Answer:
0.999958c
Explanation:
Remember that the trip involves traveling for six months at a constant velocity thus returning home at same speed . The time interval of this trip will therefore be one year.
Accurate time interval to be measured by the clock on the spaceship Δt0 = 1.0 years
Time interval as advanced on earth observed from the spacecraft Δt = 110 years
the formula for time dilation is
Δt = Δt0 /√(1-v2/c2)
or v = c*√1-(Δt0/Δt)2
=c*√[1-(1year/110years)2]
= 0.999958c
If enough heat was REMOVED from B, it would change into ____________.
The rho− meson has a charge of −e, a spin quantum number of 1, and a mass 1 507 times that of the electron. The possible values for its spin magnetic quantum number are −1, 0, and 1. Imagine that the electrons in atoms are replaced by rho− mesons. Select all of the following which are possible sets of quantum numbers (n, ℓ, mℓ, s, ms) for rho− mesons in the 3d subshell.
A. (2, 2, 1, 1, 0)
B. (3, 2, -1, 1, 1)
C. (3, 2. -1, 1, 1/2)
D. (3, 2, 0, 1, 1)
E. (3, 2, 0, 1, -1)
F. (3, 2, -1, 1, 0)
Answer:
Look up attached file
Explanation:
Two loudspeakers, 4.5 m apart and facing each other, play identical sounds of the same frequency. You stand halfway between them, where there is a maximum of sound intensity. Moving from this point toward one of the speakers, you encounter a minimum of sound intensity when you have moved 0.21 m . Assume the speed of sound is 340 m/s.
Part A
What is the frequency of the sound?
Part B
If the frequency is then increased while you remain 0.35m from the center, what is the first frequency for which that location will be a maximum of sound intensity?
Answer:
Explanation:
Given that,
Distance between speaker
L = 4.5m
Minimum intensity at L1 = 0.21m
Speed of sound is
V = 340m/s
A. Frequency of sound f?
The path difference Pd
Distance from the first speaker when you are 0.21m away
d1 = 2.25 + 0.21 = 2.46m
Distance from the second speaker when you move 0.21m closer
d2 = 2.25—0.21 = 2.04m
So, path difference is
Pd = ∆d = d1 — d2
Pd = 2.46—2.04 = 0.42m
Using the destructive interference condition
∆d = (m + ½)λ
m = 0,1,2,3....
When m= 0
∆d = ½λ
0.42 = ½λ
λ = 0.84
Then, using wave equation
v = fλ
Then, f = v / λ
f = 340 / 0.84
f = 404.76Hz
B. Incorrect question
If he is to remain at his initial positions then it is 0.21m from the center.
Then,
Constructive interference is given as
∆d = mλ
Where m = 0,1,2,3
So when m= 1
∆d = λ
And we already got the path difference to be 0.42m
So, ∆d = λ
λ = 0.42
So, applying wave equation
V = fλ
F = v/λ
F = 340/0.42
F = 809.52 Hz
But if we are to use the data given in part B
0.35m from the center..
Following the same principle as part A, the path difference will be 0.35
Therefore, since ∆d = λ
Then, λ = 0.35
So, f, = v/λ
F = 340 /0.35
F = 971.43 Hz
A straight wire carries a current of 10 A at an angle of 30° with respect to the direction of a uniform 0.30-T magnetic field. Find the magnitude of the magnetic force on a 0.50-m length of the wire.
Answer:
Explanation:
Given that,
Current in wire is
I = 10A
And the current makes an angle of 30° with respect to the magnetic field
Then, θ = 30°
And the magnetic field is
B = 0.3 T
Length of the wire is
L = 0.5m
Force on the wire F?
The force on the wire in calculated using
F = iL × B
Where
The magnitude of the cross produce of L and B is
L × B = LB•Sinθ
Then, force becomes
F = iLB•Sinθ
F = 10 × 0.5 × 0.3 × Sin30
F = 0.75 N
The force on the wire is 0.75 Newton
Answer:
The magnitude of the magnetic force on the wire is 0.75 N
Explanation:
Given;
current in the wire, I = 10 A
angle of inclination to magnetic field, θ = 30°
magnetic field strength, B = 0.30-T
length of the wire, L = 0.50-m
The magnitude of magnetic force acting on the wire is given as;
F = BILSinθ
where;
B is magnetic field strength
I is the current in the wire
L is length of the wire
θ is the angle of inclination
F = 0.3 x 10 x 0.5 x sin(30)
F = 0.3 x 10 x 0.5 x 0.5
F = 0.75 N
Therefore, the magnitude of the magnetic force on the wire is 0.75 N
A 0.62112 slug uniform disc A is pinned to a 0.12422 slug uniform rod CB which is pinned to a 0.03106 slug collar C. C slides on a smooth vertical rod and the disc rolls without slipping. Determine: a) the angular acceleration ofA and the acceleration of C when released from rest and b)The velocity of B and C when the rod is horizontal. Ans:b) vc
Answer:
Explanation: check the attached document for step by step solution.
