First, we calculate the current through the cable using given Power and Voltage. Then, we use this current to find the resistance and subsequently, the diameter of the cable. For Part B, the electric field inside a cable is zero in steady state conditions.
Explanation:To solve Part A of the question, we will use the formula Power = I2R, where I is current and R is resistance. First, we find the Current from the given power and voltage, Power = Voltage * Current. Second, we use this current to calculate the resistance, R = Power / I2. Lastly, we find the diameter using the formula for the Resistance of a cylindrical conductor, R = rho * L / A, where rho is resistivity, L is length and A is area. For Part B, the electric field inside a conductor under steady state conditions is zero.
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The diameter of the copper wire can be determined by solving an equation using its cross-sectional area. The electric field inside the copper wire is calculated by dividing the potential difference by the distance.
Explanation:The question pertains to calculating the diameter of a cylindrical copper cable and the electric field inside it given certain conditions. This involves the use of formulas related to the principles of electromagnetism, and specifically electric current and electric field.
In order to determine the diameter of the cable (Part A), we would first need to deduce the cross-sectional area using data provided. The formula AD² = 4 could be used, where A represents the cross-sectional area and D the diameter of the cable. Substituting the value found for area, we can solve for the diameter.
For Part B, the electric field inside the cable can be calculated using the formula E = V/d, where V is the potential difference and d is the distance. Given that the potential difference is 220V and the length or distance is 3.3 km, the desired electric field can be computed.
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An earthquake on the ocean floor in the Gulf of Alaska induces a tsunami (sometimes called a "tidal wave") that reaches Hilo, Hawaii, 4780 km away, after a time of 9.99 h. Tsunami’s have enormous wavelengths (100- 200 km), and for such waves the propagation speed is v ≈ qgd , where d is the average depth of the water, d is in meters, v is in m/s, and g = 9.8 m/s^2 . From the information given, find the average ocean depth between Alaska and Hawaii.
Answer:
23785m
Explanation:
V= 4780/9.9= 482.8km/h
d= v²/g = 482.8²/9.8= 23785m
To understand length contraction and time dilation. An inertial frame of reference is one in which Newton's laws hold. Any frame of reference that moves at a constant speed relative to an inertial frame of reference is also an inertial frame. The proper length l0 of an object is defined to be the length of the object as measured in the object's rest frame. If the length of the object is measured in any other inertial frame, moving with speed u relative to the object's rest frame (in a direction parallel to l0), the resulting length l is given by the length contraction equation:
Answer:
l = l0 ( 1 - u^2/v^2) ^0.5
Explanation:
This is the equation of lenght contraction of special relativity
A charge of 9.0 x 10-5C is placed in an electric field with a strength of 4.0 x 104 N. What is the electrical force acting
on the charge?
Answer:
3.6 N
Explanation:
E=F/q
so
F=qxE
F= 9x10^-5x4x10^4
F= 3.6 N
4
An 80 kg skier stands at the top of a 40 meter slope. What is her gravitational potential energy
before she skies down the slope?
O 3200
O 31,360
19.63
O 15,680
Answer:
31,360J
Explanation:
Gravitation potential energy (gpe) is calculated from the formula mgh.
That implies, gpe = mgh
Therefore substituting the values of m and h as given in the question, knowing in mine that the acceleration due to gravity( g) is 9.8 N/kg, will give 31,360J
Never forget to put your SI units, because even if your answer is numerical correct, it will be incorrect because it represents no physical quantity.
The gravitational potential energy of the skier at the top of the slope, calculated using the formula PE = mgh, is 31,360 joules.
Explanation:To find the gravitational potential energy of the skier, we use the formula PE = mgh, where PE is the potential energy, m is the mass, g is the acceleration because of gravity, and h is the height. The acceleration because gravity is 9.8 m/s².
So, we substitute these values into the formula:
PE = (80 kg) × (9.8 m/s²) × (40 m) = 31,360 joules.
Therefore, the gravitational potential energy of the skier before she skips down the slope is 31,360 joules.
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4-True-False (experiment 6) : a) 1/(f^2) vs output current is a straight line ( f is frequency) b) Magnetic field of Helmholtz coil and earth are constant during the experiment. c) The sign of B_axial in experiment 6.2 can change the results for the value of N. d) In an ideal selenoid, the sign of B_radial is independent of sign of B_axial e) In experiment 6.1, If we use a magnetic dipole with parameters 4m and 2I, the intercept of 1/(T^2) vs current stays the same since the magnetic field of earth is constant.
Answer:
A.True
B. True
C False
D. False
E. True
Explanation:
Because the direction of Earth's motion around the Sun continually changes during the year, the apparent position of a star in the sky moves in a small loop, known as the aberration of starlight. In order to better understand this phenomenon, it is sometimes helpful to use visual analogies. In these visual analogies, the car is analogous to the Earth, and the rainfall is analogous to starlight. Determine which visual analogies correspond to the following scenarios:
A) The Earth moving around the Sun and interacting with light from a distant star
B) A person on the moving Earth observing the light from a distant star
C) A person on a motionless Earth observing the light from a distant star
Answer:
B) A person on the moving Earth observing the light from a distant star
Explanation:
The aberration of starlight is a phenomenon in Astronomy used to describe the 'seeming' or apparent movement of a star around its true position. This seeming' or apparent movement of a star is caused by and dependent on the velocity of the observer. The movement of the observer relative to the star creates the notion that the star is moving. However, in reality, the star is static. The star is not moving, what is really moving is the observer; this movement of the observer is what makes it seem that the star was moving. It is not an optical illusion, although, the effect seems close enough.
