Explanation:
Consider a loop of wire, which has an area of [tex]A=14 \mathrm{cm}^{2}[/tex] and [tex]N=250[/tex] turns, it is initially placed perpendicularly in the earth magnetic field. Then it is rotated from this position to a position where its plane is parallel to the field as shown in the following figure in [tex]\Delta t=0.030[/tex] s. Given that the earth's magnetic field at the position of the loop is [tex]B=5.0 \times 10^{-5} \mathrm{T}[/tex], the flux through the loop before it is rotated is,
[tex]\Phi_{B, i} &=B A \cos \left(\phi_{i}\right)=B A \cos \left(0^{\circ}\right[/tex]
[tex]=\left(5.0 \times 10^{-5} \mathrm{T}\right)\left(14 \times 10^{-4} \mathrm{m}^{2}\right)(1)[/tex]
[tex]=7.0 \times 10^{-8} \mathrm{Wb}[/tex]
[tex]\quad\left[\Phi_{B, i}=7.0 \times 10^{-8} \mathrm{Wb}\right[/tex]
after it is rotated, the angle between the area and the magnetic field is [tex]\phi=90^{\circ}[/tex] thus,
[tex]\Phi_{B, f}=B A \cos \left(\phi_{f}\right)=B A \cos \left(90^{\circ}\right)=0[/tex]
[tex]\qquad \Phi_{B, f}=0[/tex]
(b) The average magnitude of the emf induced in the coil equals the change in the flux divided by the time of this change, and multiplied by the number of turns, that is,
[tex]{\left|\mathcal{E}_{\mathrm{av}}\right|=N\left|\frac{\Phi_{B, f}-\Phi_{B, i}}{\Delta t}\right|}{=} & \frac{1.40 \times 10^{-5} \mathrm{Wb}}{0.030 \mathrm{s}}[/tex]
[tex]& 3.6 \times 10^{-4} \mathrm{V}=0.36 \mathrm{mV}[/tex]
[tex]\mathbb{E}=0.36 \mathrm{mV}[/tex]
(a) The initial and final flux through the coil is 1.75 × 10⁻⁵ Wb and 0 Wb
(b) The induced EMF in the coil is 0.583 mV
Flux and induced EMF:Given that the coil has N = 250 turns
and an area of A = 14cm² = 1.4×10⁻³m².
It is rotated for a time period of Δt = 0.030s such that it is parallel with the earth's magnetic field that is B = 5×10⁻⁵T
(a) The flux passing through the coil is given by:
Ф = NBAcosθ
where θ is the angle between area vector and the magnetic field
The area vector is perpendicular to the plane of the coil.
So, initially, θ = 0°, as area vector and earth's magnetic field both are perpendicular to the plane of the coil
So the initial flux is:
Φ = NABcos0° = NAB
Ф = 250×1.4×10⁻³×5×10⁻⁵ Wb
Ф = 1.75 × 10⁻⁵ Wb
Finally, θ = 90°, and since cos90°, the final flux through the coil is 0
(b) The EMF induced is given by:
E = -ΔФ/Δt
E = -(0 - 1.75 × 10⁻⁵)/0.030
E = 0.583 × 10⁻³ V
E = 0.583 mV
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A resistor made of Nichrome wire is used in an application where its resistance cannot change more than 1.35% from its value at 20.0°C. Over what temperature range can it be used (in °C)?
Answer:
Pls seeattached file
Explanation:
A resistor made of Ni chrome wire is used in an application where its resistance cannot be more than 1.35 % so its temperature range will be from 33.75 to -33.75 °C.
What is Resistance?Electrical resistance, or resistance to electricity, is a force that opposes the flow of current. Ohms are used to expressing resistance values.
When there is an electron difference between two terminals, electricity will flow from high to low. In opposition to that flow is resistance. As resistance rises, the current declines. On the other side, when the resistance falls, the current rises.
According to the question,
R = R₀ (1 + α ΔT)
(1 + 0.0135)R₀ = R₀(1 + α ΔT)
ΔT = (1 + 0.0135) / α
= 0.0135 / 0.0004
= 33.75 °C.
ΔT = [(1 - 0.0135) -1]/0.004
= -33.75 °C
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Specific heat is a measurement of the amount of heat energy input required for one gram of a substance to increase its temperature by one degree Celsius. Solid lithium has a specific heat of 3.5 J/g·°C. This means that one gram of lithium requires 3.5 J of heat to increase 1°C. Plot the temperature of 1g of lithium after 3.5, 7, and 10.5 J of thermal energy are added.
