Answer:
E = 0 r <R₁
Explanation:
If we use Gauss's law
Ф = ∫ E. dA = [tex]q_{int}[/tex] / ε₀
in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.
Consequently by Gauss's law the electric field is ZERO
E = 0 r <R₁
The number of daylight hours, D, in the city of Worcester, Massachusetts, where x is the number of days after January 1 (), may be calculated by the function: What is the period of this function? N/A What is the amplitude of this function? 12 What is the horizontal shift? What is the phase shift? What is the vertical shift? How many hours of sunlight will there be on February 21st of any year?
Answer:
a. 365; b. 3; c. 78; d. 1.343 rad; e. 12; f. 10.66
Explanation:
Assume that the function is
[tex]D(x) = 3 \sin \left (\dfrac{2\pi}{365}(x - 78) \right ) + 12[/tex]
The general formula for a sinusoidal function is
y = A sin(B(x - C))+ D
|A| = amplitude
B = frequency
2π/B = period, P
C = horizontal shift (phase shift)
D = vertical shift
By comparing the two formulas, we find
|A| = 3
B = 2π/365
C = 78
D = 12
a. Period
P = 2π/B = 2π/(2π/365) = 2π × 365/2π = 365
The period is 365.
b. Amplitude
|A| = 3
The amplitude is 3.
c. Horizontal shift
C= 78
The horizontal shift is 78.
d. Phase shift (φ)
Ths phase shift is the horizontal shift expressed in radians.
φ = C × 2π/365 = 78 × 2π/365 ≈ 1.343
The phase shift is 1.343 rad.
e. Vertical shift
D = 12
The vertical shift is 12.
f. Hours of sunlight on Feb 21
Feb 21 is the 52nd day of the year, so x = 51 (the number of days after Jan 1),
[tex]\begin{array}{rcl}D(x) &=& 3 \sin \left (\dfrac{2\pi}{365}(x - 78) \right ) + 12\\\\&=& 3 \sin (0.01721(51 - 78) ) + 12\\&=& 3\sin(-0.4648) + 12\\&=& 3(-0.4482) + 12\\\&=& -1.345 + 12\\& = & \textbf{10.66 h}\\\end{array}[/tex]
There will be 10.66 h of sunlight on Feb 21 of any given year.
The figure below shows the graph of the function from 0 ≤ x ≤ 365.
At what speed, as a fraction of c, will a moving rod have a length 65% that of an identical rod at rest
Answer:
v/c = 0.76
Explanation:
Formula for Length contraction is given by;
L = L_o(√(1 - (v²/c²))
Where;
L is the length of the object at a moving speed v
L_o is the length of the object at rest
v is the speed of the object
c is speed of light
Now, we are given; L = 65%L_o = 0.65L_o, since L_o is the length at rest.
Thus;
0.65L_o = L_o[√(1 - (v²/c²))]
Dividing both sides by L_o gives;
0.65 = √(1 - (v²/c²))
Squaring both sides, we have;
0.65² = (1 - (v²/c²))
v²/c² = 1 - 0.65²
v²/c² = 0.5775
Taking square root of both sides gives;
v/c = 0.76
A velocity selector in a mass spectrometer uses a 0.100-T magnetic field. (a) What electric field strength is needed to select a speed of 4.00 . 106 m/s
Answer:
The electric field strength needed is 4 x 10⁵ N/C
Explanation:
Given;
magnitude of magnetic field, B = 0.1 T
velocity of the charge, v = 4 x 10⁶ m/s
The velocity of the charge when there is a balance in the magnetic and electric force is given by;
[tex]v = \frac{E}{B}[/tex]
where;
v is the velocity of the charge
E is the electric field strength
B is the magnetic field strength
The electric field strength needed is calculated as;
E = vB
E = 4 x 10⁶ x 0.1
E = 4 x 10⁵ N/C
Therefore, the electric field strength needed is 4 x 10⁵ N/C
The base unit prefix used for 1,000× is _____. kilo milli centi deka
Answer:
[tex]\Large \boxed{\sf kilo}[/tex]
Explanation:
kilo is a prefix that means [tex]1000[/tex] of the base unit.
Answer:
kilo is the correct answer
Explanation:
because my exam says sooo....