For example, in the analogy given:
A car moving in the rain or under the rainfall
The car is the one moving in the same way that the Earth is the one revolving. The rainfall drops vertically in the same way that the starlight is static. The movement of the car relative to the rainfall is what makes it seem that the rain is falling 'diagonally'. The same visual analogy is observe when a person on the moving Earth observing the light from a distant star
Hence, the correct option is B
A 1.005 m chain consists of small spherical beads, each with a mass of 1.00 g and a diameter of 5.00 mm, threaded on an elastic strand with negligible mass such that adjacent beads are separated by a center-to-center distance of 10.0 mm. There are beads at each end of the chain. The strand has a spring constant of 28.8 N/m. The chain is stretched horizontally on a frictionless tabletop to a length of 1.50 m, and the beads at both ends are fixed in place.
1. What is the linear mass density of the chain? What is the tension in the chain?
2. With what speed would a pulse travel down the chain?
3. The chain is set vibrating and exhibits a standing-wave pattern with four antinodes. What is the frequency of this motion?
4. If the beads are numbered sequentially from 1 to 101, what are the numbers of the five beads that remain motionless?
5. The 13th bead has a maximum speed of 7.54 m/s. What is the amplitude of that bead's motion?
6. If x0=0 corresponds to the center of the 1st bead and x101=1.50m corresponds to the center of the 101st bead, what is the position xn of the nth bead? Calculate the first four coordinates.
7. What is the maximum speed of the 30th bead?
Answer:
1) μ = 1.33 10⁻³ kg / m , F = - 14,256 , 2) v= 103.53 m/s, 3) f = 138.04 Hz , 4) 1, 25, 50, 76, 101 , 5) A = 0.00869 m , 6) # _position = (# _account-1) (1.5m / 100 accounts)
Explanation:
1) Linear density is the mass per unit length
μ = m / L
μ = 2 1 10⁻³ / 1,5
μ = 1.33 10⁻³ kg / m
this is the density when the chain is stretched, which is when the pulse occurs
we can find the tension with
F = - k (x₁-x₀)
where k is the spring constant
F = - 28.8 (1.5 -1.005)
F = - 14.256 N
the negative sign indicates that the force is restorative
2) the pulse speed is
v = √ T /μ
v = √ 14,256 / 1,33 10⁻³
v = 103.53 m / s
3) If standing waves are formed with fixed points at the ends and 4 antinodes, the wavelength is
2 λ = L
λ = L / 2
wave speed is related to frequency and wavelength
v = λ f
f = v / λ
f = v 2 / L
f = 103.53 2 / 1.5
f = 138.04 Hz
4) The marbles are numbered, the marbles that remain motionless are
the first (1) and the last (101)
Let's look for the distance to each node, for this we must observe that in each wavelength there is a node at the beginning, one in the center and one at the end, therefore the nodes are in
#_node = m λ / 2 = m L / 4
#_node position (m)
1 1.5 / 4 = 0.375
2 2 1.5 / 4 = 0.75
3 3 1.5 / 4 = 1,125
Since there are 101 marbles in the initial length, this number does not change with increasing length, so there is 101 marble in 1.5 m. Let's find with a direct proportion rule the number of marbles at these points with nodes
#_canica = 0.375 m (101 marble / 1.5 m) 0.375 67.33
# _canica = 25
#_canica = 0.75 67.33
#_canica = 50
# _canica = 1,125 67.33
#canica = 75.7 = 76
in short the number of the fixed marbles is
1, 25, 50, 76, 101 canic
5) The movement of the account is oscillatory at this point, which is why it is described by
y = A cos wt
[tex]v_{y}[/tex]= -A w sin wt
the speed is maximum for when the breast is worth ±1
v_{y} = Aw
A = v_{y} / w
angular velocity related to frequency
w = 2π f
A = v_{y} / 2πf
A = 7.54 / (2π 138.04)
A = 0.00869 m
6) for the position of each account we can use a direct proportion rule
in total there are 100 accounts distributed in the 1.50 m distance, the #_account is in the # _position. Note that it starts to be numbered 1, so this number must be subtracted from the index of the amount
# _position = (# _account-1) (1.5m / 100 accounts)
#_canic position(m)
1 0
2 0.015
3 0.045
4 0.06
7) the wave has a constant velocity, but every wave is oscillated perpendicular to this velocity, with an oscillatory movement described by the expression
y = Acos wt
the maximum speed is
[tex]v_{y}[/tex] = -Aw sin wt
speed is maximum when the sine is ±1
v_{y} = A w
to calculate the amplitude of the count we use that for a standing wave
y = 2Asin kx
y / A = 2 sin (2π /λ x)
the wavelength is
λ = 0.75 m
the position is
x (30) = 29 1.5 / 100 = 0.435 m
y (30) A = 2 sin (2pi 0.435 / 0.75)
y (30) / A = 0.96 m
The top surface of the passenger car of a train moving at a velocity of 70 km/h is 2.8 m wide and 8 m long. The top surface is absorbing solar radiation at a rate of 200 W/m2, and the temperature of the ambient air is 308C. Assuming the roof of the car to be perfectly insulated and the radiation heat exchange with the surroundings to be small relative to convection, determine the equilibrium temperature of the top surface of the car.
Answer:
33.1 degrees celcius
Explanation:
Look up attached file
The equilibrium temperature of the train’s top cannot be calculated without the specific heat transfer due to convection. The Stefan-Boltzmann law allows us to calculate heat transfer from radiation but not convection.
Explanation:The key to solving this problem is the Stefan-Boltzmann law of radiation, which describes the rate of heat transfer by emitted radiation. In brief, P = σeAT⁴, where σ = 5.67 × 10⁻⁸ J/s.m².K⁴ is the Stefan-Boltzmann constant, e is the emissivity of the body, A is the surface area of the object, and T is its temperature in Kelvins.
Given that the top surface of the car absorbs solar radiation at a rate of 200 W/m² and its dimensions, we can calculate the area of the surface. The surface, if deemed perfectly insulated, only loses heat through convection, not radiation. However, the amount of heat transfer due to convection is not provided in the question, making it impossible to calculate the exact equilibrium temperature of the top surface of the car.
