Answer:
I guess it is A. I am not sure
calculate the work function that requires a 455 nm photon to eject an electron of a value 0.73 eV
Answer:
W = 2 eV
Explanation:
Given that,
The wavelength of a photon = 455 nm
The kinetic energy of a photon, K = 0.73 eV
We need to find the work function of the electron. It can be solved using Einstein's equation such that,
[tex]W=E-K[/tex]
E is the energy of the photon
So,
[tex]W=\dfrac{hc}{\lambda}-K\\\\W=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{455\times 10^{-9}\times 1.6\times 10^{-19}}-0.73\\\\W= 2.73\ eV-0.73\ eV\\\\W=2\ eV[/tex]
So, the work function of the metal is 2 eV.
A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and speed (in m/s) when t = 5. f(t) = 18 + 48/t + 1
Answer:
The velocity of the particle = -1.92 m/s
The speed of the particle = 5.72 m/s
Explanation:
Given equation of motion;
[tex]f(t) = 18 \ + \ \frac{48}{t} \ + \ 1[/tex]
Velocity is defined as the change in displacement with time.
[tex]V = \frac{df(t)}{dt} = -\frac{48}{t^2} \\\\at \ t = 5 \ s\\\\V = -\frac{48}{5^2} = \frac{-48}{25} = - 1.92 \ m/s[/tex]
The distance traveled by the particle in 5 s:
[tex]s = f(5) = 18 + \frac{48}{5} + 1\\\\s= 28.6 \ m[/tex]
The speed of the particle when t = 5s
[tex]Speed = \frac{28.6}{5} = 5.72 \ m/s[/tex]
To a man running east at the rate of 3m/s vain appears to fall vertically with a speed of 4m/s. Find the actual speed and direction of rain...
Answer:
The actual speed of the rain is 5 m/s and its direction is -53.13°
Explanation:
The actual speed of the rain V = speed of man, v + speed of rain relative to man, v'.
V = v + v'
We add these vectorially.
Since the man's speed is 3 m/s east, in the positive x - direction, we have v = 3i and the rain's speed is falling vertically at 4 m/s, in the negative y- direction, we have v' = -4j
So, V = v + v'
V = 3i + (-4j)
V = 3i - 4j
So, the magnitude of V which is its speed is V = √(3² + (-4)²) = √(9 + 16) = √25 = 5 m/s
The direction of V, Ф = tan⁻¹(vertical component/horizontal component) = tan⁻¹(-4/3) = tan⁻¹(-4/3) = tan⁻¹(-1.333) = -53.13°
So, the actual speed of the rain is 5 m/s and its direction is -53.13°
A 2.7 kg mass is connected to a spring (k=159 N/m) and is sliding on a horizontal frictionless surface. The mass is given an initial displacement of +19 cm and released with an initial velocity of -13 cm/s. Determine the acceleration of the spring at t=3.4 seconds.
Answer:
The acceleration of the spring at [tex]t = 3.4\,s[/tex] is -29339.947 centimeters per square second.
Explanation:
The mass-spring system experiments a Simple Harmonic Motion, whose kinematic expression is the following:
[tex]x(t) = A\cdot \cos \left(\sqrt{\frac{k}{m} }\cdot t + \phi\right)[/tex] (1)
Where:
[tex]x(t)[/tex] - Position, in centimeters.
[tex]A[/tex] - Amplitude, in centimeters.
[tex]k[/tex] - Spring constant, in newtons per meter.
[tex]m[/tex] - Mass, in kilograms.
[tex]\phi[/tex] - Phase, in radians.
By Differential Calculus, we derive expression for the velocity and acceleration of the mass-spring system:
[tex]v(t) = -\sqrt{\frac{k}{m} }\cdot A\cdot \sin \left(\sqrt{\frac{k}{m} }\cdot t +\phi\right)[/tex] (2)
[tex]a(t) = -\frac{k\cdot A}{m}\cdot \cos \left(\sqrt{\frac{k}{m} }\cdot t +\phi\right)[/tex] (3)
Where [tex]v(t)[/tex] and [tex]a(t)[/tex] are the velocity and acceleration of the system, in centimeters per second and centimeters per square second, respectively.
