Answer:
The three factors that can contribute to creating a delay in advanced care for the passengers in the overturned bus include:
1. Lack of communication: Since the accident happened on the remote hillside, there is a possibility that, there would be no communication network which could have afforded them the opportunity to call medical or technical team.
2. Steep Nature of the Hill: This is another factor which will affect the care which they could have received. Steeply area tends to be difficult for climbing in or out from.
3. Thunderstorm: This factor is another reason which could contribute to delay in receiving advance care. Thunderstorm create barriers for location f the area where the bus overturned or in other situation complicate the rescue efforts of the team sent out to rescue.
Explanation:
A thermos bottle works well because:
a. its glass walls are thin
b. silvering reduces convection
c. vacuum reduces heat radiation
d. silver coating is a poor heat conductor
e. none of the above
Answer:
A thermos bottle works well because:
A) Its glass walls are thin
Answer:
A thermos bottle works well because:
C
Vacuum reduces heat radiation
Rank the following types of electromagnetic waves by the wavelength of the wave.
a. Microwaves
b. X-rays
c. Radio waves
d. Visible light
Explanation:
In order of Increasing Wavelength of the Electromagnetic Spectrum :
B) X rays
D) Visible light
A) Microwave
C) Radio Waves
Electromagnetic waves in order of decreasing wavelength is X-rays,visible light,microwaves and radio waves.
What are electromagnetic waves?The electromagnetic radiation consists of waves made up of electromagnetic field which are capable of propogating through space and carry the radiant electromagnetic energy.
The radiation are composed of electromagnetic waves which are synchronized oscillations of electric and magnetic fields . They are created due to change which is periodic in electric as well as magnetic fields.
In vacuum ,all the electromagnetic waves travel at the same speed that is with the speed of air.The position of an electromagnetic wave in an electromagnetic spectrum is characterized by it's frequency or wavelength.They are emitted by electrically charged particles which undergo acceleration and subsequently interact with other charged particles.
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When light of wavelength 233 nm shines on a metal surface the maximum kinetic energy of the photoelectrons is 1.98 eV. What is the maximum wavelength (in nm) of light that will produce photoelectrons from this surface
Answer:
λmax = 372 nm
Explanation:
First we find the energy of photon:
E = hc/λ
where,
E = Energy of Photon = ?
λ = Wavelength of Light = 233 nm = 2.33 x 10⁻⁷ m
c = speed of light = 3 x 10⁸ m/s
h = Planks Constant = 6.626 x 10⁻³⁴ J.s
Therefore,
E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2.33 x 10⁻⁷ m)
E = 8.5 x 10⁻¹⁹ J
Now, from Einstein's Photoelectric Equation:
E = Work Function + Kinetic Energy
8.5 x 10⁻¹⁹ J = Work Function + (1.98 eV)(1.6 x 10⁻¹⁹ J/1 eV)
Work Function = 8.5 x 10⁻¹⁹ J - 3.168 x 10⁻¹⁹ J
Work Function = 5.332 x 10⁻¹⁹ J
Since, work function is the minimum amount of energy required to emit electron. Therefore:
Work Function = hc/λmax
λmax = hc/Work Function
where,
λmax = maximum wavelength of light that will produce photoelectrons = ?
Therefore,
λmax = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(5.332 x 10⁻¹⁹ J)
λmax = 3.72 x 10⁻⁷ m
λmax = 372 nm
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.
Object A, with heat capacity CA and initially at temperature TA, is placed in thermal contact with object B, with heat capacity CB and initially at temperature TB. The combination is thermally isolated. If the heat capacities are independent of the temperature and no phase changes occur, the final temperature of both objects is
Answer:
d) (CATA + CBTB) / (CA + CB)
Explanation:
According to the given situation, the final temperature of both objects is shown below:-
We assume T be the final temperature
while m be the mass
So it will be represent
m CA (TA - T) = m CB (T - TB)
or we can say that
CATA - CA T = CB T - CBTB
or
(CA + CB) T = CATA + CBTB
or
T = (CA TA + CBTB) ÷ (CA + CB)
Therefore the right answer is d
The final temperature of both objects is [tex]T = \frac{C_AT_A+ C_BT_B}{C_B + C_A} \\\\[/tex].
