Answer:
Electronic data interchange
Light of wavelength 3500Å is incident on two metals A and B whose work functions are 3.2 eV and 1.9 eV respectively. Which metal will emit photoelectrons?
(a) A
(b) B
(c) Both A and B
(d) Neither A nor B
Answer:
(c) Both A and B
Explanation:
Given;
wavelength of the incident light, λ = 3500 Å = 3500 x 10⁻¹⁰ m
The energy of the incident light is given by;
E = hf
where;
h is Planck's constant = 6.626 x 10⁻³⁴ J/s
f is frequency = c / λ
[tex]E = \frac{hc}{\lambda} \\\\E = \frac{(6.626*10^{-34})(3*10^8)}{3500*10^{-10}}\\\\E = 5.68 *10^{-19} \ J\\\\E = 5.68 \ eV[/tex]
Work function is the minimum amount of energy required to liberate electrons from a metal surface.
The work function of metal A is 3.2 eV and
Th e work function of metal B is 1.9 eV
Both work functions are less than the incident energy of the light calculated as 5.68eV.
Thus, both metals will emit photoelectrons.
(c) Both A and B
what do u mean by the activity series of metals ??
Answer:
An activity series is a list of substances ranked in order of relative reactivity. For example, magnesium metal can knock hydrogen ions out of solution, so it is considered more reactive than elemental hydrogen. The activity series is a useful guide for predicting the products of metal displacement reactions.
Calculate the kinetic energy of an 88-kg scooter moving at 12 m/s .
Answer:
kinetic energy = 1/2 X m(v)^2
= 1/2 X 88 X 12
= 528 j
Your question has been heard loud and clear.
Kinetic energy formula= 1/2mv^2
Mass = 88kg (convert to grams because velocity is in m/s)
Mass in grams = 88000g or 88*10^3 g
Velocity= 12m/s
Now kinetic energy = 1/2 * 88000*12*12= 6336 joules
Kinetic energy = 6336 Joules or 6.336 Kilo joules
Thank you.
An electron is located on the x ‑axis at x0=−2.05×10−6 m . Find the magnitude and direction of the electric field at x=2.77×10−6 m on the x ‑axis due to this electron.
Answer:
The value is [tex]E = 62 \ N/C[/tex]
Explanation:
From the question we are told that
The initial position of the electron [tex]x_o = -2.05*10^{-6} \ m[/tex]
The position that is considered [tex]x= 2.77*10^{-6} \ m[/tex]
Generally the electric field is mathematically represented as
[tex]E = k * \frac{q}{r^2 }[/tex]
Here k is the coulombs constant with value [tex]k = 9*10^9 \ \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]
r is the distance covered which is mathematically represented as
[tex]r = x- x_o[/tex]
[tex]r = 2.77*10^{-6} -(-2.05*10^{-6})[/tex]
[tex]r = 4.82 *10^{-5}[/tex]
So
[tex]E = \frac{9*10^9 * 1.60*10^{-19}}{ (4.82 *10^{-6})}[/tex]
[tex]E = 62 \ N/C[/tex]
Answer:
Direction: Negative x-direction
Explanation:
The question tells you that both lie on the ‑axis, and the data indicate that the electron is in the negative ‑direction from the field point ( 0< ). Because the field produced by a negative source charge points toward the source, the field in this case points in the negative ‑direction.
Two long wires hang vertically. Wire 1 carries an upward current of 1.20 A. Wire 2, 20.0 cm to the right of wire 1, carries a downward current of 4.20 A. A third wire, wire 3, is to be hung vertically and located such that when it carries a certain current, each wire experiences no net force.
(a) Is this situation possible?
A. Yes
B. No
Is it possible in more than one way?
