The maximum kinetic energy that can be stored in the flywheel is 27,909.78 J.
Explanation:To calculate the maximum kinetic energy that can be stored in the flywheel, we need to find the moment of inertia. For a uniform solid disk, the moment of inertia is given by the formula I = 0.5 * m * r^2, where m is the mass of the disk and r is the radius. Plugging in the values, we get I = 0.5 * 67.0 kg * (1.22 m)^2 = 49.518 kg·m².
The maximum kinetic energy is equal to 0.5 * I * ω^2, where ω is the angular velocity. Since the radial acceleration of a point on the rim is given as 3510 m/s², we can calculate ω using the formula a = r * ω^2. Rearranging the formula, we get ω = √(a / r) = √(3510 m/s² / 1.22 m) = 33.24 rad/s.
Plugging in the values, the maximum kinetic energy is 0.5 * 49.518 kg·m² * (33.24 rad/s)^2 = 27,909.78 J.
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The maximum kinetic energy that can be stored in the flywheel is approximately 71,791 Joules.
To determine the maximum kinetic energy that can be stored in the flywheel, we need to use the concepts of rotational motion. The relevant formulas are:
Kinetic Energy of a rotating object: KE = 1/2 I ω², where I is the moment of inertia and ω is the angular velocity.
Moment of Inertia for a solid disk: I = 1/2 m R², where m is the mass and R is the radius.
Relation between radial acceleration and angular velocity: a = ω² R.
We are given:
Radius (R): 1.22 m
Mass (m): 67.0 kg
Maximum radial acceleration (a): 3510 m/s²
First, solve for angular velocity (ω) using the radial acceleration formula:
ω² = a / R
⇒ ω² = 3510 m/s² / 1.22 m
⇒ ω² ≈ 2877.05 s⁻²
⇒ ω ≈ 53.61 rad/s
Next, calculate the moment of inertia (I) for the flywheel:
I = 1/2 m R² => I = 0.5 × 67.0 kg × (1.22 m)²
⇒ I ≈ 49.917 kgm²
Finally, calculate the maximum kinetic energy:
KE = 1/2 I ω² => KE = 1/2 × 49.917 kgm² × (53.61 rad/s)²
⇒ KE ≈ 71,791 Joules
Thus, the maximum kinetic energy that can be stored in the flywheel is approximately 71,791 Joules.
A 191 191 kg sculpture hangs from a horizontal rod that serves as a pivot about which the sculpture can oscillate. The sculpture's moment of inertia with respect to the pivot is 17.2 17.2 kg·m2, and when it is swung at small amplitudes, it is found to oscillate at a frequency of 0.925 0.925 Hz. How far is the pivot from the sculpture's center of mass?
Answer:
r = 0.31 m
Explanation:
Given that,
Mass of the sculpture, m = 191 kg
The sculpture's moment of inertia with respect to the pivot is, [tex]I=17.2\ kg-m^2[/tex]
Frequency of oscillation, f = 0.925 Hz
Let r is the distance of the the pivot from the sculpture's center of mass. The frequency of oscillation is given by :
[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{mgr}{I}}[/tex]
[tex]r=\dfrac{4\pi^2f^2I}{mg}[/tex]
[tex]r=\dfrac{4\pi^2\times 0.925^2\times 17.2}{191\times 9.8}[/tex]
r = 0.31 m
So, the pivot is 0.31 meters from the sculpture's center of mass. Hence, this is the required solution.
A playground merry-go-round has radius 2.80 m and moment of inertia 2400 kg⋅m2 about a vertical axle through its center, and it turns with negligible friction.A child applies an 20.0 N force tangentially to the edge of the merry-go-round for 25.0 s . If the merry-go-round is initially at rest, what is its angular speed after this 25.0 s interval?
Answer:
[tex]\omega_f = 0.602\ rad/s[/tex]
Explanation:
given,
radius of merry- go- round = 2.80 m
moment of inertia = I = 2400 kg⋅m²
child apply force tangentially = 20 N
for time = 25 s
angular speed after 25 speed = ?
initial angular speed of the merry go round = 0 rad/s
we know,
torque = I α.............(1)
α is angular acceleration
and also
τ = F.r........................(2)
computing equation (1) and (2)
F . r = I α
[tex]\alpha = \dfrac{\omega_f - \omega_i}{t}[/tex]
[tex]F . r =I \times \dfrac{\omega_f - \omega_i}{t}[/tex]
[tex]20 \times 2.89 =2400\times \dfrac{\omega_f -0}{25}[/tex]
[tex]\omega_f = 0.602\ rad/s[/tex]
the angular speed of merry-go-round after 25 second is equal to [tex]\omega_f = 0.602\ rad/s[/tex]
A 3 kg steel ball falls down vertically and strikes a floor with a speed of 12 m/s. It bounces off upward with the same speed. If the ball is in contact with the floor for 0.12 s, what is the magnitude of the average force exerted on the ball by the floor?
Answer:
600N
Explanation:
If the ball bounces off upward with the same speed, the velocities before and after are equal and in opposite directions. That means the floor must have caused a momentum impact to change its direction.
By the law of momentum conservation:
[tex]mv + \Delta P_f = mV[/tex]
where m = 3 kg is the ball mass. v = -12 m/s is the velocity right before the impact. V = 12m/s is the velocity right after the impact. ΔP_f is the momentum caused by the floor.
[tex]\Delta P_f = m(V-v) = 3(12 - (-12)) = 3*24 = 72kgm/s[/tex]
Then the average force exerted during the 0.12s impact is
[tex]F = \frac{\Delta P_f}{\Delta t} = \frac{72}{0.12} = 600N[/tex]
Consider the amideanion.What is the central atom? Enter its chemical symbol.How many lone pairs are around the central atom?What is the ideal angle between the nitrogen-hydrogen bonds?Compared to the ideal angle, you would expect the actual angle between the nitrogen-hydrogen bonds to be ...
Answer:
a) The central atom of the amide ion is nitrogen
b) (NH2)-
c) There are two lone pairs around the nitrogen atom
d) The ideal angle between nitrogen hydrogen bonds is 109.5° C, typical of a tetrahedral electron density arrangement
e) I would expect the nitrogen-hydrogen bond angles in the amide ion to be less than 109.5 ° because unbonded pairs repel bonded pairs much more than other bonded pairs do. So the bonds in this ion would be pushed closer together than normal
You drop a rock off a bridge. When the rock has fallen 4 m, you drop a second rock. As the two rocks continue to fall, what happens to their velocities?
a. Both increase at the same rate.
b. The velocity of the first rock increases faster than the velocity of the second.
c. The velocity of the second rock increases faster than the velocity of the first.
d. Both velocities stay constant.
Answer:
option A.