Like a transverse wave, a longitudinal wave has
-wavelength, speed, and frequency.
-amplitude, frequency, wavelength, and speed.
-amplitude, frequency, and speed.
-amplitude, wavelength, and speed.
-amplitude, frequency, and wavelength.
Final answer:
A longitudinal wave, like a transverse wave, has properties of amplitude, frequency, wavelength, and speed. These characteristics define the wave's physical behavior and are related by the equation v = fλ, where 'v' is wave speed, 'f' is frequency, and 'λ' is wavelength. Option 2 is correct.
Explanation:
Like a transverse wave, a longitudinal wave has amplitude, frequency, wavelength, and speed. Both types of waves have these fundamental properties, but the way they propagate through mediums is different. In longitudinal waves, the particles of the medium move parallel to the wave's direction of travel, while in transverse waves, particles move perpendicular to the direction of the wave's travel. An example of a transverse wave is a wave on a string, like when playing a guitar. In contrast, sound waves in air are longitudinal waves.
The wavelength (λ) is the distance between adjacent identical parts of the wave, which can be considered from one compression to the next in the case of longitudinal waves. Wave speed (v) is the rate at which the wave travels through the medium. The frequency (f) is the number of wave cycles that pass a given point per unit time, and amplitude refers to the maximum displacement from the equilibrium position within the wave cycle.
It's important to remember that all these properties are related: the speed of a wave is given by the product of its frequency and wavelength (v = fλ).
How did Kepler’s discoveries contribute to astronomy?
Answer:They established the laws of planetary motion.(option B)
Answer:
They established the laws of planetary motion
Explanation:
i did it on edge
An object of mass m attached to a spring of force constant k oscillates with simple harmonic motion. The maximum displacement from equilibrium is A and the total mechanical energy of the system is E. Part A What is the system's potential energy when its kinetic energy is equal to 34E
The correct question is;
An object of mass m attached to a spring of force constant K oscillates with simple harmonic motion. The maximum displacement from equilibrium is A and the total mechanical energy of the system is E.
What is the system's potential energy when its kinetic energy is equal to ¾E?
Answer:
P.E = ⅛KA²
Explanation:
From conservation of energy, the total energy in the system is given as the sum of potential and kinetic energy.
Thus,
Total Energy; E = K.E.+P.E.
In simple harmonic motion, the total energy is given by;
E = ½KA²
We are told that kinetic energy is ¾E.
Thus, ½KA² = ¾(½KA²) + P.E
P.E = ½KA² - ⅜KA²
P.E = ⅛KA²
You build a grandfather clock, whose timing is based on a pendulum. You measure its period to be 2s on Earth. You then travel with the clock to the distant planet CornTeen and measure the period of the clock to be 4s. By what factor is the gravitational acceleration constant g different on planet CornTeen compared to g on Earth
Answer:
[tex]\frac{g_{2}}{g_{1}} = \frac{1}{4}[/tex]
Explanation:
The period of the simple pendulum is:
[tex]T = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex]
Where:
[tex]l[/tex] - Cord length, in m.
[tex]g[/tex] - Gravity constant, in [tex]\frac{m}{s^{2}}[/tex].
Given that the same pendulum is test on each planet, the following relation is formed:
[tex]T_{1}^{2}\cdot g_{1} = T_{2}^{2}\cdot g_{2}[/tex]
The ratio of the gravitational constant on planet CornTeen to the gravitational constant on planet Earth is:
[tex]\frac{g_{2}}{g_{1}} = \left(\frac{T_{1}}{T_{2}} \right)^{2}[/tex]
[tex]\frac{g_{2}}{g_{1}} = \left(\frac{2\,s}{4\,s} \right)^{2}[/tex]
[tex]\frac{g_{2}}{g_{1}} = \frac{1}{4}[/tex]
One end of a string is attached to a rigid wall on a tabletop. The string is run over a frictionless pulley and the other end of the string is attached to a stationary hanging mass. The distance between the wall and the pulley is 0.405 meters, When the mass on the hook is 25.4 kg, the horizontal portion of the string oscillates with a fundamental frequency of 261.6 Hz (the same frequency as the middle C note on a piano). Calculate the linear mass density of the string.
Answer:
The linear mass density is of the string [tex]\mu= 5.51*10^{-3} kg / m[/tex]
Explanation:
From the question we are told that
The distance between wall and pulley is [tex]d = 0.405m[/tex]
The mass on the hook is [tex]m = 25.4\ kg[/tex]
The frequency of oscillation is [tex]f = 261.6 Hz[/tex]
Generally, the frequency of oscillation is mathematically represented as
[tex]f = \frac{1}{2d} \sqrt{\frac{T}{\mu} }[/tex]
Where T is the tension mathematically represented as
T = mg
Substituting values
[tex]T = 25.4 *9.8[/tex]
[tex]=248.92N[/tex]
[tex]\mu[/tex] is the mass linear density
Making [tex]\mu[/tex] the subject of the formula above
[tex]\mu = \frac{T}{(2df)^2}[/tex]
Substituting values
[tex]\mu = \frac{248.92}{(2 * 0.405 * 261.6)^2}[/tex]
[tex]\mu= 5.51*10^{-3} kg / m[/tex]
Answer:
0.005550 Kg/m
Explanation:
The picture attached below shows the full explanation
A proton and a deuteron are moving with equal velocities perpendicular to a uniform magnetic field. A deuteron has the same charge as the proton but has twice its mass. The ratio of the magnetic force on the proton to that on the deuteron is:
a. 0.5.
b. 1.
c. 2.
d. There is no magnetic force in this case.
