Answer:
Explanation:
Mirror formation in mirror A
focal length f = - 10 cm ( concave mirror )
Let u be object distance , v is image distance .
In case image is virtual
v / u = 2
v = 2u
Mirror formula
[tex]\frac{1}{v} +\frac{1}{u} =\frac{1}{f}[/tex]
Putting the values
[tex]-\frac{1}{2u} +\frac{1}{u} =-\frac{1}{10}[/tex]
[tex]\frac{1}{2u}[/tex] = [tex]-\frac{1}{10}[/tex]
u = -5 cm
In case image is real
[tex]\frac{1}{v} +\frac{1}{u} =\frac{1}{f}[/tex]
[tex]\frac{1}{2u} +\frac{1}{u} =-\frac{1}{10}[/tex]
[tex]\frac{3}{2u} =-\frac{1}{10}[/tex]
u = [tex]-\frac{30}{2}[/tex]
u = - 15 cm
Wonder Woman and Superman fly to an altitude of 1610 km, carrying between them a chest full of jewels that they intend to put into orbit around Earth. They want to make this tempting treasure inaccessible to their evil enemies who are trying to gain possession of it, yet keep it available for themselves for future use when they retire and settle down. But perhaps the time to retire is now! They accidentally drop the chest, which leaves their weary hands at rest, and discover that they are no longer capable of catching it as it falls into the Pacific Ocean. At what speed vf does the chest impact the surface of the water? Ignore air resistance (in reality, it would make large difference). The radius and mass of Earth are 6370 km and 5.98×1024 kg, respectively
Answer:
5026.55 m/s
Explanation:
Gravitation potential of a body in orbit from the center of the earth is given as
Pg = -GM/R
Where G is the gravitational constant 6.67x10^-11 N-m^2kg^-2
M is the mass of the earth = 5.98x10^24 kg
R is the distance from that point to the center of the earth = r + Re
r is the distance above earth surface, Re is the earth's radius.
R = 1610 km + 6370 km = 7980 km
Pg = -(6.67x10^-11 x 5.98x10^24)/7980x10^3
Pg = -49983208.02 J/kg
The negative sign means that the gravitational potential is higher away from earth than it is at the earth's surface (it shows convention).
This indicates the kinetic energy per kilogram that the chest of jewel will fall with to earth.
Gravitation Potential on earth's surface is
Pg = -GM/Re
= -(6.67x10^-11 x 5.98x10^24)/6370x10^3 = -62616326.53 J/kg
Difference in gravity potential = -49983208.02 - (-62616326.53)
= 12633118.51 J/kg
The velocity V of the jewel chest will be
0.5v^2 = 12633118.51
V^2 = 25266237.02
V = 5026.55 m/s
A photoelectric experiment is performed where green light with a wavelength of 546.1 nm is shined on a metal plate, creating a photocurrent from it to a collector plate. When the potential difference between the metal plate and the collector is increased to a magnitude of 0.728 V, the photocurrent goes to zero—in other words, this is the stopping potential. What is the work function (in eV) for this metal?
Answer:
[tex]\phi=1.55 [eV][/tex]
Explanation:
We can use the work function equation for a photoelectric experiment:
[tex]\phi=\frac{hc}{\lambda}-K_{max}[/tex]
h is the plank constantc is the speed of lightλ is the wave lengthK is the kinetic energy (or K=eΔV)So we will have:
[tex]\phi=\frac{hc}{\lambda}-e\Delta V[/tex]
[tex]\phi=\frac{6.63*10^{-34}*3*10^{8}}{546.1*10^{-9}}-0.728eV[/tex]
[tex]\phi=3.64*10^{-19}[J]-0.728 [eV][/tex]
[tex]\phi=(3.64*10^{-19}[J]*\frac{1eV}{1.6*10^{-19}[J]})-0.728 [eV][/tex]
[tex]\phi=2.28 [eV] - 0.728 [eV][/tex]
[tex]\phi=1.55 [eV][/tex]
I hope it helps you!
Answer:
The work function is [tex]\phi = 1.544eV[/tex]
Explanation:
From the question we are told that
The wavelength of the green light is [tex]\lambda = 546.1nm[/tex]
The stopping potential is [tex]V= 0.728V[/tex]
At stopping potential the kinetic is also maximum because this where the electron(causing the current ) would flow at it highest speed before return to zero
So the maximum kinetic energy of this electron in term of electron volt is
[tex]KE_{max} = 0.728 eV[/tex]
This maximum kinetic energy is mathematically represented as
[tex]KE_{max} = \frac{hc}{\lambda} - \phi[/tex]
Where h is the planck's constant with a value [tex]h = 4.1357 *10^{-15} eV[/tex]
c is the speed of light [tex]c = 3.0 *10^8 m/s[/tex]
[tex]\phi[/tex] is the work function
Making [tex]\phi[/tex] the subject of the formula
[tex]\phi = \frac{hc}{\lambda} - KE_{max}[/tex]
Substituting values
[tex]\phi = \frac{4.1357 *10^{-15} * (3.0 *10^8)}{546.1 *10^{-9}} - 0.728[/tex]
[tex]\phi = 1.544eV[/tex]
A musician on a unicycle is riding at a steady 6.2 m/s while playing their well tuned oboe at an A above middle C (440 Hz). They then ride toward another musician sitting on a park bench also playing an A above middle C. However it appears to them that one of them is off as they hear beats. If we assume both are tuned properly and the outside temperature is 27 oC what will the beat frequency they hear be
Answer:
The frequency of the beat is [tex]f_t = 8Hz[/tex]
Explanation:
From the question we are told that
The speed of the ride [tex]v = 6,2 m/s[/tex]
The frequency of the oboe is [tex]f = 440Hz[/tex]
The temperature outside is [tex]T = 27^oC[/tex]
Generally the speed of sound generated in air is mathematically evaluated as
[tex]v_s = 331 + 0.61 T[/tex]
Substituting value
[tex]v_s = 331 + 0.61 *27[/tex]
[tex]v_s = 347.47 \ m/s[/tex]
The frequency of sound(generated by the musician on he park) getting to the musicians on the unicycle is mathematically evaluated as
[tex]f_a = \frac{v_s }{v_s - v} f[/tex]