Answer:
ΔT = 1ºC , 2ºCand 3ºC
Explanation:
In this exercise they indicate the specific heat of lithium
let's calculate the temperature increase as a function of the heat introduced
Q = m [tex]c_{e}[/tex] ΔT
ΔT = Q / m c_{e}
calculate
for Q = 3.5 J
ΔT = 3.5 / (1 3.5)
ΔT = 1ºC
For Q = 7.0 J
ΔT = 7 / (1 3.5)
ΔT = 2ºC
for Q = 10.5 J
ΔD = 10.5 / (1 3.5)
ΔT = 3ºC
we see that this is a straight line, see attached
An undiscovered planet, many light-years from Earth, has one moon, which has a nearly circular periodic orbit. If the distance from the center of the moon to the surface of the planet is 2.165×105 km and the planet has a radius of 4175 km and a mass of 6.70×1022 kg , how long (in days) does it take the moon to make one revolution around the planet? The gravitational constant is 6.67×10−11N·m2/kg2 .
Answer:
364days
Explanation:
Pls see attached file
Explanation:
The moon will take 112.7 days to make one revolution around the planet.
What is Kepler's third law?The period of the satellite around any planet only depends upon the distance between the planet's center and satellite and also depends upon the planet's mass.
Given, the distance from the moon's center to the planet's surface,
h = 2.165 × 10⁵ km,
The radius of the planet, r = 4175 km
The mass of the planet = 6.70 × 10²² kg
The total distance between the moon's center to the planet's center:
a = r +h = 2.165 × 10⁵ + 4175
a = 216500 + 4175
a = 220675
a = 2.26750 × 10⁸ m
The period of the planet can be calculated as:
[tex]T =2\pi \sqrt{\frac{a^3}{Gm} }[/tex]
[tex]T =2\3\times 3.14 \sqrt{\frac{(2.20675 \times 10^8)^3}{(6.67\times 10^{-11}).(6.70\times 10^{22})} }[/tex]
T = 9738253.26 s
T = 112.7 days
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A neutron star has a mass of between 1.4-2.8 solar masses compressed to the size of:
A. Earth
B. The state of Oregon
C. North America
D. An average city
The correct answer is D. An average city
Explanation:
A neutron star differs from others due to its massive density, this means a lot of matter is compressed in a small area. Indeed, neutron stars have a mass of around 1.4 to 2.8 times the mass of the sun. But these are considerably small as they only measure around 20 kilometers, which is the size of an average city. Additionally, neutron stars are this dense because they are the result of a regular star exploding, which leads to a super-dense core, or neutron star. In this context, the mass of a neutron star is compressed to the size of an average city.
A 0.50-T magnetic field is directed perpendicular to the plane of a circular loop of radius 0.25 m. What is the magnitude of the magnetic flux through the loop
Answer:
The magnitude of the magnetic flux through the loop is 0.0982 T.m²
Explanation:
Given;
magnitude of magnetic field, B = 0.5 T
radius of the loop, r = 0.25 m
Area of the loop is given by;
A = πr²
A = 3.142 x (0.25)²
A = 0.1964 m²
The magnitude of the magnetic flux through the loop is given by;
Ф = BA
Where;
B is the magnitude of the magnetic field
A is area of the field
Ф = 0.5 x 0.1964
Ф = 0.0982 T.m²
Therefore, the magnitude of the magnetic flux through the loop is 0.0982 T.m²
Which statement accurately describes the inner planets? Uranus is one of the inner planets. The inner planets formed when the solar system cooled. The inner planets are also called terrestrial planets. The inner planets are larger than the outer planets.
The correct answer is C. The inner planets are also called terrestrial planets.
Explanation:
Our solar system includes a total of eight planets. Additionally, planets are classified into broad categories including inner planets and outer planets. The inner planets category applies to planets such as Earth, Mercury, or Mars because these are located within the asteroid belt (region of asteroids between Mars and Jupiter). Moreover, inner planets differ from others due to their composition as they are composed of rocks and metals. Also, due to this composition, these are known as terrestrial planets. According to this, the statement that best describes inner planets is "The inner planets are also called terrestrial planets".
Answer:
The answer is c.) The inner planets are also called terrestrial planets.
Explanation:
A 26-g rifle bullet traveling 220 m/s embeds itself in a 3.8-kg pendulum hanging on a 2.7-m-long string, which makes the pendulum swing upward in an arc, Determine the vertical and horizontal component of the pendulum's maximum displacement
Answer:
displacements are 0.776m, 0.114m
Explanation:
We were given mass of 26-g rifle bullet , then we can convert to Kg since
Momentum is conserved here.