Answer:
option (b)
Explanation:
mass of proton, mp = m
mass of deuteron, md = 2m
charge on proton, qp = q
charge on deuteron, qd = q
The magnetic force on the charged particle when it is moving is given by
F = q v B Sinθ
where, θ is the angle between the velocity and magnetic field.
Here, θ = 90°
Let v is the velocity of both the particle when they enters in the magnetic field.
The force on proton is given by
Fp = q x v x B ...... (1)
The force on deuteron is
Fd = q x v x B .... (2)
Divide equation (1) by equation (2)
Fp / Fd = 1
Thus, the ratio of force on proton to the force on deuteron is 1 : 1.
Thus, option (b) is correct.
Which one of the following statements concerning the proper time interval between two events is true? a) The proper time interval is the longest time interval that any inertial observer can measure for the event. b) The proper time interval is the shortest time interval that any inertial observer can measure for the event. c) The proper time interval is the time measured by an observer who is in motion with respect to the event. d) The proper time interval depends upon the speed of the observer. e) The proper time interval depends upon the choice of reference frame.
Answer:
d) The proper time interval depends upon the speed of the observer
Explanation:
Proper time is the time as measured by a clock moving with the body in motion.
The proper time interval between two events on a world line is the change in proper time, and it depends on the events and the world line connecting them, and also on the motion of the clock between the events.
A transparent oil with index of refraction 1.28 spills on the surface of water (index of refraction 1.33), producing a maximum of reflection with normally incident orange light (wavelength 600 nm in air). Assuming the maximum occurs in the first order, determine the thickness of the oil slick. nm
To solve this problem we will apply the concepts related to the principle of destructive and constructive interference. Mathematically this expression can be given as
[tex]2nt = m\lambda[/tex]
Here,
n = Index of refraction
t = Thickness
m = Order of the reflection
[tex]\lambda[/tex] = Wavelength
We have all of this values, therefore replacing,
[tex]2(1.28)t = (1)(600nm)[/tex]
[tex]t = 233nm[/tex]
Therefore the thickness of the oil slick is 233nm
Two long, parallel wires are attracted to each other by a force per unit length of 305 µN/m. One wire carries a current of 25.0 A to the right and is located along the line y = 0.470 m. The second wire lies along the x axis. Determine the value of y for the line in the plane of the two wires along which the total magnetic field is zero.
To solve this problem we will use the concepts related to the electromagnetic force related to the bases founded by Coulumb, the mathematical expression is the following as a function of force per unit area:
[tex]\frac{F}{L} = \frac{kl_1l_2}{d}[/tex]
Here,
F = Force
L = Length
k = Coulomb constant
I =Each current
d = Distance
Force of the wire one which is located along the line y to 0.47m is [tex]305*10^{-6}N/m[/tex] then we have
[tex]l_2 = \frac{F}{L} (\frac{d}{kl_1})[/tex]
[tex]l_2 = (305*10^{-6}N/m)(\frac{0.470m}{(2*10^{-7})(25A))})[/tex]
[tex]l_2 = 28.67A[/tex]
Considering the B is zero at
[tex]y = y_1[/tex]
[tex]\frac{kI_2}{2\pi y} =\frac{kI_1}{2\pi y_1}[/tex]
[tex]\frac{(4\pi*10^{-7})(28.67)}{2\pi (y_1)} = \frac{(4\pi *10^{-7})(25)}{2\pi (0.47-y_1)}[/tex]
[tex]y_1 = 0.25m[/tex]
Therefore the value of y for the line in the plane of the two wires along which the total B is zero is 0.25m
The magnetic field at the center of a wire loop of radius , which carries current , is 1 mT in the direction (arrows along the wire represent the direction of current). For the following wires, which all also carry current , indicate the magnitude (in mT) and direction of the magnetic field at the center (red point) of each configuration.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The magnetic field is [tex]B_{net} = \frac{1}{4} * mT[/tex]
And the direction is [tex]-\r k[/tex]
Explanation:
From the question we are told that
The magnetic field at the center is [tex]B = 1mT[/tex]
Generally magnetic field is mathematically represented as
[tex]B = \frac{\mu_o I}{2R}[/tex]
We are told that it is equal to 1mT
So
[tex]B = \frac{\mu_o I}{2R} = 1mT[/tex]
From the first diagram we see that the effect of the current flowing in the circular loop is (i.e the magnetic field generated)