The net rate of heat transfer takes into account both the temperature of the object and the temperature of its surroundings. If T₁ represents the temperature of the surface to be investigated, and T₂ the ambient temperature, without the value of heat lost through convection, we cannot calculate T₁. Therefore, more information would be required to answer your question fully.
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The magnetic flux through each turn of a 110-turn coil is given by ΦB = 9.75 ✕ 10−3 sin(ωt), where ω is the angular speed of the coil and ΦB is in webers. At one instant, the coil is observed to be rotating at a rate of 8.70 ✕ 102 rev/min. (Assume that t is in seconds.) (a) What is the induced emf in the coil as a function of time for this angular speed? (Use the following as necessary: t. Do not use other variables, substitute numeric values. Assume that e m f is in volts. Do not include units
Answer:
Explanation:
Given that a coil has a turns of
N = 110 turns
And the flux is given as function of t
ΦB = 9.75 ✕ 10^-3 sin(ωt),
Given that, at an instant the angular velocity is 8.70 ✕ 10² rev/min
ω = 8.70 ✕ 10² rev/min
Converting this to rad/sec
1 rev = 2πrad
Then,
ω = 8.7 × 10² × 2π / 60
ω = 91.11 rad/s
Now, we want to find the induced EMF as a function of time
EMF is given as
ε = —NdΦB/dt
ΦB = 9.75 ✕ 10^-3 sin(ωt),
dΦB/dt = 9.75 × 10^-3•ω Cos(ωt)
So,
ε = —NdΦB/dt
ε = —110 × 9.75 × 10^-3•ω Cos(ωt)
Since ω = 91.11 rad/s
ε = —110 × 9.75 × 10^-3 ×91.11 Cos(91.11t)
ε = —97.71 Cos(91.11t)
The EMF as a function of time is
ε = —97.71 Cos(91.11t)
Extra
The maximum EMF will be when Cos(91.11t) = -1
Then, maximum emf = 97.71V
An infinitely long, straight, cylindrical wire of radius R R has a uniform current density → J = J ^ z J→=Jz^ in cylindrical coordinates. What is the magnitude of the magnetic field at some point inside the wire at a distance r i < R ri
Answer:
B(2 pi r) = uo i = uo (J pi r^2)
Magnetic field, B = μo Jr/2
Explanation:
The magnitude of the magnetic field inside the wire at a distance r1 < R inside the wire is;
B = ½μ_o × Jr
We want to find the magnitude of the magnetic field inside the wire at a distance r1 < R.
The formula for Magnetic field at a distance r located inside a conductor of radius R is given by Ampere's circuital law as;
B × 2πr = μ_o × Jπr²
Where;
B is magnitude of magnetic field
μ_o is a constant known as magnetic permeability
J is current density
Thus;
B × 2πr = μ_o × Jπr²
πr will cancel out from both sides to give;
2B = μ_o × Jr
B = ½μ_o × Jr
Thus, in conclusion them magnitude of the magnetic field is ½μ_o × Jr
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Ch 26 HW
Gedanken Conceptual Questions
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Gedanken Conceptual Questions
Einstein developed much of his understanding of relativity through the use of gedanken, or thought, experiments. In a gedanken experiment, Einstein would imagine an experiment that could not be performed because of technological limitations, and so he would perform the experiment in his head. By analyzing the results of these experiments, he was led to a deeper understanding of his theory.
In each the following gedanken experiments, Albert is in the exact center of a glass-sided freight car speeding to the right at a very high speed v relative to you.
Albert has a flashlight in each hand and directs them at the front and rear ends of the freight car. Albert switches the flashlights on at the same time.
Part A
In Albert's frame of reference, which beam of light travels at a greater speed, the one directed toward the front or the one toward the rear of the train, or do they travel at the same speed? Which beam travels faster in your frame of reference?
Enter the answers for Albert's frame of reference and your frame of reference separated by a comma using the terms front, rear, and same. For example, if in Albert's frame of reference the beam of light directed toward the front of the train travels at a greater speed and in your frame of reference the two beams travel at the same speed, then enter front,same.
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Part B
In Albert's frame of reference, which end, front or rear, is struck by light first, or are they struck at the same time? Which end is struck first in your frame of reference?
Enter the answer for Albert's frame of reference and your frame of reference separated by a comma using the terms front, rear, and same. For example, if in Albert's frame of reference the beam of light strikes the front first and in your frame of reference the two beams strike at the same time, then enter front,same.
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Now Albert directs his flashlights at the ceiling and floor of the freight car. The flashlights are located midway between the ceiling and the floor and Albert switches them on at the same time
Part C
In Albert's frame of reference, which surface, ceiling or floor, is struck by light first, or are they struck at the same time? Which surface is struck first in your frame of reference?
Enter the answer for Albert's frame of reference and your frame of reference separated by a comma using the terms ceiling, floor, and same. For example, if in Albert's frame of reference the ceiling is struck by light first and in your frame of reference the floor and the ceiling are struck by light at the same time, then enter ceiling,same.
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Albert is playing laser tag in the freight car. Two "assassins" sneak into the freight car with Albert. One is positioned against the front end and the other against the rear end. They each fire a laser at Albert. The two lasers strike Albert at the same time.
Part D
In Albert's frame of reference, who fired first, the person against the rear end or the person against the front end, or did they fire at the same time? In your frame of reference, who fired first?
Enter the answers for Albert's frame of reference and your frame of reference separated by a comma using the terms front, rear, and same. For example, if in Albert's frame of reference the person against the front end fired first and in your frame of reference both "assassins" fired at the same time, then enter front,same.
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Answer:
Please see the attached picture for the answer.
Explanation:
Final answer:
Einstein's Gedanken experiments reveal the constant speed of light and challenge intuitive notions of time and simultaneity in special relativity. In Albert's reference frame and an outside observer's frame, light beams travel at the same speed, but the perception of simultaneity can differ due to relativistic effects.