substituting values
[tex]f_a = \frac{347.47 }{347.47 - 6.2} * 440[/tex]
[tex]f_a = 448Hz[/tex]
Since the musician on the park is not moving the frequency of sound (from the musicians riding the unicycle )getting to him is = 440Hz
The beat frequency these musician here is mathematically evaluated as
[tex]f_t = f_a - f[/tex]
So [tex]f_t = 448 - 440[/tex]
[tex]f_t = 8Hz[/tex]
A uniform disk has a mass of 3.7 kg and a radius of 0.40 m. The disk is mounted on frictionless bearings and is used as a turntable. The turntable is initially rotating at 30 rpm. A thin-walled hollow cylinder has the same mass and radius as the disk. It is released from rest, just above the turntable, and on the same vertical axis. The hollow cylinder slips on the turntable for 0.20 s until it acquires the same final angular velocity as the turntable. What is the final angular momentum of the system
Answer:
1.25 kgm²/sec
Explanation:
Disk inertia, Jd =
Jd = 1/2 * 3.7 * 0.40² = 0.2960 kgm²
Disk angular speed =
ωd = 0.1047 * 30 = 3.1416 rad/sec
Hollow cylinder inertia =
Jc = 3.7 * 0.40² = 0.592 kgm²
Initial Kinetic Energy of the disk
Ekd = 1/2 * Jd * ωd²
Ekd = 0.148 * 9.87
Ekd = 1.4607 joule
Ekd = (Jc + 1/2*Jd) * ω²
Final angular speed =
ω² = Ekd/(Jc+1/2*Jd)
ω² = 1.4607/(0.592+0.148)
ω² = 1.4607/0.74
ω² = 1.974
ω = √1.974
ω = 1.405 rad/sec
Final angular momentum =
L = (Jd+Jc) * ω
L = 0.888 * 1.405
L = 1.25 kgm²/sec
To find the final angular momentum of the system, we can apply the principle of conservation of angular momentum. The initial angular momentum of the turntable is equal to its final angular momentum, which can be determined using the moment of inertia and final angular velocity of the system.
Explanation:The final angular momentum of the system can be determined by applying the principle of conservation of angular momentum. Since no external torques act on the system, the angular momentum of the system remains constant. Initially, the turntable has an angular momentum given by L1 = I1ω1, where I1 is the moment of inertia of the turntable and ω1 is its initial angular velocity. After the hollow cylinder slips on the turntable and acquires the same final angular velocity, the system's angular momentum becomes L2 = I1ω2. Setting L1 = L2 and solving for ω2 gives us the final angular velocity of the system. Finally, we can use the moment of inertia of the turntable to calculate its final angular momentum, Lf = I1ω2.
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A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 3.59 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2190 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) in the positive z direction and has a magnitude of 3.91 V/m, (b) in the negative z direction and has a magnitude of 3.91 V/m, and (c) in the positive x direction and has a magnitude of 3.91 V/m?
Answer:
a) 1.88*10^-18 N
b) 6.32*10^-19 N
c) 1.9*10^-18 N
Explanation:
The total force over the electron is given by:
[tex]\vec{F}=q\vec{E}+q\vec{v}\ X\ \vec{B}[/tex]
the first term is the electric force and the second one is the magnetic force.
You have that the velocity of the electron and the magnetic field are:
[tex]\vec{v}=2190\frac{m}{s}\ \hat{j}\\\\\vec{B}=-5.39*10^{-3}T\ \hat{i}[/tex]
by using the relation j X (-i) =- j X i = -(-k) = k, you obtain:
[tex]\vec{v} \ X\ \vec{B}=(2190m/s)(3.59*10^{-3}T)\hat{k}=7.862\ T m/s[/tex]
a) For an electric field of 3.91V/m in +z direction:
[tex]\vec{F}=q[\vec{E}+\vec{v}\ X\ \vec{B}]=(1.6*10^{-19})[3.91\hat{k}+7.862\ \hat{k}]N\\\\\vec{F}=1.88*10^{-18}N\hat {k}\\\\F=1.88*10^{-18}N[/tex]
b) E=3.91V/m in -z direction:
[tex]\vec{F}=(1.6*10^{-19})[-3.91\hat{k}+7.862\ \hat{k}]N\\\\\vec{F}=6.32*10^{-19}N\hat {k}\\\\F=6.32*10^{-19}N[/tex]
c) E=3.91 V/m in +x direction:
[tex]\vec{F}=(1.6*10^{-19})[3.91\hat{i}+7.862\ \hat{k}]N\\\\\vec{F}=[6.25*10^{-19}\ \hat{i}+1.25*10^{-18}\ \hat{k} ]N\\\\F=\sqrt{(6.25*10^{-19})^2+(1.25*10^{-18})^2}N=1.9*10^{-18}\ N[/tex]
A uniform meter stick is pivoted at the 50.00 cm mark on the meter stick. A 400.0 gram object is hung at the 20.0 cm mark on the stick and a 320.0 gram object is hung at the 75.0 cm mark. Drawing is approximate. The meter stick is unbalanced. Determine the cm-mark on the meterstick that a 400 gram object needs to be hung to achieve equilibrium. A) 10.0 B) 40.0 C) 60.0 D) 90.0 E) none of the above is within 10% of my answer
Answer:C
Explanation:
Given
mass [tex]m_1=400\ gm[/tex] is at [tex]x=20\ cm[/tex] mark
mass [tex]m_2=320\ gm[/tex] is at [tex]x=75\ cm[/tex] mark
Scale is Pivoted at [tex]x=50\ cm mark[/tex]
For scale to be in equilibrium net torque must be equal to zero
Taking ACW as positive thus
[tex]T_{net}=0.4\times g\times (0.5-0.2)-0.32\times g\times(0.75-0.50)[/tex]
[tex]T_{net}=0.12g-0.08g=0.04g[/tex]
Therefore a net torque of 0.04 g is required in CW sense which a mass [tex]400\ gm[/tex] can provide at a distance of [tex]x_o[/tex] from pivot
[tex]0.04g=0.4\times g\times x_o [/tex]
[tex]x_o=0.1\ m[/tex]
therefore in meter stick it is at a distance of [tex]x=60\ cm[/tex]
A single slit forms a diffraction pattern, with the first minimum at an angle of 40 degrees from central maximum. Monochromatic light of 410 nm wavelength is used. The same slit, illuminated by a different monochromatic light source, produces a diffraction pattern with the second minimum at a 60 degree angle from the central maximum. The wavelength of this light, in nm, is closest to:
Answer:
276.19nm
Explanation:
To find the other wavelength you use the following condition for the diffraction of both wavelengths:
[tex]m_1\lambda_1=asin\theta_1\\\\m_2\lambda_2=asin\theta_2\\\\[/tex] ( 1 )
λ1=410nm
m=1 for wavelength 1
m=2 for wavelength 2
a: width of the slit
θ1: angle of the first minimum
θ2: angle of the second minimum
you divide both equations and you obtain:
[tex]\frac{m_1\lambda_1}{m_2\lambda_2}=\frac{sin\theta_1}{sin\theta_2}\\\\\lambda_2=\frac{sin\theta_2}{sin\theta_1}\frac{m_1\lambda_1}{m_2}\\\\\lambda_2=\frac{sin60\°}{sin40\°} \frac{(1)(410nm)}{2}=276.19nm[/tex]
hence, the wavelength of the second monochromatic wave is 276.19nm
Answer:
The wavelength of the second monochromatic light is [tex]\lambda = 277 nm[/tex]
Explanation:
From the question we are told that
The angle of the first minimum is [tex]\theta_1 = 40^o[/tex]
The wavelength of the first monochromatic light is [tex]\lambda_1 = 410 \ nm[/tex]
The angle of the second minima is [tex]\theta_2 = 60^o[/tex]
For the first minima the distance of separation of diffraction patterns is mathematically represented as
[tex]a = \frac{\lambda_1 }{sin \theta_1}[/tex]
Substituting values
[tex]a = \frac{410 *10^{-9}}{sin (40) }[/tex]
[tex]a = 638 nm[/tex]
The distance between two successive diffraction is constant for the same slit
Thus the wavelength of the second light is
[tex]\lambda = \frac{a * sin (60)}{2}[/tex]
Substituting value
[tex]\lambda = \frac{638 * sin (60)}{2}[/tex]
[tex]\lambda = 277 nm[/tex]
You are working in an optical research laboratory. Your supervisor needs you to set up a doubleslit apparatus for a presentation that will provide a vertically varying pattern of red light of wavelength 693 nm on a screen. The screen of width 4.75 m at the front of the presentation room must have red fringes on either end and 21 additional red fringes between those on the end. The separation between the slits in the double slit you will use at the back of the room is 65.0 µm.
You need to determine how far away from the slits (in m) the screen must be placed. (Assume the central maximum is at the center of the screen.)
To determine the distance between the slits and the screen, we can use the formula for the separation between fringes in a double-slit interference pattern. Applying the small angle approximation, we find that the screen must be placed approximately 1.065 km away from the slits.
Explanation:To determine the distance between the slits and the screen, we can use the formula for the separation between fringes in a double-slit interference pattern:
d*sin(theta) = m*lambda
Where d is the separation between the slits (65.0 µm), lambda is the wavelength of light (693 nm), m is the order of the fringe (in this case, m = 21), and theta is the angle of the fringe from the central maximum. Since we want the central maximum to be at the center of the screen, we can consider the angle to be very small, and therefore sin(theta) ≈ theta. Rearranging the formula, we have:
theta = m*lambda / d = 21 * (693e-9 m) / (65.0e-6 m) = 2.241e-3 radians
Next, we can use the small angle approximation to find the distance from the slits to the screen. This approximation states that tan(theta) ≈ theta for small angles. Therefore, tan(theta) = y / L, where y is the distance between the center of the screen and the fringe, and L is the distance between the slits and the screen. Solving for L, we have:
L = y / tan(theta) = (4.75 m / 2) / tan(2.241e-3 radians) ≈ 1.065 km
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A 15m long wire is placed horizontally on the surface of a liquid and is gently pulled up with the force of 60N to keep the wire in equilibrium.what is the surface tension of the liquid?
Answer:
2 N/m
Explanation:
Given that
The length of a wire, l = 15 m
Force on the wire, F = 60 N
Surface tension on the liquid, S = ?
The formula to solve this problem is given as
F = S * 2 L, where
F = force on the liquid
S = surface tension of the liquid
L= length of the wire
On substituting the values of each, we have
60 = S * (2 * 15)
S = 60 / 30
S = 2 N/m
Thus, we can say then, that the surface tension of the liquid is 2 N/m
While a guitar string is vibrating, you gently touch the midpoint of the string to ensure that the string does not vibrate at that point. The lowest-frequency standing wave that could be present on the string vibrates at Group of answer choices a. twice the fundamental frequency. b. the fundamental frequency. c. three times the fundamental frequency. d. four times the fundamental frequency. e. There is not enough information to decide.
Answer:
Option A is correct - While a guitar string is vibrating, you gently touch the midpoint of the string to ensure that the string does not vibrate at that point. The lowest-frequency standing wave that could be present on the string vibrates at twice the fundamental frequency.
Explanation:
Before touching the midpoint of the string, the string vibrates with one loop.
Fundamental frequency, f1 = v/(2*L)
Now, when the midpoint of the guitar string was touched, the string vibrates with two loops.
Hence, f2 = 2*v/(2*L)
f2 = 2*f1
Therefore, compared to the fundamental frequency the frequency would be double.