Explanation:
The correct answer is option A.
Two rocks are off a bridge first rock is fallen 4 when the second rock is dropped.
Both the rock is dropped under the effect of acceleration due to gravity so, the rate of change of the velocity for both the rock particle will be the same.
Hence, the first rock will reach ground earlier than the second rock because the rate of change of both the rock is at the same rate.
At the time when the two rocks should continue to fall so their velocities should be increased.
The correct answer is option A.
Velocity:When two rocks are off a bridge so here first rock is fallen 4 at the time when the second rock should be fallen. Both the rock should be fallen under the impact of acceleration because of gravity due to which the rate of change of the velocity for both the rock particle should be similar.
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Two masses sit at the top of two frictionless inclined planes that have different angles. Which mass gets to the bottom first?
Answer:
Explanation:
Let the two inclined planes having angle of inclinations α and β.
The acceleration along the inclined plane acting on the body is gSinα and gSinβ.
If α > β
So, g Sinα > g Sinβ
So, more the inclination of the plane more be the acceleration of body and hence the time taken is less.
So, the body kept on the inclined whose inclination is more reaches at the bottom first.
The mass sitting on the plane inclined to a greater angle to the horizontal will reach the bottom first because it will experience greater acceleration.
Let the angle of inclination of the first mass to the horizontal = θLet the angle of inclination of the second mass to the horizontal = β
The net horizontal force on the masses is calculated as follows;
[tex]\Sigma F_x = 0\\\\mgsin(\theta) = ma\\\\gsin(\theta) = a\ \ \ or \ \ \ \\\\gsin(\beta ) = a[/tex]
Assuming the value of the angles of inclination of the plane is the following;
θ = 60⁰ and β = 30⁰a = gsin(60) = 0.866g
a = gsin(30) = 0.5g
Thus, we can conclude that the mass sitting on the plane inclined to a greater angle to the horizontal will reach the bottom first because it will experience greater acceleration.
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A 50-N crate is pulled up a 5-m inclined plane by a worker at constant velocity. If the plane is inclined at an angle of 37° to the horizontal and there exists a constant frictional force of 10 N between the crate and the surface, what is the force applied by the worker?
Answer:F=40.09 N
Explanation:
Given
weight of crate [tex]W=50 N[/tex]
Inclination [tex]\theta =37^{\circ}[/tex]
Frictional Force [tex]f=10 N[/tex]
as the crate is moving with constant velocity therefore net Force on crate is zero
[tex]F-50\sin (37)-f=0[/tex]
[tex]F=50\sin (37)+10[/tex]
[tex]F=30.09+10[/tex]
[tex]F=40.09 N[/tex]
What impulse occurs when an average force of 7.0 N is exerted on a cart for 2.5 s ?
Express your answer to two significant figures and include the appropriate units.
Answer:[tex]1.7\times 10[/tex]
Explanation:
Given
Force [tex]F=7 N[/tex]
time interval [tex]t=2.5 s[/tex]
Impulse is given by [tex]=Force \times time\ interval\ for\ applied\ force [/tex]
[tex]Impulse=7\times 2.5=17.5[/tex]
For two significant Figure
[tex]Impulse=1.7\times 10[/tex]
8. A gas is contained in a horizontal piston-cylinder apparatus at a pressure of 350 kPa and a volume of 0.02 m3 . Determine the work done by the piston on the gas if the cylinder volume is increased to 0.15 m3 through heating. Assume the pressure of the
Answer:
45500 J
Explanation:
Pressure, P =350 kPa = 350 x 1000 Pa
V1 = 0.02 m^3
V2 = 0.15 m^3
Work done by the piston
W = Pressure x increase in volume
W = P x (V2 - V1)
W = 350 x 1000 (0.15 - 0.02)
W = 45500 J
Thus, the work done is 45500 J.
Final answer:
The piston performs 45500 Joules of work on the gas when the volume of the gas in the piston-cylinder apparatus increases from 0.02 m³ to 0.15 m³ at a constant pressure of 350 kPa.
Explanation:
To determine the work done by the piston on the gas when the cylinder volume increases from 0.02 m³ to 0.15 m³ at a constant pressure of 350 kPa, we can use the formula for work done during a quasi-static process in thermodynamics, which is W = PΔV, where W is the work done by the gas, P is the constant pressure, and ΔV is the change in volume.
Given:
P = 350 kPa = 350000 Pa (since 1 kPa = 1000 Pa)
ΔV = 0.15 m³ - 0.02 m³ = 0.13 m³
We can now calculate the work done:
W = PΔV
W = 350000 Pa × 0.13 m³
W = 45500 J
Therefore, the piston performs 45500 Joules of work on the gas when the volume increases to 0.15 m³.
Imagine that you have obtained spectra for several galaxies and have measured the observed wavelength of a hydrogen emission line that has a rest wavelength of 656.3 nanometers. Here are your results:Galaxy 1: Observed wavelength of hydrogen line is 660.0 nanometers.Galaxy 2: Observed wavelength of hydrogen line is 665.8 nanometers.Galaxy 3: Observed wavelength of hydrogen line is 682.7 nanometers.Calculate the redshift, z, for each of these galaxies.
Answer:
Galaxy 1:
z = 0.0056
Galaxy 2:
z = 0.014
Galaxy 3:
z = 0.040
Explanation:
Spectral lines will be shifted to the blue part of the spectrum¹ 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 source in this particular case is represented for each of the galaxies of interest.
Hence, the redshift represents this shift of the spectral lines to red part in the spectrum of a galaxy or any object which is moving away. That is a direct confirmation of how the universe is in an expanding accelerated motion.
The redshift can be defined in analytic way by means of the 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]
[tex]\frac{v}{c} = \frac{\lambda_{measured}-\lambda_{0}}{\lambda_{0}}[/tex]
[tex]z = \frac{\lambda_{measured}-\lambda_{0}}{\lambda_{0}}[/tex] (2)
Where z is the redshift.
For the case of Galaxy 1:
Where [tex]\lambda_{measured} = 660.0 nm[/tex] and [tex]\lambda_{0} = 656.3 nm[/tex].
[tex]z = \frac{\lambda_{measured}-\lambda_{0}}{\lambda_{0}}[/tex]
[tex]z = (\frac{660.0 nm - 656.3 nm}{656.3 nm})[/tex]
[tex]z = 0.0056[/tex]
For the case of Galaxy 2:
Where [tex]\lambda_{measured} = 665.8 nm[/tex] and [tex]\lambda_{0} = 656.3 nm[/tex].
[tex]z = \frac{665.8 nm - 656.3 nm}{656.3 nm}[/tex]
[tex]z = 0.014[/tex]
For the case of Galaxy 3:
Where [tex]\lambda_{measured} = 682.7 nm[/tex] and [tex]\lambda_{0} = 656.3 nm[/tex].