How wide is the central diffraction peak on a screen 2.30 m behind a 0.0368-mm-wide slit illuminated by 558-nm light
Answer:
The value [tex]y = 0.0349 \ m[/tex]
Explanation:
From the question we are told that
The distance of the screen is [tex]D = 2.30 \ m[/tex]
The width of the slit is [tex]d = 0.0368 \ nm = 0.0368 *10^{-3} \ m[/tex]
The wavelength is [tex]\lambda = 558 \ nm = 558 *10^{-9} \ m[/tex]
The width of the central diffraction peak is mathematically represented as
[tex]k = 2 * y[/tex]
Where y is the distance from the center to the high peak which is mathematically represented as
[tex]y = \frac{\lambda * D }{d }[/tex]
substituting values
[tex]y = \frac{ 558 *10^{-8} * 2.30 }{0.0368 *10^{-3} }[/tex]
[tex]y = 0.0349 \ m[/tex]
Adjust the mass of the refrigerator by stacking different objects on top of it. If the mass of the refrigerator is increased (with the Applied Force held constant), what happens to the acceleration
Answer:
The acceleration of the refrigerator together with the objects decreases.
Explanation:
If the mass of the refrigerator is increased by stacking more masses (objects) on it,
and the force applied remains constant, then we know from
F = ma
where
F is the applied force
m is the total mass of the refrigerator and the objects
a is the acceleration of the masses.
If F is constant, and m is increased, the acceleration will decrease
Answer:
The acceleration decreases.
Explanation:
its right
Suppose you drop paperclips into an open cart rolling along a straight horizontal track with negligible friction. As a result of the accumulating paper clips, explain whether the momentum and kinetic energy increase, decrease, or stay the same.
Answer:
Stay the same
Explanation:
Since, friction is negligible:
Initial Momentum = Final Momentum
Initial KE = Final KE
m1 * v1 = m2 * v2
When m increases v decreases.
The momentum and kinetic energy remain the same if you drop paper clips into an open cart rolling along a straight horizontal track with negligible friction.
What is friction?Between two surfaces that are sliding or attempting to slide over one another, there is a force called friction. For instance, friction makes it challenging to push a book down the floor. Friction always moves an object in a direction that is counter to the direction that it is traveling or attempting to move.
Given:
The paperclips into an open cart rolling along a straight horizontal track with negligible friction,
Calculate the momentum, Since friction is negligible,
Initial Momentum = Final Momentum
Initial Kinetic Energy = Final Kinetic Energy
m₁ × v₁ = m₁ × v₂
When m increases, v decreases,
Thus, momentum will remain the same.
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Categorize each ray tracing statement as relating to ray 1, ray 2, or ray 3.
A. Drawn from the top of the object so that it passes through the center of the lens at the optical axis.
B. Drawn from the top of the object so that it passes through the focal point on the same side of the lens as the object.
C. Drawn parallel to the optical axis from the top of the object.
D. Ray bends parallel to the optical axis.
E. Ray bends so that it passes through the focal point on the opposite side of the lens as the object.
F. Ray does not bend.
Answer:
statement 1 with answer C
statement 2 with answer F
statement 3 with answer B
Statement 1 with E
Statement 2 with A
Statement 3 with D
Explanation:
In this exercise you are asked to relate each with the answers
In general, in the optics diagram,
* Ray 1 is a horizontal ray that after stopping by the optical system goes to the focal point
* Ray 2 is a ray that passes through the intercept point between the optical axis and the system and does not deviate
* Ray 3 is a ray that passes through the focal length and after passing the optical system, it comes out horizontally.
With these statements, let's review the answers
statement 1 with answer C
statement 2 with answer F
statement 3 with answer B
Statement 1 with E
Statement 2 with A
Statement 3 with D
Four charges each of magnitude 15 µC are arranged on the corners of a square of side 5 cm. What is the total potential energy of the system?
Answer:
-105J
Explanation:
See attached file
Problem 25.40 What is the energy (in eV) of a photon of visible light that has a wavelength of 500 nm
Answer:
E = 2.48 eV
Explanation:
The energy of a photon is given by the following formula:
E = hυ
where,
E = Energy of Photon = ?