[tex]\frac{\mu_o I}{2R} = 1mT[/tex]
This implies that the effect of a current flowing in the smaller semi-circular loop is (i.e the magnetic field generated)
[tex]B_1 = \frac{1}{2} \frac{\mu_o I}{2R}[/tex]
and for the larger semi-circular loop is
[tex]B_2 = \frac{1}{2} \frac{\mu_o I}{2 * (2R)}[/tex]
Now a closer look at the second diagram will show us that the current in the semi-circular loop are moving in the opposite direction
So the net magnetic field would be
[tex]B_{net} = B_1 - B_2[/tex]
[tex]= \frac{1}{2} \frac{\mu_o I}{2R} -\frac{1}{2} \frac{\mu_o I}{2 * (2R)}[/tex]
[tex]=\frac{\mu_o I}{4R} -\frac{\mu_o I }{8R}[/tex]
[tex]=\frac{\mu_o I}{8R}[/tex]
[tex]= \frac{1}{4} \frac{\mu_o I}{2R}[/tex]
Recall [tex]\frac{\mu_o I}{2R} = 1mT[/tex]
So
[tex]B_{net} = \frac{1}{4} * mT[/tex]
Using the Right-hand rule we see that the direction is into the page which is [tex]-k[/tex]
A girl on a bike is moving at a speed of 1.40 m/s at the start of a 2.45 m high and 12.4 m long incline. The total mass is 60.0 kg, air resistance and rolling resistance can be modeled as a constant friction force of 41.0 N, and the speed at the lower end of the incline is 6.70 m/s. Determine the work done (in J) by the girl as the bike travels down the incline.
Answer:
Explanation:
Given that,
Initial speed of the girl is
u = 1.4m/s
Height she is going is
H = 2.45m
Incline plane she will pass to that height
L = 12.4m
Mass of girl and bicycle is
M=60kg
Frictional force that oppose motion is
Fr = 41N
Speed at lower end of inclined plane
V2 = 6.7m/s
Work done by the girl when the car travel downward
Using conservation of energy
K.E(top) + P.E(top) + work = K.E(bottom) + P.E(bottom) + Wfr
Where Wfr is work done by friction
Wfr = Fr × d
P.E(bottom) is zero, sicne the height is zero at the ground
K.E is given as ½mv²
Then,
½M•u² + MgH + W = ½M•V2² + 0 + Fr×d
½ × 60 × 1.4² + 60×9.8 × 2.45 + W = ½ × 60 × 6.7² + 41 × 12.4
58.8 + 1440.5 + W = 1855.1
W = 1885.1 —58.8 —1440.5
W = 355.8 J
An object of mass M is dropped near the surface of Earth such that the gravitational field provides a constant downward force on the object. Which of the following describes what happens to the center of mass of the object-Earth system as the object falls downward toward Earth? a. It moves toward the center of Earth. b. It moves toward the object.c. It does not move. d. The answer cannot be determined without knowing the mass of Earth and the distance between the object and Earth’s center.
Answer:
The answer is: c. It does not move
Explanation:
Because the gravitational force is characterized by being an internal force within the Earth-particle system, in this case, the object of mass M. And since in this system there is no external force in the system, it can be concluded that the center of mass of the system will not move.
As the object is dropped at a constant downward force, the center mass of the object-Earth system does not move.
In the absence of an external force, the center mass of the object-Earth system will remain constant as the object fall to the ground.
The force of attraction on the object above the surface of the Earth is given as;
[tex]F = \frac{Gm_1 m_2}{R^2}[/tex]
where;
G is gravitational constantm is massR is the distance of the object from the center of the earthThus, as the object is dropped at a constant downward force, the center mass of the object-Earth system does not move.
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Why might these "Mental Maps" be inaccurate or differ between different people?
Answer:
Catastrophic events of weather related outcomes.
Explanation:
The mental map of a person from a certain place may change due to long periods of time outside, since this gives the mind time to forget certain important details, they may also change according to people's experience and perception, places, regions and environments since these places are also changing and the perception we had is no longer the same. A mental map is a first person perspective of an area that an individual possesses. This type of subconscious map shows a person how a place looks and how to interact with it.
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Sketch both, the time domain AM signal and its frequency spectrum and explain what you see in terms of themodulation property of the Fourier transform.(Hint: How is the frequency spectrum of the message signal(co-sinusoid of 880 Hz) plus a DC component in base band, i.e. before modulation?)
