypotenuse being the length of the arm. By using the sine function, we can find the horizontal distance traveled by the orangutan in one swing.
Using the formula sin(20°) = horizontal distance / 0.90 m, we can rearrange the equation to find the horizontal distance traveled:
horizontal distance = 0.90 m * sin(20°)
Since the orangutan takes one swing immediately after another, the time taken for one swing is negligible. Therefore, the speed at which the orangutan is moving can be calculated by dividing the horizontal distance by the time taken for one swing. As the time taken is not given in the question, we cannot provide an exact value for the speed.
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To estimate the speed of an orangutan swinging, we use the physics formula for the speed of an object in pendulum motion. Under the given conditions of a 20° swing and arm length of 0.9 m, the orangutan's speed is approximately 2.54 m/s.
Explanation:The behavior you're describing, where an orangutan swings like a pendulum beneath handholds, involves the study of pendulum motion, which is a concept in physics. Under simplified physics assumptions, we can model the swinging motion using the formula for the speed (v) of an object in pendulum motion: v = √(2*g*L(1-cos(θ))), where g is the acceleration due to gravity (9.8 m/s^2), L is the length of the pendulum (which we can approximate as the length of an orangutan's arm, 0.9 m), and θ is the angle swung through (20° or 0.349 radians after conversion).
Plugging the given values into the formula gives us: v = √(2*9.8*0.9(1-cos(0.349))), so the speed of an orangutan swinging would be approximately 2.54 m/s.
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A bowling ball encounters a 0.760-m vertical rise on the way back to the ball rack, as the drawing illustrates. Ignore frictional losses and assume that the mass of the ball is distributed uniformly. The translational speed of the ball is 8.57 m/s at the bottom of the rise. Find the translational speed at the top.
To find the translational speed of the bowling ball at the top of the vertical rise, we can use the principle of conservation of mechanical energy.
Explanation:To find the translational speed of the bowling ball at the top of the vertical rise, we can use the principle of conservation of mechanical energy. Initially, the ball has kinetic energy due to its translational speed. As it moves up the vertical rise, its gravitational potential energy increases. Since there is no friction, we can assume there is no loss of energy.
Using the equation for conservation of mechanical energy:
Initial kinetic energy = final potential energy
0.5 * m * v^2 = m * g * h
Where m is the mass of the ball, v is the translational speed at the bottom, g is the acceleration due to gravity, and h is the height of the vertical rise.
Plugging in the given values:
0.5 * m * (8.57 m/s)^2 = m * 9.8 m/s^2 * 0.760 m
Simplifying and solving for the mass:
m = (9.8 m/s^2 * 0.760 m) / (0.5 * (8.57 m/s)^2)
Finally, we can plug the mass back into the initial equation to find the translational speed at the top of the rise.
A 0.20 kg object, attached to a spring with spring constant k = 10 N/m, is moving on a horizontal frictionless surface in simple harmonic motion of amplitude of 0.080 m. What is its speed at the instant when its displacement is 0.040 m? (Hint: Use conservation of energy.)
The question involves using the Conservation of Energy to find the speed of an object in simple harmonic motion. The concept relies on the fact that the total energy in the harmonic system remains constant, allowing us to determine the kinetic energy of the object and consequently its speed.
Explanation:This question can be answered using the Conservation of Energy principle.
In the given problem, the object attached to the spring performs simple harmonic motion, meaning that the total energy of the system (kinetic energy + potential energy) remains constant.
At any instant in time, the total mechanical energy of the object will be:
E = 1/2 k A²
Where.
E is total energy, k is the spring constant, and A is the system's amplitude.
When the displacement of the object is at x = 0.040 m, the potential energy of the object (U) will be:
U = 1/2 k x²
At this point, the kinetic energy (K) can be found by subtracting the potential energy (U) from the total energy (E).
K = E - U
Knowing that the kinetic energy (K) is also given by 1/2 m v² (where m is mass and v is speed), we can rearrange this equation to find the speed of the object:
v = sqrt((2K)/m)
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The speed of the object at a displacement of 0.040 m is 0.49 m/s.
Calculating Speed in Simple Harmonic Motion Using Conservation of Energy
To find the speed of a 0.20 kg object attached to a spring (k = 10 N/m) at a displacement of 0.040 m, we can use the principle of conservation of energy.