[tex]z = \frac{682.7 nm - 656.3 nm}{656.3 nm}[/tex]
[tex]z = 0.040[/tex]
Key terms:
¹Spectrum: Decomposition of light in its characteristic colors (wavelengths).
A conducting sphere of radius r1= 0.38m has a total charge of Q=.75 uC. A second uncharged conducting sphere of radius r2= 0.28m is then connected to the first by a thin conducting wire. The spheres are separated by a very large distance to their size. What is the total charge on sphere two, Q2, in C?
Answer:
The Q₂ is 0.318 μC
Explanation:
The charge flows is the same on both, then:
[tex]V_{1} =\frac{kQ_{1} }{r_{1} } \\V_{2} =\frac{kQ_{2} }{r_{2} } \\Q_{2} =Q-Q_{1} \\\frac{kQ_{1} }{r_{1} } =\frac{k*(Q-Q_{1}) }{r_{2} } \\Q_{1} =\frac{\frac{Q}{r_{2} } }{(1/r_{1})+(1/r_{2}) }[/tex]
But:
[tex]\frac{1}{r_{1} } +\frac{1}{r_{2} }=\frac{1}{0.38} +\frac{1}{0.28} =6.2[/tex]
Q = = 0.75 μC
Replacing:
[tex]Q_{1} =\frac{\frac{0.75}{0.28} }{6.2} =0.432\mu C[/tex]
The Q₂ is equal:
Q₂ = 0.75 - 0.432 = 0.318 μC
The second conducting sphere, once connected and then disconnected from the first, ends up with the same charge as the first sphere, 0.75 µC or 0.75 x 10^-6 C.
Explanation:When two conducting spheres are connected by a thin conducting wire, the charge distributes itself evenly across both spheres, assuming both spheres are identical in size. In the case of spheres with different radii, the amount of charge on each sphere once they are separated again will still be proportional to the original amount. Since the two spheres in the question are described to be separated by a large distance relative to their size, the thin wire connecting them would create an equipotential surface, allowing the charge to redistribute.
The total charge Q is conserved in the system, thus the charge on the second sphere Q2 after disconnecting the wire would be the same as the initial total charge Q since it was not charged before. Therefore, the total charge on sphere two, Q2, will also be 0.75 μC or 0.75 x 10^{-6} C.
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A 3.1 g spider is dangling at the end of a silk thread. You can make the spider bounce up and down on the thread by tapping lightly on his feet with a pencil. You soon discover that you can give the spider the largest amplitude on his little bungee cord if you tap exactly once every second.What is the spring constant of the silkthread?
Answer:0.122 N/m
Explanation:
Given
Maximum Amplitude can be obtained when tapped after every second
i.e. Time Period [tex]T=1 s[/tex]
mass of spider [tex]m=3.1 gm[/tex]
and we know
[tex]T\cdot \omega =2\pi [/tex]
where [tex]\omega [/tex]=natural frequency of oscillation
T=time Period
[tex]\omega [/tex] is also given by [tex]\omega =\sqrt{\frac{k}{m}}[/tex]
[tex]1\cdot \sqrt{\frac{k}{m}}=2\pi [/tex]
[tex]1\cdot \sqrt{\frac{k}{3.1\times 10^{-3}}}=2\pi [/tex]
[tex]k=(2\pi )^2\times 3.1\times 10^{-3}[/tex]
[tex]k=0.122 N/m[/tex]
The answer is: [tex]k \approx 0.12 \text{ N/m}.[/tex]
To determine the spring constant of the silk thread, we need to consider the spider's motion as simple harmonic motion (SHM). The frequency of the motion is given by the tapping rate, which is once every second, so the frequency [tex]\( f \)[/tex]is 1 Hz. The mass [tex]\( m \)[/tex]of the spider is 3.1 g, which we need to convert to kilograms for consistency with standard units. Since 1 g = 0.001 kg, we have:
[tex]\[ m = 3.1 \text{ g} = 3.1 \times 10^{-3} \text{ kg} \][/tex]
The period [tex]\( T \)[/tex] of the motion is the reciprocal of the frequency:
[tex]\[ T = \frac{1}{f} = \frac{1}{1 \text{ Hz}} = 1 \text{ s} \][/tex]
For SHM, the period [tex]\( T \)[/tex] is related to the mass[tex]\( m \)[/tex] and the spring constant [tex]\( k \)[/tex] by the following equation:
[tex]\[ T = 2\pi \sqrt{\frac{m}{k}} \][/tex]
Solving for [tex]\( k \),[/tex] we get:
[tex]\[ k = \frac{4\pi^2 m}{T^2} \][/tex]
Substituting the known values for [tex]\( m \), \( T \)[/tex], and using[tex]\( \pi \approx 3.14159 \)[/tex], we have:
[tex]\[ k = \frac{4\pi^2 \times 3.1 \times 10^{-3} \text{ kg}}{(1 \text{ s})^2} \][/tex]
[tex]\[ k = 4\pi^2 \times 3.1 \times 10^{-3} \text{ N/m} \][/tex]
[tex]\[ k = 4 \times (3.14159)^2 \times 3.1 \times 10^{-3} \text{ N/m} \][/tex]
[tex]\[ k \approx 4 \times 9.8696 \times 3.1 \times 10^{-3} \text{ N/m} \][/tex]
[tex]\[ k \approx 120.47 \times 10^{-3} \text{ N/m} \][/tex]
[tex]\[ k \approx 0.12047 \text{ N/m} \][/tex]
Therefore, the spring constant[tex]\( k \)[/tex]of the silk thread is approximately[tex]\( 0.12047 \text{ N/m} \).[/tex]
A rope is tied to a tree 4.5 feet from the ground and then run through a pulley hooked to a vehicle 33 feet from the tree. If a person stands at the tree and pulls ten feet of the rope (after all slack is taken up) how far will the vehicle move?
If a person pulls ten feet of the rope tied to a vehicle through a pulley, the vehicle will move the same distance, which is ten feet assuming ideal conditions with no slack or stretch in the system.
Explanation:The question is asking about how far a vehicle moves when a rope attached to it and run through a pulley is pulled a certain distance. This scenario can be understood using the principles of distance and leverage as applied in physics. When you pull a rope through a pulley, the movement or distance covered by the object attached to the rope is equivalent to the length of the rope pulled.
In this case, if a person pulls ten feet of the rope, the vehicle will also move ten feet. It's important to note that this is applicable under ideal conditions where the rope and pulley do not stretch or bend and the there is no slack in the system.
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A car drives at steady speed around a perfectly circular track.
a. The car's acceleration is zero.
b. Both the acceleration and net force on the car point outward.
c. The net force on the car is zero.
d. If there is no friction, the acceleration is outward.
e. Both the acceleration and net force on the car point inward.