Answer:
see the attachment
Explanation:
In frequency spectrum there is no change in value of frequency of signal. However, The amplitude of signal after modulation increases. The fourier transform of sinwct is
1/2(F(w+wc) - F(w-wc))
For 2sinwct, the fourier transform is,
(F(w+wc) - F(w-wc))
In frequency domain, in AM only amplitude changes. Frequency remains same
A 150kg person stands on a compression spring with spring constant 10000n/m and nominal length of 0.50.what is the total length of the loaded spring
Answer:
The total length of the spring would be 0.65 m
Explanation:
The Concept
Hooke's law evaluates the increment of spring in relation to the force acting on the body. Hooke's law states that for a spring undergoing deformation, the force applied is directly proportional to the deformation experienced by the spring. Hooke's law is represented thus;
F = k x ..................1
where F is the force applied to the spring
k is the spring constant
x is the spring stretch or extension
Step by Step Calculations
We have to obtain x before adding it to the nominal length, We make x the subject formula in equation 1;
x = F/k
but F = m x g
so, x = (m x g)/k
given that, the mass of the person m =150 kg
g is the acceleration due to gravity = 9.81 m/[tex]s^{2}[/tex]
k is the spring constant = 10000 N/m
then x = (9.81 m/[tex]s^{2}[/tex] x 150 kg)/10000 N/m
x = 0.14715 m
the extension experienced by the spring after the compression is 0.14715 m
The total length of the spring would be;
L = 0.14715 m + 0.5 m = 0.64715
L ≈ 0.65 m
Therefore the total length of the spring would be 0.65 m
Electric potential is associated with both electric fields due to static charges and induced electric fields. Electric potential is associated with magnetic fields but not with electric fields due to static charges. Electric potential is associated with both electric fields due to static charges and magnetic fields. Electric potential is associated with electric fields due to static charges but not with induced electric fields. Electric potential is associated with induced electric fields but not with electric fields due to static charges.
Answer:
Electric potential is associated with electric fields due to static charges but not with induced electric fields.
Explanation:
An electric potential is the amount of work needed to move a unit charge from a reference point to a specific point inside the field without producing an acceleration. Typically, the reference point is the Earth or a point at infinity. Induced electricity is as a result of changing magnetic flux linkage (no charge is involved)
Electric potential is related to the electrostatic field created by static charges and is given by V = kQ/r; it is not related to induced electric fields or magnetic fields because they are nonconservative.
Electric potential is associated with electric fields due to static charges but not with induced electric fields.
This is because electric potential is defined for conservative fields, such as those produced by static charges, where the potential difference is related to the work done in moving a charge between two points in the field.
For point charges, the electric potential is given by the equation V = kQ/r, which clearly illustrates that electric potential depends on the charge and the distance from it.
On the other hand, induced electric fields are produced by changing magnetic fields and are nonconservative.
Nonconservative fields do not have a well-defined electric potential because the work done to move a charge can vary depending on the path taken, unlike the work done in a conservative electrostatic field.
Therefore, electric potential cannot be associated with induced electric fields or magnetic fields.
Because all blocks weigh the same, their masses must also be equal. Density is defined as mass per unit volume, so we know the densities of the red and blue blocks are different because their volumes are not the same. Since the volume of a blue block is one-half the volume of a red block, the density of a blue block is ________ the density of a red block.
a.half
b.equal to
Answer:
the density of a blue block is ___twice _____ the density of a red block.
Explanation:
If both blocks have the same mass, then the block with the lesser volume will be more densely packed when compared to the block with the larger volume. This is because the molecules are more closely packed.
Check the image below for detailed calculations for proof
If the blue block and red block have the same mass but different volumes - specifically, the blue block's volume is half that of the red block - then the density of the blue block is twice that of the red block.
Explanation:Given the blocks weigh the same, their masses are equal. Density is defined as mass per unit volume (mass/volume), meaning the density is influenced by both the mass and the volume of an object. As the red and blue blocks have different volumes, their densities must be different if their masses are the same.
In the case of the blue block and the red block, the blue block has half the volume of the red block but the same mass. Thus, because the blue block's volume is half that of the red block but the mass is the same, the density of the blue block is twice the density of a red block, not half. This is because the proportion of mass to volume in the blue block is larger than in the red block. In other words: Density of blue block = 2 * Density of red block.
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g If 6.35 moles of a monatomic ideal gas at a temperature of 320 K are expanded isothermally from a volume of 1.45 L to a volume of 3.95 L, calculate: a) the change in the internal energy of the gas. b) the work done by the gas. c) the heat flow into or out of the gas.
Answer:
(a) change in the internal energy of the gas is zero
(b) the work done by the gas is 16.93 kJ
(c) the heat flow is 16.93 kJ, which is into the gas
Explanation:
Given;
number of moles of gas, n = 6.35 moles
temperature of the gas, T = 320 K
initial volume of the gas, V₁ = 1.45 L
final volume of the gas, V₂ = 3.95 L
Part (a)
For isothermal expansion, temperature is constant and internal energy will also be constant.
Therefore, change in the internal energy of the gas is zero since the gas expanded isothermally (constant temperature).
ΔU = Q - W
where;
ΔU is change in internal energy
Q is heat transferred to the system
W is the work done by the system
Thus, Q = W
ΔU = 0
Part (b)
the work done by the gas
[tex]W = nRTln{[\frac{V_2}{V_1}][/tex]
where;
R is gas constant = 8.314 J/mol.K
[tex]W = (6.35)(8.314)(320)ln{[\frac{3.95}{1.45}]}\\\\W =16930.4\ J\\\\W = 16.93\ kJ[/tex]
Part (c)
the heat flow into or out of the gas
Q = ΔU + W
Q = 0 + 16.93 kJ
Q = 16.93 kJ
Since the heat flow is positive, then it is heat flow into the gas.