Option A is correct - While a guitar string is vibrating, you gently touch the midpoint of the string to ensure that the string does not vibrate at that point. The lowest-frequency standing wave that could be present on the string vibrates at twice the fundamental frequency.
The ball rolling along a straight and level path. The ball is rolling at a constant
speed of 2.0 m/s. The mass of the ball is 3.0 kg. What is the kinetic energy of the
ball? (KE=0.5*m*v)
Answer:6 joules
Explanation:
Mass(m)=3kg
Velocity(v)=2m/s
Kinetic energy=0.5 x m x v^2
Kinetic energy=0.5 x 3 x 2^2
Kinetic energy=0.5 x 3 x 2 x 2
Kinetic energy=6
An airplane is traveling at 150 m/s in level flight. In order to make a change in direction, the airplane travels in a horizontal curved path. To fly in the curved path, the pilot banks the airplane at an angle such that the lift has a horizontal component that provides the horizontal radial acceleration to move in a horizontal circular path. If the airplane is banked at an angle of 12.0 degrees, then the radius of curvature of the curved path of the airplane is
Given Information:
Velocity = v = 150 m/s
angle = θ = 12°
Required Information:
Radius of curvature = R = ?
Answer:
Radius of curvature = [tex]R = 1.079 \times 10^{4} \: m[/tex]
Explanation:
Please refer to the attached diagram,
[tex]Fcos(\theta) = m \cdot g\\Fcos(12^{\circ}) = m \cdot g \:\:\:\:\:\:eq. 1[/tex]
Where m is the mass of the plane and g is the gravitational acceleration.
[tex]Fsin(\theta) = \frac{mv^{2}}{R}\\Fsin(12^{\circ}) = \frac{m \cdot v^{2}}{R}\:\:\:\:\:\:eq. 2[/tex]
Where v is the velocity of the plane and R is the radius of curvature of the curved path of the airplane.
Dividing eq. 2 by eq. 1 yields,
[tex]tan(12^{\circ}) = \frac{v^{2}}{R\cdot g }[/tex]
[tex]since \: \frac{sin(\theta)}{cos(\theta)} = tan(\theta)[/tex]
[tex]tan(12^{\circ}) = \frac{v^{2}}{R\cdot g }\\\\R = \frac{v^{2}}{g\cdot tan(12^{\circ}) }\\\\R = \frac{150^{2}}{9.81\cdot 0.212 }\\\\R = 10793\\R = 1.079 \times 10^{4} \: m[/tex]
Therefore, the radius of curvature of the curved path of the airplane is 1.079×10⁴ m
Final answer:
The radius of curvature of an airplane's horizontal circular path can be determined by the centripetal force, provided by the horizontal component of lift when the airplane is banked at a specific angle.
Explanation:
The question involves an airplane traveling in a horizontal curved path and banking at an angle to achieve this motion. To find the radius of curvature for the airplane's path, we need to consider the centripetal force necessary for circular motion, which is supplied by the horizontal component of the lift force when the airplane is banked. The lift force itself acts perpendicular to the wings. By using trigonometry and the principles of circular motion, we can relate the banking angle, the airplane's velocity, and the force required for level flight to find the radius of curvature. Notably, this type of problem is a common application of Newton's laws of motion in a circular motion context.
Where does sheering often occur
A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.30 Ω is in a 1.0 mT magnetic field, with the coil oriented for maximum flux. The coil is connected to an uncharged 3.0 μF capacitor rather than to a current meter. The coil is quickly pulled out of the magnetic field.
Afterward, what is the voltage across the capacitor?
Answer:
The voltage across the capacitor = 0.8723 V
Explanation:
From the question, it is said that the coil is quickly pulled out of the magnetic field. Therefore , the final magnetic flux linked to the coil is zero.
The change in magnetic flux linked to this coil is:
[tex]\delta \phi = \phi _f - \phi _i[/tex]
= 0 - BA cos 0°
[tex]\delta \phi =-BA \\ \\ \delta \phi =-B( \pi r^2) \\ \\ \delta \phi =(1.0 \ mT)[ \pi ( \frac{d}{2} )^2][/tex]
[tex]\delta \phi =(1.0 \ mT)[ \pi ( \frac{0.01}{2} )^2][/tex]
[tex]\delta \phi = -7.85*10^8 \ Wb[/tex]
Using Faraday's Law; the induced emf on N turns of coil is;
[tex]\epsilon = N |\frac{ \delta \phi}{\delta t} |[/tex]
Also; the induced current I = [tex]\frac{ \epsilon}{R}[/tex]
[tex]\frac{ \delta q}{ \delta t} = \frac{1}{R} (N|\frac{\delta \phi}{\delta t } |)[/tex]
[tex]\delta q = \frac{1}{R} N|\delta \phi |[/tex]
[tex]\delta q = \frac{1}{0.30} (10)| -7.85*10^8 \ Wb |[/tex]
[tex]\delta q = 2.617*10^{-6} \ C[/tex]
The voltage across the capacitor can now be determined as:
[tex]\delta \ V =\frac{ \delta q}{C}[/tex]
= [tex]\frac{ 2.617*10^{-6} \ C}{3.0 \ *10^{-6} F}[/tex]
= 0.8723 V
A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.80 rad????s. Its total moment of inertia is 1360 kg ???? m2. Four people standing on the ground, each of mass 65 kg, suddenly step onto the edge of the merry-go-round. (a) What is the angular velocity of the merry-go-round now? (b) What if the people were on it initially and then jumped off in a radial direction (relative to the merry-go-round)
This question is about the principle of the conservation of angular momentum and how changes in moment of inertia affect angular velocity. When people step onto the merry-go-round, its moment of inertia increases causing the angular velocity to decrease, whereas when they jump off, the moment of inertia decreases, thus increasing the angular velocity.