A. Yes
B. No
(b) Describe the position of wire 3.
distance ________ cm
direction: left of wire 1
(c) Describe the magnitude and direction of the current in wire 3.
magnitude
direction
down
Answer:
Explanation:
A, yes
b,
The force(per unit length) on wire 1 by 2 is to the left and
= μo*I1*I2/2πr
= 2*10^-7 * 1.2 * 4.2/0.20
= 5.04*10^-6N/m
From third law we have the force on wire2 by wire 1 is 5.04*10^-6N/m and to the right
So the magnitude of the force on each wire by wire 3 must be 4.2*10^-6N/m
Since the current in wire 2 is > wire 1 then wire 3 must be closer to wire 1 than 2
and wire 3 must be to the left of wire 1
Let x be the distance from wire 3 to wire 1 so x + 0.20 = distance from wire 3 to wire 2
Now μo*I1*I3/2πx = μo*I3*I2/2π(x + 0.20)
so solving for x we get
I1/x = I2/(x+0.20)
or x +0.20 = 4x
so x = 0.20/3 = 0.0667m
Now μo*I1*I3/2πx = 4.2*10^-6
so I3 = 4.2x10^-6 * 0.0667/(2.0x10^-7 * 1.2) = 1.167 A and its direction is down
An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.200 rev/s . The magnitude of the angular acceleration is 0.883 rev/s2 . Both the the angular velocity and angular accleration are directed counterclockwise. The electric ceiling fan blades form a circle of diameter 0.760 m .A. Compute the fan's angular velocity magnitude after time 0.204 s has passed.Express your answer numerically in revolutions per second.B. Through how many revolutions has the blade turned in the time interval 0.204 s from Part A?Express the number of revolutions numerically.C. What is the tangential speed vt of a point on the tip of the blade at time t = 0.204 s?Express your answer numerically in meters per second.D. Calculate the magnitude atat of the tangential acceleration of a point on the tip of the blade at time t = 0.204 s.Express the acceleration numerically in meters per second squared.
Answer:
(A). the angular velocity of fan is 0.380 rev/s.
(B). The number of revolution is 0.059 rad.
(C). The tangential speed of the blade is 0.907 m/s.
(D). The tangential acceleration of the blade is 2.10 m/s²
Explanation:
Given that,
Initial angular velocity = 0.200 rev/s
Angular acceleration = 0.883 rev/s²
Diameter = 0.760 m
Time = 0.204 s
(A). We need to calculate the angular velocity of fan
Using equation of angular motion
[tex]\omega_{f}=\omega_{i}+\alpha t[/tex]
Put the value into the formula
[tex]\omega_{f}=0.200+0.883\times0.204[/tex]
[tex]\omega_{f}=0.380\ rev/s[/tex]
(B). We need to calculate the number of revolution
Using formula of angular displacement
[tex]\theta=\omega_{i}t+\dfrac{1}{2}\alpha t^2[/tex]
Put the value into the formula
[tex]\theta=0.200\times0.204+\dfrac{1}{2}\times0.883\times(0.204)^2[/tex]
[tex]\theta=0.059\ rad[/tex]
(C). We need to calculate the tangential speed of the blade
Using formula of speed
[tex]v=\dfrac{d}{2}\omega_{f}[/tex]
Put the value into the formula
[tex]v=\dfrac{0.760}{2}\times0.380\times2\pi[/tex]
[tex]v=0.907\ m/s[/tex]
(D). We need to calculate the tangential acceleration of the blade
Using formula of tangential acceleration
[tex]a_{t}=\alpha r[/tex]
Put the value into the formula
[tex]a_{t}=0.883\times0.38\times2\pi[/tex]
[tex]a_{t}=2.10\ m/s^2[/tex]
Hence, (A). the angular velocity of fan is 0.380 rev/s.
(B). The number of revolution is 0.059 rad.
(C). The tangential speed of the blade is 0.907 m/s.
(D). The tangential acceleration of the blade is 2.10 m/s²
The gravitational pull of the sun on Earth keeps Earth orbiting around the sun. Which statement is correct about the force that Earth exerts on the sun?
A. Earth pulls the sun away from itself with a force equal to the ratio of the mass of the sun to the mass of Earth.
B. Earth pulls the sun toward itself with a force equal to the ratio of the mass of the sun to the mass of Earth.
C. Earth pulls the sun away from itself with an equal force.
D. Earth pulls the sun toward itself with an equal force,
Answer:
Earth pulls the sun towards itself with a force equal to the ratio of the mass of the sun to the mass of Earth
The Earth pulls the sun toward itself with an equal force. (Option D)
According to the 3rd Newton's Law, forces existing in the universe are shown in pairs, that is to say, for all action force, there is a corresponding reaction force of equal magnitude and opposite direction. By the Newton's Law of Gravitation we understand that gravitation between the Sun and the Earth is done because of a gravitational field created by mutual interaction, in terms of their masses and distance among them. The pair of forces is not based on direct contact.