If we know [tex]m = 2.7\,kg[/tex], [tex]k = 159\,\frac{N}{m}[/tex], [tex]t = 0\,s[/tex], [tex]x(t) = 19\,cm[/tex] and [tex]v(t) = -13\,\frac{cm}{s}[/tex], then we have the following system of nonlinear equations:
[tex]A \cdot \cos \phi = 19[/tex] (1)
[tex]-7.674\cdot A \cdot \sin \phi = - 13[/tex] (2)
If we divide (2) by (1):
[tex]-7.674\cdot \tan \phi = -0.684[/tex]
[tex]\tan \phi = 0.089[/tex]
[tex]\phi = \tan^{-1} 0.089[/tex]
[tex]\phi \approx 0.028\,rad[/tex]
By (1), we get the value of the amplitude:
[tex]A = \frac{19}{\cos \phi}[/tex]
[tex]A = 19.075\,cm[/tex]
If we know that [tex]A = 19.075\,cm[/tex], [tex]k = 159\,\frac{N}{m}[/tex], [tex]m = 2.7\,kg[/tex], [tex]t = 3.4\,s[/tex] and [tex]\phi \approx 0.028\,rad[/tex], then the acceleration of the spring is:
[tex]a(t) = -29339.947\,\frac{cm}{s^{2}}[/tex]
The acceleration of the spring at [tex]t = 3.4\,s[/tex] is -29339.947 centimeters per square second.
hoW DO YOU SPELL THIS?
Answer:
Spell what?
Can u please mention that too?
Answer:
what
Explanation: how do you spell what please include this in your next post
The position of an object of
mass 5 kg as a function of
time is giving by r = (20m/s4)t4
i + (12 m/s3)t3 j. Find the
force acting on the object as a
function of time. Express the
force in unit vectors. Hint:
Remember that Newton's
second Law relates the force
to the acceleration
Answer:
[tex]F=5(240t^2i+72tj)\ N[/tex]
Explanation:
Given that,
The mass of the object, m = 5 kg
The position vector is, [tex]r=20t^4i+12t^3j[/tex]
Velocity, [tex]v=\dfrac{dr}{dt}=80t^3i+36t^2j[/tex]
Acceleration, [tex]a=\dfrac{dv}{dt}=240t^2i+72tj[/tex]
Newton's second law of motion is given as follows:
F = ma
Put all the values,
[tex]F=5(240t^2i+72tj)\ N[/tex]
Hence, this is the required solution.
Part A What will be the equilibrium temperature when a 227 g block of copper at 283 °C is placed in a 155 g aluminum calorimeter cup containing 844 g of water at 14.6°C?
Answer:
T = 20.84°C
Explanation:
From the law of conservation of energy:
Heat Lost by Copper Block = Heat Gained by Aluminum Calorimeter + Heat Gained by Water
[tex]m_cC_c\Delta T_c = m_wC_w\Delta T_w + m_aC_a\Delta T_a[/tex]
where,
[tex]m_c[/tex] = mass of copper = 227 g
[tex]m_w[/tex] = mass of water = 844 g
[tex]m_a[/tex] = mass of aluminum = 155 g
[tex]C_c[/tex] = specific heat capacity of calorimeter = 385 J/kg.°C
[tex]C_w[/tex] = specific heat capacity of water = 4200 J/kg.°C
[tex]C_a[/tex] = specific heat capacity of aluminum = 890 J/kg.°C
[tex]\Delta T_c[/tex] = change in temperature of copper = 283°C - T
[tex]\Delta T_w[/tex] = change in temperature of water = T - 14.6°C
[tex]\Delta T_a[/tex] = change in temperature of aluminum = T - 14.6°C
T = equilibrium temperature = ?
Therefore,
[tex](227\ g)(385\ J/kg.^oC)(283^oC-T)=(844\ g)(4200\ J/kg.^oC)(T-14.6^oC)+(155\ g)(890\ J/kg.^oC)(T-14.6^oC)\\\\24732785\ J - (87395\ J/^oC) T = (3544800\ J/^oC) T - 51754080\ J+ (137950\ J/^oC) T-2014070\ J\\\\24732785\ J +51754080\ J+2014070\ J = (3544800\ J/^oC) T+(137950\ J/^oC+(87395\ J/^oC) T\\\\78560935\ J = (3770145\ J/^oC) T\\\\T = \frac{78560935\ J}{3770145\ J/^oC}[/tex]
T = 20.84°C
HELP ME PLEASE!!!
Which 2 statements are true about this chemical reaction that forms acid rain?
Answer:
B.