Explanation:This question falls into the realm of rotational dynamics, specifically dealing with the conservation of angular momentum and the effect of moment of inertia on angular velocity. Let's break it down:
(a) Four people step onto the merry-go-round
When the four people step onto the merry-go-round, they increase its moment of inertia, thus reducing its angular velocity according to the principle of conservation of angular momentum. We can find the new angular velocity using the equation L_initial = L_final, where L is the angular momentum. Hence, we have I_initial * ω_initial = I_final * ω_final. Rewriting for ω_final gives us ω_final = (I_initial * ω_initial / I_final).
(b) People jump off the merry-go-round
On the other hand, when people jump off the merry-go-round, they are reducing its moment of inertia, thus increasing the angular velocity under the conservation of angular momentum. Hence, a similar equation would apply, only that the final angular velocity would be higher than the initial angular velocity because the final moment of inertia is lower due to the people jumping off.
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Your friend is trying to construct a clock for a craft show and asks you for some advice. She has decided to construct the clock with a pendulum. The pendulum will be a very thin, very light wooden bar with a thin, but heavy, brass ring fastened to one end. The length of the rod is 80 cm and the diameter of the ring is 10 cm. She is planning to drill a hole in the bar to place the axis of rotation 15 cm from one end. She wants you to tell her the period of this pendulum.
Answer:
The period of the pendulum is [tex]T = 1.68 \ sec[/tex]
Explanation:
The diagram illustrating this setup is shown on the first uploaded image
From the question we are told that
The length of the rod is [tex]L = 80 \ cm[/tex]
The diameter of the ring is [tex]d = 10 \ cm[/tex]
The distance of the hole from the one end [tex]D = 15cm[/tex]
From the diagram we see that point A is the center of the brass ring
So the length from the axis of rotation is mathematically evaluated as
[tex]AP = 80 + 10 -5 -15[/tex]
[tex]AP = 70 \ cm = \frac{70}{100} = 0.7 \ m[/tex]
Now the period of the pendulum is mathematically represented as
[tex]T = 2 \pi \sqrt{\frac{AP}{g} }[/tex]
[tex]T = 2 \pi \sqrt{\frac{0.7}{9.8 } }[/tex]
[tex]T = 1.68 \ sec[/tex]
Select the correct answer from each drop-down menu. Various plant and animal species are on the verge of extinction. Because all organisms are [BLANK] to their native habitat, human activities that result in [BLANK] are the major reason for their extinction.
Answer:
Because all organisms are
adapted
to their native habitat, human activities that result in
habitat destruction
are the major reason for their extinction.
Explanation:
Answer:
Adapted
and
habitat destruction
A 22-turn circular coil of radius 3.00 cm and resistance 1.00 Ω is placed in a magnetic field directed perpendicular to the plane of the coil. The magnitude of the magnetic field varies in time according to the expression B = 0.010 0t + 0.040 0t2, where B is in teslas and t is in seconds. Calculate the induced emf in the coil at t = 4.60 s.
Answer:
23.5 mV
Explanation:
number of turn coil 'N' =22
radius 'r' =3.00 cm=> 0.03m
resistance = 1.00 Ω
B= 0.0100t + 0.0400t²
Time 't'= 4.60s
Note that Area'A' = πr²
The magnitude of induced EMF is given by,
lƩl =ΔφB/Δt = N (dB/dt)A
=N[d/dt (0.0100t + 0.0400 t²)A
=22(0.0100 + 0.0800(4.60))[π(0.03)²]
=0.0235
=23.5 mV
Thus, the induced emf in the coil at t = 4.60 s is 23.5 mV
Pls help ASAP. This is astronomy but there isn't an option :)
Order the hierarchy of astronomy from smallest to largest.
Universe, Galaxy, Moons, Stars, Planet
From largest to smallest they are: Universe, galaxy, solar system, star, planet, moon and asteroid.
Explanation:Let's describe them from smallest to largest. In fact the size order is not exact as there are exceptions.An asteroid is a rocky body which lies in the asteroid belt between Mars and Jupiter. They are typically quite small object. The largest asteroid Ceres has been reclassified as a dwarf planet.A moon is typically a rocky body which is in orbit around a planet. Some moons such as our Moon are quite large and are typically bigger than asteroid. Some moons can actually be smaller than some asteroids.A planet is a nearly spherical body which is in orbit around the Sun. Planets are larger than moons.A star is what planets orbit around. It is the source of light and heat. Our Sun is a star which is many times bigger than all of the planets.A solar system is a star and all of its planets, asteroids, comets and other bodies. It is significantly bigger than a star.A galaxy, such as our Milky Way Galaxy, is a collection of solar systems orbiting around a central core. Most galaxies have a supermassive black hole at their centres.Galaxies also form clusters which are large scale structures.The universe is everything. It contains billions of galaxies. Lots of information RIGHT!!!!
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Answer:
Universe, galaxy, solar system, star, planet, moon and asteroid. Hope this helped!
Sandy and Chris stand on the edge of a cliff and throw identical mass rocks at the same speed. Sandy throw her rock horizontally while Chris throws his upward at an angle of 45° to the horizontal. (i) Which one has a higher speed : Sandy’s thorwn rock at the launch or Chris’s rock when it reaches its highest point in its flight? (ii) Which one (sandy’s rock or Chris’s rock has the higher speed right before hitting the ground? : Explain carefully your answer.
(i) Sandy's rock has a higher speed at the launch than Chris's rock at its peak. (ii) Both rocks will have the same speed just before hitting the ground due to the conservation of energy.
Explanation:The subject of this question is physics, specifically pertaining to projectile motion and velocity. To answer this, we need to use the principles of the conservation of energy and the behaviour of objects in projectile motion.
(i) At the point of release, both Sandy's and Chris's rocks have the same initial speed since they were thrown with the same force. However, when Chris's rock reaches the highest point in its flight, its vertical velocity component becomes zero (it stops moving upwards). Despite this, it will retain a horizontal component of velocity, which is equivalent to the initial horizontal speed of the rock thrown by Sandy. Therefore, Sandy's rock will have a higher speed at the point of launch compared to Chris’s rock at its peak.