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
υ = frequency of photon = c/λ
Therefore,
E = hc/λ
where,
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of light = 500 nm = 5 x 10⁻⁷ m
Therefore,
E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(5 x 10⁻⁷ m)
E = (3.97 x 10⁻¹⁹ J)(1 eV/1.6 x 10⁻¹⁹ J)
E = 2.48 eV
A photon of visible light that has a wavelength of 500 nm, has an energy of 2.48 eV.
We can calculate the energy (E) of a photon with a wavelength (λ) of 500 nm using the Planck's-Einstein relation.
[tex]E = \frac{h \times c}{\lambda } = \frac{(6.63 \times 10^{-34}J.s ) \times (3.00 \times 10^{8}m/s )}{500 \times 10^{-9}m } = 3.98 \times 10^{-19} J[/tex]
where,
h: Planck's constantc: speed of lightWe can convert 3.98 × 10⁻¹⁹ J to eV using the conversion factor 1 J = 6.24 × 10¹⁸ eV.
[tex]3.98 \times 10^{-19} J \times \frac{6.24 \times 10^{18} eV }{1J} = 2.48 eV[/tex]
A photon of visible light that has a wavelength of 500 nm, has an energy of 2.48 eV.
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A fireworks rocket is launched vertically upward at 40 m/s. At the peak of its trajectory, it explodes into two equal-mass fragments. One reaches the ground t1 = 2.71s after the explosion.When does the second reach the ground?t=?
Answer:
6.13 seconds
Explanation:
At the peak of the fireworks trajectory, the velocity of the firework would be zero. Using equation of motion, we have:
v² = u² + 2gh
0 = 40² - (2)(9.81)(h)
0 = 1600 - 19.62h
19.62h = 1600
h = 1600/19.62
h = 81.55 m
Now during the process of explosion, the two parts gained equal vertical momentum but in opposite directions.
We are told the first piece lands in a time of 2.71 s,
Using 3rd equation of motion, we have;
h = ut + ½gt²
81.55 = u(2.71) + ½(9.81 × 2.71²)
81.55 = 2.71u + 36.0228
2.71u = 81.55 - 36.0228
2.71u = 45.5272
u = 45.5272/2.71
u = 16.8 m/s
The time it takes a projectile to return back to its original launch point assuming the projectile was launched
vertically with speed u = 16.8 m/s is;
t = 2u/g
t = (2 × 16.8)/9.81
t = 3.43 s
Thus total time it takes the second mass to reach the ground = 3.43 + 2.71 = 6.13 seconds
An insect 1.1 mm tall is placed 1.0 mm beyond the focal point of the objective lens of a compound microscope. The objective lens has a focal length of 14 mm , the eyepiece a focal length of 21 mm .
A) Where is the image formed by the objective lens? Give your answer as the distance from the image to the lens. Express your answer using two significant figures.
B) How tall is the image mentioned in part A? Express your answer using two significant figures.
C) If you want to place the eyepiece so that the image it produces is at infinity, how far should this lens be from the image produced by the objective lens? Express your answer using two significant figures.
D) Under the conditions of part C, find the overall magnification of the microscope. Express your answer using two significant figures.
Answer:
Explanation:
For image formation in objective lens
object distance u = 14 +1 = 15 mm
focal length f = 14 mm .
image distance v = ?
lens formula
[tex]\frac{1}{v} -\frac{1}{u} =\frac{1}{f}[/tex]
Putting the values
[tex]\frac{1}{v} +\frac{1}{15} =\frac{1}{14}[/tex]
v = 210 mm .
B )
magnification = v / u
= 210 / 15
= 14
size of image = 14 x 1.1 mm
= 15.4 mm
= 15 mm approx
C )
For final image to be at infinity , image produced by objective lens must fall at the focal point of eye piece . so objective lens's distance from the image formed by objective must be equal to focal length of eye piece that is 21 mm .
21 mm is the answer .