Calculate the total mechanical energy (E) in the system using the amplitude (A):
E = 1/2 x k x [tex]A^2[/tex]
E = 1/2 x 10 N/m x [tex](0.080 m)^2[/tex]
E = 0.032 J
Determine the potential energy (U) at the displacement (x = 0.040 m):
U = [tex]1/2 \times k \times x^2[/tex]
U = [tex]1/2 \times 10 N/m \times (0.040 m)^2[/tex]
U = 0.008 J
Find the kinetic energy (K) at that displacement:
K = E - U
K = 0.032 J - 0.008 J
K = 0.024 J
Calculate the speed (v) using the kinetic energy:
K = 1/2 x m x [tex]v^2[/tex]
[tex]0.024 J = 1/2 \times 0.20 kg \times v^2[/tex]
[tex]v^2 = 0.24 m^2/s^2[/tex]
v = 0.49 m/s
Therefore, the speed of the object at a displacement of 0.040 m is 0.49 m/s.
If you could shine a very powerful flashlight beam toward the Moon, estimate the diameter of the beam when it reaches the Moon. Assume that the beam leaves the flashlight through a 7.0-cm aperture, that its white light has an average wavelength of 550nm, and that the beam spreads due to diffraction only. The distance from the Earth to the Moon is 384x10^3km.
To develop this problem it is necessary to apply the Rayleigh Criterion (Angular resolution)criterion. This conceptos describes the ability of any image-forming device such as an optical or radio telescope, a microscope, a camera, or an eye, to distinguish small details of an object, thereby making it a major determinant of image resolution. By definition is defined as:
[tex]\theta = 1.22\frac{\lambda}{d}[/tex]
Where,
[tex]\lambda[/tex]= Wavelength
d = Width of the slit
[tex]\theta[/tex]= Angular resolution
Through the arc length we can find the radius, which would be given according to the length and angle previously described.
The radius of the beam on the moon is
[tex]r = l\theta[/tex]
Relacing [tex]\theta[/tex]
[tex]r = l(\frac{1.22\lambda}{d})[/tex]
[tex]r = 1.22\frac{l\lambda}{d}[/tex]
Replacing with our values we have that,
[tex]r = 1.22*(\frac{(384*10^3km)(\frac{1000m}{1km})(550*10^{-9}m)}{7*10^{{-2}}})[/tex]
[tex]r = 3680.91m[/tex]
Therefore the diameter of the beam on the moon is
[tex]d = 2r[/tex]
[tex]d = 2 * (3690.91)[/tex]
[tex]d = 7361.8285m[/tex]
Hence, the diameter of the beam when it reaches the moon is 7361.82m
Final answer:
Using diffraction formulas, it is estimated that a flashlight beam with an aperture of 7.0cm and an average wavelength of 550nm would have a diameter of approximately 3.7 km when it reaches the Moon.
Explanation:
To estimate the diameter of a flashlight beam when it reaches the Moon, we'll use the concept of diffraction spreading, as the beam is assumed to spread due to diffraction only. According to the diffraction formula for a circular aperture, θ = 1.22 λ/D. The angle θ is the angle of the spread, λ is the average wavelength of the light, and D is the diameter of the aperture through which the light passes.
Using the provided average wavelength of λ = 550nm and an aperture diameter of D = 7.0cm for the flashlight, we can calculate the minimum angular spread of the beam, θ.
θ = 1.22 × (550 x 10^-9 m) / (7.0 x 10^-2 m) = 0.0000095714 radians.
Then, to find out how broad the beam would be when it reaches the Moon, we use the angular spread to calculate the diameter of the beam at the distance to the Moon, which is 384,000 km (or 384 x 10^6 m):
Beam Diameter on the Moon = θ × Distance to the Moon = 0.0000095714 radians × 384 x 10^6 m = 3677.388 m.
Therefore, the flashlight's beam would have an estimated diameter of approximately 3.7 km when it reaches the Moon, considering only the diffraction spreading.
A 47.0 kg uniform rod 4.25 m long is attached to a wall with a hinge at one end. The rod is held in a horizontal position by a wire attached to its other end. The wire makes an angle of 30.0 ∘ with the horizontal, and is bolted to the wall directly above the hinge.
If the wire can withstand a maximum tension of 1350 N before breaking, how far from the wall can a 69.0 kg person sit without breaking the wire?
Answer:
2.79 m
Explanation:
Use the static equilibrium condition, net torque actin on the system is zero.