Which of the following devices can be used to measure force?
A. Bathroom scale
(b) spring scale
c. Force sensor
d. All of the above.
Answer:
Explanation:
d.All of the above
Final answer:
All the listed devices -- bathroom scale, spring scale, and force sensor -- can be used to measure force. Bathroom and spring scales measure the weight of an object, which is a force, and display it typically in kilograms after calibration while force sensors provide direct measurement in newtons.
Explanation:
Devices that can be used to measure force include a bathroom scale, a spring scale, and a force sensor. All of these devices measure force, which can be the weight of an object (the force due to gravity acting on the object). A bathroom scale, for example, measures the normal force exerted by a person standing on it and gives a reading in kilograms by dividing the force in newtons by the acceleration due to gravity (9.80 m/s2). However, it is calibrated to display mass. A spring scale uses the extension of a spring under load to measure force which is typically indicated in newtons or pounds. Lastly, a force sensor is a more general device that can directly measure the force exerted on it in newtons and is often used for more precise scientific measurements.
If you stood on a bathroom scale in an elevator, the reading would change depending on the elevator's motion. The scale would display a higher value when the elevator starts moving upward (accelerating) due to the increase in the normal force. However, when the elevator moves at a constant speed, the scale would read your normal weight as no additional normal force is required once you're moving at constant velocity.
A 3.0-Ω resistor is connected in parallel with a 6.0-Ω resistor. This combination is then connected in series with a 4.0-Ω resistor. The resistors are connected across an ideal 12-volt battery. How much power is dissipated in the 3.0-Ω resistor? Group of answer choices
a. 12 w
b. 2.7 w
c. 6.0 w
d. 5.3 w
To solve this problem we must find the values of the equivalent resistances in both section 1 and section 2. Later we will calculate the total current and the total voltage. With the established values we can find the values of the currents in the 3 Ohms resistance and the power there.
The equivalent resistance in section 1 would be
[tex]R_{eq1} = \frac{(3\Omega)(6\Omega)}{(3+6)\Omega}[/tex]
[tex]R_{eq1} = 2\Omega[/tex]
The equivalent resistance in section 2 would be
[tex]R_{eq2} = R_{eq1} +4\Omega[/tex]
[tex]R_{eq2} = 6\Omega[/tex]
Now the total current will be,
[tex]I_t = \frac{V_t}{R_{eq2}}[/tex]
[tex]I_t = \frac{12V}{6\Omega}[/tex]
[tex]I_t = 2.0A[/tex]
Finally the total Voltage will be,
[tex]V = IR_{eq1}[/tex]
[tex]V = (2.0A)(2.0\Omega)[/tex]
[tex]V = 4V[/tex]
Since the voltage across the 3 and 6 Ohms resistor is the same, because they are in parallel, the current in section 3 would be
[tex]I_{3.0\Omega} = \frac{V}{R}[/tex]
[tex]I_{3.0\Omega} = \frac{4.0V}{3.0\Omega}[/tex]
[tex]I_{3.0\Omega} = 1.3A[/tex]
Finally the power ratio is the product between the current and the voltage then,
[tex]P_{3.0\Omega} = I_{3.0\Omega} V[/tex]
[tex]P_{3.0\Omega} = (1.3A)(4.0V)[/tex]
[tex]P_{3.0\Omega} = 5.3W[/tex]
Therefore the correct answer is D.
Final answer:
To find the power dissipated in the 3.0-Ω resistor, calculate the total resistance and current, then apply Ohm's law to find the voltage across and current through the resistor. Finally, use the power formula P = V²/R, resulting in a power dissipation of (a) 12 W for the 3.0-Ω resistor.
Explanation:
The question deals with finding the power dissipated in the 3.0-Ω resistor. To answer this, we first need to find the total resistance of the circuit and the current through the circuit. We then apply this current to the parallel resistors to find the voltage across and the current through the 3.0-Ω resistor before calculating its power dissipation.
First, calculate the resistance of the resistors in parallel. Using the formula for resistors in parallel:
1/Rparallel = 1/R₁ + 1/R₂
1/Rparallel = 1/3.0 + 1/6.0
1/Rparallel = 1/3 + 1/6 = 2/6 + 1/6 = 3/6
1/Rparallel = 1/2
Rparallel = 2.0 Ω
Now, add the series resistor to find the total resistance:
Rtotal = Rparallel + Rseries
Rtotal = 2.0 + 4.0 = 6.0 Ω
Then, using Ohm's law, calculate the total current from the battery:
I = V/Rtotal
I = 12V/6.0Ω
I = 2.0 A
The current through the 3.0-Ω resistor in parallel is the same as the total current, so we use Ohm's law V = IR to find the voltage across the 3.0-Ω resistor:
V3.0-Ω = 2.0 A × 3.0 Ω = 6.0 V
Finally, calculate the power dissipated by the 3.0-Ω resistor:
P = V2/R
P = (6.0 V)2/3.0 Ω
P = 36 W/3.0 Ω
P = 12 W
Therefore, the power dissipated in the 3.0-Ω resistor is 12 W, which corresponds to choice (a).