D )
overall magnification =
[tex]\frac{210}{15} \times \frac{D}{f_e}[/tex]
D = 25 cm , f_e = focal length of eye piece
= 14 x 250 / 21
= 166.67
= 170 ( in two significant figures )
(a) The distance of the image v=220mm
(b) SIze of the image 15 mm
(c) Distance of lens be from the image produced by the objective lens 21 mm
(d) overall magnification of the microscope 170
What is objective lens?The objective lens of a microscope is the one at the bottom near the sample. At its simplest, it is a very high-powered magnifying glass, with very short focal length. This is brought very close to the specimen being examined so that the light from the specimen comes to a focus inside the microscope tube
For image formation in objective lens
object distance u = 14 +1 = 15 mm
focal length f = 14 mm .
image distance v = ?
By using lens formula
[tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}[/tex]
Putting the values
[tex]\dfrac{1}{v}+\dfrac{1}{15}=\dfrac{1}{14}[/tex]
v = 210 mm .
B ) Magnification is the ratio of the size of the image to the size of the an object.
[tex]\rm magnification = \dfrac{v} { u}[/tex]
[tex]M= \dfrac{210} { 15}[/tex]
M= 14
size of image = 14 x 1.1 mm
= 15.4 mm
= 15 mm approx
C )
For final image to be at infinity , image produced by objective lens must fall at the focal point of eye piece . so objective lens's distance from the image formed by objective must be equal to focal length of eye piece that is 21 mm .
21 mm is the answer .
D )
overall magnification =
[tex]\dfrac{210}{15}\times \dfrac{D}{f_e}[/tex]
D = 25 cm , f_e = focal length of eye piece
[tex]= 14 \times \dfrac{ 250} { 21}[/tex]
= 166.67
= 170 ( in two significant figures )
Hence all the answers are:
(a) The distance of the image v=220mm
(b) SIze of the image 15 mm
(c) Distance of lens be from the image produced by the objective lens 21 mm
(d) overall magnification of the microscope 170
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How to do this question
Answer:
(a) 10 m/s
(b) 22.4 m/s
Explanation:
(a) Draw a free body diagram of the car when it is at the top of the loop. There are two forces: weight force mg pulling down, and normal force N pushing down.
Sum of forces in the centripetal direction (towards the center):
∑F = ma
mg + N = mv²/r
At minimum speed, the normal force is 0.
mg = mv²/r
g = v²/r
v = √(gr)
v = √(10 m/s² × 10.0 m)
v = 10 m/s
(b) Energy is conserved.
Initial kinetic energy + initial potential energy = final kinetic energy
½ mv₀² + mgh = ½ mv²
v₀² + 2gh = v²
(10 m/s)² + 2 (10 m/s²) (20.0 m) = v²
v = 22.4 m/s
A) Hooke's law is described mathematically using the formula Fsp = -ku. Which statement is correct about the spring force, Fsp?
A.It is a vector quantity
B.It is the force doing the push or pull,
C.It is always a positive force.
D.It is larger than the applied force.
1. Which example best describes a restoring force?
B) the force applied to restore a spring to its original length
2. A spring is compressed, resulting in its displacement to the right. What happens to the spring when it is released?
C) The spring exerts a restoring force to the left and returns to its equilibrium position.
3. A 2-N force is applied to a spring, and there is displacement of 0.4 m. How much would the spring be displaced if a 5-N force was applied?
D) 1 m
4. Hooke’s law is described mathematically using the formula Fsp=−kx. Which statement is correct about the spring force, Fsp?
D)It is a vector quantity.
5. What happens to the displacement vector when the spring constant has a higher value and the applied force remains constant?
A) It decreases in magnatude.
Hope this Helps!! Sorry its late
The main purpose of a written report may be to _____. A. revise a hypothesis B. summarize other scientists' results C. design a procedure for an experiment D. analyze data without drawing conclusions
PLZZZ HURRY TIMED MARK BRAINLIEST
Answer:
analyze data without drawing conclusions
Explanation:
Research reports are written in order to communicate clearly, information obtained primarily from research and analysis of data.
Typical reports of scientific research endeavours are written in such a way that they convey the research process succinctly without excessive extraneous information. A report is typically made up of; summary of the contents, introduction/ background, methods, results, discussion, conclusion and recommendations.
Hence a report does not really make inferences from the research findings.
The square armature coil of an alternating current generator has 200 turns and is 20.0 cm on side. When it rotates at 3600 rpm, its peak output voltage is 120 V.
A) What is the frequency of the output voltage?
B) What is the strength of the magnetic field in which the coil is turning?