[tex]\sum \tau= 0[/tex]
[tex]T(Lsin\theta)- Mg\frac{L}{2}-mgx=0[/tex]
solve for the distance of the person from the wall x
[tex]x= \frac{TLsin\theta-Mg(L/2)}{Mg}[/tex]
now putting the values we get
[tex]x= \frac{1350\times4.25sin30-47\times9.81\times(4.25/2)}{69\times9.80}[/tex]
= 2.79 m
Water at 298 K discharges from a nozzle and travels horizontally hitting a flat, vertical wall. The nozzle diameter is 12 mm and the water exits the nozzle with a flat velocity profile at a velocity of 6.0 m/s. Use macroscopic balance for momentum to compute the force [N] on the wall neglecting gravitational and frictional effects
The force on the wall can be calculated using the macroscopic balance for momentum. We need to calculate the mass flow rate of water and use the equation: mass flow rate = density × velocity × cross-sectional area of the nozzle. With the given values, we can find the force on the wall.
Explanation:The force on the wall can be computed using the macroscopic balance for momentum. Since water discharges from a nozzle and travels horizontally, the initial momentum of the water is zero. The change in momentum of the water is equal to the force exerted on the wall. To calculate the force, we need to find the mass flow rate of the water. Assuming the water is an ideal fluid with a flat velocity profile, we can use the equation: mass flow rate = density × velocity × cross-sectional area of the nozzle. Plugging in the given values, we can then calculate the force on the wall.
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Final answer:
To calculate the force exerted on the wall by the water discharging from the nozzle, we use the mass flow rate and velocity of the water to find the momentum change. This calculation applies the macroscopic momentum balance principle and yields the force in Newtons.
Explanation:
The student is asking to calculate the force on a wall when water from a nozzle hits it horizontally. To find this force, we can use the macroscopic momentum balance principle which states that the rate of change of momentum of the fluid is equal to the sum of the external forces applied to the fluid.
As gravity and frictional effects are negligible, the only external force is the force on the wall exerted by the water. Key information given includes the diameter of the nozzle (12 mm) and the velocity of the water (6.0 m/s).
To calculate the force, we need to find the mass flow rate of the water, which can be determined by multiplying the density of water (ρ = 1000 kg/m³ at standard conditions) by the cross-sectional area of the nozzle (A) and the velocity (v) of the water.
The cross-sectional area, A, of the nozzle is A = π * (d/2)², where d is the diameter of the nozzle. Substituting the given values, the area A is π * (0.012/2)² m². The mass flow rate (ṭ) then is ρ * A * v.
With the mass flow rate known, the force (F) exerted on the wall can be calculated by taking the rate of change of momentum as ṭ * v, because the water's velocity is reduced to zero upon hitting the wall, thus the change in momentum is just the initial momentum. Therefore, F = ṭ * v.
After calculating the mass flow rate and applying the equation for force, the student will get the desired force in Newtons (N) on the wall.
An electric motor has an eccentric mass of 10 kg (10% of the total mass) and is set on two identical springs (k=3200 N/m). The motor runs at 1750 rpm, and the mass eccentricity is 100 mm from the center. The springs are mounted 250 mm apart with the motor shaft in the center. Neglect damping and determine the amplitude of vertical vibration.
Answer:
The amplitude is 0.010 m
Solution:
As per the question:
Eccentric Mass, [tex]m_{e} = 10\ kg[/tex]
[tex]m_{e}[/tex] = 10%M kg
0.1M = 10
Total mass, M = 100 kg
Spring constant, k = 3200 N/m
Eccentric center, e = 100 mm = 0.1 m
Speed of motor, N = 1750 rpm
Distance between two springs, d = 250 mm = 0.25 m
Now,
Angular velocity, [tex]\omega = \frac{2\pi N}{60} = \frac{2\pi \times 1750}{60} = 183.26\ rad/s[/tex]
For the vertical vibrations:
[tex]\omega_{n} = \sqrt{\frac{2k}{M}} = \sqrt{\frac{6400}{100}} = 8\ rad/s[/tex]
Now, the frequency ratio is given by:
r = [tex]\frac{\omega}{\omega_{n}} = \frac{183.26}{8} = 22.91[/tex]
Thus the amplitude is given by:
A = [tex]e^{\frac{m_{e}}{M}}.\frac{r^{2}}{|1 - r^{2}|}[/tex]
A = [tex]e^{\frac{10}{100}}.\frac{22.91^{2}}{|1 - 22.91^{2}|} = 0.010 m[/tex]
Final answer:
The amplitude of vertical vibration for the electric motor mounted on springs is 100 mm or 0.1 meters, considering the eccentric mass and neglecting damping.
Explanation:
The subject of this question is Physics, and it pertains to a college-level study of mechanics in the context of oscillations and waves, specifically relating to a system experiencing simple harmonic motion (SHM). The motor with an eccentric mass which causes vertical vibrations can be modeled as a mass-spring system, characteristic of simple harmonic oscillators.