Explanation:
HNO2 is less stable thus dissociates easily to HNO3 + NO + H2O while HNO3 is a strong acid. Thus when they react with H2O they form acid rain
Answer:
B
Explanation:
dont have one just trust me
Answer the following questions
1. Heat in liquid travels from
a) bottom to top
b) top to bottom
c) left to right
d) right to left
2. The direction of flow of heat is
a) always from a cooler body to a hotter body
b) always from a hotter body to cooler body
c) always from a body at a lower temperature to a body at a higher temperature
d) all the above
3. A cold steel spoon is dipped in a cup of hot milk. The steel spoon transfer the heat to its other end by the process of
a) convection
b) conduction
c) radiation
d) none of the above
what is the resistance of a bulb of 40w connected in a line of 70v
Answer:
122.5 ohm
Explanation:
Given :
P=40 w
V= 70 V
R=?
Resistance can be calculated as :
[tex]P=\frac{V^{2} }{R} \\40=\frac{(70)^{2} }{R}\\40=\frac{4900}{R} \\R=\frac{4900}{40} \\R=122.5 ohm[/tex]
Therefore, resistance of the bulb will be 122.5 ohm
a brick of mass 0.8 kg is accidentally dropped from a high scaffolding. it reaches the ground with a kinetic energy of 240 J. How high is scaffolding ?(Take acceleration due to gravity g be 10 m s-¹)
Answer:
30 m
General Formulas and Concepts:
Energy
Gravitational Potential Energy: [tex]\displaystyle U_g = mgh[/tex]
m is mass (in kg)g is gravityh is height (in m)Kinetic Energy: [tex]\displaystyle KE = \frac{1}{2}mv^2[/tex]
m is mass (in kg)v is velocity (in m/s²)Law of Conservation of Energy
Explanation:
Step 1: Define
Identify variables
[Given] m = 0.8 kg
[Given] g = 10 m/s²
[Given] U = 240 J
[Solve] h
Step 2: Solve for h
[LCE] Substitute in variables [Gravitational Potential Energy]: (0.8 kg)(10 m/s²)h = 240 JMultiply: (8 kg · m/s²)h = 240 JIsolate h [Cancel out units]: h = 30 mWhich of the following is a good example of a contact force?
ОА.
Earth revolving around the Sun
OB.
a bridge suspended by cables
OC.
a ball falling downward a few seconds after being thrown upward
OD. electrically charged hairs on your head repelling each other and standing up
Answer:
A bridge suspended by cables
Explanation:
Both objects represent a contact force (in this case, normal force) acting on each other. The force occurs since both objects are in direct physical contact.
Question 1 of 25
Which equation is an example of a synthesis reaction?
A. HNO3 + KOH → KCI + H20
B. 2Li+ CaCl2 - 2LiCl + Ca
O C. S+ 02 - S02
7
O D. CH4 + 202 - 2H2O + CO2
Answer:
C. S + 02 → S02
Explanation:
A synthesis or combination reaction is that reaction involving two elements as reactants to form a single compound as product.
In the reaction given below;
S + 02 → S02
Sulphur and oxygen are elemental substances that combine to synthesize sulfur IV oxide (SO2), and hence it is an example of synthesis reaction.
A swift blow with the hand can break a pine board. As the hand hits the board, the kinetic energy of the hand is transformed into elastic potential energy of the bending board; if the board bends far enough, it breaks. Applying a force to the center of a particular pine board deflects the center of the board by a distance that increases in proportion to the force. Ultimately the board breaks at an applied force of 870 N and a deflection of 1.4 cm.
a. To break the board with a blow from the hand, how fast must the hand be moving? Use 0.50 kg for the mass of the hand.
b. If the hand is moving this fast and comes to rest in a distance of 1.2 cm, what is the average force on the hand?
Answer:
A. The hand must move with a velocity of 6.98 m/s to break the board.
B. Average force on the hand = 1025 N
Explanation:
A.To determine the speed the hand must move with to break the board, the force workdone in breaking the board is found first.
Workdone = force × distance
Minimum force required = 870 N;
Distance moved by board/Deflection in order to break = 1.4 cm = 0.014 M
Workd done = 870 N × 0.014 m = 12.18 Nm or 12.18 J
This work done = Kinetic energy of the hand
Kinetic energy = mv²/2 ; where m is mass and v is velocity
Mass of hand = 0.50 Kg, velocity = ?, K.E. = 12.18 J
v² = 2 KE/m
v = √2KE/m
v = √(2 × 12.18/0.50)
v = 6.98 m/s
Therefore, the hand must move with a velocity of 6.98 m/s to break the board.