Hence, we conclude that the Earth pulls the sun toward itself with an equal force.
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Explain how your weight can change even though your mass remains the same.
Answer:
See explanation below
Explanation:
Recall that Weight is a Force that results from the action of a gravitational field on our body mass. Therefore if one changes locations to places where the acceleration of gravity is different, our weight can change although the mass of our body stays the same.
For example, going to a high mountain, where the acceleration of gravity is a little smaller than at sea level, will produce such change. Also, going to another planet with different gravitational field, or going to the Moon (where the acceleration of gravity is about 1/6 of that on Earth which therefore will reduce our weight without reducing our mass)
Since weight is calculated with the equation:
Q=mg
the weight can also change when the gravitational acceleration (g) changes.
g can change at example when you travel to a different planet or (slightly) when gaining attitude on Earth (e.g. when climbing a mountain or flying on a plane you get lighter).
Describe the motion of an object:
a. for which the direction of the acceleration is the same as the direction of motion of the object.
b. for which the direction of the acceleration is opposite to the direction of motion of the object.
c. for which the change in velocity is zero.
d. for which the initial velocity is zero but the acceleration is not zero.
Answer:
A. Here object is moving with increasing velocity
B. Here the object is moving with decreasing velocity that is it is decelerating
C. Here the initial and final velocity are the same so the object moves at constant velocity
D. Here the object started from rest but reached a certain velocity with time so a non zero acceleration was attained
Answer:
Answer:
A. Here object is moving with increasing velocity
B. Here the object is moving with decreasing velocity that is it is decelerating
C. Here the initial and final velocity are the same so the object moves at constant velocity
D. Here the object started from rest but reached a certain velocity with time so a non zero acceleration was attained
Explanation:
A bus took 8 hours to travel 639 km. For the first 5 hours, it
travelled at an average speed of 72 km/h. What was the
average speed for the remaining time of the journey?
Answer:
93
Explanation:
A spaceship is traveling through deep space to a space station and needs to make a course correction to go around a nebula. The captain orders the ship to travel 2.3 106 kilometers before turning 70° and traveling 1.5 106 kilometers to reach the space station. If the captain had not ordered a course correction, what would have been the magnitude and direction of the path of the spaceship if it had traveled to the space station through the nebula?
Answer:
Explanation:
Using the given values
Let
Px= 2.3 x 10^6, Py=0, Qx=1.5x10^6cos(70),
Qy= 1.5x10^6sin(70)
So let
Hx= Px+Qx=2.813 x 10^6, Hy=Py+Qy=1.401 x 10^6
Magnitude = √ ((2.813 x 10^6)^2+(1.401 x 10^6)^2)= 3.14 x 10^6
To find direction
tan စ = Hy/Hx = 1.401/2.813= 0.4980
စ = 26.5°
The magnitude and direction of the path of the spaceship if it had traveled to the space station through the nebula are 3.14 x 10⁶ m and 26.5°.
What is nebula?A nebula is a huge cloud consisting of dust and gas in space. Some nebula come from the gas and dust thrown out by the explosion of a dying star known as a supernova.
The captain orders the ship to travel in horizontal direction, Px= 2.3 x 10⁶ m, Py=0,
After turning 70° and it travels to reach the space station, Qx = 1.5x10⁶ x cos70° in horizontal direction.
and Qy= 1.5x10⁶sin70°
So, total horizontal distance moved is
Hx= Px+Qx = 2.813 x 10⁶
Total vertical distance moved is
Hy=Py+Qy = 1.401 x10⁶
Magnitude of the path = sq rt [(2.813 x 10⁶)²+(1.401 x 10⁶)²]
Path = 3.14 x 10⁶ m
Direction of the path,
tan α = Hy/Hx
tan α = 1.401/2.813
tan α = 0.4980
α = 26.5°
Thus, the magnitude and direction of the path of spaceship is 3.14 x 10⁶ m and 26.5°.