The initial momentum before impact = (Mi * Vi)
Where Mi= initial given mass
Vi=initial velocity given
= 0.026 * 220 = 5.72 kgm/s
The final momentum after impact is (Mf * Vf )
Mf= final mass
5.72=( 3.82* Vf )
= 5.72/ 3.82
= 1.497 m/s
the speed of the pendulum bob with bullet afterwards= 1.497 m/s
the total energy after the collision is the addition of the kinetic energy of the bob+bullet and the potential energy of the bob and bullet, potential energy can be taken as zero.
M = 3.82 kg the mass of the bob containing the bullet
E(total) = ¹/₂MV² = 1/2 * (3.82kg)*(1.497m/s)² = 4.280J
When the Bob got to highest point the kinetic energy is zero and the potential energy is due to the increase in height of the bob, and the addition of the potential and kinetic energies still equal the total energy from before
E(total) = Mgh + 0 = Mgh = 4.280J
solving for h and substituting,
h = 4.280 J/(9.8m/s^2*3.82kg) = 0.114 m
Since the height is found,we the angle of the pendulum at the top of the swing can also be determined
A = arccos[(2.7 - 0.114) / 2.7] or A = 16.71degrees
Since A is known, the displacement along the horizontal axis can be calculated as
x = 2.7* sin(A) = 0.776m
therefore, displacement is 0.776m, 0.114m
the vertical and horizontal component of the pendulum's maximum displacement are displacement is 0.776m, 0.114m
Explain why the two plates of a capacitor are charged to the same magnitude when a battery is connected to the capacitor.
Answer:
This is because the same electron removed from the positively charged plate is what is taken to the negatively charged plate, maintaining the same amount of electron according to the conservation of charge in an electric circuit.
Explanation:
In any circuit, electrons are neither created nor destroyed according to the laws of conservation of charge, but are transferred from one point to another on the circuit. When the plates of a capacitor are connected to battery, the battery pushes the electron to move due to its potential difference. Electrons are then moved from the positive plate, at a steady rate, to the negative plate. The removal of electrons from the positive plate is what leaves it positively charged from deficiency of electrons, and the addition of electrons at the negatively charged plate is what leaves the plate negatively charge from excess of electrons. From this, we can see that the same electrons removed from the positively charged plate are taken to the negatively charged plate.
A screen is placed a distance dd to the right of an object. A converging lens with focal length ff is placed between the object and the screen. In terms of f, what is the smallest value d can have for an image to be in focus on the screen?
Answer:
2f
Explanation:
The formula for the object - image relationship of thin lens is given as;
1/s + 1/s' = 1/f
Where;
s is object distance from lens
s' is the image distance from the lens
f is the focal length of the lens
Total distance of the object and image from the lens is given as;
d = s + s'
We earlier said that; 1/s + 1/s' = 1/f
Making s' the subject, we have;
s' = sf/(s - f)
Since d = s + s'
Thus;
d = s + (sf/(s - f))
Expanding this, we have;
d = s²/(s - f)
The derivative of this with respect to d gives;
d(d(s))/ds = (2s/(s - f)) - s²/(s - f)²
Equating to zero, we have;
(2s/(s - f)) - s²/(s - f)² = 0
(2s/(s - f)) = s²/(s - f)²
Thus;
2s = s²/(s - f)
s² = 2s(s - f)
s² = 2s² - 2sf
2s² - s² = 2sf
s² = 2sf
s = 2f
Which best identifies the requirements for work to be performed? an object that has a force acting on it an object that is moving and has no net force a force acting on a motionless object a force that moves an object
Answer:
a force that moves an object
Explanation:
the formula for work is force * distance
This question involves the concepts of work, force, and displacement.
The statement that best identifies the requirements for work to be performed is "a force that moves an object".
Work is defined as the product of force applied on an object and the distance moved by the object. Mathematically,
Work = (Force)(Displacement)
Hence, both the applied force and the displacement of the object as a result of the application of the force is necessary for the work to be done. If any one of these values becomes zero, the work automatically becomes zero, which means no work is performed.
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An emf is induced in response to a change in magnetic field inside a loop of wire. Which of the following changes would increase the magnitude of the induced emf? A. Straighten the wire out to be flat B. Reduce the resistance of the wire of which the loop is made C. Turning the plane of the loop to be parallel to the magnetic field D. Reducing the diameter of the loop
Answer:
changing the magnetic field more rapidly
Explanation:
According to Faraday's law, whenever there is a change in the magnetic lines of force, it leads the production of induced emf. The magnitude of induced emf is proportional to to the rate of change of flux.