Answer:
e. Both the acceleration and net force on the car point inward.
Explanation:
If no net force acts on the car, the car must drive in a straight line, at constant speed.
As the acceleration is defined as the rate of change of the velocity vector, this means that it can produce either a change in the magnitude of the velocity (the speed) or in the direction.
In order to the car can follow a circular trajectory, it must be subjected to an acceleration, that must go inward, trying to take the car towards the center of the circle.
The net force that causes this acceleration, aims inward, and is called the centripetal force.
It is not a different type of force, it can be a friction force, a tension force, a normal force, etc., as needed.
The net force on a car driving around a circular track at steady speed is zero, resulting in zero acceleration. Hence, the correct option is c.
The net force on the car is zero. When a car drives at a steady speed around a circular track, the centripetal force required to keep it moving in a circle balances out the outward inertia. Therefore, there is no net force on the car. This leads to zero acceleration (choice c).
Glycerin (C3H8O3) is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 154 g of glycerin to 316 mL of H2O at 40.0°C? The vapor pressure of pure water at 40.0°C is 55.32 torr and its density is 0.992 g/cm3.
Answer:
[tex]P_{sol}=50.4\ mm.Hg[/tex]
Explanation:
According to given:
molecular mass of glycerin, [tex]M_g=3\times 12+8+3\times 16=92\ g.mol^{-1}[/tex]molecular mass of water, [tex]M_w=2+16=18\ g.mol^{-1}[/tex]∵Density of water is [tex]0.992\ g.cm^{-3}= 0.992\ g.mL^{-1}[/tex]∴mass of water in 316 mL, [tex]m_w=316\times 0.992=313.5 g[/tex]mass of glycerin, [tex]m_g=154\ g[/tex]pressure of mixture, [tex]P_x=55.32\ torr= 55.32\ mm.Hg[/tex]temperature of mixture, [tex]T_x=40^{\circ}C[/tex]Upon the formation of solution the vapour pressure will be reduced since we have one component of solution as non-volatile.
moles of water in the given quantity:
[tex]n_w=\frac{m_w}{M_w}[/tex]
[tex]n_w=\frac{313.5}{18}[/tex]
[tex]n_w=17.42 moles[/tex]
moles of glycerin in the given quantity:
[tex]n_g=\frac{m_g}{M_g}[/tex]
[tex]n_g=\frac{154}{92}[/tex]
[tex]n_g=1.674 moles[/tex]
Now the mole fraction of water:
[tex]X_w=\frac{n_w}{n_w+n_g}[/tex]
[tex]X_w=\frac{17.42}{17.42+1.674}[/tex]
[tex]X_w=0.912[/tex]
Since glycerin is non-volatile in nature so the vapor pressure of the resulting solution will be due to water only.
[tex]\therefore P_{sol}=X_w\times P_x[/tex]
[tex]\therefore P_{sol}=0.912\times 55.32[/tex]
[tex]P_{sol}=50.4\ mm.Hg[/tex]
A rectangular coil of dimensions 5.40cm x 8.50cm consists of25 turns of wire. The coil carries a current of 15.0 mA.
a) Calculate the magnitude of its magnetic moment
b) Suppose a uniform magnetic field of magnitude of 0.350 T isapplied parallel to the plane of the loop. What is the magnitude ofthe torque acting on the loop?
Answer:
(a) Magnetic moment will be [tex]17.212\times 10^{-4}A-m^2[/tex]
(b) Torque will be [tex]6.024\times 10^{-4}N-m[/tex]
Explanation:
We have given dimension of the rectangular 5.4 cm × 8.5 cm
So area of the rectangular coil [tex]A=5.4\times 8.5=45.9cm^2=45.9\times 10^{-4}m^2[/tex]
Current is given as [tex]i=15mA=15\times 10^{-3}A[/tex]
Number of turns N = 25
(A) We know that magnetic moment is given by [tex]magnetic\ moment=NiA=25\times 45.9\times 10^{-4}\times 15\times 10^{-3}=17.212\times 10^{-4}A-m^2[/tex]
(b) Magnetic field is given as B = 0.350 T
We know that torque is given by [tex]\tau =BINA=0.350\times 15\times 10^{-3}\times 25\times 45.9\times 10^{-4}=6.024\times 10^{-4}N-m[/tex]
Final answer:
a) The magnitude of the magnetic moment is 0.0172 A·m². b) The magnitude of the torque on the loop is 0 N·m.
Explanation:
a) To calculate the magnitude of the magnetic moment, we can use the formula μ = nIA, where μ is the magnetic moment, n is the number of turns, I is the current, and A is the area of the coil. In this case, n = 25, I = 15.0 mA = 0.015 A, and A = (5.40 cm)(8.50 cm) = 45.9 cm² = 0.00459 m². Plugging these values into the formula, we get μ = (25)(0.015 A)(0.00459 m²) = 0.0172 A·m².
b) To calculate the magnitude of the torque, we can use the formula τ = μBsinθ, where τ is the torque, μ is the magnetic moment, B is the magnetic field strength, and θ is the angle between μ and B. In this case, μ = 0.0172 A·m² (calculated in part a),
B = 0.350 T, and since the magnetic field is applied parallel to the plane of the loop, θ = 0°.
Plugging these values into the formula, we get τ = (0.0172 A·m²)(0.350 T)(sin 0°)
= 0 N·m.
006 10.0 points Two pucks of equal mass collide. One puck (puck A) is stationary before the collision. After the collision the puck that was initially stationary moves along in the same direction in which the puck that hit it (puck B) was initially moving. Puck B Puck A Which of the following statements is correct if the total energy of the system is conserved during the collision (elastic collision)? 1. There isn't enough information to answer the question. 2. Both pucks slide past the point of colli- sion. 3. The puck B remains at the point of colli- sion. 4. The puck B rebounds.
Answer: 3. The puck B remains at the point of collision.
Explanation:
Assuming no external forces acting during the collision, total momentum must be conserved.
The initial momentum is due only to puck B, as Puck A is at rest.
The final momentum is given by the sum of the momenta of both pucks, so we can write the following equation:
mA*viA = (mA * vfA) + (mB * vfB)
As mA = mB = m, we can simplify the former equation as follows:
viA = vfA + vfB (1)
Now, we also know, that the collision was an elastic collision, so total kinetic energy must be conserved too:
½ m viA²= ½ m vfA² + ½ m vfB²
Simplifying on both sides, we finally have:
viA² = vfA² + vfB² (2)
Now, if we square both sides of (1), we get:
viA² = (vfA + vfB)²= vfA² + 2* vfA * vfB +vfB² (3)
As the right side in (2) and (3) must be equal each other (as the left sides do), the only choice is that either vfA or vfB, be zero.
