Explanation:
Dark Energy. Dark Energy is a hypothetical form of energy that exerts a negative, repulsive pressure, behaving like the opposite of gravity. It has been hypothesised to account for the observational properties of distant type Ia supernovae, which show the universe going through an accelerated period of expansion
. If you live in a region that has a particular TV station, you can sometimes pick up some of its audio portion on your FM radio receiver. Explain how this is possible. Does it imply that TV audio is broadcast as FM
Answer:
Please see below as the answer is self-explanatory.
Explanation:
The low band of the VHF TV Spectrum, spans channels 2-6, from 54 to 88 Mhz.
In the analog TV, in the Americas, the total bandwidth of any channel is 6 Mhz, with the visual carrier modulated in VSS (Vestigial Side Band) at 1.25 Mhz from the lowest frequency of the channel.
The aural carrier is located at 4.5 Mhz from the visual carrier, and is FM modulated.
For Channel 6, which spans between 82 and 88 Mhz, the visual carrier is at 83.25 Mhz, so the aural carrier is at 87.75 Mhz, which falls within the FM Band, so it is possible to listen the audio part of this channel in a FM radio receiver, even at a lower volume, due to the FM radio has a greater deviation than TV aural carrier.
The reason why it is possible for TV station to sometimes pick up some of the audio portion on your FM radio receiver is because; TV waves can sometimes deviate into the FM radio frequency range.
Let us start with explaining the waves of TV and radio.
The frequency range utilized by TV stations is either the range 54 MHz to 88 MHz or 174 MHz to 222 MHz. In contrast, the frequency range utilized by FM Radio band is between 88 MHz and 174 MHz.
Now, in some cases, it is possible that the TV signal may deviate into the range of the FM Radio and as such in that case, the TV signal will pick the audio portion of an FM Radio. These TV waves are very high frequency waves.
Finally, it does not imply that the TV wave is broadcasting as an FM because it only deviated a bit from the TV range and not like that is where it is made to operate.
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The phenomenon of magnetism is best understood in terms ofA) the existence of magnetic poles.B)the magnetic fields associated with the movement of charged particles.C)gravitational forces between nuclei and orbital electrons.D) electrical fluid
Answer:
A) the existence of magnetic poles.Explanation:
Magnetism is defined as the ability of a magnet to attract magnetic substance to itself. Such magnet has the ability of being magnetized. A magnet is known to possess poles which are the north poles and south poles. The presence of this poles is what makes them possess the properties of a magnet. An ordinary steel bar doesn't have the properties of a magnet unless it is magnetized and when you are trying to magnetize a steel bar, you are invariably introducing the magnetic poles.
According to the law of magnetism, like poles repel but unlike poles attract. From the above explanation, it can be concluded that the phenomenon of magnetism is best understood interns of existence of magnetic poles. This poles are called the north and the south poles.
g A particle (charge = +40 mC) is located on the x axis at the point x = -20 cm, and a second particle (charge = -50 mC) is placed on the x axis at x = +30 cm. What is the magnitude of the total electrostatic force on a third particle (charge = -4.0 mC) placed at the origin (x = 0)? Group of answer choices
Answer:
Explanation:
We shall find electric field at origin due to two given charges sitting on the either side of origin .
Total field will add up due to their same direction .
Field due to a charge Q
= 9 x 10⁹ x Q / R² ; R is distance of point , Q is charge
Field due to first charge
= 9 x 10⁹ x 40 x 10⁻³ / 2² x 10⁻⁴
= 90 x 10¹⁰ N/C
Field due to second charge
= 9 x 10⁹ x 50 x 10⁻³ / 2² x 10⁻⁴
= 112.5 x 10¹⁰ N/C
Total field
= 202.5 x 10¹⁰ N/C
Force on given charge at origin
= charge x field
= 4 x 10⁻³ x 202.5 x 10¹⁰
= 810 x 10⁷ N .
The rotor of a gas turbine is rotating at a speed of 7000 rpm when the turbine is shut down. It is observed that 3.5 minutes is required for the rotor to coast to rest. Assuming uniformly accelerated motion, determine the number of revolutions that the rotor executes before coming to rest. Hint: there will be a large number of rotations.