Hence if the magnetic field inside a loop of wire is changed rapidly, the magnitude of induced emf increases in accordance with Faraday's law of electromagnetic induction stated above when the magnetic field is changed more rapidly, hence the answer.
A double-slit experiment is performed with light of wavelength 620 nm. The bright interference fringes are spaced 2.3 mm apart on the viewing screen. What will the fringe spacing be if the light is changed to a wavelength of 360 nm?
Answer:
1.34 mm
Explanation:
A double slit experiment is conducted with a light which has a wavelength of 620 nm
The fringes are separated 2.3 mm apart
The light is changed to a wavelength length of 360 nm
Let x represent the fringe spacing as a result of the change in wavelength
Therefore,the fringe spacing can be calculated as follows
2.3mm/x= 620nm/360nm
Multiply both sides
x × 620= 2.3×360
620x= 828
x= 828/620
x= 1.34 mm
Improved balance is a primary benefit of regular cardiovascular exercise .
Answer:
Cardiovascular exercise is the activity use that aerobic metabolism and cellular reaction.
Explanation:
Cardiovascular exercise is activity increase heart rate and raises oxygen large muscle group of the body.Cardiovascular exercise is that contain cardio improve to the health mental health, heart health.Cardiovascular exercise such as walking, swimming, running is that exercise is benefit to the health.Cardiovascular exercise to the internal body organs that the healthy heart for the function and performance of the heart.Cardiovascular exercise that having involve feet of the ground this type of activity is called high impact of cardio.Cardio is a good and maintaining exercise for the lungs and heart or healthy bones.Cardio exercise is performed that to a water in reduce to the gravity of that pull on the body weight.Cardiovascular daily to build the stronger muscle and that control the blood pressure.A car moving east at 45 km/h turns and travels west at 30 km/h. What is the
magnitude and direction of the change in velocity?
mahalle 1.11
Explanation:
Change in Velocity = final velocity - initial velocity
Change in velocity = 30km/h - (- 45km/h )
= 75 km/h due west
At what speed (in m/s) will a proton move in a circular path of the same radius as an electron that travels at 7.45 ✕ 106 m/s perpendicular to the Earth's magnetic field at an altitude where the field strength is 1.10 ✕ 10−5 T
Answer:
The speed of the proton is 4059.39 m/s
Explanation:
The centripetal force on the particle is given by;
[tex]F = \frac{mv^2}{r}[/tex]
The magnetic force on the particle is given by;
[tex]F = qvB[/tex]
The centripetal force on the particle must equal the magnetic force on the particle, for the particle to remain in the circular path.
[tex]\frac{mv^2}{r} = qvB\\\\r = \frac{mv^2}{qvB} \\\\r = \frac{mv}{qB}[/tex]
where;
r is the radius of the circular path moved by both electron and proton;
⇒For electron;
[tex]r = \frac{(9.1*10^{-31})(7.45*10^6)}{(1.602*10^{-19})(1.1*10^{-5})}\\\\r = 3.847 \ m[/tex]
⇒For proton
The speed of the proton is given by;
[tex]r = \frac{mv}{qB}\\\\mv = qBr\\\\v = \frac{qBr}{m} \\\\v = \frac{(1.602*10^{-19})(1.1*10^{-5})(3.847)}{1.67*10^{-27}} \\\\v = 4059.39 \ m/s[/tex]
Therefore, the speed of the proton is 4059.39 m/s
Suppose a 58-turn coil lies in the plane of the page in a uniform magnetic field that is directed into the page. The coil originally has an area of 0.150 m2. It is stretched to have no area in 0.100 s. What is the magnitude (in V) and direction (as seen from above) of the average induced emf if the uniform magnetic field has a strength of 1.10 T? magnitude V direction ---Select--- †\
Answer:
95.7v
Explanation
Using Faraday's law of electromagnetic induction we know that rate of change in magnetic flux will induce EMF in closed loop
So it is given as
E= Ndစ/dt
E= N BA-0/ deta t
Given that
N = 58turns
B = 1.10T
A = 0.150m^²
Deta t= 0.1s
now we have
E = 58(1.10x0.150)/0.1
= 95.7v
Magnetic flux is decreasing, so the direction of the current will be to aid the decreasing flux $decrease= CLOCKWISE
Explanation:
A Galilean telescope adjusted for a relaxed eye is 36.2 cm long. If the objective lens has a focal length of 39.5 cm , what is the magnification
Answer:
The magnification is [tex]m = 12[/tex]
Explanation:
From the question we are told that
The object distance is [tex]u = 36.2 \ cm[/tex]
The focal length is [tex]v = 39.5 \ cm[/tex]
From the lens equation we have that
[tex]\frac{1}{f} = \frac{1}{u} + \frac{1}{v}[/tex]
=> [tex]\frac{1}{v} = \frac{1}{f} - \frac{1}{u}[/tex]
substituting values
[tex]\frac{1}{v} = \frac{1}{39.5} - \frac{1}{36.2}[/tex]
[tex]\frac{1}{v} = -0.0023[/tex]
=> [tex]v = \frac{1}{0.0023}[/tex]
=> [tex]v =-433.3 \ cm[/tex]
The magnification is mathematically represented as
[tex]m =- \frac{v}{u}[/tex]
substituting values
[tex]m =- \frac{-433.3}{36.2}[/tex]
[tex]m = 12[/tex]
light of wavelength 550 nm is incident on a diffraction grating that is 1 cm wide and has 1000 slits. What is the dispersion of the m = 2 line?