Explanation:
Gedanken Experiments and Einstein's Theory of Relativity
Albert Einstein used Gedanken's experiments to explore the implications of relativity on the concepts of time and simultaneity. These thought experiments challenged intuitive notions by demonstrating how the speed of light remains constant in all frames of reference, a cornerstone of special relativity. Let's address the student's questions based on this principle:
Part A: In both Albert's frame of reference and your frame of reference, the beams of light travel at the same speed since the speed of light is constant regardless of the observer's motion relative to the source.
Part B: Albert sees both beams of light strike the ends of the freight car simultaneously. However, due to relativistic effects, a stationary observer might see the light traveling in the direction of the freight car's motion strike first.
Part C: Both in Albert's and your frame of reference, the light strikes the ceiling and floor simultaneously because there is no relative motion along the direction of the light beams' propagation.
Part D: In Albert's frame of reference, both lasers strike him simultaneously. For a stationary observer, depending on the direction of the freight car's motion, one assassin may seem to fire before the other due to relativistic effects.
Einstein's ingenious analytical methods demonstrated through Gedanken experiments, led to the conclusion that many seemingly obvious predictions must be reconsidered when dealing with the postulates of relativity, which have been confirmed through meticulous experimentation.
According to the theory, the average force Favg, applied to the mass multiplied by the duration the force is applied should equal the mass change in momentum. For each run calculate the change in momentum of the mass using the average force, then find the % difference between that and change in momentum using the change in velocity.
Answer:
The change in momentum for average force is: Δ[tex]P_{average force} = F_{avg}[/tex]Δt and it is calculated below.
For thick and thin spring, the %difference between change in momentum for average force and change in momentum for velocities is; 178.27% and 159.72% respectively
Explanation:
The average force applied to an object can be calculated by multiplying the average net external force by the time interval over which the force acts. The change in momentum can be calculated using the average force, as well as the change in velocity. The percentage difference between the two values can be found by dividing the difference by the change in momentum using the change in velocity and then multiplying by 100.
Explanation:The average force applied to an object can be calculated using the average net external force multiplied by the time interval over which the force acts. This relationship is known as the impulse-momentum theorem, which states that the change in momentum is equal to the impulse. Impulse is the product of the average force and the time interval.
To calculate the change in momentum of an object using the average force, you can multiply the average force by the duration the force is applied. This will give you the change in momentum caused by the force. You can then calculate the change in momentum using the change in velocity by multiplying the mass of the object by the change in velocity. Finally, you can find the percentage difference between the two values by dividing the difference between the two values by the change in momentum using the change in velocity and then multiplying by 100.
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If the wave speed is 20 m/s and the wave length is 2 meters find the frequency ?
Answer:
10
Explanation:
v=f×lander
20=f2
f=10
A coil of 43 turns in the shape of a rectangle with width 5 cm and length 10 cm is dropped from a position where magnetic field is 0 T to a position where the field is 0.5 T and is directed perpendicular to the plane of the coil. The displacement occurs in a time of 0.431 s. Calculate the resulting average emf induced in the coil
Answer:
Explanation:
Given that,
Number of turns
N = 43 turns
Dimensions of rectangle
Length L = 10cm = 0.1m
Width W = 5cm = 0.05
Then, area of rectangle
A = L×W = 0.1 × 0.05 = 0.005m²
Initial magnetic field
Bi = 0T
Final magnetic field
Bf = 0.5T
The displacement time
∆t = 0.431s
Induce emf ε?
EMF is given as
ε = —NdΦ/dt
Where magnetic flux
Φ = BA
Note A is constant
Then,
ε = —NdΦ/dt = —Nd(BA)/dt
ε = —N•A•dB/dt
ε = —N•A•(∆B/∆t)
ε = —N•A•(Bf - Bi) / (∆t)
ε = —43 × 0.005 × (0.5 - 0) /0.431
ε = —43 × 0.005 × 0.5 /0.431
ε = —0.249 V
The average induced EMF is 0.249V
A velocity selector, in which charged particles of a specific speed pass through undeflected while those of greater or lesser speeds are deflected either to the left or to the right. After leaving the velocity selector, particles enter a purely magnetic field. A particle's radius of motion is then used to find the mass-to-charge ratio of the particle, which in turn can identify it. Taken altogether, this device is called a mass spectrometer. The illustrated electric field is directed to the right with magnitude 1.95 times 10^5 V/m and the magnetic field is directed into the screen with magnitude 0.555 T.
a. Determine the speed v of the undeflected charged particle.
b. After passing through the velocity selector, the charged particle moves in a circular path with a radius of r = 6.61 mm. Determine the particle's mass-to-charge ratio. m/q = kg/C
Answer:
a) 351351.35m/s
b) 1.044*10^{-8}kg/C
Explanation:
a) Electric force and magnetic force over the charge must have the same magnitude. From there we can compute the seep of the charge.
[tex]F_E=F_B\\\\qE=qvB\\\\v=\frac{E}{B}=\frac{1.95*10^{5}V/m}{0.555T}=351351.35\frac{m}{s}[/tex]
b) the mass-charge ratio is given by:
[tex]\frac{m}{q}=\frac{rB}{v}=\frac{(6.61*10^{-3}m)(0.555T)}{351351.35m/s}=1.044*10^{-8}\frac{kg}{C}[/tex]
hope this helps!!
Light of wavelength 199 nm shines on a metal surface. 4.08 eV is required to eject an electron. What is the kinetic energy of
(a) the fastest and
(b) the slowest ejected electrons?
(c) What is the stopping potential for this situation?
(d) What is the cutoff wavelength for this metal?