Answer:
The rotor of the gas turbine rotates 12250 revolutions before coming to rest.
Explanation:
Given that rotor of gas turbine is decelerating at constant rate, it is required to obtained the value of angular acceleration as a function of time, as well as initial and final angular speeds. That is:
[tex]\dot n = \dot n_{o} + \ddot n \cdot t[/tex]
Where:
[tex]\dot n_{o}[/tex] - Initial angular speed, measured in revolutions per minute.
[tex]\dot n[/tex] - Final angular speed, measured in revolutions per minute.
[tex]t[/tex] - Time, measured in minutes.
[tex]\ddot n[/tex] - Angular acceleration, measured in revoiutions per square minute.
The angular acceleration is now cleared:
[tex]\ddot n = \frac{\dot n - \dot n_{o}}{t}[/tex]
If [tex]\dot n_{o} = 7000\,\frac{rev}{min}[/tex], [tex]\dot n = 0\,\frac{rev}{min}[/tex] and [tex]t = 3.5\,min[/tex], the angular acceleration is:
[tex]\ddot n = \frac{0\,\frac{rev}{min}-7000\,\frac{rev}{min} }{3.5\,min}[/tex]
[tex]\ddot n = -2000\,\frac{rev}{min^{2}}[/tex]
Now, the final angular speed as a function of initial angular speed, angular acceleration and the change in angular position is represented by this kinematic equation:
[tex]\dot n^{2} = \dot n_{o}^{2} + 2\cdot \ddot n \cdot (n-n_{o})[/tex]
Where [tex]n[/tex] and [tex]n_{o}[/tex] are the initial and final angular position, respectively.
The change in angular position is cleared herein:
[tex]n-n_{o} = \frac{\dot n^{2}-\dot n_{o}^{2}}{2\cdot \ddot n}[/tex]
If [tex]\dot n_{o} = 7000\,\frac{rev}{min}[/tex], [tex]\dot n = 0\,\frac{rev}{min}[/tex] and [tex]\ddot n = -2000\,\frac{rev}{min^{2}}[/tex], the change in angular position is:
[tex]n-n_{o} = \frac{\left(0\,\frac{rev}{min} \right)^{2}-\left(7000\,\frac{rev}{min} \right)^{2}}{2\cdot \left(-2000\,\frac{rev}{min^{2}} \right)}[/tex]
[tex]n-n_{o} = 12250\,rev[/tex]
The rotor of the gas turbine rotates 12250 revolutions before coming to rest.
A physics professor wants to perform a lecture demonstration of Young's double-slit experiment for her class using the 633-nm light from a He-Ne laser. Because the lecture hall is very large, the interference pattern will be projected on a wall that is 4.0 m from the slits. For easy viewing by all students in the class, the professor wants the distance between the m
What slit separation is required in order to produce the desired interference pattern?
d=________m
Note: if the professor wants the distance between the m = 0 and m = 1 maxima to be 25 cm
Answer:
d = 1.0128×10⁻⁵m
Explanation:
given:
length L = 4.0m
maximum distance between m = 0 and m = 1 , y = 25cm = 0.25m
wavelength λ = 633nm = 633×10⁻⁹m
note:
dsinθ = mλ (constructive interference)
where d is slit seperation, θ is angle of seperation , m is order of interference , and λ is wavelength
for small angle
sinθ ≈ tanθ
[tex]d (\frac{y}{L}) =[/tex] mλ
[tex]d (\frac{y}{L}) = (1)(633nm)[/tex]
[tex]d(\frac{0.25}{4} ) = (1)(633nm)[/tex]
d = 1.0128×10⁻⁵m
The AC voltage source supplies an rms voltage of 146 V at frequency f. The circuit has R = 110 Ω, XL = 210 Ω, and XC = 110 Ω. At the instant the voltage across the generator is at its maximum value, what is the magnitude of the current in the circuit?