Answer:
The dispersion is [tex]D = 2.01220 *10^{5} \ rad/m[/tex]
Explanation:
From the question we are told that
The wavelength of the light is [tex]\lambda = 550 \ = 550 *10^{-9} \ n[/tex]
The width of the grating is[tex]k = 1\ cm = 0.01 \ m[/tex]
The number of slit is N = 1000 slits
The order of the maxima is m = 2
Generally the spacing between the slit is mathematically represented as
[tex]d = \frac{k}{N}[/tex]
substituting values
[tex]d = \frac{ 0.01}{1000}[/tex]
[tex]d = 1.0 *10^{-5} \ m[/tex]
Generally the condition for constructive interference is
[tex]d\ sin(\theta ) = m * \lambda[/tex]
substituting values
[tex]1.0 *10^{-5} sin (\theta) = 2 * 550 *10^{-9}[/tex]
[tex]\theta = sin^{-1} [\frac{ 2 * 550 *10^{-9}}{ 1.0 *10^{-5}} ][/tex]
[tex]\theta = 6.315^o[/tex]
Generally the dispersion is mathematically represented as
[tex]D = \frac{ m }{d cos(\theta )}[/tex]
substituting values
[tex]D = \frac{ 2 }{ 1.0 *10^{-5} cos(6.315 )}[/tex]
[tex]D = 2.01220 *10^{5} \ rad/m[/tex]
The near point (the smallest distance at which an object can be seen clearly) and the far point (the largest distance at which an object can be seen clearly) are measured for six different people.
Near Point(cm) Far Point(cm)
Avishka 40 [infinity]
Berenice 30 300
Chadwick 25 500
Danya 25 [infinity]
Edouard 80 200
Francesca 50 [infinity]
Of the farsighted people, rank them by the power of the lens needed to correct their hyperopic vision. Rank these from largest to smallest power required.
1. Berenice
2. Avishka
3. Francesca
4. Edouard
Answer:
1. Berenice = 0.67 D
2. Avishka = 1.50 D
3. Francesca = 2.00 D
4. Edouard = 2.75 D
Explanation:
The farsighted people are those with near point greater than 25 cm.
They include Avishka, Berenice, Edouard and Francisca.
A converging lens is needed to correct farsightedness, or hyperopia, therefore, the focal length, f, is positive. The image formed is virtual and on the same side of the lens. Thus the image distance is negative
From the lens formula, 1/f = 1/v 1/u; but v is negative
Therefore, 1/f = 1/u - 1/v
But, power of a lens = 1/f in meters.
Therefore, P = 1/u - 1/v
For Avishka, u = 25 cm or 0.25 m, v = 0.4 m
P = 1/ 0.25 - 1/0.4 = 1.5 D
For Berenice, u = 0.25 m, v = 0.3 m
P = 1/0.25 - 1/0.30 =0.7 D
For Edouard, u = 0.25 m, v = 0.80 m
P = 1/0.25 - 1/0.80 =2.75 D
For Francesca, u = 0.25 m, v = 0.50 m
P = 1/0.25 - 1/0.5 = 2.0 D
When The farsighted people are those with a near point greater than 25cm.
Berenice is = 0.67 D
Avishka is = 1.50 D
Francesca is = 2.00 D
Edouard is = 2.75 D
What is Hyperopic Vision?
When The farsighted people are those with a near point greater than 25 cm. Then, They include Avishka, Berenice, Edouard, and also Francisca.