The given parameters;
heat capacity of object A = CAinitial temperature of object A = TAheat capacity of object B = CBinitial temperature of object B = TBThe final temperature of both objects is calculated as follows;
heat lost by object A is equal to heat gained by object B
[tex]mC_A (T_A - T) = mC_B(T- T_B)\\\\C_AT_A-C_AT = C_BT - C_BT_B\\\\C_BT+C_AT = C_AT_A+ C_BT_B\\\\T(C_B + C_A) = C_AT_A+ C_BT_B \\\\T = \frac{C_AT_A+ C_BT_B}{C_B + C_A} \\\\[/tex]
Thus, the final temperature of both objects is [tex]T = \frac{C_AT_A+ C_BT_B}{C_B + C_A} \\\\[/tex].
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A charged capacitor and an inductor are connected in series. At time t = 0, the current is zero, but the capacitor is charged. If T is the period of the resulting oscillations, the next time, after t = 0 that the energy stored in the magnetic field of the inductor is a maximum is
Answer:
t = T / 2 all energy is stored in the inductor
Explanation:
The circuit described is an oscillating circuit where the charge of the condensation stops the inductor and vice versa, in this system the angular velocity of the oscillation is
w = √1/LC
2π / T =√1 / LC
T = 2π √LC
The energy is constant and for the initial instant it is completely stored in the capacitor
Uc = Q₀² / 2C
In the process, the capacitor is discharging and the energy is stored in the inductor until when the charge in the capacitors zero, all the energy is stored in the inductor
U = L I² / 2
in the intermediate instant the energy is stored in the two elements.
Since the period of the system is T for time t = 0 all energy is stored in the capacitor and for t = T / 2 all energy is stored in the inductor
After t = 0 the maximum energy stored in the magnetic field of the inductor is equal to [tex]U'=\dfrac{L I^{2}}{2}[/tex] for the time period, half of period of oscillation (t = T/2).
The given problem is based on the charging and discharging concepts of capacitor. An oscillating circuit is a circuit where the charge of the capacitor stops the inductor and vice versa, in this system the angular frequency of the oscillation is given as,
[tex]\omega =\dfrac{1}{\sqrt{LC}}\\\\\\\dfrac{2 \pi}{T} =\dfrac{1}{\sqrt{LC}}\\\\\\T = 2\pi \times \sqrt{LC}[/tex]
here, T is the period of oscillation.
Also, the energy stored in the capacitor is constant and for the initial instant it is completely stored in the capacitor. So, the energy stored is given as,
[tex]U =\dfrac{Q^{2}}{2C}[/tex]
here, C is the capacitance.
In the process, the capacitor is discharging and the energy is stored in the inductor until when the charge in the capacitors zero, all the energy is stored in the inductor. So, the expression for the energy stored in the inductor is,
[tex]U'=\dfrac{L I^{2}}{2}[/tex]
here, L is the inductance and I is the current.
Note :- The period of the system is T for time t = 0 all energy is stored in the capacitor and for t = T / 2 all energy is stored in the inductor.
Thus, we conclude that after t = 0 the maximum energy stored in the magnetic field of the inductor is equal to [tex]U'=\dfrac{L I^{2}}{2}[/tex] for the time period, half of period of oscillation (t = T/2).
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A simple arrangement by means of which e.m.f,s. are compared is known
Answer:
A simple arrangement by means of which e.m.f,s. are compared is known as?
(a)Voltmeter
(b)Potentiometer
(c)Ammeter
(d)None of the above
Explanation:
Light of wavelength 500 nm falls on two slits spaced 0.2 mm apart. If the spacing between the first and third dark fringes is to be 4.0 mm, what is the distance from the slits to a screen?
Answer:
L = 0.8 m
Explanation:
Since, the distance between first and third dark fringes is 4 mm. Therefore, the fringe spacing between consecutive dark fringes will be:
Δx = 4 mm/2 = 2 mm = 2 x 10⁻³ m
but,
Δx = λL/d
λ = wavelength of the light = 500 nm = 5 x 10⁻⁷ m
d = slit spacing = 0.2 mm = 0.2 x 10⁻³ m
L = Distance between slits and screen = ?
Therefore, using the values, we get:
2 x 10⁻³ m = (5 x 10⁻⁷ m)(L)/(0.2 x 10⁻³)
L = (2 x 10⁻³ m)(0.2 x 10⁻³ m)/(5 x 10⁻⁷ m)
L = 0.8 m
g In the atmosphere, the shortest wavelength electromagnetic waves are called A. infrared waves. B. ultraviolet waves. C. X-rays. D. gamma rays. E.