As we are told that puck A (initially stationary) after the collision, moves, the only possible choice is that puck B remain at rest in the point of collision, after the collision, exchanging his speed with puck A.
In an elastic collision where the total energy of the system is conserved, Puck B rebounds. This is because kinetic energy, along with momentum, is conserved in such a collision, causing the moving puck to rebound.
Explanation:Considering this scenario from the lens of physics, the principle of conservation of momentum and kinetic energy comes into play, which means the total momentum and kinetic energy before and after the collision should remain constant. This is consistent with the properties of an elastic collision. Both pucks have the same mass and initially, Puck B is in motion while Puck A is stationary. Hence, all of the initial energy and momentum are with Puck B.
Upon collision, the energy is transferred, causing Puck A to move in the same direction that Puck B was initially moving. The correct statement is that if the collision is elastic and the total energy of the system is conserved, 'Puck B rebounds'. This rebounding comes from the conservation of kinetic energy principle which holds in an elastic collision. That is, the total initial kinetic energy equals the total final kinetic energy.
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Hoover Dam on the Colorado River is the highest dam in the United States at 221m, with a power output of 680 MW. The dam generates electricity by flowing water down to a point 150 m below the stop, at an average flow rate of 650 m3/s.
Part (a) Calculate the power in this flow in watts.
Answer:
Power = 9.75 ×10^8[tex]\frac{kgm^2}{s^3}[/tex]
Explanation:
Power is rate of change of energy.Here gravitational energy is transferred to kinetic energy of water at a definite rate.For one second 650m^3 of water flows out down to 150m oh depth.
So, the energy at a height of 150m is transformed to kinetic energy.
for a second,
650m^3 of water flows down ⇒ (1000kg/m^3 × 650m^3) = 6.5×10^5kg of warer flos down.
The total gravitational potential energy stored in water is
= mass of water × height× gravity
= 6.5 ×10^5 × 150 × 10 = 9.75 ×10^8[tex]\frac{kgm^2}{s^2}[/tex]
As it is transformed in a second it is also equal to Power.
Answer:
power = 1407.77 MW
Explanation:
The basic principle of a hydro electric station is the conversion of potential energy to electrical energy. Here, water is allowed to fall from a height which will increase its the kinetic energy. This high speed flowing water is used to rotate the shaft of a turbine which will in turn produce electrical energy.
So here,
power = rate of change of potential energy with respect to time
power = [tex]\frac{d(mgh)}{dt}[/tex]
where,
m = mass of water
g = acceleration due to gravity
h = height through which the water falls
here h and g are constants ( h is the total height through which the water falls and it doesn't change with time). Therefore we can take them out of differentiation.
thus,
power = gh[tex]\frac{d(m)}{dt}[/tex]
now,
m = ρV
where,
ρ = density
V = volume
substituting this in the above equation we get,
power = gh[tex]\frac{d(ρV)}{dt}[/tex]
again ρ is a constant. Thus,
power = ρgh[tex]\frac{d(V)}{dt}[/tex]
Given that,
h = 221 m
[tex]\frac{d(V)}{dt}[/tex] = 650 [tex]m^{3}[/tex]/s
g = 9.8 m/[tex]s^{2}[/tex]
ρ = 1000 kg/[tex]m^{3}[/tex]
substituting these values in the above equation
power = 1000 x 9.8 x 221 x 650
power = 1407.77 MW
A 200-Ω resistor is connected in series with a 10-µF capacitor and a 60-Hz, 120-V (rms) line voltage. If electrical energy costs 5.0¢ per kWh, how much does it cost to leave this circuit connected for 24 hours?
Answer:
Cost to leave this circuit connected for 24 hours is $ 3.12.
Explanation:
We know that,
[tex]\mathrm{X}_{\mathrm{c}}=\frac{1}{2 \pi \mathrm{fc}}[/tex]
f = frequency (60 Hz)
c= capacitor (10 µF = [tex]10^-6[/tex])
[tex]\mathrm{X}_{\mathrm{c}}=\text { Capacitive reactance }[/tex]
Substitute the given values
[tex]\mathrm{X}_{\mathrm{c}}=\frac{1}{2 \times 3.14 \times 10 \times 10^{-6} \times 60}[/tex]
[tex]\mathrm{X}_{\mathrm{c}}=\frac{1}{3.768 \times 10^{-3}}[/tex]
[tex]\mathrm{x}_{\mathrm{c}}=265.39 \Omega[/tex]
Given that, R = 200 Ω
[tex]X^{2}=R^{2}+X c^{2}[/tex]
[tex]X^{2}=200^{2}+265.39^{2}[/tex]
[tex]X^{2}=40000+70431.85[/tex]
[tex]X^{2}=110431.825[/tex]
[tex]x=\sqrt{110431.825}[/tex]
X = 332.31 Ω
[tex]\text { Current }(I)=\frac{V}{R}[/tex]
[tex]\text { Current }(I)=\frac{120}{332.31}[/tex]
Current (I) = 0.361 amps
“Real power” is only consumed in the resistor,
[tex]\mathrm{I}^{2} \mathrm{R}=0.361^{2} \times 200[/tex]
[tex]\mathrm{I}^{2} \mathrm{R}=0.1303 \times 200[/tex]
[tex]\mathrm{I}^{2} \mathrm{R}=26.06 \mathrm{Watts} \sim 26 \mathrm{watts}[/tex]
In one hour 26 watt hours are used.
Energy used in 54 hours = 26 × 24 = 624 watt hours
E = 0.624 kilowatt hours
Cost = (5)(0.624) = 3.12
n this example we will look at the periodic motion of an ultrasonic transducer, a kind of ultrasonic loudspeaker used for medical diagnosis. The transducer is oscillating at a frequency of 6.7 MHz (6.7×106Hz). How much time does each oscillation take, and what is the angular frequency?
Answer:
Explanation:
frequency, f = 6.7 MHz = 6.7 x 10^6 Hz
Time period is defined as the time taken by the oscillating body to complete one oscillation.
the formula for the time period is
T = 1/ f = 1 / (6.7 x 10^6) = 1.5 x 10^-7 second
The angular frequency is given by
ω = 2 π f = 2 x 3.14 x 6.7 x 10^6
ω = 4.2 x 10^7 rad/s
The time for each oscillation is approximately 1.49 x 10^-7 seconds, and the angular frequency is approximately 4.21 x 10^7 rad/s.
Explanation:To find the time for each oscillation, we can use the formula T = 1/f, where T represents the period and f represents the frequency. In this case, the frequency is given as 6.7 MHz, which is equivalent to 6.7 x 10^6 Hz. So, substituting the value of the frequency into the formula, we get: T = 1 / (6.7 x 10^6) = 1.49 x 10^-7 seconds. Therefore, each oscillation takes approximately 1.49 x 10^-7 seconds.