The pressure exerted by a phonograph needle on a record is surprisingly large, due to the very small width of the needle. show answer Incorrect Answer If the equivalent of 0.95 g is supported by a needle, the tip of which is a circle 0.205 mm in radius, what pressure is exerted on the record, in pascals
Answer:
Explanation:
Given that,
Mass support is
M = 0.95g= 0.95/1000 = 0.00095kg
Radius of circle R = 0.205mm
r = 0.205/1000 = 0.000205m
Then, area of the circle can be determined using
A = πr²
A = π × 0.000205²
A = 1.32 × 10^-7 m²
From pressure definitions
Pressure = Force / Area
The force is perpendicular to the area
Force = weight = mg
F = mg = 0.00095 × 9.8
F = 9.31 × 10^-3 N
Then,
Pressure = Force / Area
P = F/A
P = 9.31 × 10^-3 / 1.32 × 10^-7
P = 70,530.30 N/m²
Since 1 pascal = 1 N/m²
Then,
P = 70,530.30 Pascals
To find the pressure exerted by the phonograph needle, multiply the weight of the needle by the gravitational constant to get the force, calculate the area of the needle tip using the given radius, and divide the force by the area using the pressure formula P = F / A.
Explanation:The pressure exerted by the needle on the record can be calculated using the formula for pressure P = F / A, where F is the force (in this case the weight of the needle) and A is the area over which the force is applied (in this case the area of the needle tip).
First, you need to convert the weight of the needle into a force. Since weight is a force caused by gravity acting on a mass, you can find it by multiplying the mass of the needle by the acceleration due to gravity (g = 9.8 m/s²). So, F = 0.95 g * 9.8 m/s². However, you need to convert grams to kilograms (as 1g = 0.001kg). Hence, F = 0.95 * 0.001 kg * 9.8 m/s².
Next, you find the area of the very tip of the needle. Since the tip is circular, we use the formula for the area of a circle, A = πr², where r is the radius of the needle tip. Substituting r = 0.205 mm = 0.205 * 10^-3 m (since 1mm = 10^-3m) into the formula will give you the area.
Finally, substitute F and A into the formula P = F / A to find the pressure.
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Light from an LED with a wavelength of 4.90 ✕ 102 nm is incident on (and perpendicular to) a pair of slits separated by 0.310 mm. An interference pattern is formed on a screen 2.20 m from the slits. Find the distance (in mm) between the first and second dark fringes of the interference pattern.
Answer:
Δx = 3.477 x 10⁻³ m = 3.477 mm
Explanation:
The distance between two consecutive dark fringes is given by the following formula, in Young's Double Slit experiment:
Δx = λL/d
where,
Δx = distance between two consecutive dark fringes = ?
λ = wavelength of light = 4.9 x 10² nm = 4.9 x 10⁻⁷ m
L = Distance between slits and screen = 2.2 m
d = slit separation = 0.31 mm = 0.31 x 10⁻³ m
Therefore,
Δx = (4.9 x 10⁻⁷ m)(2.2 m)/(0.31 x 10⁻³ m)
Δx = 3.477 x 10⁻³ m = 3.477 mm
The distance (in mm) between the first and second dark fringes of the
interference pattern is 3.477 mm
This is calculated by using the formula in Young's Double Slit experiment:
Δx = λL/d
where,
Δx = distance between two consecutive dark fringes which is unknown
λ which is wavelength of light = 4.9 x 10² nm = 4.9 x 10⁻⁷ m
L which is distance between slits and screen = 2.2 m
d which is slit separation = 0.31 mm = 0.31 x 10⁻³ m
We then substitute them into the equation
Δx = (4.9 x 10⁻⁷ m ×2.2 m)/(0.31 × 10⁻³ m)
Δx = 3.477 x 10⁻³ m = 3.477 mm
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Lightning bolts can carry currents up to approximately 20 kA. We can model such a current as the equivalent of a very long, straight wire. If you were unfortunate enough to be 4.9 m away from such a lightning bolt, how large a magnetic field would you experience
Answer:
how large a magnetic field would you experience = 8.16 x 10∧-4T
Explanation:
I = 20KA = 20,000A
r = 4.9 m
how large a magnetic field would you experience = u.I/2πr
how large a magnetic field would you experience = (4π x10∧-7) × 20000/2π × 4.9
how large a magnetic field would you experience = 8.16 x 10∧-4T
Answer: 8.16*10^-4 T
Explanation:
Given
Current of the lightening bolt, I = 20 kA
Distance from the strike of the lightening bolt, r = 4.9 m
To solve, we use the formula
B = [μ(0) * I] / 2πr, where
B = magnetic field of the lightening
μ = permeability constant = 4π*10^-7 N/A²
I = current of the lightening
r = distance from the lightening strike
B = [(4 * 3.142*10^-7) * 20*10^3] / (2 * 3.142 * 4.9)
B = (12.568*10^-7 * 20*10^3) / 6.284 * 4.9
B = 0.025 / 30.79
B = 8.16*10^-4 T
The magnetic field to be experienced would be 8.16*10^-4 T large
A vertical wire carries current in the upward direction. An electron is traveling parallel to the wire. What is the angle ααalpha between the velocity of the electron and the magnetic field of the wire? Express your answer in degrees. View Available Hint(s) ααalpha = nothing degreesdegrees
Answer:
First of all note that The magnetic field produced by the vertical wire will be into on the right hand side and it will be out of the page on the left hand side
Assuming that the electron beam is coming from the right hand side of the page parallel to the wire, The direction of the velocity vector(V) is left and the direction of magnetic field due the wire(B) is into the page. If u use right hand rule, you will get the direction downwards but as the formula also depends on q , the charge on electron is negative .Therefore the direction will be inverted i.e Upwards.