Answer:gamma ray
Explanation:
Light with an intensity of 1 kW/m2 falls normally on a surface and is completely absorbed. The radiation pressure is
Answer:
The radiation pressure of the light is 3.33 x 10⁻⁶ Pa.
Explanation:
Given;
intensity of light, I = 1 kW/m²
The radiation pressure of light is given as;
[tex]Radiation \ Pressure = \frac{Flux \ density}{Speed \ of \ light}[/tex]
I kW = 1000 J/s
The energy flux density = 1000 J/m².s
The speed of light = 3 x 10⁸ m/s
Thus, the radiation pressure of the light is calculated as;
[tex]Radiation \ pressure = \frac{1000}{3*10^{8}} \\\\Radiation \ pressure =3.33*10^{-6} \ Pa[/tex]
Therefore, the radiation pressure of the light is 3.33 x 10⁻⁶ Pa.
A body is thrown vertically upwards with a speed of 95m / s and after 7s it reaches its maximum height. How fast does it reach its maximum height? What was the maximum height reached?
Explanation:
u = 95 m/sec ( Initial speed)
t = 7 sec ( Time of ascent)
According to Equations of Motion :
[tex]s = ut - \frac{1}{2} g {t}^{2} [/tex]
Max. Height = 95 * 7 - 4.9 * 49 = 424. 9 = 425 m
Answer:
332.5 m
Explanation:
At the maximum height, the velocity is 0.
Given:
v₀ = 95 m/s
v = 0 m/s
t = 7 s
Find: Δy
Δy = ½ (v + v₀) t
Δy = ½ (0 m/s + 95 m/s) (7 s)
Δy = 332.5 m
A pair of narrow, parallel slits separated by 0.230 mm is illuminated by green light (λ = 546.1 nm). The interference pattern is observed on a screen 1.50 m away from the plane of the parallel slits.
A) Calculate the distance from the central maximum to the first bright region on either side of the central maximum.
B) Calculate the distance between the first and second dark bands in the interference pattern.
Answer:
A) y = 3.56 mm
B) y = 3.56 mm
Explanation:
A) The distance from the central maximum to the first bright region can be found using Young's double-slit equation:
[tex] y = \frac{m\lambda L}{d} [/tex]
Where:
λ: is the wavelength = 546.1 nm
m: is first bright region = 1
L: is the distance between the screen and the plane of the parallel slits = 1.50 m
d: is the separation between the slits = 0.230 mm
[tex] y = \frac{m\lambda L}{d} = \frac{1*546.1 \cdot 10^{-9} m*1.50 m}{0.230 \cdot 10^{-3} m} = 3.56 \cdot 10^{-3} m [/tex]
B) The distance between the first and second dark bands is:
[tex] \Delta y = \frac{\Delta m*\lambda L}{d} [/tex]
Where:
[tex] \Delta m = m_{2} - m_{1} = 2 - 1 = 1 [/tex]
[tex] \Delta y = \frac{1*546.1 \cdot 10^{-9} m*1.50 m}{0.230 \cdot 10^{-3} m} = 3.56 \cdot 10^{-3} m [/tex]
I hope it helps you!
The following equation is an example of
decay.
181
185
79
Au →
4
2
He+
Answer:
Alp decay.
Explanation:
From the above equation, the parent nucleus 185 79Au produces a daughter nuclei 181 77 Ir.
A careful observation of the atomic mass of the parent nucleus (185) and the atomic mass of the daughter nuclei (181) shows that the atomic mass of the daughter nuclei decreased by a factor of 4. Also, the atomic number of the daughter nuclei also decreased by a factor of 2 when compared with the parent nucleus as shown in the equation given above.
This simply means that the parent nucleus has undergone alpha decay which is represented with a helium atom as 4 2He.
Therefore, the equation is an example of alpha decay.
What is the de Broglie wavelength of an object with a mass of 2.50 kg moving at a speed of 2.70 m/s? (Useful constant: h = 6.63×10-34 Js.)
Answer:
9.82 × [tex]10^{-35}[/tex] Hz
Explanation:
De Broglie equation is used to determine the wavelength of a particle (e.g electron) in motion. It is given as:
λ = [tex]\frac{h}{mv}[/tex]
where: λ is the required wavelength of the moving electron, h is the Planck's constant, m is the mass of the particle, v is its speed.