Answer:
1.03A
Explanation:
For computing the magnitude of the current in the circuit we need to do the following calculations
LCR circuit impedance
[tex]Z = \sqrt{R^2 + (X_L - X_c)^2} \\\\ = \sqrt{110^2 + (210 - 110)^2}[/tex]
= 148.7Ω
Now the phase angle is
[tex]\phi = tan^{-1} (\frac{X_L - X_C}{R}) \\\\ = tan^{-1} (\frac{210 - 110}{110})\\\\ = 42.3^{\circ}[/tex]
Now the rms current flowing in the circuit is
[tex]I_{rms} = \frac{V_{rms}}{Z} \\\\ = \frac{146}{148.7}[/tex]
= 0.98 A
The current flowing in the circuit is
[tex]I = I_{rms}\sqrt{2} \\\\ = (0.98) (1.414)[/tex]
= 1.39 A
And, finally, the current across the generator is
[tex]I'= I cos \phi[/tex]
[tex]= (1.39) cos 42.3^{\circ}[/tex]
= 1.03A
Hence, the magnitude of the circuit current is 1.03A
Based on the graph below, what prediction can we make about the acceleration when the force is 0 newtons? A. It will be 0 meters per second per second. B. It will be 5 meters per second per second. C. It will be 10 meters per second per second. D. It will be 15 meters per second per second.
Answer:
Option A
Explanation:
From the graph, we came to know that Force and acceleration are in direct relationship.
Also,
Force = 0 when Acceleration = 0
Because Both are 0 at the origin.
Answer:
A. It will be 0 meters per second per second.
Explanation:
The force and acceleration is in a proportional relationship, that means the line goes through the origin.
On the graph, when the force is at 0, the acceleration is 0. The line passes through the origin.
If a bicycle starts from rest and is pedaled normally until the bike is moving at 6 meters per second across level ground, what kinds of energy have its tires been given? (Select all that apply) g
Answer: Translational Kinetic Energy
Rotational Kinetic Energy
Explanation:
An object has translational kinetic energy when it is undergoing through a linear displacement.
Rotational energy is kinetic energy due to the rotation of an object .
Here the wheel of bicycle undergoes both translational and rotational kinetic energy has it moves with linear displacement with rotation in it.
Hence, the tires have been two kinds of energy : translational and rotational kinetic energy
What is the wave length if the distance from the central bright region to the sixth dark fringe is 1.9 cm . Answer in units of nm.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The wavelength is [tex]\lambda = 622 nm[/tex]
Explanation:
From the question we are told that
The distance of the slit to the screen is [tex]D = 5 \ m[/tex]
The order of the fringe is m = 6
The distance between the slit is [tex]d = 0.9 \ mm = 0.9 *10^{-3} \ m[/tex]
The fringe distance is [tex]Y = 1.9 \ cm = 0.019 \ m[/tex]
Generally the for a dark fringe the fringe distance is mathematically represented as
[tex]Y = \frac{[2m - 1 ] * \lambda * D }{2d}[/tex]
=> [tex]\lambda = \frac{Y * 2 * d }{[2*m - 1] * D}[/tex]
substituting values
=> [tex]\lambda = \frac{0.019 * 2 * 0.9*10^{-3} }{[2*6 - 1] * 5}[/tex]
=> [tex]\lambda = 6.22 *10^{-7} \ m[/tex]
[tex]\lambda = 622 nm[/tex]
The 2-Mg truck is traveling at 15 m/s when the brakes on all its wheels are applied, causing it to skid for 10 m before coming to rest. The total mass of the boat and trailer is 1 Mg. Determine the constant horizontal force developed in the coupling C, and the friction force developed between the tires of the truck and the road during this time.
Answer:
constant horizontal force developed in the coupling C = 11.25KN
the friction force developed between the tires of the truck and the road during this time is 33.75KN
Explanation:
See attached file
The friction force between the tires of the truck and the road is 22500 N.
Calculating the friction force:It is given that a 2 Mg truck ( m = 2000 Kg) is initially moving with a speed of u = 15 m/s.
Distance traveled before coming to rest, s = 10m
The final velocity of the truck will be zero, v = 0
When the breaks are applied, only the frictional force is acting on the truck and it is opposite to the motion of the truck.
The frictional force is given by:
f = -ma
the acceleration of the truck = -a
The negative sign indicates that the acceleration is opposite to the motion.