Given the mass and the spring constant, we can calculate the natural angular frequency (not represented in this problem due to lack of damping) of this system. The driving frequency can also be converted to an angular frequency to check against the natural frequency. However, the question directly asks for the amplitude of vertical vibration, which in the undamped case of SHM, depends on the initial displacement of the mass from its equilibrium position. Since the question neglects damping, the amplitude can be considered equal to the eccentricity of the mass in the static case. Therefore, the amplitude of vertical vibration is 100 mm, or 0.1 meters.
The force exerted by an electric charge at the origin on a charged particle at a point (x, y, z) with position vector r = x, y, z is F(r) = Kr/|r|3 where K is a constant. Find the work done as the particle moves along a straight line from (2, 0, 0) to (2, 4, 5).
Answer:
Explanation:
F(r) = Kr/|r|3
= k / r²
Work done
= ∫F(r) dr
= ∫ k / r² dr
limit from
r₁ = 2 to r₂ = √ ( 2² + 4² + 5² ) = 6.7
Work done
∫ k / r² dr
[ - k / r ]
Taking limit from 2 to 6.7
Work done = K ( - 1 / 6.7 + 1 / 2 )
= 0.35 K
The work done as a particle moves along a straight line from one point to another in the presence of an electric field is 0.
To find the work done as a particle moves along a straight line from one point to another in the presence of an electric field, you can use the following formula for the work done:
[tex]\[W = \int_{\mathbf{r_1}}^{\mathbf{r_2}} \mathbf{F} \cdot d\mathbf{r}\][/tex]
Where:
- [tex]\(W\)[/tex] is the work done.
- [tex]\(\mathbf{F}\)[/tex] is the force acting on the particle.
- [tex]\(d\mathbf{r}\)[/tex] is the infinitesimal displacement vector along the path of the particle.
- [tex]\(\mathbf{r_1}\)[/tex] and [tex]\(\mathbf{r_2}\)[/tex] are the initial and final positions of the particle.
In this case, the force [tex]\(\mathbf{F}\)[/tex] is given as [tex]\(F(\mathbf{r})[/tex] = [tex]\frac{K\mathbf{r}}{|\mathbf{r}|^3}\)[/tex], and the particle moves along a straight line from [tex]\(\mathbf{r_1} = (2, 0, 0)\)[/tex] to [tex]\(\mathbf{r_2} = (2, 4, 5)[/tex].
We can rewrite [tex]\(\mathbf{F}[/tex] as a vector:
[tex]\[\mathbf{F} = \frac{K}{|\mathbf{r}|^3}\mathbf{r}\][/tex]
Now, let's calculate the differential displacement [tex]\(d\mathbf{r}\)[/tex] along the path from [tex]\(\mathbf{r_1}\)[/tex] to [tex]r_2[/tex]:
[tex]\[d\mathbf{r} = (dx, dy, dz)\][/tex]
Now, let's calculate the dot product [tex]\(\mathbf{F} \cdot d\mathbf{r}\)[/tex]:
[tex]\[\mathbf{F} \cdot d\mathbf{r} = \frac{K}{|\mathbf{r}|^3}\mathbf{r} \cdot (dx, dy, dz)\][/tex]
Since the path is a straight line, [tex]\(\mathbf{r}\) and \(d\mathbf{r}\)\\[/tex] are parallel, so[tex]\(\mathbf{r} \cdot d\mathbf{r} = |\mathbf{r}||d\mathbf{r}|\cos(0^\circ) = |\mathbf{r}||d\mathbf{r}|\)[/tex].
Therefore, we have:
[tex]\[\mathbf{F} \cdot d\mathbf{r} = \frac{K}{|\mathbf{r}|^3}|\mathbf{r}||d\mathbf{r}| = \frac{K}{|\mathbf{r}|^2}|d\mathbf{r}|\][/tex]
Now, we can integrate [tex]\(\mathbf{F} \cdot d\mathbf{r}\)[/tex] from [tex]\(\mathbf{r_1}\)[/tex] to [tex]\(\mathbf{r_2}\)[/tex] along the straight line path:
[tex]\[W = \int_{\mathbf{r_1}}^{\mathbf{r_2}} \frac{K}{|\mathbf{r}|^2}|d\mathbf{r}|\][/tex]
The integration limits and the path are straight, so this simplifies to:
[tex]\[W = K \int_{2}^{2} \frac{1}{(2^2+4^2+5^2)^{3/2}}|d\mathbf{r}|\][/tex]
Since the particle moves from [tex]\(\mathbf{r_1}\) to \(\mathbf{r_2}\)[/tex], the displacement [tex]\(|d\mathbf{r}| = |\mathbf{r_2} - \mathbf{r_1}| = |(0, 4, 5)| = \sqrt{0^2 + 4^2 + 5^2} = \sqrt{41}\)[/tex].