Given that: h = 6.63 ×[tex]10^{-34}[/tex] Js, m = 2.50 kg, v = 2.70 m/s, the wavelength, λ, can be determined as follows;
λ = [tex]\frac{h}{mv}[/tex]
= [tex]\frac{6.63*10^{-34} }{2.5*2.7}[/tex]
= [tex]\frac{6.63 * 10^{-34} }{6.75}[/tex]
= 9.8222 × [tex]10^{-35}[/tex]
The wavelength of the object is 9.82 × [tex]10^{-35}[/tex] Hz.
For exercise, an athlete lifts a barbell that weighs 400 N from the ground to a height of 2.0 m in a time of 1.6 s. Assume the efficiency of the human body is 25%, and that he lifts the barbell at a constant speed. Show all work and include proper unit for your final answer.
a) In applying the energy equation (ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W) to the system consisting of the earth, the barbell, and the athlete,
1. Which terms (if any) are positive?
2. Which terms (if any) are negative?
3. Which terms (if any) are zero?
b) Determine the energy output by the athlete in SI unit.
c) Determine his metabolic power in SI unit.
d) Another day he performs the same task in 1.2 s.
1. Is the metabolic energy that he expends more, less, or the same?
2. Is his metabolic power more, less, or the same?
Answer:
Explanation:
(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)
ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .
ΔK = 0
ΔUg is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .
ΔUg = positive
ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .
ΔUs = 0
ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence
ΔEth = negative .
b )
work done by athlete
= 400 x 2 = 800 J
energy output = 800 J
c )
It is 25% of metabolic energy output of his body
so metalic energy output of body
= 4x 800 J .
3200 J
power = energy output / time
= 3200 / 1.6
= 2000 W .
d )
1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .
2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .
A) Applying the energy equation
The positive terms is : ΔUg The negative terms is : ΔEth The zero term are : ΔK and ΔUsB) The energy output by the athlete is ; 800 Joules
C) The metabolic power is : 2000 w
D) When he performs the task in 1.2 s
The metabolic energy he expends is : the same His metabolic power is : moreGiven data :
Weight of barbell = 400 N
Height = 2.0 m
Time = 1.6 secs
efficiency of the human body = 25%
Speed = constant
A) From the energy equation the ΔK is zero because the athlete is lifting the barbell at a constant speed. ΔUg is positive because as the weight is lifted its potential energy increases. ΔEth ( change in energy of earth ) is negative because it exerts a force in opposite direction to displacement
B) Determine the energy output of the athlete
weight of barbell * Height = 400 * 2 = 800 J
C) Determine the metabolic power
Metabolic power = energy output / Time
where ; energy output = 4 * 800 = 3200
∴ Metabolic power = 3200 / 1.6
= 2000 w
D) When performs same task at 1.2 s
The metabolic energy he expends is the same and His metabolic power is more
Hence we can conclude that the answers to your questions are as listed above
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The switch on the electromagnet, initially open, is closed. What is the direction of the induced current in the wire loop (as seen from the left)?
Answer:
The induced current is clockwise
Krishna and Seldon now try a homework problem. A policeman sitting in his unmarked police car sees an approaching motorcyclist go through a red light two blocks away. He turns on his siren at a frequency of 1000 Hz as the motorcyclist heads directly toward him at 61 mph (27.27 m/s). What frequency does the motorcyclist hear? (Enter your answer to at least the nearest integer. Assume the speed of sound in air is 331 m/s.) Hz What frequency does the motorcyclist hear when stopped with the police car approaching at 61 mph (27.27 m/s)? (Enter your answer to at least the nearest integer. Assume the speed of sound in air is 331 m/s.) Hz
Answer:
Explanation:
We shall apply formula of Doppler's effect
Here source is fixed and observer is approaching the source
f = f₀ x [(V + v ) / V ]
f₀ is original and f is apparent frequency , V is velocity of sound and v is velocity of motorcyclist .
f = 1000 x [(331 + 27.27 ) / 331 ]
= 1082 .4 Hz
This is the frequency heard by motorcyclist .
When police car is approaching him when he is stopped
f = f₀ x [V /(V - v ) ]
v is velocity of police car .