If you assume the electron beam coming from left hand side.Then also u will get the same answer.
So, the angle α between the velocity of the electron and the magnetic field of the wire is 90°.
Explanation:
For an electron traveling parallel to a wire carrying an upward direction current, the angle between its velocity and the magnetic field of the wire, according to the right-hand rule, is 90 degrees.
Explanation:The setting of this problem involves a vertical wire carrying an upward direction current and an electron traveling parallel to it. According to the right-hand rule in magnetism, which states that if your thumb points in the direction of the current, then your fingers will curl in the direction of the magnetic field, the magnetic field of the wire would form concentric circles around the wire.
For an electron traveling parallel to the wire, according to this rule, it would be always at a right angle or 90 degrees to the magnetic field as it moves along the circumference of these imaginary concentric circles. Therefore, the angle alpha (ααalpha) between the velocity of the electron and the magnetic field of the wire is 90 degrees.
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A string of 18 identical Christmas tree lights are connected in series to a 130 V source. The string dissipates 61 W.
What is the equivalent resistance of the light string?
Answer in units of Ω.
What is the resistance of a single light? Answer in units of Ω.
How much power is dissipated in a single light?
Answer in units of W.
One of the bulbs quits burning. The string has a wire that shorts out the bulb filament when it quits burning, dropping the resistance of that bulb to zero. All the rest of the bulbs remain burning.
What is the resistance of the light string now?
Answer in units of Ω.
How much power is dissipated by the string now?
Answer in units of W.
Answer:
(a) 277.05 Ω
(b) 15.39 Ω
(c) 3.76 W
Explanation:
(a)
Applying,
P = V²/R.......................... Equation 1
Where P = Power dissipated by the string. V = Voltage source, R = equivalent resistance of the light string
Make R the subject of the equation
R = V²/P................... Equation 2
Given: V = 130, P = 61 W
Substitute into equation 2
R = 130²/61
R = 277.05 Ω
(b) The resistance of a single light is given as
R' = R/18 (since the light are connected in series and the are identical)
Where R' = Resistance of the single light.
R' = 277.05/18
R' = 15.39 Ω
(c)
Heat dissipated in a single light is given as
P' = I²R'..................... Equation 3
Where P' = heat dissipated in a single light, I = current flowing through each light.
We can calculate for I using
P = VI
make I the subject of the equation
I = P/V
I = 61/130
I = 0.469 A.
Also given: R' = 15.39 Ω
Substitute into equation 3
P' = 0.496²(15.39)
P' = 3.76 W
(a)The equivalent resistance of the light string is 277.05 Ω
(b)The power is dissipated in a single ligh15.39 Ω
(c)The resistance of the light string now3.76 W
Calculation of Power is dissipates
(a) P is = V²/R.......................... Equation 1
Where P is = Power dissipated by the string.
Then V = Voltage source,
After that R is = the equivalent resistance of the light string
Now Make R the subject of the equation
R is = V²/P................... Equation 2
Then Given: V = 130,
P = 61 W
After that Substitute into equation 2
Then R = 130²/61
Therefore, R = 277.05 Ω
(b) When The resistance of a single light is given as
R' is = R/18 (since the light are connected in series and are identical)
Now Where R' is = Resistance of the single light.
R' is = 277.05/18
Therefore, R' is = 15.39 Ω
(c) When Heat dissipated in a single light is given as
P' is = I²R'..................... Equation 3
Where P' is = heat dissipated in a single light,
Then I = current flowing through each light.
Now We can calculate for I using
P is = VI
Now we make I the subject of the equation
After that I = P/V
Then I = 61/130
I is = 0.469 A.
Also given: R' is = 15.39 Ω
Then Substitute into equation 3
P' is = 0.496²(15.39)
Therefore, P' is = 3.76 W
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