So, the work done is:
[tex]\[W = K \int_{2}^{2} \frac{1}{(2^2+4^2+5^2)^{3/2}}\sqrt{41} dr\][/tex]
Since the limits of integration are the same, the integral evaluates to zero. Therefore, the work done as the particle moves along a straight line from (2, 0, 0) to (2, 4, 5) is zero.
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A voltmeter and an ammeter are used to respectively monitor the voltage across and the current through a resistor connected to an AC source. If the resistor has a value of 10.0 Ω and the ammeter reads 8.02 A, determine the following.
(a) the rms voltage (in V) across the resistor(b) the peak voltage (in V) of the source(c) the maximum current (in A) in the resistor(d) the average power (in W) delivered to the resistor
Final answer:
To determine the values requested - (a) the rms voltage across the resistor is 80.2 V, (b) the peak voltage of the source is 113.4 V, (c) the maximum current in the resistor is 8.02 A, and (d) the average power delivered to the resistor is 641 W.
Explanation:
In order to determine the values requested, we can use Ohm's law which states that V = IR, where V is the voltage, I is the current, and R is the resistance. Let's go through each part:
(a) To find the rms voltage across the resistor, we use the formula V = IR, where I is the ammeter reading and R is the resistance. Plugging in the values, we get V = (8.02 A) x (10.0 Ω) = 80.2 V.
(b) The peak voltage of the source can be found by multiplying the rms voltage by √2. So the peak voltage is (80.2 V) x √2 = 113.4 V.
(c) The maximum current in the resistor is equal to the ammeter reading, which is 8.02 A.
(d) The average power delivered to the resistor can be calculated using the formula P = IV, where I is the rms current and V is the rms voltage. Since the ammeter reading is the rms current, we can plug in the values to get P = (8.02 A) x (80.2 V) = 641 W.
Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edge of the tires. If the car takes 6 seconds to come to a stop, calculate the average angular acceleration of a point on the outer edge of the tires.
To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.
From the perspective of angular movement, we find the relationship with the tangential movement of velocity through
[tex]\omega = \frac{v}{R}[/tex]
Where,
[tex]\omega =[/tex]Angular velocity
v = Lineal Velocity
R = Radius
At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is
[tex]\alpha = \frac{\omega}{t}[/tex]
Where
[tex]\alpha =[/tex]Angular acceleration
[tex]\omega =[/tex] Angular velocity
t = Time
Our values are
[tex]v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})[/tex]
[tex]v = 16.67m/s[/tex]
[tex]r = 0.25m[/tex]
[tex]t=6s[/tex]
Replacing at the previous equation we have that the angular velocity is
[tex]\omega = \frac{v}{R}[/tex]
[tex]\omega = \frac{ 16.67}{0.25}[/tex]
[tex]\omega = 66.67rad/s[/tex]
Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s
At the same time the angular acceleration would be
[tex]\alpha = \frac{\omega}{t}[/tex]
[tex]\alpha = \frac{66.67}{6}[/tex]
[tex]\alpha = 11.11rad/s^2[/tex]
Therefore the angular acceleration of a point on the outer edge of the tires is [tex]11.11rad/s^2[/tex]
A skateboarder is attempting to make a circular arc of radius r = 11 m in a parking lot. The total mass of the skateboard and skateboarder is m = 87 kg. The coefficient of static friction between the surface of the parking lot and the wheels of the skateboard is μs = 0.62 .
(a) What is the maximum speed, in meters per second, he can travel through the arc without slipping?
Answer:
V = 8.26 m/s
Explanation:
The sum of forces on the centripetal-axis is:
[tex]Ff = m*a_c[/tex]
Maximum speed happens when friction force is maximum, so:
[tex]\mu*N = m*V^2/R[/tex]
[tex]\mu*m*g = m*V^2/R[/tex]
[tex]V=\sqrt{R*\mu*g}[/tex]
V = 8.26 m/s
The maximum speed that he can travel through the arc without slipping is mathematically given as
v = 9.93m/s
The maximum speedQuestion Parameters:
Generally the equation for the Maximum velocity is mathematically given as
[tex]\frac{mv^2}{r} = \mu_s mg[/tex]
Therefore
[tex]v = \sqrt{(0.63)(9.8)(16)}[/tex]
v = 9.93m/s
Therefore, the maximum speed that he can travel through the arc without slipping is
v = 9.93m/s
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