= 1000 x 331 / (331 - 27.27)
= 1090 Hz
A velocity selector can be used to measure the speed of a charged particle. A beam of particles is directed along the axis of the instrument. A parallel plate capacitor sets up an electric field E which is oriented perpendicular to a uniform magnetic field B. If the plates are separated by 3 mm and the value of the magnetic field is 0.3 T, what voltage between the plates will allow particles of speed 5 x 105 m/s to pass straight through without deflection? A. 70 V B. 140 V C. 450 V D. 1,400 V E. 2,800 V
Answer:
C. 450v
Explanation:
Using
Voltage= B*distance of separation*velocity
3mm x 0.3T x 5E5m/s
= 450v
PLEASE HELP FAST WILL GIVE BRAINLIEST The sentence, "The popcorn kernels popped twice as fast as the last batch," is a(n) _____. 1.experiment 2.hypothesis 3.observation 4.control
The answer is 3. Observation
Explanation:
The sentence "The popcorn kernels popped twice as fast as the last batch" is the result of observing or measuring the time popcorn kernels require to pop. In this context, the sentence best matches the word "observation" which the term used in the Scientific method to refer to statements that are the result of studying a phenomenon, either through the senses such as sight or through precise instruments that allow scientists to understand numerically variables such as time, speed, temperature, etc.
Five wheels are connected as shown in the figure. Find the velocity of the block “Q”, if it is known that: RA= 5 [m], RB= 10 [m], RD= 6 [m], RE=12 [m].
Answer:
-5 m/s
Explanation:
The linear velocity of B is equal and opposite the linear velocity of E.
vB = -vE
vB = -ωE rE
10 m/s = -ωE (12 m)
ωE = -0.833 rad/s
The angular velocity of E is the same as the angular velocity of D.
ωE = ωD
ωD = -0.833 rad/s
The linear velocity of Q is the same as the linear velocity of D.
vQ = vD
vQ = ωD rD
vQ = (-0.833 rad/s) (6 m)
vQ = -5 m/s
If the magnetic field of an electromagnetic wave is in the +x-direction and the electric field of the wave is in the +y-direction, the wave is traveling in the
Answer:
The wave is travelling in the ±z-axis direction.
Explanation:
An electromagnetic wave has an oscillating magnetic and electric field. The electric and magnetic field both oscillate perpendicularly one to the other, and the wave travels perpendicularly to the direction of oscillation of the electric and magnetic field.
In this case, if the magnetic field is in the +x-axis direction, and the electric field is in the +y-axis direction, we can say with all assurance that the wave will be travelling in the ±z-axis direction.
A thick wire with a radius of 4.0 mm carries a uniform electric current of 1.0 A, distributed uniformly over its cross-section. At what distance from the axis of the wire, and greater than the radius of the wire, is the magnetic field strength equal to that at a distance 2.0 mm from the axis. distance
Answer:
8 mm
Explanation:
From the information given:
The Ampere circuital law can be used to estimate the magnetic field strength at two points when the distance is less than the radius and when the distance is greater than the radius.
when the distance is less than the radius ; we have:
[tex]B_1 = \dfrac{ \mu_o \times I \times r}{2 \times \pi \times \ d^2}[/tex]
when the distance is greater than the radius; we have:
[tex]B_2 = \dfrac{\mu_o \ \times I}{2 \ \times \pi \times R}[/tex]
Equating both equations together ; we have :
[tex]\dfrac{ \mu_o \times I \times r}{2 \times \pi \times \ d^2} = \dfrac{\mu_o \ \times I}{2 \ \times \pi \times R}[/tex]
[tex]\dfrac{1}{R}= \dfrac{r}{d^2}[/tex]
[tex]R= \dfrac{d^2}{r}[/tex]
where; d = radius of the wire and r = distance;
[tex]R =\dfrac{4^2}{2}[/tex]
[tex]R =\dfrac{16}{2}[/tex]
R = 8 mm
Ellen says that whenever the acceleration is directly proportional to the displacement of an object from its equilibrium position, the motion of the object is simple harmonic motion. Mary says this is true only if the acceleration is opposite in direction to the displacement. Which one, if either, is correct
Answer:
Both Ellen and Mary are correct.
Explanation:
Both are correct, it's just different ways of saying the same thing.
When the acceleration is always opposite in direction to the displacement, then, the acceleration is directly proportional to the displacement of an object from its equilibrium position
A single-slit diffraction pattern is formed on a distant screen. Assume the angles involved are small. Part A By what factor will the width of the central bright spot on the screen change if the wavelength is doubled
Answer:
If the wavelength is doubled, the width of the central bright spot on the screen will increase by a factor of 2 (that is, it will also double).