Also, A converging lens is needed to correct farsightedness, or hyperopia, thus, When the focal length, f, is positive. Also, The image formed is virtual and also on the same side of the lens. hence the image distance is negative.
Also, From the lens formula, 1/f = 1/v 1/u; but v is negative
Thus, 1/f = 1/u - 1/v
But, when the power of a lens = 1/f in meters.
Thus, P = 1/u - 1/v
For Avishka, u = 25 cm or 0.25 m, v = 0.4 m
After that, P = 1/ 0.25 - 1/0.4 = 1.5 D
Then, For Berenice, u = 0.25 m, v = 0.3 m
Now, P = 1/0.25 - 1/0.30 =0.7 D
For Edouard, u = 0.25 m, v is = 0.80 m
Then, P = 1/0.25 - 1/0.80 is =2.75 D
Now, For Francesca, u = 0.25 m, v is = 0.50 m
Therefore, P = 1/0.25 - 1/0.5 = 2.0 D
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A charming friend of yours who has been reading a little bit about astronomy accompanies you to the campus observatory and asks to see the kind of star that our Sun will ultimately become, long, long after it has turned into a white dwarf. Why is the astronomer on duty going to have a bit of a problem satisfying her request? a. All the old stars in our Galaxy are located in globular clusters and all of these are too far away to be seen with the kind of telescope a college or university campus would have. b. After being a white dwarf, the Sun will explode, and there will be nothing left to see. c. The universe is not even old enough to have produced any white dwarfs yet d. Astronomers only let people with PhD's look at these stellar corpses; it's like an initiation rite for those who become astronomers. e. After a white dwarf cools off it becomes too cold and dark to emit visible light
Answer:
b
Explanation:
Vector has a magnitude of 6.0 m and points 30° north of east. Vector has a magnitude of 4.0 m and points 30° east of north. The resultant vector + is given by
Answer:
The resultant vector is [tex]\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j[/tex].
Explanation:
First, each vector is determined in terms of absolute coordinates:
6-meter vector with direction: 30º north of east.
[tex]\vec A = (6\,m)\cdot (\cos30^{\circ} \,i + \sin 30^{\circ}\,j)[/tex]
[tex]\vec A = 5.196\,i + 3\,j[/tex]
4-meter vector with direction: 30º east of north.
[tex]\vec B = (4\,m)\cdot (\cos 60^{\circ}\,i + \sin 60^{\circ}\,j)[/tex]
[tex]\vec B = 2\,i + 3.464\,j[/tex]
The resultant vector is obtaining by sum of components:
[tex]\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j[/tex]
The resultant vector is [tex]\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j[/tex].
A 0.500 H inductor is connected in series with a 93 Ω resistor and an ac source. The voltage across the inductor is V = −(11.0V)sin[(500rad/s)t]. What is the voltage across the resistor at 2.09 x 10-3 s? Group of answer choices 205 V 515 V 636 V 542 V
Answer:
205 V
V[tex]_{R}[/tex] = 2.05 V
Explanation:
L = Inductance in Henries, (H) = 0.500 H
resistor is of 93 Ω so R = 93 Ω
The voltage across the inductor is
[tex]V_{L} = - IwLsin(wt)[/tex]
w = 500 rad/s
IwL = 11.0 V
Current:
I = 11.0 V / wL
= 11.0 V / 500 rad/s (0.500 H)
= 11.0 / 250
I = 0.044 A
Now
V[tex]_{R}[/tex] = IR
= (0.044 A) (93 Ω)
V[tex]_{R}[/tex] = 4.092 V
Deriving formula for voltage across the resistor
The derivative of sin is cos
V[tex]_{R}[/tex] = V[tex]_{R}[/tex] cos (wt)
Putting V[tex]_{R}[/tex] = 4.092 V and w = 500 rad/s
V[tex]_{R}[/tex] = V[tex]_{R}[/tex] cos (wt)
= (4.092 V) (cos(500 rad/s )t)
So the voltage across the resistor at 2.09 x 10-3 s is which means
t = 2.09 x 10⁻³
V[tex]_{R}[/tex] = (4.092 V) (cos (500 rads/s)(2.09 x 10⁻³s))
= (4.092 V) (cos (500 rads/s)(0.00209))
= (4.092 V) (cos(1.045))
= (4.092 V)(0.501902)
= 2.053783
V[tex]_{R}[/tex] = 2.05 V
Which of the following explains why a “control” is important in a case-control study of a disease? The researchers need to control the bias that those who contracted the disease may create when they talk to others. The researchers need to compare those who contracted the disease to those who did not. The researchers need to compare those who contracted the disease to those who contracted previous diseases. The researchers need to control the disease so that it is not spread further.