B. Average force on the hand
This can be determined using the equation of motion, v² = u² + 2as to find acceleration, since force = mass × acceleration
From the equation of motion, a = v² - u²/2s
At rest, v = 0, u = 6.98, s = 1.2 cm = 0.012 m
a = 0² - 6.98²/ 2 × 0.012
a = -2030 m/s²
Force = 2030 m/s² × 0. 50 kg = 1015 N
Therefore, Average force on the hand = 1025 N
In addition to acceleration, what else will be a maximum at the amplitude for SHM?
A. Potential energy
B. Kinetic energy
C. Nuclear energy
D. Chemical energy
It is Potential energy's
An empty parallel plate capacitor is connected between the terminals of a 8.85-V battery and charges up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor
Answer:
The new voltage is 17.7 V.
Explanation:
Voltage, V = 8.85 V
The spacing is doubled.
When it is disconnected, the charge remains same,
q = C V ..... (1)
where, C is the capacitance, V is the voltage.
The capacitance is inversely proportional to the distance between the two plates.
So, when the spacing is doubled, the capacitance is halved.
Let the new voltage is V'.
C V = C' V'
C x 8.85 = C/2 x V'
V' = 17.7 V
You are packing for a trip to another star, and on your journey you will be traveling at a speed of 0.99c. Can you sleep in a smaller cabin than usual, because you will be shorter when you lie down? Explain your answer.
Explanation:
No. From your own reference frame, nothing will change. Everything will look the same to you even if you're traveling at 99% speed of light. However, an outside observer will see you shrink to about 14% of your original length along the direction of your motion according to the Lorentz contraction predicted by special relativity:
[tex]L=L_{0} \sqrt{1 - \frac{v^2 }{c^2 }}[/tex]
Một ô tô khối lượng một tấn chuyển động trên một đường nằm ngang. Hệ số ma sát giữa bánh ô tô và mặt đường là 0,07. Gia tốc trọng trường g=9,8m/s2
a) vẽ và xác định tên các lực tác động lên vật. Viết phương trình chuyển động của vật.
b) nếu ô tô chuyển động đều, xuống dốc có độ dốc 5%. Tính lực kéo của ô tô.
c) nếu ô tô chuyển động đều. Lên dốc có độ dốc 5%. Tính lực kéo của động cơ ô tô
A 2.5-kg rock is dropped off a 32-m cliff and hits a spring, compressing it 57 cm. What is the spring constant? Round your answer to two significant figures.
The spring constant, k, is
StartFraction N over m EndFraction.
Answer: 4800 N/m
Explanation:
Given
mass of rock [tex]m=2.5\ kg[/tex]
Height of cliff [tex]h=32\ m[/tex]
compression in the spring [tex]x=57\ cm[/tex]
Here, potential energy is converted into kinetic energy which in turn converts to elastic potential energy of the spring
[tex]\Rightarrow mgh=\dfrac{1}{2}kx^2\\\\\Rightarrow k=\dfrac{2mgh}{x^2}\\\\\Rightarrow k=\dfrac{1568}{0.3249}\\\\\Rightarrow k=4826.100\approx 4800\ N/m[/tex]
Answer:
480
Explanation:
2021
A car on the roller coaster begins with 0j of kinetic energy and 12,928j of potential energy and finishes the track with 3,715j of potential energy. How much kinetic energy dose the car finish with????
Answer:
9213 J
Explanation:
Change in Kinetic energy = Change in Potential energy
= 12,928J - 3715J
=9213 J
For more assistance: +1 (304) 223-3136
Alice's friends Bob and Charlie are having a race to a distant star 10 light years away. Alice is the race official who stays on Earth, and her friend Darien is stationed on the star where the race ends. Bob is in a rocket that can travel at 0.7c; whereas Charlie's rocket can reach a speed of 0.866c. Bob and Charlie start at the same time. Draw space- diagrams from each perspective.
Required:
Estimate how long it takes Bob and Charlie to finish the race from each perspective.
Solution :
The distance between the starting point and the end point, [tex]L_0[/tex] = 10 light years
But due to the relativistic motion of Bob and Charlie, the distance will be reduced following the Lorentz contraction. The contracted length will be different since they are moving with different speeds.