Answer:
a) = 6.23eV
b) = 0eV
c) = 6.23V
d) = 304.12nm
Explanation:
λ = 199nm
E = 4.08eV = Φ
hf = K(max) + Φ
Φ = work function of the target material
hf = photon energy
K(max) = k.e of the metal
K(max) = [(6.626*10⁻³⁴) * (3.0*10⁸)] / 199*10⁻⁹
K(max) = 9.99*10⁻¹⁹J
1.602*10⁻¹⁹J = 1eV
9.99*10⁻¹⁹ = 6.23eV
K(max) = 6.23eV
b). The slowest moving electron has a kinetic energy of zero
c). The stopping potential is the potential difference required to stop the fastest electron.
eV = K(max)
V = K(max) / e
V = 6.23eV / e
V = 6.23V
d)
The cut- off wavelength occurs when the maximum K.E is zero.
Φ = hc / λ
λ₀ = hc / Φ
λ₀ = (6.626*10⁻³⁴ * 3.0*10⁸) / (4.08 * 1.602*10⁻¹⁹)
λ₀ = 1.9878*10⁻²⁵ / 6.5361*10⁻¹⁹
λ₀ = 3.04*10⁻⁷m
λ₀ = 304.12nm
7. A pendulum of mass 5.0 kg hangs in equilibrium. A frustrated student walks up to it and kicks the bob with a horizontal force of 30.0 ???? applied over 0.30 seconds. What is the length of the pendulum if it has a period of 5.0 seconds? What is the maximum angle of displacement of the swinging pendulum?
Answer:
Explanation:
For pendulum , time period is
T = 2π √ l / g
T² = 4π² l/ g
l = / g T² / 4π²
= 9.8 x 5² / 4π²
= 6.21 m
Angular impulse = Torque x time
= 30 x 6.21 x .3 = increase in angular momentum
30 x 6.21 x .3 = I ω ; I is moment of inertia and ω is angular velocity .
I ω = 55.89
I = 1/3 m l²
= 1/3 x 5 x 6.21²
= 64.27
Angular kinetic energy = 1/2 I ω²
= 1/2(I ω)²/ I
= .5x 55.89²/ 64.27
= 24.30 J
Due to it , the centre of mass of pendulum increases by height h so that
h = l/2 ( 1 - cos θ )
1/2 I ω² = mg l/2 ( 1 - cos θ)
24.3 = 5 x 9.8 x 3.1 ( 1 - cos θ)
( 1 - cos θ) = .16
cos θ = .84
θ = 33 degree.
The length of the pendulum is about 6.25m and the maximum angle of displacement would occur when the pendulum is kicked with the maximum force of 30N.
Explanation:The length of a pendulum is determined by the formula T=2π√(L/g). Given that the period T is 5.0 seconds and g (acceleration due to gravity) is 9.8 m/s², we can rearrange the equation to solve for L (length), resulting in L = (T² * g) / (4π²) = (5.0)² * 9.8 / (4π²) ≈ 6.25m. This regard to the maximum angle of displacement, when a pendulum is in simple harmonic motion, which is typically for displacements less than 15 degrees, the displacement is directly proportional to its force. Therefore, the maximum angle would occur when the force is the greatest, in this case, when the student kicks the pendulum with a force of 30 N.
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A ventriloquist is able to convince you that words are coming from his dummy’s mouth because the dummy’s mouth movements match the timing of the words actually coming from the ventriloquist. The ventriloquist is making use of the perceptual principle of
The ventriloquist uses the perceptual principle known as the McGurk Effect to create the illusion that words are coming from their dummy's mouth. This involves syncing their speech with the movements of the dummy's mouth and mastering aspects of speech production. We perceive the words to be coming from the dummy due to the combined audiovisual cues.
Explanation:A ventriloquist convinces us that words are coming from their dummy's mouth by making use of the perceptual principle known as the McGurk Effect. This audiovisual speech perception demonstrates that we don't just hear speech but also 'see' it. The ventriloquist creates the illusion by carefully syncing the opening and closing of the dummy's mouth with their own speech. This overlaid speech and vision information persuades us to perceive the sound as coming from the dummy.
Human speech itself is produced by shaping the cavity formed by the throat and mouth, the vibration of vocal cords, and using the tongue to adjust the fundamental frequency and combination of those sounds. Thus, excellent ventriloquists master these aspects of speech production without visibly moving their own mouth or lips, reinforcing the illusion.
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An alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 4.29 cm in a uniform magnetic field with B = 1.53 T. Calculate (a) its speed, (b) its period of revolution, (c) its kinetic energy, and (d) the potential difference through which it would have to be accelerated to achieve this energy.
Answer:
A) speed = 3144287.425 m/s
B) period = 8.57x10^-8 s
C) KE = 3.302x10^-14 J
D) potential difference = 103187.5 V
Explanation:
Detailed explanation and calculation is shown in the image below.
A leaf spring in an off-road truck with a spring constant k of 87.6 kN/m (87,600 N/m) is compressed a distance of 6.2 cm (0.062 m) from its original unstretched position. What is the increase in potential energy of the spring (in kJ)?
The increase in potential energy of the spring when compressed a distance of 0.062m from its unstretched position, with a spring constant of 87600N/m, is approximately 0.168 kJ.
Explanation:The increase in potential energy of the spring can be calculated using the formula U = ½kx². In this problem, k is the spring constant which is 87,600 N/m and x is the displacement of the leaf spring when it is compressed, which is 0.062 m. Plugging in these values, we get U = ½ * (87,600 N/m) * (0.062 m)² = 167.5424 Joules. This value is in Joules, so to convert it to kilojoules (kJ), we divide by 1000. So, the increase in potential energy of the spring is approximately 0.168 kJ.
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A uniform electric field, with a magnitude of 650 N/C, is directed parallel to the positive x-axis. If the potential at x = 3.0 m is 1 700 V, what is the potential at x = 1.0 m?
The potential at x = 1.0 m in the given uniform electric field, with the given parameters, would be 400 V.