The researchers need to compare those who contracted the disease to those who did not.
A bucket filled with water has a mass of 23 Kg and is attached to a rope, which in turn is wound around a 0.050 m radius cylinder at the top of a well. What torque does the weight of water and bucket produce on the cylinder if the cylinder is ont permitted to rotate? (g= 9.8 m/s2)
Answer:
The torque is 11.27 N m
Explanation:
Recall that torque is the vector product of the force times the distance to the pivoting point. So in our case, the distance to the pivoting point is the radius of the cylinder (0.05 m), and the force is given by the weight of the bucket full of water (W = 9.8 * 23 N = 225.4 N)
Then the torque is: 0.05 * 225.4 N m = 11.27 N m
The following situation will be used for the next three problems: A rock is projected upward from the surface of the moon, at time t = -0.0s, with a velocity of 30m/s. The acceleration due to gravity at the surface of the moon is 1.62m/s2 the time when the rock is ascending at a height of 180m is closest to:______.
a. 8s .
b. 12s.
c. 17s.
d. 23s.
e. 30s
For the previous situation, the height of the rock when it is descending with a velocity of 20m/s is closest to:_____.
A. 115m.
B. 125m.
C. 135m.
D. 145m
E. 155m.
Explanation:
Given that,
Initial speed of the rock, u = 30 m/s
The acceleration due to gravity at the surface of the moon is 1.62 m/s².
We need to find the time when the rock is ascending at a height of 180 m.
The rock is projected from the surface of the moon. The equation of motion in this case is given by :
[tex]h=ut-\dfrac{1}{2}gt^2\\\\180=30t-\dfrac{1}{2}\times 1.62t^2[/tex]
It is a quadratic equation, after solving whose solution is given by:
t = 7.53 s
or
t = 8 seconds
(e)If it is decending, v = -20 m/s
Now t' is the time of descending. So,
[tex]v=-u+gt\\\\t=\dfrac{v+u}{g}\\\\t=\dfrac{20+30}{1.62}\\\\t=30.86\ s[/tex]
Let h' is the height of the rock at this time. So,
[tex]h'=ut-\dfrac{1}{2}gt^2\\\\h'=30\times 30.86-\dfrac{1}{2}\times 1.62\times 30.86^2\\\\h'=154.40\ m[/tex]
or
h' = 155 m
Two long parallel wires are a center-to-center distance of 1.30 cm apart and carry equal anti-parallel currents of 2.40 A. Find the magnitude of the magnetic field at the point P which is equidistant from the wires. (R = 5.00 cm).
Image is missing, so i have attached it
Answer:
19.04 × 10⁻⁴ T in the +x direction
Explanation:
We are told that the point P which is equidistant from the wires. (R = 5.00 cm). Thus distance from each wire to O is R.
Hence, the magnetic field at P from each wire would be; B = μ₀I/(2πR)
We are given;
I = 2.4 A
R = 5 cm = 0.05 m
μ₀ is a constant = 4π × 10⁻⁷ H/m
B = (4π × 10⁻⁷ × 2.4)/(2π × 0.05)
B = 9.6 × 10⁻⁴ T
To get the direction of the field from each wire, we will use Flemings right hand rule.
From the diagram attached:
We can say the field at P from the top wire will point up/right
Also, the field at P from the bottom wire will point down/right
Thus, by symmetry, the y components will cancel out leaving the two equal x components to act to the right.
If the mid-point between the wires is M, the the angle this mid point line to P makes with either A or B should be same since P is equidistant from both wires.
Let the angle be θ
Thus;
sin(θ) = (1.3/2)/5
θ = sin⁻¹(0.13) = 7.47⁰
The x component of each field would be:
9.6 × 10⁻⁴cos(7.47) = 9.52 × 10⁻⁴ T
Thus, total field = 2 × 9.52 × 10⁻⁴ = 19.04 × 10⁻⁴ T in the +x direction
The magnitude of the magnetic field at the point P will be "9.6 × 10⁻⁴ T".
Magnetic fieldThe region of the environment close to something like a magnetic entity or a current-carrying body wherein this same magnetic forces caused by the body as well as a current might well be sensed.