Answer:
A) 60 Hz
B) 0.04 T
Explanation:
Given that.
Number of turns, N = 200
Length of the side, l = 20 cm = 0.2 m
Speed if rotation, w = 3600 rpm
Voltage, V = 120 V
First, we try to convert the speed from rpm to rad/s
3600 * (2π/60)
3600 * 0.10473
3600 rpm = 377 rad/s
Now, we use that as our w, speed of rotation
Frequency of output, f =
w/2π
f = 377 / 6.284
f = 59.99 Hz or approximately, 60 Hz.
B
Strength of the magnetic field in which the coil is turning
E• = NABw
Where, A = l² = 0.2² = 0.04, on substituting the values to the equation, we have
120 = 200 * 0.04 * 377 * B
120
Making B subject of formula,
B = 120/ 3016
B = 0.04 T..
The frequency of the output voltage is 60 Hz and the strength of the magnetic field is 0.04 T
Which is a dopant for a p-type semiconductor? arsenic indium phosphorus antimony
Answer:
As opposed to n-type semiconductors, p-type semiconductors have a larger hole concentration than electron concentration.
Explanation:
In p-type semiconductors, holes are the majority carriers and electrons are the minority carriers. A common p-type dopant for silicon is boron or gallium. hope this you :)
Answer:
IndiumHere are notes I took on semiconductor conductivity :
________________________________________________________
-A p-type semiconductor is made of a material in which electrical conduction is due to the movement of a positive charge.
-Examples of p-type dopants - boron, aluminum, gallium, indium, and thallium
Explanation:
In this case, indium only correct option being a dopant of a P-type semiconductor. Other options are N-type dopants.
Hopefully its correct !! <3
An electric train operates on 800 V. What is its power consumption when the current flowing through the train's motor is 2,130 A?
Answer:
1704 kWExplanation:
To solve for the power consumed by the trains motor we have to employ the formula for power which is
Power= current * voltage
Given that
voltage V= 800 V
current I= 2130 A
Substituting in the formula for power we have
Power= 2130*800= 1704000 watt
Power = 1704 kW
This is the amount of energy consumed, transferred or converted per unit of time
Hence the power consumed by the trains motor is 1704 kW
Air at 27oC and 1 atm flows over a flat plate 40 cm in length and 1 cm in width at a speed of 2 m/s. The plate is heated over its entire length to a temperature of 600C. Calculate the heat transferred from the plate.
Answer:
Heat transferred = 22.9 watt
Explanation:
Given that:
[tex]T_1[/tex] = 27°C = (273 + 27) K = 300 K
[tex]T_2[/tex]= 600°C = (600 +273) K = 873 K
speed v = 2 m/s
length x = 40 cm = 0.4 cm
width = 1 cm = 0.001 m
The heat transferred from the plate can be calculate by using the formula:
Heat transferred = h×A ×ΔT
From the tables of properties of air, the following values where obtained.
[tex]k = 0.02476 \ W/m.k \\ \\ \rho = 1.225 \ kg/m^3 \\ \\ \mu = 18.6 \times 10^{-6} \ Pa.s \\ \\ c_p = 1.005 \ kJ/kg[/tex]
To start with the reynolds number; the formula for calculating the reynolds number can be expressed as:
reynolds number = [tex]\dfrac{\rho \times v \times x }{\mu}[/tex]
reynolds number = [tex]\dfrac{1.225 \times 2 \times 0.4}{18.6 \times 10^{-6}}[/tex]
reynolds number = [tex]\dfrac{0.98}{18.6 \times 10^{-6}}[/tex]
reynolds number = 52688.11204
Prandtl number = [tex]\dfrac{c_p \mu}{k}[/tex]
Prandtl number = [tex]\dfrac{1.005 \times 18.6 \times 10^{-6} \times 10^3}{0.02476}[/tex]
Prandtl number = [tex]\dfrac{0.018693}{0.02476}[/tex]
Prandtl number = 0.754963
The nusselt number for this turbulent flow over the flat plate can be computed as follows:
Nusselt no = [tex]\dfrac{hx}{k} = 0.0296 (Re) ^{0.8} \times (Pr)^{1/3}[/tex]
[tex]\dfrac{h \times 0.4}{0.02476} = 0.0296 (52688.11204) ^{0.8} \times (0.754968)^{1/3}[/tex]
[tex]\dfrac{h \times 0.4}{0.02476} =161.4252008}[/tex]
[tex]h =\dfrac{161.4252008 \times 0.02476}{ 0.4}[/tex]
h = 9.992 W/m.k
Recall that:
The heat transferred from the plate can be calculate by using the formula:
Heat transferred = h×A ×ΔT
Heat transferred = [tex]h\times A \times (T_2-T_1)[/tex]
Heat transferred = 9.992 × (0.4 × 0.01) ×(873-300)
Heat transferred = 22.9 watt
A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotates horizontally about one end while the pilot is seated at the other end. In order to achieve a radial acceleration of 34.1 m/s2 with a beam of length 5.55 m , what rotation frequency is required?