(ii) Just before hitting the ground, both rocks would have the same speed. This is because the final velocity of a freely falling object is independent of the direction of launch, assuming the object falls to the same height it was thrown from. Conservation of energy ensures that each rock will have the same speed it started with. This remains true irrespective of air resistance.
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(i) Sandy's thrown rock at the launch has a higher speed than Chris's rock when it reaches its highest point in its flight.
(ii) Sandy's rock has the higher speed right before hitting the ground.
(i) At the launch, both Sandy and Chris throw their rocks with the same initial speed [tex]\( v_0 \)[/tex]. Sandy throws her rock horizontally, so its initial vertical speed is 0, and its initial horizontal speed is [tex]\( v_0 \)[/tex]. Chris throws his rock at a 45° angle to the horizontal, which means the initial speed is divided equally between the horizontal and vertical components due to the symmetry of the 45° angle. Therefore, both the horizontal and vertical initial speeds of Chris's rock are [tex]\( \frac{v_0}{\sqrt{2}} \)[/tex].
At the highest point of Chris's rock's trajectory, the vertical component of its velocity becomes 0 because it momentarily stops ascending before starting to fall back down. However, it still has a horizontal component of velocity equal to [tex]\( \frac{v_0}{\sqrt{2}} \)[/tex]. Since [tex]\( \frac{v_0}{\sqrt{2}} \)[/tex] is less than [tex]\( v_0 \)[/tex], Sandy's rock has a higher speed at the launch than Chris's rock at its highest point.
(ii) When both rocks are about to hit the ground, they will both have a vertical component of velocity due to the acceleration of gravity. For Chris's rock, the horizontal component of velocity remains constant throughout the flight (assuming no air resistance), which is [tex]\( \frac{v_0}{\sqrt{2}} \)[/tex]. The vertical component of velocity for Chris's rock will be equal in magnitude to the horizontal component at the moment of release but will have changed direction due to gravity, resulting in a net speed of [tex]\( v_0 \)[/tex] it hits the ground.
Sandy's rock, on the other hand, has been accelerating downwards under gravity while maintaining its initial horizontal speed. Just before hitting the ground, the vertical component of Sandy's rock's velocity will be greater than [tex]\( \frac{v_0}{\sqrt{2}} \)[/tex] due to the acceleration of gravity. The total speed of Sandy's rock just before impact is calculated by combining the horizontal and vertical components using the Pythagorean theorem. Since the horizontal component is [tex]\( v_0 \)[/tex] and the vertical component is greater than [tex]\( \frac{v_0}{\sqrt{2}} \)[/tex], the total speed of Sandy's rock will be greater than [tex]\( v_0 \)[/tex].
Therefore, right before hitting the ground, Sandy's rock has a higher speed than Chris's rock, which has a speed of [tex]\( v_0 \)[/tex]. Sandy's rock's speed is greater because it has both the initial horizontal speed [tex]\( v_0 \)[/tex] and an additional vertical speed component due to the acceleration of gravity.
interdependence between plants and animals
Answer:
plants depend on animals for CO2 (to use during photosynthesis) while animals depend on plants for food (consumation)
Explanation:
Answer:
plants depend on animals for CO2 (to use during photosynthesis) while animals depend on plants for food (consumation)
Explanation:
You are suffering through a court case arising from a recent car accident that you were in. You were traveling West, alone in your car, which has a mass of 1000 kg, through an intersection when another driver in a small car (mass 850 kg) traveling South crashed into your passenger side at the center of the intersection. The two cars became stuck together and skidded off the road. For this problem, choose the positive x-axis to point to the East (right) and the positive y-axis to point North (up). You recall that you had just looked at the speedometer before the accident and were traveling at 20 m/s. What was your initial momentum vector?
Answer:
p₀ = (- 20,000 i - 17,000 j) kg m / s
vₓ = -10.81 m / s , v_{y} = - 9,189 m / s
Explanation:
This exercise can be solved with the conservation of the moment.
The system is formed by the two cars, so the forces during the crash are internal and the moment is conserved
initial moment. Just before the crash
p₀ = M v₁ + m v₂
where the masses M = 1000 kg with a velocity of v₁ = - 20 i m / s for traveling west, the second vehicle has a mass m = 850 kg and a velocity of v₂ = - 20 i m / s, we substitute
p₀ = - 1000 20 i - 850 20 j
p₀ = (- 20,000 i - 17,000 j) kg m / s
final moment. Right after the crash
how the cars fit together
[tex]p_{f}[/tex] = (M + m) [tex]v_{f}[/tex]
the final speed is shaped
v_{f} = vₓ i + v_{y} j
we substitute
p_{f} = (M + m) (vₓ i + v_{y} j)
p_{f} = (1000 + 850) (vₓ i + [tex]v_{y}[/tex]j)
p₀ = p_{f}
(- 20000 i - 17000 j) = 1850 (vₓ i + v_{y} j)
we solve for each component
vₓ = -20000/1850
vₓ = -10.81 m / s
v_{y} = -17000/1850
v_{y} = - 9,189 m / s
Suppose you defined ""positive"" motion as motion westward. What would be the persons total velocity?
Answer: I think that you are asking about an interactive reader which talks about an exercise which asked for the person's velocity who were on the bus.
The answer is 14 m/s
Explanation: When you have to do an exercise about velocity, you can do this exercise knowing the velocity from the bus in relative the street and you ALWAYS have to define your positive or negative motion about the direction in which the person were walking to know if you have to add or subtract
In this case, the bus were travelling east at 15 m/s and the person walked toward the back of the bus at 1 m/s.
It doesn't matter if the positive is the west or the east. While you only know the positive and knowing that the person was walking toward back the bus you can deduce this:
The person's total velocity is (+15 m/s) - (+1 m/s) = +14 m/s
If positive motion is defined as motion westward, the total velocity would be the individual's speed in the westward direction. It includes both the speed and the direction, which makes velocity a vector quantity.