For Bob,
Speed of Bob's rocket with respect to Alice, [tex]L_b = 0.7 \ c[/tex]
So the distance appeared to Bob due to the length contraction,
[tex]$L_b=L_0\sqrt{1-\frac{V_b^2}{c^2}$[/tex]
[tex]$L_b=10\times \sqrt{1-0.49} \ Ly$[/tex]
[tex]$=7.1 \ Ly$[/tex]
Therefore, the time required to finish the race by Bob is
[tex]$t_b = \frac{L_b}{V_b}$[/tex]
[tex]$=\frac{7.1 \ c}{0.7 \ c}$[/tex]
= 10.143 year
For Charlie,
Speed of Charlie's rocket with respect to Alice, [tex]L_c = 0.866 \ c[/tex]
So the distance appeared to Charlie due to the length contraction,
[tex]$L_b=L_0\sqrt{1-\frac{V_c^2}{c^2}$[/tex]
[tex]$L_b=10\times \sqrt{1-0.75} \ Ly$[/tex]
[tex]$=5 \ Ly$[/tex]
The time required to finish the race by Charlie is
[tex]$t_b = \frac{L_c}{V_c}$[/tex]
[tex]$=\frac{5 \ c}{0.866 \ c}$[/tex]
= 5.77 year
You take your pulse and observe 80 heartbeats in a minute. What is the period of your heartbeat? What is the frequency of your heartbeat?
Answer:
120 beats per minute.
Explanation:
If I take your pulse and observe 80 heartbeats in a minute. Then the period of your heartbeat is 0.8 s and frequency is 1.3Hz.
What is Heartbeat ?A pulse is the term used in medicine to describe the tactile arterial palpation of the cardiac cycle (heartbeat) by skilled fingertips. Any location where an artery can be compressed close to the surface of the body, such as the carotid artery in the neck, the radial artery in the wrist, the femoral artery in the groyne, the popliteal artery behind the knee, the posterior tibial artery near the ankle joint, and on the foot, can be used to palpate the pulse (dorsalis pedis artery). Heart rate may be determined by monitoring pulse, or the number of arterial pulses per minute. Auscultation, which is the process of counting the heartbeats while listening to the heart using a stethoscope, is another way to determine the heart rate. Typically, three fingers are used to gauge the radial pulse.
Given,
heart beat = 80 beats/min = 1.3 beats/s
Frequency is nothing but how much beats is heart having in one second and that is 1.3 beats/s. Hence frequency of heart is 1.3Hz.
The Period is reciprocal of frequency,
T = 1/f = 0.8 s
To know more about frequency :
https://brainly.com/question/29739263
#SPJ2.
When you release some of the paint from a spray paint can and the can remains at the same temperature, which gas law does this represent?
Answer:
Boyle's law.
Explanation:
Robert Boyle was an Irish chemist and is famously referred to as the first modern chemist. He was born on the 25th of January, 1627 in Lismore, Ireland and died on the 31st, December 1691, London, United Kingdom.
Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.
Mathematically, Boyles law is given by;
[tex] PV = K[/tex]
[tex] P_{1}V_{1} = P_{2}V_{2} [/tex]
Where;
P1 is the original pressure.
P2 is the final pressure.
V1 is the original volume.
V2 is the final volume.
Hence, when you release some of the paint from a spray paint can by applying an amount of pressure and the can remains at the same temperature, the gas law which this represent is Boyle's law.
If distance between two charges increased by 2 times then force
Explanation:
The size of the force varies inversely as the square of the distance between the two charges. Therefore, if the distance between the two charges is doubled, the attraction or repulsion becomes weaker, decreasing to one-fourth of the original value.
what is effort arm
don't say the answer of gogle
Answer:
effort arm mean the use of any work by using your hand force motion or by hand power
help plzzzzzzzzzzzz ?