Learn more about nebula.
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URGENT!!!!!! Assume that a wire has 1.5 ohms of resistance. If the wire is connected to two batteries with a total voltage of 3.0 V, how much current will flow through the wire? 3.0 amps 2.3 amps 2.0 amps 1.0 amps
Answer:
2.0 amps
Explanation:
Current is the ratio of voltage to resistance:
I = V/R = (3.0)/(1.5) = 2.0
The current in the wire is 2.0 amps.
A billiard ball moves with 3 kg⋅m/s of momentum and strikes three other billiard balls. What is the total momentum of the balls after the collision? A. less than 3 kg⋅m/s B. It depends on the mass of each ball. C. 3 kg⋅m/s D. more than 3 kg⋅m/s
Answer:
It should be C
Explanation:
It depends on the mass of each ball
Remember the formula for momentum, mass X velocity
Momentum depends on the mass of the object
The momentum of the billiard balls after the collision is 3 Kgm/s.
Let the billiard ball be B.Given the following data:
Momentum B = 3 Kgm/sTo find the momentum of the balls after the collision:
Momentum is simply a multiplication (product) of the mass of an object with its velocity. Also, momentum is a vector quantity because it has both magnitude and direction.
Mathematically, momentum is given by the formula;
[tex]Momentum = Mass[/tex] × [tex]Velocity[/tex]
[tex]Total \; momentum = Momentum \; B + Momentum \;of \;the \;other\; balls.[/tex]
Since the other billiard balls were at rest (not experience any motion), there velocities would be zero (0) m/s.
Substituting the values, we have:
[tex]Total \; Momentum = 3 + 0[/tex]
Total momentum = 3 Kgm/s.
Therefore, the momentum of the billiard balls after the collision is 3 Kgm/s.
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What are the magnitude and direction of the net force on the third spider (S3) when it is resting at the origin
Complete Question
The complete question is shown on the first uploaded image
Answer:
The magnitude is [tex]F_3 =0.62\ N[/tex]
Explanation:
From the question we are told that
The distance between the spiders is [tex]a =0.59\ m[/tex]
The charge on the [tex]S_1\ and\ S_3[/tex] is [tex]q_{1}=q_{3}=q =+4.9\mu \ C[/tex]
The charge on [tex]S_2[/tex] is [tex]q_2 = 4.9\mu \ C[/tex]
Given that the charge are arranged in an equilateral triangle, it implies that the angle at each vertices of the triangle is 60°
Generally the magnitude of net force on [tex]S_3[/tex] is mathematically represented as
[tex]F_3= F_{13} cos(60)+ F_{21} cos(60)[/tex]
So
[tex]F_{13} =\frac{k* q^2 }{a^2}[/tex]
Here k is the columns constant with value [tex]k = 9*10^9 \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]
So
[tex]F_{13}=\frac{9*10^9*((4.9*10^{-6})^2)}{0.59^2}[/tex]
=> [tex]F_{13} = 0.62 \ N[/tex]
From the diagram we see that [tex]F_{13}= F_{23} = 0.62\ N[/tex]
So
[tex]F_3=0.62cos(60)+0.62cos(60)[/tex]
[tex]F_3 =0.62\ N[/tex]
Now from the diagram the ls [tex]270^o[\tex] counterclockwise from x-axis
Help me please !!!!!
Answer:
confounding cause they had exposure to many programmes
Answer:
Confounding should be your answer please mark brainliest
what are the x and y components of the vector that has a magnitude of 4N and a direction of 30 degrees
Answer:
We are given
vector as 4N and direction 30°
So since X-component is
adjacent /hypotenuse which is cosine we have
= 4 cos 30 = 3.46 N
And Y-component which is opposite/hypotenuse we have
= 4 sin 30 = 2N
explain the different methods that can be used to model the motion of an object.
Answer:Even things that appear to be at rest move.
When we describe the motion of one object with
respect to another, we say that the object is moving
relative to the other object.
• A book that is at rest, relative to the table it lies
on, is moving at about 30 kilometers per second
relative to the sun.