Explanation:In the context of the question, if positive motion is defined as motion westward, then the individual's total velocity would be their speed in the westward direction with the magnitude and direction incorporated. Velocity is a vector quantity because it encompasses both magnitude (how fast an object is moving) and direction (where it is going). So if the person is walking at a speed of 3 km/h, in the westward (positive) direction, then we can say the person's total velocity is 3 km/h west.
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A coil with an inductance of 2.3 H and a resistance of 14 Ω is suddenly connected to an ideal battery with ε = 100 V. At 0.13 s after the connection is made, what is the rate at which (a) energy is being stored in the magnetic field, (b) thermal energy is appearing in the resistance, and (c) energy is being delivered by the battery?
Given Information:
Resistance = R = 14 Ω
Inductance = L = 2.3 H
voltage = V = 100 V
time = t = 0.13 s
Required Information:
(a) energy is being stored in the magnetic field
(b) thermal energy is appearing in the resistance
(c) energy is being delivered by the battery?
Answer:
(a) energy is being stored in the magnetic field ≈ 219 watts
(b) thermal energy is appearing in the resistance ≈ 267 watts
(c) energy is being delivered by the battery ≈ 481 watts
Explanation:
The energy stored in the inductor is given by
[tex]U = \frac{1}{2} Li^{2}[/tex]
The rate at which the energy is being stored in the inductor is given by
[tex]\frac{dU}{dt} = Li\frac{di}{dt} \: \: \: \: eq. 1[/tex]
The current through the RL circuit is given by
[tex]i = \frac{V}{R} (1-e^{-\frac{t}{ \tau} })[/tex]
Where τ is the the time constant and is given by
[tex]\tau = \frac{L}{R}\\ \tau = \frac{2.3}{14}\\ \tau = 0.16[/tex]
[tex]i = \frac{110}{14} (1-e^{-\frac{t}{ 0.16} })\\i = 7.86(1-e^{-6.25t})\\\frac{di}{dt} = 49.125e^{-6.25t}[/tex]
Therefore, eq. 1 becomes
[tex]\frac{dU}{dt} = (2.3)(7.86(1-e^{-6.25t}))(49.125e^{-6.25t})[/tex]
At t = 0.13 seconds
[tex]\frac{dU}{dt} = (2.3) (4.37) (21.8)\\\frac{dU}{dt} = 219.11 \: watts[/tex]
(b) thermal energy is appearing in the resistance
The thermal energy is given by
[tex]P = i^{2}R\\P = (7.86(1-e^{-6.25t}))^{2} \cdot 14\\P = (4.37)^{2}\cdot 14\\P = 267.35 \: watts[/tex]
(c) energy is being delivered by the battery?
The energy delivered by battery is
[tex]P = Vi\\P = 110\cdot 4.37\\P = 481 \: watts[/tex]
Find the magnitude of the average force ⟨Fx⟩⟨Fx⟩ in the x direction that the particle exerts on the right-hand wall of the container as it bounces back and forth. Assume that collisions between the wall and particle are elastic and that the position of the container is fixed. Be careful of the sign of your answer. Express the magnitude of the average force in terms of mmm, vxvxv_x, and LxLxL_x.
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
The attached file has a detailed solution of the given problem.
Newton's second law allows us to find the average force for the impact of a particle against the wall of a container is:
[tex]F = \frac{m v_x^2}{L_x}[/tex]
Newton's second law is the change of the momentum with respect to time and the momentum is defined as the product of the mass and the velocity of the body.
[tex]F = \frac{dp}{dt} \\p = m v[/tex]
where F is the force, p the moment, m the mass, t the time and v is the velocity.
Ask for the average force, therefore we change the differentials for variations.
[tex]<F> = m \frac{\Delta v}{\Delta t}[/tex]
They indicate that the velocity in the direction of the wall is vₓ and the mass of the container is much greater than the mass of the particle, they also indicate that the collision is elastic, therefore the speed of the particle before and after the collision is equal , but its address changes.
Δv = vₓ - (-vₓ)
Δv = 2 vₓ
The change in velocity occurs during the collision, in the rest of the motion the particle has a constant velocity, using the uniform motion relation.
[tex]v = \frac{d}{t} \\t = \frac{d}{v}[/tex]
The particle travels a distance Lₓ from inside the container to the wall and bounces, we can find the total time for the particle where the distance of the entire route is:
d = 2 Lₓ
t = [tex]\frac{2L_x}{v_x}[/tex]
Let's substitute in Newton's second law.
[tex]<F>= \frac{m \ 2v_x}{ \frac{2L_x}{v_x} } \\<F>= \frac{m v_x^2}{L_x}[/tex]
In conclusion, using Newton's second law we can find the average force for the collision of a particle against the wall of a container is:
[tex]<F> = \frac{m \ v_x^2}{L_x}[/tex]
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We find that 2N current loops are coplanar and coaxial. The first has radius a and current I. The second has radius 2a and current 2I, and the pattern is repeated up to the Nth, which has radius Na and current NI. The current in the loops alternates in direction from loop to loop as seen from above. Thus the current in the first loop is counterclockwise, in the next clockwise, up to the last loop where it is again clockwise. The magnitude of the magnetic field at the center of the loops is:_________
a. 0
b. µoNI/2a
c. µoI/Na
d. µoNI/a
e. µI/2Na
Final answer:
The magnitude of the magnetic field at the center of the loops can be determined using Ampere's law. It is given by µ₀NI/(2a), where µ₀ is the permeability of free space, N is the number of turns, I is the current, and a is the radius of the loop.
Explanation:
The magnitude of the magnetic field at the center of the loops can be determined using Ampere's law. According to the law, the magnetic field inside a current-carrying wire of radius a is given by µ₀NI/(2a), where µ₀ is the permeability of free space, N is the number of turns, I is the current, and a is the radius of the loop.
In this case, the radius of the first loop is a and the current is I. Since the pattern is repeated up to the Nth loop, the radius of the Nth loop is Na and the current is NI. Therefore, the magnitude of the magnetic field at the center of the loops is µ₀NI/(2a).