Explanation:The subject question pertains to the concept of electric potential in a uniform electric field. The potential difference, V, in an uniform electric field is given by the equation V = Ed, where E is the field strength and d is the distance. Given an electric field strength, E, of 650 N/C and a potential, V, at x = 3.0 m of 1 700 V, we seek the potential at x = 1.0m. Since the electric field is uniform, we know that the change in potential is linear with distance. Because the distance decreases by 2 meters (from x = 3.0 m to x = 1.0 m), the potential will decrease by E*2, which is 650 N/C * 2 m = 1300 V. So, the potential at x = 1.0 m is 1 700 V - 1 300 V = 400 V.
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To find the electric potential at x = 1.0 m in a uniform electric field directed along the x-axis, subtract the potential difference due to the field over the distance moved from the initial potential; the answer is 400 V.
The question is asking to calculate the electric potential at a certain point given a uniform electric field and an initial electric potential at a different point. To find the potential at x = 1.0 m, we can use the relationship between electric field (E), electric potential (V), and distance (d), with the following formula: V = V_0 - E * d, where V is the potential at the point of interest, V_0 is the initial potential, E is the electric field strength, and d is the distance between the points.
Gor the given electric field of 650 N/C directed parallel to the positive x-axis, and with the potential at x = 3.0 m being 1700 V, the potential difference \\Delta V caused by moving from x = 3.0 m to x = 1.0 m can be calculated by multiplying the electric field strength by the distance over which the field is applied, which is 2.0 m (3.0 m - 1.0 m). Hence, \\Delta V = E * d = 650 N/C * 2.0 m = 1300 V.
Therefore, the potential at x = 1.0 m is V = 1700 V - 1300 V = 400 V.
You are asked to construct a mobile with four equal m = 165 kg masses, and three light rods of negligible mass and equal lengths. The rods are of length 55 cm. (a) At what location on the level 1 rod should the free end of rod 2 be attached? L1 = (b) At what location on the level 2 rod should the free end of rod 3 be attached? L2 = (c) At what location on the level 3 rod should the whole assembly be suspended from so that the mobile is in equilibrium? L3 =
Answer:
Explanation:
the picture attached shows the whole explanation
) Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, the first having a mass of 1.50 × 105 kg and a velocity of 0.30 m/s to the right, and the second having a mass of 1.10 × 105 kg and a velocity of 0.12 m/s to the left. What is their final velocity?
Answer:
Their final velocity is 0.122m/s
Explanation:
Using the law of conservation of momentum which states that the sum of momentum of bodies before collision is equal to the sum of their momentum after collision. The bodies move with a common velocity after collision.
Momentum = mass × velocity
BEFORE COLLISION
momentum of the first train
= 1.5×10^5 × 0.3
= 0.45×10^5kgm/s
Momentum of second train
= 1.1×10^5 × (-0.12) (velocity is negative since the body is moving towards the left)
= -0.132 × 10^5kgm/s
AFTER COLLISION
Momentum of both bodies is given as:
(1.5×10^5+1.1×10^5)V
= 2.6×10^5V
V is their common velocity
According to the law:
0.45×10^5+(-0.132×10^5) =2.6×10^5V
0.45×10^5 - 0.132×10^5 = 2.6×10^5V
0.318×10^5 = 2.6×10^5V
0.318 = 2.6V
V = 0.318/2.6
V = 0.122m/s
Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum that has a wavelength of 700 nm and a peak electric field magnitude of 3.5 V/m. (Use the following as necessary: t and x. Assume that E is in volts per meter, B is in teslas, t is in seconds, and x is in meters. Do not include units in your answer. Assume that E = 0 and B = 0 when x = 0 and t = 0.) E(x, t) = B(x, t) =
Answer:
[tex]E=3.5(8.98*10^{6}x-2.69*10^{15}t)[/tex]
[tex]B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)[/tex]
Explanation:
The electric field equation of a electromagnetic wave is given by:
[tex]E=E_{max}(kx-\omega t)[/tex] (1)
E(max) is the maximun value of E, it means the amplitude of the wave.k is the wave numberω is the angular frequencyWe know that the wave length is λ = 700 nm and the peak electric field magnitude of 3.5 V/m, this value is correspond a E(max).
By definition:
[tex]k=\frac{2\pi}{\lambda}[/tex]
[tex]k=8.98*10^{6} [rad/m][/tex]
And the relation between λ and f is:
[tex]c=\lambda f[/tex]
[tex]f=\frac{c}{\lambda}[/tex]
[tex]f=\frac{3*10^{8}}{700*10^{-9}}[/tex]
[tex]f=4.28*10^{14}[/tex]
The angular frequency equation is:
[tex]\omega=2\pi f[/tex]
[tex]\omega=2\pi*4.28*10^{14}[/tex]
[tex]\omega=2.69*10^{15} [rad/s][/tex]
Therefore, the E equation, suing (1), will be:
[tex]E=3.5(8.98*10^{6}x-2.69*10^{15}t)[/tex] (2)
For the magnetic field we have the next equation:
[tex]B=B_{max}(kx-\omega t)[/tex] (3)
It is the same as E. Here we just need to find B(max).
We can use this equation:
[tex]E_{max}=cB_{max}[/tex]
[tex]B_{max}=\frac{E_{max}}{c}=\frac{3.5}{3*10^{8}}[/tex]
[tex]B_{max}=1.17*10^{-8}T[/tex]
Putting this in (3), finally we will have:
[tex]B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)[/tex] (4)
I hope it helps you!
Final answer:
To find the equations of the fields for an electromagnetic wave in the red spectrum with given parameters, calculate the frequency using the speed of light and wavelength, then apply it to the sinusoidal equations representing the electric and magnetic fields.