Explanation:
1. First, let's find the total resistance of the circuit. We begin by combining [tex]R_{4}[/tex], [tex]R_{5}[/tex] and [tex]R_{6}[/tex]:
[tex]R_{456}=R_{4} + \dfrac{R_{5}R_{6}}{R_{5} + R_{6}}[/tex]
[tex]= 6\:Ω + \dfrac{(3\:Ω)(5\:Ω)}{3\:Ω+5\:Ω} = 7.9\:Ω[/tex]
Now time to combine [tex]R_{2}[/tex] and [tex]R_{3}[/tex] and they are connected in series so
[tex]R_{23} =R_{2} + R_{3} = 17\:Ω[/tex]
Note that [tex]R_{23}[/tex] and [tex]R_{456}[/tex] are connected in parallel so
[tex]R_{23456} = \dfrac{R_{23}R_{456}}{R_{23}+R_{456}}=5.4\:Ω[/tex]
Finally, [tex]R_{23456}[/tex] is connected in series with [tex]R_{1}[/tex] so the total resistance [tex]R_{T}[/tex] is
[tex]R_{T} = R_{1} + R_{23456} = 10\:Ω + 5.4\:Ω = 15.4\:Ω[/tex]
2. The total current in the circuit is
[tex]I_{T} = \dfrac{V}{R_{T}} = \dfrac{20\:V}{15.4\:Ω} = 1.3\:A[/tex]
3. The voltage drop across [tex]R_{1},\:V_{1}[/tex] is
[tex]V_{1} = I_{T}R_{1} = (1.3\:A)(10\:Ω) = 13\:V[/tex]
4. We can see that [tex]I_{T} = I_{1} + I_{2}[/tex]. To solve for [tex]I_{1}[/tex], we need [tex]V_{23}[/tex], which is just [tex]V_{T} - V_{1} = 20\:V - 13\:V = 7\:V[/tex] , which gives us
[tex]I_{1} = \dfrac{V_{23}}{R_{23}} = \dfrac{7\:V}{17\:Ω} = 0.4\:A[/tex]
5. From #2 & #4, [tex]I_{2} = 1.3\:A - 0.4\:A = 0.9\:A[/tex] and we also know that the voltage drop across [tex]R_{456}[/tex] is 7 V, the same as that of [tex]R_{23}[/tex]. The voltage drop across [tex]R_{4}[/tex] is
[tex]V_{4} = I_{2}R_{4} =(0.9\:A)(6\:Ω) = 5.4\:V[/tex]
This means that the voltage drop across [tex]R_{6}[/tex] is 7 V - 5.4 V = 1.6 V. Knowing this, the current through [tex]R_{6}[/tex] is
[tex]I_{6} = \dfrac{1.6\:V}{5\:Ω} = 0.3\:A[/tex]
Hey, can a physics major help me?
I have been wondering about the exact difference between theories laws facts and hypothosis.
I know the general layout but I am still a bit confused.
100 points for answering and brainly if it is a good one.
Answer:
A hypothesis is a limited explanation of a phenomenon; a scientific theory is an in-depth explanation of the observed phenomenon. A law is a statement about an observed phenomenon or a unifying concept
Answer:
Explanation:
will try 2 explain fact, hypothesis, theory n law
fact is the starting pt: e.g. apple falls from tree
hypothesis tries 2 explain a fact: e.g. there is a force pulling down apple
theory is a complete explanation w/ equations n stuff: e.g. Newton came up w/ theory of gravitational attraction force
law is a theory dat has been proven right through tests n experiments: Newton's gravity theory had been proven right in many many tests.
who won the battle of Buxar
Answer:
British East India Company
The battle was fought at Buxar, a "small fortified town" within the territory of Bihar, located on the banks of the Ganga river about 130 kilometres (81 mi) west of Patna; it was a decisive victory for the British East India Company.
A conductor is placed in an external electrostatic field. The external field is uniform before the conductor is placed within it. The conductor is completely isolated from any source of current or charge.
1. Which of the following describes the electricfield inside this conductor?
a. It is in thesame direction as the original external field.
b. It is in theopposite direction from that of the original externalfield.
c. It has adirection determined entirely by the charge on itssurface.
d. It is alwayszero.
2. The charge density inside theconductor is:
a. 0
b. non-zero;but uniform
c. non-zero;non-uniform
d. infinite
Answer:pp
Explanation:
ii
Ocean waves of wavelength 22 m are moving directly toward a concrete barrier wall at 4.6 m/s . The waves reflect from the wall, and the incoming and reflected waves overlap to make a lovely standing wave with an antinode at the wall. (Such waves are a common occurrence in certain places.) A kayaker is bobbing up and down with the water at the first antinode out from the wall.
Required:
a. How far from the wall is she?
b. What is the period of her up-and-down motion?
Answer:
a) [tex]d=11m[/tex]
b) [tex]T=4.68s[/tex]
Explanation:
From the question we are told that:
Wavelength [tex]\lambda=22m[/tex]
Velocity [tex]v=4.6m/s[/tex]
a)
Generally the equation for distance between her and the wall d is mathematically given by
Since
The First Anti node distance is [tex]\frac{\lambda}{2}[/tex]
Therefore
[tex]d= \frac{\22}{2}[/tex]
[tex]d=11m[/tex]
b)
Generally the equation for her up-and-down motion is mathematically given by
[tex]T=\frac{22}{4.7}[/tex]
[tex]T=4.68s[/tex]