Answer:
f = 0.4 Hz
Explanation:
The frequency of rotation of an object in order to achieve required centripetal or radial acceleration, can be found out by using the following formula:
f = (1/2π)√(ac/r)
where,
f = frequency of rotation = ?
ac = radial acceleration = 34.1 m/s²
r = radius = length of beam = 5.55 m
Therefore,
f = (1/2π)√[(34.1 m/s²)/(5.55 m)]
f = 0.4 Hz
A car is travelling west at 22.2 m/s when it accelerated for 0.80 s to the west at 2.68 m/s2. Calculate the car's final velocity. Show all your work.
Answer:
24.34 m/s
Explanation:
recall that one of the equations of motions takes the form:
v = u + at
where,
v = final velocity
u = initial velocity (given as 22.2 m/s)
a = acceleration (given as 2.68m/s²)
t = time elapsed during acceleration (given as 0.80s)
since we are told that the the acceleration is in the direction of the intial velocity, we can simply substitute the known values into the equation above:
v = u + at
v = 22.2 + (2.68) (0.8)
v = 24.34 m/s
please help !!!!! please note that two images are there................ i am urgently needs this question
Answer:
can you tell me about this property
If the terminals of a battery with zero internal resistance are connected across two identical resistors in series, the total power delivered by the battery is 8.00 W. If the same battery is connected across the same resistors in parallel, what is the total power delivered by the battery
Answer:
24W
Explanation:
The series connection has a resistance of 2R
The parallel connection has a resistance of R/2 .. the resistance has decreased by a factor 4
Assuming the battery still provides the same pd .. the current increases by a factor of 4 .. increasing the power output by a factor of 4 also (P = V x A)
Power output = 4 x 8W .. .. So P = 24 W
Metal 1 has a larger work function than metal 2. Both are illuminated with the same short-wavelength ultraviolet light.
Do electrons from metal 1 have a higher speed, a lower speed, or the same speed as electrons from metal 2? Explain.
Answer:
a lower speed
Explanation:
Let us look closely at the Einstein's photoelectric equation;
KE= E-Wo
Where;
KE= kinetic energy of the emitted photoelectron
E= energy of the incident photon
Wo= work function of the metal
Hence,where Wo for metal 1 > Wo for metal 2, it follows that KE for metal 1 must also be less than KE for metal 2.
This is because the difference between E and Wo for metal 1 is smaller than the same difference for metal 2 hence the answer.
How many turns of wire are needed in a circular coil 13 cmcm in diameter to produce an induced emf of 5.6 VV
Answer:
Number of turns of wire(N) = 3,036 turns (Approx)
Explanation:
Given:
Diameter = 13 Cm
emf = 5.6 v
Note:
The given question is incomplete, unknown information is as follow.
Magnetic field increases = 0.25 T in 1.8 (Second)
Find:
Number of turns of wire(N)
Computation:
radius (r) = 13 / 2 = 6.5 cm = 0.065 m
Area = πr²
Area = (22/7)(0.065)(0.065)
Area = 0.013278 m²
So,
emf = (N)(A)(dB / dt)
5.6 = (N)(0.013278)(0.25 / 1.8)
5.6 = (N)(0.013278)(0.1389)
N = 3,036.35899
Number of turns of wire(N) = 3,036 turns (Approx)
The filament in the bulb is moving back and forth, first pushed one way and then the other. What does this imply about the current in the filament
Answer:
energy carried by the current is given by the pointyng vector
Explanation:
The current is defined by
i = dQ / dt
this is the number of charges per unit area over time.