Explanation:
For a single-slit diffraction, diffraction patterns are found at angles θ for which
w sinθ = mλ
where w is the width
λ is wavelength
m is an integer, m = 1,2,3, ....
From the equation, w sinθ = mλ
For the first case, where nothing was changed
w₁ = mλ₁ / sinθ
Now, If the wavelength is doubled, that is, λ₂ = 2λ₁
The equation becomes
w₂ = mλ₂ / sinθ
Then, w₂ = m(2λ₁) / sinθ
w₂ = 2(mλ₁) / sinθ
Recall that, w₁ = mλ₁ / sinθ
Therefore, w₂ = 2w₁
Hence, If the wavelength is doubled, the width of the central bright spot on the screen will increase by a factor of 2 (that is, it will also double).
This problem explores the behavior of charge on conductors. We take as an example a long conducting rod suspended by insulating strings. Assume that the rod is initially electrically neutral. For convenience we will refer to the left end of the rod as end A, and the right end of the rod as end B. In the answer options for this problem, "strongly attracted/repelled" means "attracted/repelled with a force of magnitude similar to that which would exist between two charged balls.A. A small metal ball is given a negative charge, then brought near (i.e., within about 1/10 the length of the rod) to end A of the rod. What happens to end A of the rod when the ball approaches it closely this first time?
What happens to end A of the rod when the ball approaches it closely this first time?a. It is strongly repelled.b. It is strongly attracted.c. It is weakly attracted.d. It is weakly repelled.e. It is neither attracted nor repelled.
Answer:
e. It is neither attracted nor repelled.
Explanation:
Electrostatic attraction or repulsion occurs between two or more charged particles or conductors. In this case, if the negatively charged ball is brought close to the neutral end A of the rod, there would be no attraction or repulsion between the rod end A and the negatively charged ball. This is because a charged particle or conductor has no attraction or repulsion to a neutral particle or conductor.
If Superman really had x-ray vision at 0.12 nm wavelength and a 4.1 mm pupil diameter, at what maximum altitude could he distinguish villains from heroes, assuming that he needs to resolve points separated by 5.4 cm to do this?
Answer:
Maximum altitude to see(L) = 1.47 × 10⁶ m (Approx)
Explanation:
Given:
wavelength (λ) = 0.12 nm = 0.12 × 10⁻⁹ m
Pupil Diameter (d) = 4.1 mm = 4 × 10⁻³ m
Separation distance (D) = 5.4 cm = 0.054 m
Find:
Maximum altitude to see(L)
Computation:
Resolving power = 1.22(λ / d)
D / L = 1.22(λ / d)
0.054 / L = 1.22 [(0.12 × 10⁻⁹) / (4 × 10⁻³ m)]
0.054 / L = 1.22 [0.03 × 10⁻⁶]
L = 0.054 / 1.22 [0.03 × 10⁻⁶]
L = 0.054 / [0.0366 × 10⁻⁶]
L = 1.47 × 10⁶
Maximum altitude to see(L) = 1.47 × 10⁶ m (Approx)
An unstable particle at rest spontaneously breaks into two fragments of unequal mass. The mass of the first fragment is 3.00 10-28 kg, and that of the other is 1.86 10-27 kg. If the lighter fragment has a speed of 0.844c after the breakup, what is the speed of the heavier fragment
Answer: Speed = [tex]3.10^{-31}[/tex] m/s
Explanation: Like in classical physics, when external net force is zero, relativistic momentum is conserved, i.e.:
[tex]p_{f} = p_{i}[/tex]
Relativistic momentum is calculated as:
p = [tex]\frac{mu}{\sqrt{1-\frac{u^{2}}{c^{2}} } }[/tex]
where:
m is rest mass
u is velocity relative to an observer
c is light speed, which is constant (c=[tex]3.10^{8}[/tex]m/s)
Initial momentum is zero, then:
[tex]p_{f}[/tex] = 0
[tex]p_{1}-p_{2}[/tex] = 0
[tex]p_{1} = p_{2}[/tex]
To find speed of the heavier fragment:
[tex]\frac{mu_{1}}{\sqrt{1-\frac{u^{2}_{1}}{c^{2}} } }=\frac{mu_{2}}{\sqrt{1-\frac{u^{2}_{2}}{c^{2}} } }[/tex]
[tex]\frac{1.86.10^{-27}u_{1}}{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }=\frac{3.10^{-28}.0.844.3.10^{8}}{\sqrt{1-\frac{(0.844c)^{2}}{c^{2}} } }[/tex]