Explanation:
To determine the equations for the electric and magnetic fields (E and B, respectively) of an electromagnetic wave in the visible red spectrum with a wavelength (λ) of 700 nm and a peak electric field magnitude (Ē) of 3.5 V/m, we first need to calculate the frequency (f) of the wave.
Since the speed of light (c) is approximately 3 × 108 m/s, the frequency can be calculated by using the relationship c = f × λ. After finding the frequency, we can then write down the equation for the electric field E(x, t). Assuming that the wave is propagating in the +x direction and that the electric field oscillates in the y-direction, with no initial phase change, the equation for the electric field is:
E(x, t) = Ē × sin(2π(ft - x/λ))
To find the associated magnetic field B, we use the fact that the magnitudes of E and B are related by the speed of light c, such that B = E/c. Since electromagnetic waves have their electric and magnetic fields perpendicular to each other, if E is in the y-direction, B will be in the z-direction. The equation for the magnetic field is:
B(x, t) = B × sin(2π(ft - x/λ))
Where B is the peak magnetic field strength.
ii. At the instant shown in Figure 3, the blocks are moving toward each other with the same speed of 0.35m/s. The blocks collide 0.50 seconds later. What is the speed of the two-block system’s center of mass just before the blocks collide?
Answer:
2.9 m/s
Explanation:
The speed of the two-block system center of mass will not be changed by the collision, since it is an internal force. The only force acting upon the two-block system, then, will be gravity. We can calculate the effects of gravity upon the speed of the blocks along the slope by taking the sine of the angle of the slope (37°) and multiplying it by the magnitude of Fg in the vertical direction. sin(37°)*(3.92 N) = 2.36 N. This force acts upon the system and accelerates it down the slope, which can be modeled with the equation 2.36 N = (0.40 kg)*a. Solving for a, we find that gravity accelerates the block at 5.9 m/s2 along the slope of the block. Using this figure, we can find the speed of the system with the equation v = v0 + at. Initial velocity of the system is zero, as both blocks are moving towards each other at equal speed, so v = (5.9 m/s2)*(0.50 s). Velocity of the system after 0.50 seconds = 2.9 m/s.
The center-of-mass of an isolated two-block system moving towards each other at the same speed remains the same before the collision; therefore, the center-of-mass velocity is also 0.35 m/s just before the blocks collide.
Explanation:In this scenario, we are dealing with the principle of conservation of momentum in an isolated system. In two-dimensional collisions involving identical masses moving toward each other at the same speed, the center-of-mass velocity does not change due to the collision.
To find the speed of the two-block system's center of mass, we can use the formula for center-of-mass velocity, that is the weighted average of the individual velocities. Since the blocks have identical mass, the center-of-mass velocity is just the average of their individual velocities. Given both blocks are moving toward each other at a speed of 0.35 m/s, the speed of the center-of-mass of the system before the blocks collide is also 0.35 m/s.
Given two identical blocks moving towards each other at a speed of 0.35 m/s, we can denote the velocity of each block as +0.35 m/s and -0.35 m/s. The positive and negative signs apply due to the blocks moving towards each other.The center-of-mass velocity (V_CM) is found by using the formula: V_CM = (m1*v1 + m2*v2) / (m1 + m2) where m1 and m2 are the masses of the blocks and v1 and v2 are their respective velocities.Since both blocks have the same mass and velocity, except for direction, the center-of-mass velocity is 0.Learn more about Center-of-Mass Velocity here:https://brainly.com/question/32099134
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HELP ASAP PLEASE!!!
The earth is:
A. a perfect sphere
B. thinner at the equator
C. slightly flatter at the poles
D. made of mostly carbon and iron
The Earth is slightly flatter at the poles, making the correct answer option C. It is an oblate spheroid with a denser core made primarily of iron and nickel.
The correct answer to the question, 'The earth is:', is C: slightly flatter at the poles. This is because the Earth's shape is an oblate spheroid, meaning it's somewhat squished at the poles and bulges a bit at the equator due to its rotation. The Earth's equatorial radius is greater than its polar radius, which results in this slight flattening at the poles. Also, the Earth's core is composed predominantly of iron and nickel, making it very dense.
While it is true that the Earth is not a perfect sphere, it is also not thinner at the equator as option B suggests. Option D mentions that Earth is made of mostly carbon and iron, which is incorrect; the Earth is composed primarily of iron, oxygen, silicon, and other elements, but not mostly carbon.
An AC power source has an rms voltage of 120 V and operates at a frequency of 60.0 Hz. If a purely inductive circuit is made from the power source and a 46.8 H inductor, determine the inductive reactance and the rms current through the inductor.(a) the inductive reactance (in Ω)
(b) the rms current through the inductor (in A)
Answer:
(a) 17634.24 Ω
(b) 0.0068 A
Explanation:
(a)
The formula for inductive inductance is given as
X' = 2πFL................... Equation 1
Where X' = inductive reactance, F = frequency, L = inductance
Given: F = 60 Hz, L = 46.8 H, π = 3.14
Substitute into equation 1
X' = 2(3.14)(60)(46.8)
X' = 17634.24 Ω
(b)
From Ohm's law,
Vrms = X'Irms
Where Vrms = Rms Voltage, Irms = rms Current.
make Irms the subject of the equation
Irms = Vrms/X'...................... Equation 2
Given: Vrms = 120 V, X' = 17634.24 Ω
Substitute into equation 2
Irms = 120/17634.24
Irms = 0.0068 A
Global Precipitation Measurement (GPM) is a tool scientist use to forecast weather. Which statements describe GPM
Answer:
Correct answers:
Explanation:
• It is a satellite that collects data about rain and snow.
GPM is part of NASA's Earth Systematic Missions program and its function is to track precipitation from space with satellites and to provide accurate information of the time, place an amount of rain or snow anywhere in the world.
• Its orbit covers 90 percent of Earth's surface.
GPM has unique perspective on measuring precipitation from space because of their collected satellite retrievals.