The angular frequency can be calculated using the formula ω = 2πf, where ω represents the angular frequency and f represents the frequency. Substituting the given value of the frequency into the formula, we get: ω = 2π x 6.7 x 10^6 = 4.21 x 10^7 rad/s. Therefore, the angular frequency is approximately 4.21 x 10^7 rad/s.
An applied force of 122 N is exerted horizontally on an 28 kg box of dogs in Yulin. The coefficient of kinetic friction between the floor and the box is 0.35. If the box is initially at rest, what is the final kinetic energy after it has been moved 9.46 m across the floor?
Answer:
The final kinetic energy is 242 J.
Explanation:
Hi there!
According to the work-energy theorem, the total work done on the box is equal to its change in kinetic energy:
W = ΔKE
Where:
W = work done on the box.
ΔKE = change in kinetic energy (final KE - initial KE).
The only forces that do work in this case are the applied force and the friction force because the box moves only horizontally.
The equation of work is the following:
W = F · s
Where:
F = force.
s = traveled distance.
Then, the work done by the applied force is:
W = 122 N · 9.46 m = 1.15 × 10³ J
To calculate the work done by friction, we have to find the friction force:
Fr = N · μ
Where:
Fr = friction force.
N = normal force.
μ = coefficient of kinetic friction.
The box does not have a net vertical acceleration. It means that the sum of the vertical forces acting on the box is zero:
∑Fy = 0
In this case, the only vertical forces are the weight of the box and the normal force. Then:
Weight + N = 0
N = - Weight
The weight of the box is calculated as follows:
Weight = m · g
Where:
m = mass of the box.
g = acceleration due to gravity.
Then:
-Weight = N = 28 kg · 9.8 m/s² = 274.4 N
Now, we can calculate the friction force:
Fr = N · μ
Fr = 274.4 N · 0.35 = 96 N
The work done by the friction force will be:
W = Fr · s
W = 96 N · 9.46 m = 908 J
Since the work done by friction opposes to the sense of movement, the work is negative.
Now, we can calculate the total work done on the box:
W total = W applied forece + W friction force
W = 1.15 × 10³ J - 908 J = 242 J
Applying the work-energy theorem:
W = final KE - initial KE
Since the box is initially at rest the initial kinetic energy is zero. Then:
W = final KE - 0
W = final KE
Final KE = 242 J
The final kinetic energy is 242 J.
Suppose that a simple pendulum consists of a small 81 g bob at the end of a cord of negligible mass. If the angle θ between the cord and the vertical is given by θ = (0.068 rad) cos[(9.2 rad/s) t + φ], what are (a) the pendulum's length and (b) its maximum kinetic energy?
Answer:
(a) 0.115 m
(b) 2.08 x 10^-5 J
Explanation:
mass of bob, m = 81 g = 0.081 kg
The equation of oscillation is given by
θ = 0.068 Cos {9.2 t + Ф}
Now by comparison
The angular velocity
ω = 9.2 rad/s
(a) [tex]\omega^{2} =\frac{g}{L}[/tex]
where, L be the length of the pendulum
[tex]L =\frac{g}{\omega ^{2}}[/tex]
[tex]L =\frac{9.8}{9.2 \times 9.2}[/tex]
L = 0.115 m
(b) A = L Sinθ
A = 0.115 x Sin 0.068
A = 7.8 x 10^-3 m
Maximum kinetic energy
K = 0.5 x mω²A²
K = 0.5 x 0.081 x 9.2 x 9.2 x 7.8 x 7.8 x 10^-6
K = 2.08 x 10^-5 J
A 5 kg ball takes 13.3 seconds for one revolution around the circle. What's the magnitude of the angular velocity of this motion?
Answer: 0.47 rad/sec
Explanation:
By definition, the angular velocity is the rate of change of the angle traveled with time, so we can state the following:
ω = ∆θ/ ∆t
Now, we are told that in 13.3 sec, the ball completes one revolution around the circle, which means that, by definition of angle, it has rotated 2 π rad (an arc of 2πr over the radius r), so we can find ω as follows:
ω = 2 π / 13.3 rad/sec = 0.47 rad/sec
Answer:
0.472rad/s
Explanation:
Angular velocity = 2πf where f = frequency and frequency is the number of revolution per second and 2π represent a cycle of revolution. The mass of the body was 5kg, the time taken to complete a cycle was 13.3 s.
Frequency = 1/period where period is the time it takes to complete a revolution.
F = 1/13.3 = 0.075hz
Angular velocity = 2* 3.142* 0.075 = 0.472rad/s
You are lying in bed and want to shut your bedroom door. You have a bouncy "superball" and a blob of clay, both with the same mass. Which one would be more effective to throw at your door to close it?
a. The superball.
b. Neither will work.
c. The blob of clay.
d. Both the same.
Answer:
Choice A. The bouncy "superball" will exert approximately twice as much force than the blob of clay when tossed out at the same speed.
Explanation:
The momentum of an object is equal to the product of its mass and its velocity. The momentum of the system is conserved even if the collision is inelastic.
When the bouncy ball or the blob of clay hits the door, their momentum changes. Since momentum is conserved, the door would move in the opposite direction such that the total momentum stays at zero.
Let the mass of each object be [tex]m[/tex]. Suppose that both objects hit the door at a speed of [tex]v[/tex] in the same direction.
The bouncy ball will likely bounce back at about the same speed (in the opposite direction, away from the door.) The magnitude of the change in velocity will be at most [tex]2v[/tex]. The magnitude of the change in momentum will be at most [tex]2 \,m\cdot v[/tex].However, when the blob of clay hits the door, its shape would change (and it would potentially stick to the door.) In other words, its velocity becomes almost [tex]0[/tex] after the collision. The magnitude of the change in its momentum will only be around [tex]m \cdot v[/tex], which is only about [tex]\dfrac{1}{2}[/tex] the momentum change due to the bouncy ball.By Newton's Second Law, the net force on the door is proportion to the rate of change in its momentum. Assume that the two objects are in contact with the door for the same amount of time, the bouncy ball would exert about twice as much force on the door as the clay would. Hence it is "more effective" for closing the door.
The clay is more effective than the superball for shutting the door because upon impact the clay sticks to the door and transfers all of its momentum to it, while the superball bounces back and doesn't transfer all of its momentum.
Explanation:The most effective object to throw at your door to shut it would be c. The blob of clay. This is because of the principle of conservation of momentum. When the clay hits the door, it sticks to it, thus transferring all of its momentum to the door. This momentum is what causes the door to move and potentially close. On the other hand, when the superball hits the door, it bounces back, meaning not all of its momentum is transferred to the door. Hence, the clay, by virtue of its ability to stick and transfer its full momentum, would be more effective.