Applying the third equation of motion we get:
v² = u² -2as
0 = 15² - 2×a×10
225 = 20a
a = 11.25 m/s²
So the magnitude of frictional force is:
f = ma = 2000 × 11.25 N
f = 22500 N
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Force and distance are used to calculate work. Work is measured in which unit? joules watts newtons meters
Answer:
The unit of work is joules
Force and displacement are used to calculate the work done by an object. This work is measured in the units of Joules. Thus, the correct option is A.
What is Work?Work can be defined as the force that is applied on an object which shows some displacement. Examples of work done include lifting an object against the Earth's gravitational force, and driving a car up on a hill. Work is a form of energy. It is a vector quantity as it has both the direction as well as the magnitude. The standard unit of work done is the joule (J). This unit is equivalent to a newton-meter (N·m).
The nature of work done by an object can be categorized into three different classes. These classes are positive work, negative work and zero work. The nature of work done depends on the angle between the force and displacement of the object. Positive work is done if the applied force displaces the object in its direction, then the work done is known as positive work. Negative work is opposite of positive work as in this work, the applied force and displacement of the object are in opposite directions to each other and zero work is done when there is no displacement.
Therefore, the correct option is A.
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A wheel rotating about a fixed axis has a constant angular acceleration of 4.0 rad/s2. In a 4.0-s interval the wheel turns through an angle of 80 radians. Assuming the wheel started from rest, how long had it been in motion at the start of the 4.0-s interval
Answer:
The time interval is [tex]t = 3 \ s[/tex]
Explanation:
From the question we are told that
The angular acceleration is [tex]\alpha = 4.0 \ rad/s^2[/tex]
The time taken is [tex]t = 4.0 \ s[/tex]
The angular displacement is [tex]\theta = 80 \ radians[/tex]
The angular displacement can be represented by the second equation of motion as shown below
[tex]\theta = w_i t + \frac{1}{2} \alpha t^2[/tex]
where [tex]w_i[/tex] is the initial velocity at the start of the 4 second interval
So substituting values
[tex]80 = w_i * 4 + 0.5 * 4.0 * (4^2)[/tex]
=> [tex]w_i = 12 \ rad/s[/tex]
Now considering this motion starting from the start point (that is rest ) we have
[tex]w__{4.0 }} = w__{0}} + \alpha * t[/tex]
Where [tex]w__{0}}[/tex] is the angular velocity at rest which is zero and [tex]w__{4}}[/tex] is the angular velocity after 4.0 second which is calculated as 12 rad/s s
[tex]12 = 0 + 4 t[/tex]
=> [tex]t = 3 \ s[/tex]
Following are the response to the given question:
Given:
[tex]\to \alpha = 4.0 \ \frac{rad}{s^2}\\\\[/tex]
[tex]\to \theta= 80\ radians\\\\\to t= 4.0 \ s\\\\ \to \theta_0=0\\[/tex]
To find:
[tex]\to \omega=?\\\\\to t=?\\\\[/tex]
Solution:
Using formula:
[tex]\to \theta- \theta_0 = w_{0} t+ \frac{1}{2} \alpha t^2\\\\ \to 80-0= \omega_{0}(4) + \frac{1}{2} (4)(4^2)\\\\ \to 80= \omega_{0}(4) + \frac{1}{2} (4)(16)\\\\\\to 80= \omega_{0}(4) + (4)(8)\\\\\to 80= \omega_{0}(4) + 32\\\\\to 80-32 = \omega_{0}(4) \\\\\to \omega_{0}(4)= 48 \\\\\to \omega_{0}= \frac{48}{4} \\\\ \to \omega_{0} = 12 \frac{rad}{ s} \\\\[/tex]
It would be the angle for rotation at the start of the 4-second interval.
This duration can be estimated by leveraging the fact that the wheel begins from rest.
[tex]\to \omega = \omega_{0} + \alpha t\\\\\to 12 = 0 +4(t) \\\\\to 12 = 4(t) \\\\ \to t=\frac{12}{4}\\\\\to t= 3\ s[/tex]
Therefore, the answer is "[tex]12\ \frac{rad}{s}[/tex] and [tex]3 \ s[/tex]".