The movement of the charge carriers (electrons) is governed by the applied potential difference, when the filament has a movement the drag speed of these moving electrons should change slightly.
But the energy carried by the current is given by the pointyng vector of the electromagnetic wave
S = 1 / μ₀ EX B
It moves at the speed of light and its speed depends on the properties of the doctor and is not disturbed by small changes in speed, therefore the current in the circuit does not change due to this movement
what path would an object have to take to have the distance and the displacement to be equal
Answer:
When an object move in a straight line without moving back.
Explanation:
Distance is covered by an object is the magnitude of length from one position to the another. It is a scalar quantity.
While displacement is the distance covered in a specific direction. Displacement is a vector quantity. It has both magnitude and direction.
If an object move in a straight path without going back, then, the magnitude of distance will be the same with the magnitude of displacement.
Both distance and displacement are measured in the same unit which is metres.
Therefore, an object have to take a straight path without going back to have the distance and the displacement equal.
A deep-space vehicle moves away from the Earth with a speed of 0.870c. An astronaut on the vehicle measures a time interval of 3.10 s to rotate her body through 1.00 rev as she floats in the vehicle. What time interval is required for this rotation according to an observer on the Earth
Answer:
t₀ = 1.55 s
Explanation:
According to Einstein's Theory of Relativity, when an object moves with a speed comparable to speed of light, the time interval measured for the event, by an observer in motion relative to the event is not the same as measured by an observer at rest.
It is given as:
t = t₀/[√(1 - v²/c²)]
where,
t = time measured by astronaut in motion = 3.1 s
t₀ = time required according to observer on earth = ?
v = relative velocity = 0.87 c
c = speed of light
3.1 s = t₀/[√(1 - 0.87²c²/c²)]
(3.1 s)(0.5) = t₀
t₀ = 1.55 s
Answer:
The time interval required for this rotation according to an observer on the Earth = [tex]6.29sec[/tex]Explanation:
Time interval required for this rotation according to an observer on the Earth is given as [tex]\delta t[/tex]
where,
[tex]t_o = 3.1\\\\v = 0.87[/tex]
[tex]\delta t = \frac{t_o}{\sqrt{1-\frac{v^2}{c^2}}}\\\\\delta t = \frac{3.1}{\sqrt{1-(\frac{0.87c}{c})^2}}\\\\\delta t = 6.29sec[/tex]
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The distance from the center of a lens to the location where parallel rays converge or appear to converge is called the _____ length.
Answer:
FOCALExplanation:
The center of a lens is known as its optical center. All light rays incident on a particular lens converges at a points a point known as the principal focus or the focal point after reflecting. Note that all light incident on a reflecting surface must all converge at this focal point after reflection.
The distance measured from the center of this lens to its principal focus (otherwise known as focal point) is known as the focal length of the lens.
Based on the explanation above, it cam be concluded that the distance from the center of a lens to the location where parallel rays converge or appear to converge is called the FOCAL length.
Answer:
X and Y are two uncharged metal spheres on insulating stands, and are in contact with each other. A positively charged rod R is brought close to X as shown in Figure (a).
The figure shows two spheres on stands and the positively charged rod. The sphere on the left is marked X. The sphere on the right is marked Y. The spheres are in contact with each other. The rod is marked R and it is located to the left of sphere X.
Sphere Y is now moved away from X , as in Figure (b).
The figure shows two spheres on stands and the positively charged rod. The sphere on the left is marked X. The sphere on the right is marked Y and it is moved away from sphere X. The rod is marked R and it is located to the left of sphere X.
What are the final charge states of X and Y?
Both X and Y are neutral.
X is neutral and Y is positive.
X is positive and Y is neutral.
X is negative and Y is positive.
Both X and Y are negative.
Explanation:
Light passes through a single slit. If the width of the slit is reduced, what happens to the width of the central bright fringe
Explanation:
In Single Slit Experiment:
The width of the central diffraction maximum is inversely proportional to the width of the slit.
Therefore, if we make the slit width smaller, the angle T(representing the angle between the wave ray to a point on the screen and the normal line between the slit and the screen) increases, giving a wider central band.