[tex]\frac{1.86.10^{-27}u_{1}}{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }=1.42.10^{-19}[/tex]
[tex]1.86.10^{-27}u_{1} = 1.42.10^{-19}.{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }[/tex]
[tex](1.86.10^{-27}u_{1})^{2} = (1.42.10^{-19}.{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } })^{2}[/tex]
[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38}.(1-\frac{u_{1}^{2}}{9.10^{16}} )[/tex]
[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38} -[2.02.10^{-38}(\frac{u_{1}^{2}}{9.10^{16}} )][/tex]
[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38} -2.24.10^{-23}.u^{2}_{1}[/tex]
[tex]3.46.10^{-54}.u_{1}^{2}+2.24.10^{-23}.u^{2}_{1} = 2.02.10^{-38}[/tex]
[tex]2.24.10^{-23}.u^{2}_{1} = 2.02.10^{-38}[/tex]
[tex]u^{2}_{1} = \frac{2.02.10^{-38}}{2.24.10^{-23}}[/tex]
[tex]u_{1} = \sqrt{9.02.10^{-62}}[/tex]
[tex]u_{1} = 3.10^{-31}[/tex]
The speed of the heavier fragment is [tex]u_{1} = 3.10^{-31}[/tex]m/s.
an electromagnetic wave propagates in a vacuum in the x-direction. In what direction does the electric field oscilate
Answer:
The electric field can either oscillates in the z-direction, or the y-direction, but must oscillate in a direction perpendicular to the direction of propagation, and the direction of oscillation of the magnetic field.
Explanation:
Electromagnetic waves are waves that have an oscillating magnetic and electric field, that oscillates perpendicularly to one another. Electromagnetic waves are propagated in a direction perpendicular to both the electric and the magnetic field. If the wave is propagated in the x-direction, then the electric field can either oscillate in the y-direction, or the z-direction but must oscillate perpendicularly to both the the direction of oscillation of the magnetic field, and the direction of propagation of the wave.
1. Suppose that a solid ball, a solid disk, and a hoop all have the same mass and the same radius. Each object is set rolling without slipping up an incline with the same initial linear (translational) speed. Which goes farthest up
the incline?
a. the ball
b. the disk
c. the hoop
d. the hoop and the disk roll to the same height, farther
than the ball
e. they all roll to the same height
2. Suppose that a solid ball, a solid disk, and a hoop all have the same mass and the same radius. Each object is set rolling with slipping up an incline with the same initial linear (translational) speed. Which goes farthest up
the incline?
a. the ball
b. the disk
c. the hoop
d. the hoop and the disk roll to the same height, farther
than the ball
e. they all roll to the same height
Answer:
The hoop
Explanation:
Because it has a smaller calculated inertia of 2/3mr² compares to the disc
In a double-slit experiment the distance between slits is 5.0 mm and the slits are 1.4 m from the screen. Two interference patterns can be seen on the screen: one due to light of wavelength 450 nm, and the other due to light of wavelength 590 nm. What is the separation in meters on the screen between the m = 5 bright fringes of the two interference patterns?
Answer:
Δy = 1 10⁻⁴ m
Explanation:
In double-slit experiments the constructive interference pattern is described by the equation
d sin θ = m λ
In this case we have two wavelengths, so two separate patterns are observed, let's use trigonometry to find the angle
tan θ = y / L
as the angles are small,
tan θ = sin θ / cos θ = sin θ
substituting
sin θ = y / L
d y / L = m λ
y = m λ / d L
let's apply this formula for each wavelength
λ = 450 nm = 450 10⁻⁹ m
m = 5
d = 5.0 mm = 5.0 10⁻³ m
y₁ = 5 450 10⁻⁹ / (5 10⁻³ 1.4)
y₁ = 3.21 10⁻⁴ m
we repeat the calculation for lam = 590 nm = 590 10⁻⁹ m
y₂ = 5 590 10⁻⁹ / (5 10⁻³ 1.4)
y₂= 4.21 10⁻⁴ m
the separation of these two lines is
Δy = y₂ - y₁
Δy = (4.21 - 3.21) 10⁻⁴ m
Δy = 1 10⁻⁴ m