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An insulating hollow sphere has inner radius a and outer radius b. Within the insulating material the volume charge density is given by rho(r)=αr,where α is a positive constant.
A). What is the magnitude of the electric field at a distance r from the center of the shell, where a
Express your answer in terms of the variables α, a, r, and electric constant ϵ0.
B) .A point charge
q is placed at the center of the hollow space, at r=0. What value must q have (sign and magnitude) in order for the electric field to be constant in the region a
Express your answer in terms of the variables α, a, and appropriate constants.
C). What then is the value of the constant field in this region?
Express your answer in terms of the variable αand electric constant ϵ0.
Answer:
E = α/2∈₀ [ 1 - a²/r² ]
Ф = α/2∈₀
Explanation:
Using Gauss Law:
ρ(r) = a/r, dA
= 4 π r²d r
Ф = [tex]\int\limits^r_a[/tex] ρ(r')dA
Ф[tex]_{encl}[/tex] = [tex]\int\limits^r_a[/tex] ρ(r')dA
= 4πα [tex]\int\limits^r_a[/tex] r'dr'
Ф[tex]_{encl}[/tex] = 4 π α 1/2(r²-a²)
E(4πr²) = [tex]2\pi\alpha (r^{2}-a^{2} )/[/tex]∈₀
= [tex]2\pi\alpha (r^{2}-a^{2} )/[/tex]∈₀(4πr²)
= α (r² - a²) / 2 ∈₀ (r²)
= α/2∈₀ [ r²/r² - a²/r² ]
E = α/2∈₀ [ 1 - a²/r² ]
Electric field of the point charge:
E[tex]_{q}[/tex] = q / 4π∈₀r²
[tex]E_{total}[/tex] = α / 2 ∈₀ - (α / 2 ∈₀ )(a² / r²) + q / 4 π ∈₀ r²
For [tex]E_{total}[/tex] to be constant:
- (αa²/ 2 ∈₀ ) + q / 4 π ∈₀ = 0 and q = 2παa²
-> α / 2 ∈₀ - αa²/ 2 ∈₀ + 2παa² / 4 π ∈₀
= α - αa² + αa² / 2 ∈₀
= α /2 ∈₀
Hence:
Ф = α/2∈₀
A 285-kg object and a 585-kg object are separated by 4.30 m. (a) Find the magnitude of the net gravitational force exerted by these objects on a 42.0-kg object placed midway between them.
Answer:
The magnitude of the net gravitational force exerted by these objects on a 42.0-kg object is 1.818 x 10⁻⁷ N
Explanation:
Given;
first object with mass, m₁ = 285 kg
second object with mass, m₂ = 585 kg
distance between the two objects, r = 4.3 m
The midpoint between the two objects = r/₂ = 4.3 /2 = 2.15 m
Gravitational force between the first object and the 42 kg object;
[tex]F = \frac{GMm}{r^2}[/tex]
where;
G = 6.67 x 10⁻¹¹ Nm²kg⁻²
[tex]F = \frac{6.67*10^{-11} *285*42}{2.15^2} \\\\F = 1.727*10^{-7} \ N[/tex]
Gravitational force between the second object and the 42 kg object
[tex]F = \frac{6.67*10^{-11} *585*42}{2.15^2} \\\\F = 3.545*10^{-7} \ N[/tex]
Magnitude of net gravitational force exerted on 42kg object;
F = 3.545x 10⁻⁷ N - 1.727 x 10⁻⁷ N
F = 1.818 x 10⁻⁷ N
Therefore, the magnitude of the net gravitational force exerted by these objects on a 42.0-kg object is 1.818 x 10⁻⁷ N
A 1.8 kg microphone is connected to a spring and is oscillating in simple harmonic motion up and down with a period of 3s. Below the microphone is 1.8 hz, calculate the spring constant
Answer:
230N/m
Explanation:
Pls see attached file
A computer has a mass of 3 kg. What is the weight of the computer?
A. 288 N.
B. 77.2 N
C. 3N
D. 29.4 N
Answer:
29.4 NOption D is the correct option.
Explanation:
Given,
Mass ( m ) = 3 kg
Acceleration due to gravity ( g ) = 9.8 m/s²
Weight ( w ) = ?