Final answer:
The net magnetic field at the center of a series of alternating coaxial current loops is zero due to the cancellation of the magnetic fields of each pair of loops with opposite currents.
Explanation:
The question refers to a configuration of coaxial current loops whose radii and currents increase with each subsequent loop, and where the direction of current alternates with each loop. To find the magnetic field at the center due to each loop, we use the formula for the magnetic field (B) at the center of a current loop: B = (µ0I)/(2r), where µ0 is the permeability of free space, I is the current, and r is the radius of the loop.
For each pair of consecutive loops, the magnetic fields will be in opposite directions due to the alternating current direction. Thus, the magnetic fields generated by each pair of loops (except for the last one if N is odd) will cancel each other out. If N is even, all fields cancel out and the net magnetic field is zero; if N is odd, the net magnetic field is only due to the largest loop. In both cases, the net magnetic field at the center will be zero.
Therefore, the correct answer is (a) 0.
A fan is designed to last for a certain time before it will have to be replaced (planned obsolescence). The fan only has one speed (at a maximum of 700 rpm), and it reaches the speed in 4.0 s (starting from rest). It takes the fan 11.0 s for the blade to stop once it is turned off. The manufacturer specifies that the fan will operate up to 1 billion rotations. Andre lives in a hot climate, works outside of the home from approximately 8:00 am to 5:00 pm, Monday through Friday, does not own an air conditioner, and can't sleep with the fan running. Estimate how many hot days ????hot Andre will be able to use the fan, rounded to the nearest day. ????hot= days
Answer:
1355 days
Explanation:
Find the given attachment
Andre will not be able to use the fan at all during his working hours.
Explanation:To estimate how many hot days Andre will be able to use the fan, we need to calculate the number of rotations the fan will make during his working hours and subtract the number of rotations it takes for the fan to stop once turned off. We know that the fan reaches a speed of 700 rpm in 4.0 s from rest. This means its angular velocity is 700 * 2π/60 = 73.33 rad/s. So, in 8 hours (Andre's working hours), the fan will rotate for 73.33 * 3600 * 8 = 210,624 rotations. Subtracting the rotations it takes for the fan to stop (700 * 11 = 7700 rotations), Andre will be able to use the fan for approximately 210,624 - 7700 = 202,924 rotations. Since the fan is specified to operate up to 1 billion rotations, Andre will be able to use the fan for approximately 202,924 / 1,000,000,000 = 0.0002029 (or about 0.02%) of its lifespan. Rounded to the nearest day, this is 0 days.
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What is the function of a power source in a circuit
Answer:
Electrical energy
Explanation:
A power source, such as a battery, provides electrical energy for the circuit. The power source is a device or system that converts another form of energy to electrical energy.
The function of a power source in a circuit is Electrical energy.
Electrical energy:Electrical energy seems to be a sort of energy (Kinetic) that would be generated by transporting or shifting electric charges. This same quantity of electricity relies somewhat on the velocity of the charges – therefore more electrical energy they represent, the quicker they accelerate.
Examples of Electrical energy are:
Lightning.Batteries.Electric eels.Thus the response above is the appropriate one.
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An ammeter is connected in series with a resistor of unknown resistance R and a DC power supply of known emf e. A student uses the ammeter to measure the current I and calculates the power P dissipated by the resistor using I and e. The student later discovers that the ammeter is not an ideal ammeter. Which of the following correctly compares the actual power dissipated by the resistor to the value of power P calculated by the student and explains why? (A) The actual power dissipated by the resistor is greater than P because the ammeter had some resistance (B) The actual power dissipated by the resistor is less than P because the ammeter had some resistance. (C) The actual power dissipated by the resistor is greater than P because the ammeter should have been connected in parallel to the resistor. (D) The actual power dissipated by the resistor is less than P because the ammeter should have been connected in parallel to the resistor. (E) The actual power dissipated by the resistor is equal to P because the ammeter was connected properly in series.
Answer:
B:The actual power dissipated by the resistor is less than P because the ammeter had some resistance.
Explanation:
Here,power has been calculated using current I and total EMF \ε . So,P=EMF*current= ε I will represent total power dissipated in resistor and ammeter.
Now, this total power P has been dissipated in both resistor and ammeter. So, power dissipated in resistor must be less than P as some power is also dissipated in ammeter because it has non-zero resistance.
So, the answer is B:The actual power dissipated by the resistor is less than P because the ammeter had some resistance.
Note that option A,C and E are ruled out as they state power dissipated by resistor is greater than or equal to P which is false.
Also,option D is ruled out as ammeter is connected in series.
A solid sphere of uniform density has a mass of 3.0 × 104 kg and a radius of 1.0 m. What is the magnitude of the gravitational force due to the sphere on a particle of mass 9.3 kg located at a distance of (a) 1.4 m and (b) 0.21 m from the center of the sphere? (c) Write a general expression for the magnitude of the gravitational force on the particle at a distance r ≤ 1.0 m from the center of the sphere.
The gravitational force on a particle due to a sphere can be calculated using Newton's Law of gravitation. Different distances result in different force magnitudes.
Explanation:The magnitude of the gravitational force (F) on a particle due to a sphere can be calculated by using Newton's Law of Gravitation, F = G*(m1*m2)/r², where G is the gravitational constant (6.67 × 10-¹¹ N·m²/kg²), m1 and m2 are the masses of the two bodies, and r is the distance between their centers.
(a) For a distance of 1.4 m, F = G*(3.0 × 10⁴ kg*9.3 kg) / (1.4 m)².
(b) For a distance of 0.21 m, F = G*(3.0 × 10⁴ kg*9.3 kg) / (0.21 m)².
(c) For a general expression for the magnitude of the gravitational force on the particle at a distance r ≤ 1.0 m from the center of the sphere, the formula changes to F = G*(m1*m2)/r², where r ≤ 1.0 m.
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