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The auditory canal behaves like a resonant tube to aid in hearing. One end terminated at the eardrum, while the other opens to the outside. Typically, the canal is about 2.4 cm long. At which frequency would it resonate in its first harmonic?
4.2 kHz
3.6 kHz
2.9 kHz
5.7 kHz
Answer:
3.6 kHz
Explanation:
The pipes behave like a closed pipe . The end open is the end of the air canal outside the ear and the closed end is the eardrum.
The first harmonic will be as seen in the figure attached.
The length of the first harmonic will be λ/4.
λ/4=2.4 cm
λ=2.4 * 4=9.6 cm 0.096 m
Speed of Sound- 344 m/s(in air)
velocity(v) * Time Period(T) = Wavelength (λ)
Also, Time Period(T)= \frac{\textrm{1}}{\textrm{Frequency(f)}}
\frac{\textrm{Velocity}}{\textrm{Wavelength}}=\frac{\textrm{1}}{\textrm{Time Period}} =Frequency
Plugging in the values into the equation,
Frequency = [tex]\frac{344}{0.096}[/tex] Hz
= 3583.3 Hz≈3600 Hz= 3.6 kHz
Frequency= 3.6 kHz
Answer:
3.6 kHz
Explanation:
The auditory canal is a closed pipe because it has one closed end, the end terminated at the eardrum.
The length of the first harmonic of a closed pipe is given as;
L = λ/4 -------------------- (i)
where L = Length
and λ = wavelength
2.4 = λ/4
λ = 2.4 x 4 = 9.6 cm
Also, v = fλ ------------------ (ii)
where v = speed of sound in air = 344 m/s
f = frequency of wave in Hertz
f = v/λ ------------------ (iii)
convert 9.6 cm to m = 0.096 m
substitute for λ and v in (iii)
[tex]f = \frac{344}{0.096} = 3583.33[/tex]
3583.33 Hz = 3600 Hz = 3.6 kHz
Four students use different instruments to measure the length of the same pen. Which measurement implies the greatest precision?
a. 16.0 cm.
b. 0.160 m.
c. 0.00016 km.
d. Need more information.
e. 160.0 mm
Answer:
e. 160.0 mm
Explanation:
Precision is defined as the accuracy in measurement that the measuring instrument measure any given specimen. The instrument having the lowest least count will be more precise than the instrument with higher least count.
Also precision of the reading is given by the number of the significant figures in the given sample of reading.
Option (e) that is 160.0 has a total number of significant figure of 4, which is the largest total significant figures among the other options.
Also the measurement reads 160.0 mm which measure to the lowest possible measurement of the given pen. Thus this reading is more precise.
Hence the correct option is --- Option (e) 160.0 mm
The colors of the stars in the sky range from red to blue. Assuming that the color indicates the frequency at which the star radiates the maximum amount of electromagnetic energy, estimate the surface temperature of red, yellow, white, and blue stars.
A) Estimate the surface temperature of red star. Assume that red color corresponds to wavelengths of approximately 650 nm.B) Estimate the surface temperature of yellow star. Assume that yellow color corresponds to wavelengths of approximately 570 nm.C) Estimate the surface temperature of blue star. Assume that blue color corresponds to wavelengths of approximately 470 nmD) Estimate the surface temperature of white star. Since green is in the middle of the visible spectrum, choose 520 nm wavelength for white star.
Answer:
a) 4458K b) 5048K, c) 6166K, d) 5573K
Explanation:
The temperature of the stars and many very hot objects can be estimated using the Wien displacement law
[tex]\lambda_{max}[/tex] T = 2,898 10⁻³ [m K]
T = 2,898 10⁻³ / [tex]\lambda_{max}[/tex]
a) indicate that the wavelength is
Lam = 650 nm (1 m / 109 nm) = 650 10⁻⁹ m
Lam = 6.50 10⁻⁷ m
T = 2,898 10⁻³ / 6.50 10⁻⁷
T = 4,458 10³ K
T = 4458K
b) lam = 570 nm = 5.70 10⁻⁷ m
T = 2,898 10⁻³ / 5.70 10⁻⁷
T = 5084K
c) lam = 470 nm = 4.70 10⁻⁷ m
T = 2,898 10⁻³ / 4.7 10⁻⁷
T = 6166K
d) lam = 520 nm = 5.20 10⁻⁷ m
T = 2,898 10⁻³ / 5.20 10⁻⁷
T = 5573K
A ) the surface temperature of red star is about 4500 K
B ) the surface temperature of yellow star is about 5100 K
C ) the surface temperature of blue star is about 6200 K
D ) the surface temperature of white star is about 5600 K
[tex]\texttt{ }[/tex]
Further explanationLet's recall the Wien's Displacement Law as follows:
[tex]\boxed {\lambda_{max}\ T = 2.898 \times 10^{-3} \texttt{ m.K}}[/tex]
where:
λ_max = the wavelength of the maximum radiation energy ( m )
T = surface temperature of the star ( K )
Let us now tackle the problem!
[tex]\texttt{ }[/tex]
Given:
wavelength of red light = λ_r = 650 nm = 650 × 10⁻⁹ m
wavelength of yellow light = λ_y = 570 nm = 570 × 10⁻⁹ m
wavelength of blue light = λ_b = 470 nm = 470 × 10⁻⁹ m
wavelength of white light = λ_w = 520 nm = 520 × 10⁻⁹ m
Asked:
A ) the surface temperature of red star = T_r = ?
B ) the surface temperature of yellow star = T_y = ?
C ) the surface temperature of blue star = T_b = ?
D ) the surface temperature of white star = T_w = ?
Solution:
Part A) :[tex]T_r = ( 2.898 \times 10^{-3} ) \div \lambda_r[/tex]
[tex]T_r = ( 2.898 \times 10^{-3} ) \div ( 650 \times 10^{-9} )[/tex]
[tex]\boxed {T_r \approx 4500 \texttt{ K} }[/tex]
[tex]\texttt{ }[/tex]
Part B) :[tex]T_y = ( 2.898 \times 10^{-3} ) \div \lambda_y[/tex]
[tex]T_y = ( 2.898 \times 10^{-3} ) \div ( 570 \times 10^{-9} )[/tex]
[tex]\boxed {T_y \approx 5100 \texttt{ K} }[/tex]
[tex]\texttt{ }[/tex]
Part C) :[tex]T_b = ( 2.898 \times 10^{-3} ) \div \lambda_b[/tex]
[tex]T_b = ( 2.898 \times 10^{-3} ) \div ( 470 \times 10^{-9} )[/tex]
[tex]\boxed {T_b \approx 6200 \texttt{ K} }[/tex]
[tex]\texttt{ }[/tex]
Part D) :[tex]T_w = ( 2.898 \times 10^{-3} ) \div \lambda_w[/tex]
[tex]T_w = ( 2.898 \times 10^{-3} ) \div ( 520 \times 10^{-9} )[/tex]
[tex]\boxed {T_w \approx 5600 \texttt{ K} }[/tex]
[tex]\texttt{ }[/tex]
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Answer detailsGrade: High School
Subject: Mathematics
Chapter: Energy
Moe is carrying and supporting a large crate by exerting 1000 N. He is walking at constant velocity across a level, horizontal floor. If he covers a distance of 10 meters, how much work does he do on the crate? Use g = 10m/s2 if needed. Ignore friction between Moe and the crate.