• The book moves even faster relative to the center
of our galaxy.
Explanation:
If one has twice the length of the other, what is the ratio of the diameter of the longer wire to the diameter of the shorter wire
Explanation:
The resistance of a wire is given by the formula as follows :
[tex]R=\rho \dfrac{l}{A}[/tex]
l is length of wire
A is area of cross section, [tex]A=\pi r^2[/tex]
Since, r=d/2, d = diameter
[tex]A=\pi \dfrac{d^2}{4}[/tex]
It is given in the problem that the resistance of two aluminum wires is same.
[tex]R_1=R_2\\\\\rho\dfrac{L_1}{\pi d_1^2/4}=\rho\dfrac{L_2}{\pi d_2^2/4}\\\\\dfrac{L_1}{L_2}=\dfrac{d_1^2}{d_2^2}[/tex]
We have, L₁=2L₂
So,
[tex]\dfrac{2L_2}{L_2}=\dfrac{d_1^2}{d_2^2}\\\\\dfrac{d_1^2}{d_2^2}=2\\\\\dfrac{d_1}{d_2}=\sqrt{2}[/tex]
So, the ratio of the diameter of the longer wire to the diameter of the shorter wire is [tex]\sqrt2:1[/tex]
Approximately how long will it take light to travel to 1.7 AU in minutes? Use 3 × 108 m/s for the speed of light. unanswered
Answer:
14.1mins
Explanation:
We know that
Time =Distance/Speed
So
= 1.7 au/ (3 E8 m /s )
= 1.7 x 1.496 x 10^11 m /(3 x 10^8 m /s
Converting Au to meters (1.7 x 1.496 x 10^11 m)
= 847.7 s
=14.1 mins
Answer:
Time = 847.73 s = 14.12 min
Explanation:
The distance traveled by an object travelling at a constant speed is given by the following formula:
Distance = (Speed)(Time)
Time = Distance/Speed
where,
Distance = 1.7 AU
converting it into meters:
Distance = (1.7 AU)(1.496 x 10¹¹ m/1 AU) = 2.543 x 10¹¹ m
Speed = 3 x 10⁸ m/s
Therefore, using these values in the equation we get:
Time = (2.543 x 10¹¹ m)/(3 x 10⁸ m/s)
Time = 847.73 s = 14.12 min
According to Newton's Third Law, if someone jumps off a skateboard, the
skateboard will roll backwards because
Answer:
every action has a same an equal reaction
If someone jumps off a skateboard, the skateboard will roll backwards because action and reaction are equal and opposite.
Newton's third law states that action and reaction are equal and opposite. That is, for every action there is a resulting reaction equal in magnitude to the action but opposite in direction.
If someone jumps off a skateboard, he will exert an action force downwards while the skate board exerts a reaction force upwards on him. This causes the skateboard to move backwards while he moves upwards.
Thus, if someone jumps off a skateboard, the skateboard will roll backwards because action and reaction are equal and opposite.
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A stretched string is 2.11 m long and has a mass of 19.5 g. When the string oscillates at 440 Hz , which is the frequency of the standard A pitch, transverse waves with a wavelength of 15.3 cm travel along the string. Calculate the tension T in the string.
Answer:
The tension of the string is 41.876 N
Explanation:
Given;
length of the string, L = 2.11 m
mass of the string, m = 19.5 g = 0.0195 kg
frequency of the wave, f = 440 Hz
wavelength, λ = 15.3 cm = 0.153 m
The velocity of the wave is given by;
v = fλ
v = 440 x 0.153
v = 67.32 m/s
Also the velocity of the wave is given by
[tex]v = \sqrt{\frac{T}{\mu} }[/tex]
where;
μ is mass per unit length = 0.0195 / 2.11 = 0.00924 kg/m
T is the tension of the string
T = v²μ
T = (67.32)²(0.00924)
T = 41.876 N
Therefore, the tension of the string is 41.876 N
What will be the force if the particle's charge is tripled and the electric field strength is halved? Give your answer in terms of F
Answer:
1.5F
Explanation:
Using
E= F/q
Where F= force
E= electric field
q=charge
F= Eq
So if qis tripled and E is halved we have
F= (E/2)3q
F= 1.5Eq=>> 1.5F
While John is traveling along a straight inter-
state highway, he notices that the mile marker
reads 241 km. John travels until he reaches
the 145 km marker and then retraces his path
to the 174 km marker.