Now, let's find the weight :
[tex]w \: = \: m \times g[/tex]
plug the values
[tex] = 3 \times 9.8[/tex]
Multiply the numbers
[tex] = 29.4 \: [/tex] Newton
Hope this helps!!
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A flat loop of wire consisting of a single turn of cross-sectional area 8.20 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 2.60 T in 1.02 s. What is the resulting induced current if the loop has a resistance of 2.70
Answer:
The induced current is [tex]I = 6.25*10^{-4} \ A[/tex]
Explanation:
From the question we are told that
The number of turns is [tex]N = 1[/tex]
The cross-sectional area is [tex]A = 8.20 cm^2 = 8.20 * 10^{-4} \ m^2[/tex]
The initial magnetic field is [tex]B_i = 0.500 \ T[/tex]
The magnetic field at time = 1.02 s is [tex]B_t = 2.60 \ T[/tex]
The resistance is [tex]R = 2.70\ \Omega[/tex]
The induced emf is mathematically represented as
[tex]\epsilon = - N * \frac{ d\phi }{dt}[/tex]
The negative sign tells us that the induced emf is moving opposite to the change in magnetic flux
Here [tex]d\phi[/tex] is the change in magnetic flux which is mathematically represented as
[tex]d \phi = dB * A[/tex]
Where dB is the change in magnetic field which is mathematically represented as
[tex]dB = B_t - B_i[/tex]
substituting values
[tex]dB = 2.60 - 0.500[/tex]
[tex]dB = 2.1 \ T[/tex]
Thus
[tex]d \phi = 2.1 * 8.20 *10^{-4}[/tex]
[tex]d \phi = 1.722*10^{-3} \ weber[/tex]
So
[tex]|\epsilon| = 1 * \frac{ 1.722*10^{-3}}{1.02}[/tex]
[tex]|\epsilon| = 1.69 *10^{-3} \ V[/tex]
The induced current i mathematically represented as
[tex]I = \frac{\epsilon}{ R }[/tex]
substituting values
[tex]I = \frac{1.69*10^{-3}}{ 2.70 }[/tex]
[tex]I = 6.25*10^{-4} \ A[/tex]
a figures skater rotating at 5 rads with arms extended has a moment of inertia of 2.25 kg. if the arms are pulled in so the moment of inertia decrease to 1.8 what is the final angular speed
Answer:
The final angular speed is 6.25 rad/s
Explanation:
Given;
initial angular speed, ω₁ = 5 rad/s
initial moment of inertia, I₁ = 2.25 kg.m²
Final moment of inertia, I₂ = 1.8 kg.m²
final angular speed, ω₂ = ?
Based on conservation of angular momentum, we will have the following expression;
ω₁I₁ = ω₂I₂
ω₂ = (ω₁I₁ ) / I₂
ω₂ = (5 x 2.25) / 1.8
ω₂ = 6.25 rad/s
Therefore, the final angular speed is 6.25 rad/s
When the magnet falls toward the copper block, the changing flux in the copper creates eddy currents that oppose the change in flux. The resulting braking force between the magnet and the copper block always opposes the motion of the magnet, slowing it as it falls. The rate of the fall produces a rate of flux change sufficient to produce a current that provides the braking force. If the copper is cooled with liquid nitrogen, the resistivity of the copper drops dramatically. How will this affect the speed at which the magnet falls toward the copper
Answer:
The speed at which the magnet falls through the copper block will be reduced dramatically.
Explanation:
Eddy's current are loops of electrical current induced within conductors by a changing magnetic field in the conductor. Eddy's current is proportional to the the magnetic field strength, the rate of change of flux, the area of the loop, and is inversely proportional to the resistivity of the material. Eddy currents flows perpendicularly to the magnetic field, and in closed loops within conductors.
Reducing the resistivity of the copper will increase the Eddy current on the copper, which will in turn increase the opposition to the action producing the flux change (the falling magnet through the copper block). The result is that the speed at which the magnet falls through the copper block will be reduced dramatically.