According to the question,
Current, I = 2.4 A
Radius, R = 5 cm or,
= 0.05 m
Constant, μ₀ = 4π × 10⁻⁷ H/m
We know the relation,
The magnetic field, B = [tex]\frac{\mu_0 I}{2 \pi R}[/tex]
By substituting the values in the above relation, we get
= [tex]\frac{4 \pi\times 10^{-7}\times 2.4}{2 \pi\times 0.05}[/tex]
= 9.6 × 10⁻⁴ T
Thus the above answer is appropriate.
Find out more information about magnetic field here:
https://brainly.com/question/14411049
The metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off. This simple
process is which kind of a change?
OA a physical change
OB. a chemical change
OC. a nuclear change
OD
an ionic change
B. A chemical change
Explanation:
I'm guessing ?
An electron is trapped between two large parallel charged plates of a capacitive system. The plates are separated by a distance of 1 cm and there is vacuum in the region between the plates. The electron is initially found midway between the plates with a kinetic energy of 11.2 eV and with its velocity directed toward the negative plate. How close to the negative plate will the electron get if the potential difference between the plates is 100 V? (1 eV = 1.6 x 10-19 J)
Answer:
The electron will get at about 0.388 cm (about 4 mm) from the negative plate before stopping.
Explanation:
Recall that the Electric field is constant inside the parallel plates, and therefore the acceleration the electron feels is constant everywhere inside the parallel plates, so we can examine its motion using kinematics of a constantly accelerated particle. This constant acceleration is (based on Newton's 2nd Law:
[tex]F=m\,a\\q\,E=m\,a\\a=\frac{q\,E}{m}[/tex]
and since the electric field E in between parallel plates separated a distance d and under a potential difference [tex]\Delta V[/tex], is given by:
[tex]E=\frac{\Delta\,V}{d}[/tex]
then :
[tex]a=\frac{q\,\Delta V}{m\,d}[/tex]
We want to find when the particle reaches velocity zero via kinematics:
[tex]v=v_0-a\,t\\0=v_0-a\,t\\t=v_0/a[/tex]
We replace this time (t) in the kinematic equation for the particle displacement:
[tex]\Delta y=v_0\,(t)-\frac{1}{2} a\,t^2\\\Delta y=v_0\,(\frac{v_0}{a} )-\frac{a}{2} (\frac{v_0}{a} )^2\\\Delta y=\frac{1}{2} \frac{v_0^2}{a}[/tex]
Replacing the values with the information given, converting the distance d into meters (0.01 m), using [tex]\Delta V=100\,V[/tex], and the electron's kinetic energy:
[tex]\frac{1}{2} \,m\,v_0^2= (11.2)\,\, 1.6\,\,10^{-19}\,\,J[/tex]
we get:
[tex]\Delta\,y= \frac{1}{2} v_0^2\,\frac{m (0.01)}{q\,(100)} =11.2 (1.6\,\,10^{-19})\,\frac{0.01}{(1.6\,\,10^{-19})\,(100)}=\frac{11.2}{10000} \,meters=0.00112\,\,meters[/tex]Therefore, since the electron was initially at 0.5 cm (0.005 m) from the negative plate, the closest it gets to this plate is:
0.005 - 0.00112 m = 0.00388 m [or 0.388 cm]
Light of wavelength 520 nm is incident a on a diffraction grating with a slit spacing of 2.20 μm , what is the angle from the axis for the third order maximum?
Answer:
θ = 45.15°
Explanation:
We need to use the grating equation in this question. The grating equation is given as follows:
mλ = d Sin θ
where,
m = order number = 3
λ = wavelength of light = 520 nm = 5.2 x 10⁻⁷ m
d = slit spacing = 2.2 μm = 2.2 x 10⁻⁶ m
θ = angle from the axis = ?
Therefore,
(3)(5.2 x 10⁻⁷ m) = (2.2 x 10⁻⁶ m) Sin θ
Sin θ = (3)(5.2 x 10⁻⁷ m)/(2.2 x 10⁻⁶ m)
Sin θ = 0.709
θ = Sin⁻¹(0.709)
θ = 45.15°
The bar magnet is pushed toward the center of a wire loop. Looking down from the top view (would appear the magnet is coming up toward the observer); Which is true? A. There is no induced current in the loop B. There is a counterclockwise induced current in the loop C. There is not enough information to correctly answer the question D. There is a clockwisee induced current in the loop
Answer:
Explanation:
B. There is a counterclockwise induced current in the loop
Explanation:
This in line with the right hand grip rule,
The right hand rule states that: to determine the direction of the magnetic force on a positive moving charge, ƒ, point the thumb of the right hand in the direction of v, the fingers in the direction of B, and a perpendicular to the palm points in the direction of F.