What is John's resultant displacement from
the 241 km marker?
Answer in units of km.
Answer:
His displacement is 260 – 169 = 91 km
His distance traveled is (260 – 134) + ( 169 – 134 ) = 161 km
Explanation:
What is the power in D of the eyes of a woman who can see an object clearly at a distance of only 9.5 cm
Answer:
Explanation:
It is the far point an myopic person . For it she uses concave lens . For her ,
object distance u = infinity
image v = - 9.5 cm
f = ?
form the lens formula
[tex]\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\[/tex]
[tex]\frac{1}{f} = \frac{1}{-9.5} - 0 \\[/tex]
f = - 9.5 cm = - 9.5 / 100 m
P = 1 /f = - 100 / 9.5 = - 10.52 D .
Why are experimental investigations the best type of scientific investigation o demonstrat cause-and-effect relationships?
Answer:
Isolation of Variables.
Explanation:
I am just taking a guess but Experimental investigations allow you to isolate variables (making most variables constant, allowing only the independent variable to have an impact on the dependent variable). With other scientific investigations, if they don't isolate the variables it is very possible that an unknown third variable could be skewing the results (which will result in correlation but not causation).
Answer:
Experimental investigations are the best type of scientific investigation to demonstrate cause-and-effect relationships because they allow the investigator to actively manipulate variables and control conditions. The independent variable can be isolated and manipulated, producing a change (or no change) in the dependent variable. This change can be measured to gather evidence to support or refute the cause-and-effect relationship. Experimental investigations also use a control group (a group that does not receive the factor being tested) for comparison.
Explanation:
Automated-response....it's right
The horizontal and vertical components of the initial velocity of a projectile are 50 m/s and 120 m/s respectively. What is the magnitude of the initial velocity
Answer:
50 m/s
Explanation:
In a projectile motion, acceleration is only on the vertical plane, that is, the vertical velocity is the only velocity component that undergoes acceleration under gravitational force. The horizontal component of velocity dos not change with time, and hence is the same as the initial velocity of the projectile.
A car travels in a straight line with an average velocity of 80 km/h for 2.5 h and then with an average velocity of 40 km/h for 1.Sh. What is the total displacement for the 4h trip and what is the average velocity for the total trip
Answer:
260 km
65 km/hr
Explanation:
The displacement of an object is the distance moved by that object in a particular direction.
velocity = displacement / time, therefore,
Displacement = velocity * time.
Total displacement for the 4 HR trip = displacement for 2.5 hr + displacement for 1.5 hr
total displacement = (80 * 2.5) + (40 * 1.5)
Total displacement = 200 + 60
Total displacement = 260 Km
average velocity for the total trip = total displacement / total time taken
average velocity = 260 km/ 4 hr
average velocity = 65 km/hr
"If we increase the amplitude of transverse waves on a string, what happens to the speed of a wave along the string?"
Answer:
It is increased
Explanation:
If the amplitude of a transverse wave in a string is increased, then also, the speed of that wave along the string also increases.
In transverse waves, the particles of the medium does move vertically, m either up or down and also moves at right angles to the direction of the wave. It should be noted also, that the wave amplitude of a transverse wave is defined as the difference in height between the crest(high) and the resting position(low).
The highest point particles of the medium reach is often tagged as the crest, while the lowest point is tagged as the resting position. The higher the height of the crests are, the greater the amplitude of the wave will be.
Two students measure the speed of light. One obtains (3.001 0.001)10^8 m/s; the other obtains (2.999 0.006)10^8 m/s.Required:Which is more precise?
Answer:
(30010001) 10^8 m/s
Explanation:
In a particular wire in a circuit, conventional current flows in the z direction. What is the direction of electron current in the wire
Answer:
Electron current will flow in the negative z axis
Explanation:
This is because flow of conventional current is opposite that if electron current so since conventional is in +z axis electron will be -z axis