Answer:
he did not work on the create
Explanation:
Honestly I have no idea it makes a lot of sense to use the work equation so that was my first answer but for some reason on UT Quest it said that was the wrong and moe not doing any work on the create was the right answer
Final answer:
The work done by Moe on the crate is 10,000 J.
Explanation:
The work done by Moe on the crate can be calculated using the formula:
Work = Force x Distance x Cosine(theta)
Since Moe is exerting a force of 1000 N and the distance covered is 10 meters, we can plug these values into the formula:
Work = 1000 N x 10 m x Cosine(0°)
The angle theta is 0° because Moe is exerting a force in the same direction as the displacement.
Cosine(0°) is equal to 1, so the work done by Moe on the crate is:
Work = 1000 N x 10 m x 1 = 10,000 J
A clear sheet of polaroid is placed on top of a similar sheet so that their polarizing axes make an angle of 30° with each other. The ratio of the intensity of emerging light to incident unpolarized light is:
Answer:
[tex]\dfrac{I}{I_0}=\dfrac{3}{8}[/tex]
Explanation:
Given that
Angle ,θ = 30°
From Malus law,Intensity given as
[tex]I=\dfrac{I_0}{2}cos^2\theta[/tex]
Io=Intensity of unpolarized light
I=Intensity of emerging light
Now by putting the value of angle
[tex]I=\dfrac{I_0}{2}cos^2\theta[/tex]
[tex]I=\dfrac{I_0}{2}cos^230^{\circ}[/tex]
We know that
[tex]cos30^{\circ}=\dfrac{\sqrt{3}}{2}[/tex]
[tex]I=\dfrac{I_0}{2}\times \dfrac{3}{4}[/tex]
[tex]\dfrac{I}{I_0}=\dfrac{3}{8}[/tex]
Therefore ratio will be [tex]\dfrac{3}{8}[/tex]
Answer:
Ratio of the intensity of emerging light[tex]$\frac{I}{I_{0}}=\frac{3}{8}$[/tex]
Explanation:
Given:
Angle,[tex]$\theta=30^{\circ}$[/tex]
Step 1:
According to Malus law,
Intensity,
[tex]$I=\frac{I_{0}}{2} \cos ^{2} \theta$[/tex]
[tex]l_{0} =[/tex]Intensity of unpolarized light
[tex]l=[/tex]Intensity of emerging light
Step 2:
Put the value of angle
[tex]$I=\frac{I_{0}}{2} \cos ^{2} \theta$[/tex]
[tex]$I=\frac{I_{0}}{2} \cos ^{2} 30^{\circ}$[/tex]
We know,
[tex]$\cos 30^{\circ}=\frac{\sqrt{3}}{2}$[/tex]
[tex]$I=\frac{I_{0}}{2} \times \frac{3}{4}$[/tex]
So, the intensity of the ratio
[tex]$\frac{I}{I_{0}}=\frac{3}{8}$[/tex]
Therefore, the ratio of the intensity of light is [tex]$\frac{3}{8}$[/tex]
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A mass of 0.56 kg is attached to a spring and set into oscillation on a horizontal frictionless surface. The simple harmonic motion of the mass is described by x(t) = (0.42 m)cos[(6 rad/s)t]. Determine the following. (a) amplitude of oscillation for the oscillating mass m (b) force constant for the spring N/m (c) position of the mass after it has been oscillating for one half a period m (d) position of the mass one-third of a period after it has been released m (e) time it takes the mass to get to the position x = −0.10 m after it has been released
Answer:
(a) 0.42 m
(b) 20.16 N/m
(c) - 0.42 m
(d) - 0.21 m
(e) 17.3 s
Solution:
As per the question:
Mass, m = 0.56 kg
x(t) = (0.42 m)cos[cos(6 rad/s)t]
Now,
The general eqn is:
[tex]x(t) = Acos\omega t[/tex]
where
A = Amplitude
[tex]\omega[/tex] = angular frequency
t = time
Now, on comparing the given eqn with the general eqn:
(a) The amplitude of oscillation:
A = 0.42 m
(b) Spring constant k is given by:
[tex]\omega = \sqrt{k}{m}[/tex]
[tex]\omega^{2} = \frac{k}{m}[/tex]
Thus
[tex]k = m\omega^{2} = 0.56\times 6^{2} = 20.16\ N/m[/tex]
(c) Position after one half period:
[tex]x(t) = 0.42cos\pi = - 0.42\ m[/tex]
(d) After one third of the period:
[tex]x(t) = 0.42cos(\frac{2\pi}{3}) = - 0.21\ m[/tex]
(e) Time taken to get at x = - 0.10 m:
[tex]-0.10 = 0.42cos6t[/tex]
[tex]6t = co^{- 1} \frac{- 0.10}{0.42}[/tex]
t = 17.3 s
The amplitude of oscillation is 0.42 m. The force constant of the spring is approximately 0.03 N/m. The position of the mass after one half a period, one-third of a period, and at x = -0.10 m can be determined using the given equation x(t).
Explanation:The amplitude of oscillation is the maximum displacement from the equilibrium position. In this case, the amplitude is given as 0.42 m.
The force constant (k) of the spring can be determined using Hooke's Law, which states that the force exerted by a spring is proportional to the displacement from equilibrium. In this case, the force constant is calculated by dividing the mass (0.56 kg) by the square of the angular frequency (6 rad/s) squared, which gives a force constant of approximately 0.03 N/m.
The position of the mass after it has been oscillating for one half a period can be found by substituting the value of time (T/2) into the equation x(t), which gives a position of approximately -0.42 m.
The position of the mass one-third of a period after it has been released can be determined by substituting the value of time (T/3) into the equation x(t), which gives a position of approximately 0.33 m.
The time it takes the mass to get to the position x = -0.10 m after it has been released can be found by rearranging the equation x(t) and solving for time. The time is approximately 0.14 seconds.
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