• The sensors measure microwaves.
GPM Microwave Imager (GMI) is a sensor that observes the microwave energy emitted by the Earth and atmosphere, so it improves the accuracy of rain and snowfall estimates.
Airplanes are launched from aircraft carriers by means of a steam catapult. The catapult is a well-insulated cylinder that contains steam, and is fitted with a frictionless piston. The piston is connected to the airplane by a cable. As the steam expands, the movement of the piston causes movement of the plane. A catapult design calls for 270 kg of steam at 15 MPa and 450°C to be expanded to 0.4 MPa. How much work can this catapult generate during a single stroke? Compare this to the energy required to accelerate a 30,000 kg aircraft from rest to 350 km per hour.
Find the screenshots in the attachment for complete solution. Follow the sequence
The catapult can generate 216,900,000 J of work during a single stroke, which is significantly more energy than the 142,222,500 J required to accelerate the aircraft.
To calculate the work done during the expansion of the steam in the catapult, we can use the first law of thermodynamics, which is the energy balance equation for a closed system:
ΔU = Q - W
Where:
ΔU is the change in internal energy of the system.
Q is the heat added to the system.
W is the work done by the system.
In this case, the steam in the catapult is the system, and we want to find the work done (W) during the expansion.
We can calculate the change in internal energy (ΔU) using the following equation:
ΔU = m * (u_final - u_initial)
Where:
m is the mass of the steam.
u_final is the specific internal energy of the steam at the final state.
u_initial is the specific internal energy of the steam at the initial state.
Given that the steam expands from 15 MPa and 450°C to 0.4 MPa, we need to find the specific internal energies at these two states. You'll typically need steam tables or thermodynamic software for precise values, but I'll provide approximate values for illustration:
For the initial state (15 MPa and 450°C), the specific internal energy might be approximately u_initial = 3270 kJ/kg.
For the final state (0.4 MPa), the specific internal energy might be approximately u_final = 2480 kJ/kg.
The mass (m) of the steam is given as 270 kg.
Now, calculate the change in internal energy:
ΔU = 270 kg * (2480 kJ/kg - 3270 kJ/kg)
ΔU = -216,900 kJ
The negative sign indicates a decrease in internal energy during the expansion.
Now, let's calculate the work done (W). Since this is an expansion, the work done by the system is positive:
W = -ΔU
W = -(-216,900 kJ)
W = 216,900 kJ
Now, let's compare this to the energy required to accelerate a 30,000 kg aircraft from rest to 350 km per hour. To calculate this energy, we can use the kinetic energy formula:
KE = 0.5 * m * v^2
Where:
KE is the kinetic energy.
m is the mass of the aircraft.
v is the velocity.
Converting 350 km per hour to m/s:
350 km/h * (1000 m/km) / (3600 s/h) ≈ 97.22 m/s
Now, calculate the kinetic energy:
KE = 0.5 * 30,000 kg * (97.22 m/s)^2
KE ≈ 142,222,500 J (joules)
So, the energy required to accelerate the aircraft is approximately 142,222,500 joules.
Comparing the work done by the catapult (216,900 kJ or 216,900,000 J) to the energy required to accelerate the aircraft (142,222,500 J), we can see that the catapult generates significantly more energy during a single stroke than is required to accelerate the aircraft. The catapult can generate much more work due to the expansion of steam, making it suitable for launching an aircraft.
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A single-slit diffraction pattern is formed on a distant screen. If the width of the single slit through which light passes is reduced, what happens to the width of the central bright fringe? Assume the angles involved remain small. A single-slit diffraction pattern is formed on a distant screen. If the width of the single slit through which light passes is reduced, what happens to the width of the central bright fringe? Assume the angles involved remain small. The central bright fringe becomes narrower. The effect cannot be determined unless the distance between the slit and the screen is known. The central bright fringe becomes wider. The central bright fringe remains the same size.
Answer:
Explanation:
General guidance
Concepts and reason
The concept used to solve this problem is slit width condition for maximum diffraction in case of single slit diffraction experiment.
Initially, use the condition for diffraction maximum in the case of single slit diffraction to find the inapplicable given options.
Finally, use the condition for diffraction maximum in the case of single slit diffraction to find the applicable given options.
Fundamentals
The condition for diffraction maximum in the case of single slit diffraction is as follows:
sin Θ=λ/α
Here, the angle situated in the first dark fringe on each side of the central bright fringe isΘ , slit width is α, and the wavelength is λ .
The incorrect options are as follows:
• The central bright fringe remains in same size.
The width of the central bright fringe is inversely proportional to the slit width. Therefore, the central fringe cannot remain in the same size. Hence, it is incorrect.
• The effect cannot be determined unless the distance between the slit and screen is known.
Without knowing the distance between the slit and screen, the effect can be experienced. Therefore, this option is incorrect.
• The central bright fringe becomes narrower.
Due to the inverse proportion of central bright fringe width and slit width, the central bright fringe becoming narrower is incorrect.The central bright fringe width is directly proportional to the slit width.
If the width of the slit increases, then the central bright fringe width also increases.
Due to the inverse proportion of central bright fringe width and slit width, the central bright fringe becomes wider when the width of the single slit is reduced.
The condition for diffraction maximum is as follows:
sin Θ=λ/α
The slit width is inversely proportional to the angle of the first dark fringe on either side of the central bright fringe.
Therefore,
• The central bright fringe becomes wider is correct.
The applicable option when the width of the slit reduces is the central bright fringe becoming wider.
Slit width is inversely proportional to the angle of the first dark fringe on either side of the central bright fringe.
The central bright fringe width is directly proportional to the angle of the first dark fringe on either side of the central bright fringe.
If the central bright fringe becomes wider, then the angle of the first dark fringe on either side of the central bright fringe will be larger.
Answer
The applicable option when the width of the slit reduces is the central bright fringe becoming wider.