If the current flowing through a circuit of constant resistance is doubled, the power dissipated by that circuit will
A charge is placed on a spherical conductor of radius r1. This sphere is then connected to a distant sphere of radius r2 (not equal to r1) by a conducting wire. After the charges on the spheres are in equilibrium:__________.
1. the electric fields at the surfaces of the two spheres are equal.
2. the amount of charge on each sphere is q/2.
3. both spheres are at the same potential. the potentials are in the ratio V2/V1 = q2/q1.
4. the potentials are in the ratio V2/V1 = r2/r1 .
Answer:
Option 3 = both spheres are at the same potential.
Explanation:
So, let us complete or fill the missing gap in the question above;
" A charge is placed on a spherical conductor of radius r1. This sphere is then connected to a distant sphere of radius r2 (not equal to r1) by a conducting wire. After the charges on the spheres are in equilibrium BOTH SPHERES ARE AT THE SAME POTENTIAL"
The reason both spheres are at the same potential after the charges on the spheres are in equilibrium is given below:
=> So, if we take a look at the Question again, the kind of connection described in the question above (that is a charged sphere, say X is connected another charged sphere, say Y by a conducting wire) will eventually cause the movement of charges(which initially are not of the same potential) from X to Y and from Y to X and this will continue until both spheres are at the same potential.
An electron traveling with a speed v enters a uniform magnetic field directed perpendicular to its path. The electron travels for a time t0 along a half-circle of radius R before leaving the magnetic field traveling opposite the direction it initially entered the field. Which of the following quantities would change if the electron had entered the field with a speed 2v? (There may be more than one correct answer.)
a. The radius of the circular path the electron travels
b. The magnitude of the electron's acceleration inside the field
c. The time the electron is in the magnetic field
d. The magnitude of the net force acting on the electron inside the field
Answer:
Explanation:
For circular path in magnetic field
mv² / R = Bqv ,
m is mass , v is velocity , R is radius of circular path , B is magnetic field , q is charge on the particle .
a )
R = mv / Bq
If v is changed to 2v , keeping other factors unchanged , R will be doubled
b )
magnitude of acceleration inside field
= v² / R
= Bqv / m
As v is doubled , acceleration will also be doubled
c )
If T be the time inside the magnetic field
T = π R / v
= π / v x mv / Bq
= π m / Bq
As is does not contain v that means T remains unchanged .
d )
Net force acting on electron
= m v² / R = Bqv
Net force = Bqv
As v becomes twice force too becomes twice .
So a . b , d are correct answer.
In a polar coordinate system, the velocity vector can be written as . The term theta with dot on top is called _______________________ angular velocity transverse velocity radial velocity angular acceleration
Answer:
I believe it's called rapid growth
Explanation:
that is my answer no matter what
The length of your eye decreases slightly as you age, making the lens a bit closer to the retina. Suppose a man had his vision surgically corrected at age 30. At age 70, once his eyes had decreased slightly in length, what condition would he have?
A. Nearsightedness
B. Farsightedness
C. Neither nearsightedness nor farsightedness
Answer:
A. NearsightednessExplanation:
A nearsightedness is an eye defect that occurs when someone is only able to see close ranged object but not far distance object. According to the question, if the length of my eye decreases slightly as I age, this means there is a possibility that I will find it difficult to view a far distance object as I age.
At 70, once my eyes had decreased slightly in length, this means I will only be able to see close ranged object but not far distant object, showing that I am now suffering from nearsightedness according to its definition above.
Two people, one of mass 85 kg and the other of mass 50 kg, sit in a rowboat of mass 90 kg. With the boat initially at rest, the two people, who have been sitting at opposite ends of the boat, 3.5 m apart from each other, now exchange seats. How far does the boat move?
Answer:
0.11m
Explanation:
let's assume the boat is of uniform construction
Ignoring friction losses
Also assume the origin is at the end of the boat originally with the heavier person
the center of mass of the whole system will not change relative to the water when the two swap ends
Originally, the center of mass is
85[0] + 90[3.5/2] + 50[3.5] / (85 + 90 + 50) = 1.14m from the origin
after the swap, the center of mass is
50[0] + 90[3.5/2] + 85[3.5] / (85 + 90+ 50) = 1.030m from the origin
The center of mass has shifted
1.14-1.030 = 0.11m
as no external force acted on the system, the center of mass relative to the water will not change. The boat will therefore shift towards the end where the heavier person originally sat
A thin film with an index of refraction of 1.60 is placed in one of the beams of a Michelson interferometer. If this causes a shift of 8 bright fringes in the pattern produced by light of wavelength 580 nm, what is the thickness of the film
Answer:
3.867 μm
Explanation:
The index of refraction, μ = 1.6
Wavelength of the light, λ = 580 nm
N2 - N1 = (2L / λ) (n2 - n1), Making L subject of formula, we have
(N2 - N1) λ = 2L (n2 - n1)
L = [(N2 - N1) * λ] / 2(n2 - n1)
L = (8 * 580) / 2(1.6 - 1.0)
L = 4640 nm / 1.2
L = 3867 nm or 3.867 μm
Therefore we can come to the conclusion that the thickness of the film is 3.867 nm
Two spherical objects at the same altitude move with identical velocities and experience the same drag force at a time t. If Object 1 has twice (2x) the diameter of Object 2, which object has the larger drag coefficient? Explain your answer using the drag equation.
Answer:
The object with the twice the area of the other object, will have the larger drag coefficient.
Explanation:
The equation for drag force is given as
[tex]F_{D} = \frac{1}{2}pu^{2} C_{D} A[/tex]
where [tex]F_{D}[/tex] IS the drag force on the object
p = density of the fluid through which the object moves
u = relative velocity of the object through the fluid
p = density of the fluid
[tex]C_{D}[/tex] = coefficient of drag
A = area of the object
Note that [tex]C_{D}[/tex] is a dimensionless coefficient related to the object's geometry and taking into account both skin friction and form drag. The most interesting things is that it is dependent on the linear dimension, which means that it will vary directly with the change in diameter of the fluid
The above equation can also be broken down as
[tex]F_{D}[/tex] ∝ [tex]P_{D}[/tex] A
where [tex]P_{D}[/tex] is the pressure exerted by the fluid on the area A
Also note that [tex]P_{D}[/tex] = [tex]\frac{1}{2}pu^{2}[/tex]
which also clarifies that the drag force is approximately proportional to the abject's area.
In this case, the object with the twice the area of the other object, will have the larger drag coefficient.
. A 24-V battery is attached to a 3.0-mF capacitor and a 100-ohm resistor. If the capacitor is initially uncharged, what is the voltage across the capacitor 0.16 seconds after the circuit is connected to the battery
Answer:
The voltage is [tex]V_c = 9.92 \ V[/tex]
Explanation:
From the question we are told that
The voltage of the battery is [tex]V_b = 24 \ V[/tex]
The capacitance of the capacitor is [tex]C = 3.0 mF = 3.0 *10^{-3} \ F[/tex]
The resistance of the resistor is [tex]R = 100\ \Omega[/tex]
The time taken is [tex]t = 0.16 \ s[/tex]
Generally the voltage of a charging charging capacitor after time t is mathematically represented as
[tex]V_c = V_o (1 - e^{- \frac{t}{RC} })[/tex]
Here [tex]V_o[/tex] is the voltage of the capacitor when it is fully charged which in the case of this question is equivalent to the voltage of the battery so
[tex]V_c = 24 (1 - e^{- \frac{0.16}{100 * 3.0 *10^{-1}} })[/tex]
[tex]V_c = 9.92 \ V[/tex]
The number of neutrons in the nucleus of zinc 65 Zn 30 is:
35
Need more data to answer
65
30
Explanation:
proton number + neutron number = atomic mass
30 + 35 = 65
If the person knows that the monkey is going to drop from the tree at the same instant that the person launches the food, how should the person aim the arrow containing the food?
Answer:
The options are
A. He should aim it at the monkey
B. He should aim it below the monkey
C. He should aim it above the monkey
D. None of the above
The answer is A. He should aim it at the monkey
This is because the monkey has a large surface area and a bigger body mass. This will make aiming the food at the monkey feasible in it getting it as it could use other parts of the body to get the food aimed at it. The monkey won’t reach the food when falling if it is aimed above it. It also won’t get to the it when it is shot at below it.