Answer:
The maximum wavelength of light that could liberate electrons from the aluminum metal is 303.7 nm
Explanation:
Given;
wavelength of the UV light, λ = 248 nm = 248 x 10⁻⁹ m
maximum kinetic energy of the ejected electron, K.E = 0.92 eV
let the work function of the aluminum metal = Ф
Apply photoelectric equation:
E = K.E + Ф
Where;
Ф is the minimum energy needed to eject electron the aluminum metal
E is the energy of the incident light
The energy of the incident light is calculated as follows;
[tex]E = hf = h\frac{c}{\lambda} \\\\where;\\\\h \ is \ Planck's \ constant = 6.626 \times 10^{-34} \ Js\\\\c \ is \ speed \ of \ light = 3 \times 10^{8} \ m/s\\\\E = \frac{(6.626\times 10^{-34})\times (3\times 10^8)}{248\times 10^{-9}} \\\\E = 8.02 \times 10^{-19} \ J[/tex]
The work function of the aluminum metal is calculated as;
Ф = E - K.E
Ф = 8.02 x 10⁻¹⁹ - (0.92 x 1.602 x 10⁻¹⁹)
Ф = 8.02 x 10⁻¹⁹ J - 1.474 x 10⁻¹⁹ J
Ф = 6.546 x 10⁻¹⁹ J
The maximum wavelength of light that could liberate electrons from the aluminum metal is calculated as;
[tex]\phi = hf = \frac{hc}{\lambda_{max}} \\\\\lambda_{max} = \frac{hc}{\phi} \\\\\lambda_{max} = \frac{(6.626\times 10^{-34}) \times (3 \times 10^8) }{6.546 \times 10^{-19}} \\\\\lambda_{max} = 3.037 \times 10^{-7} m\\\\\lambda_{max} = 303.7 \ nm[/tex]
How do the magnitude and direction of the electric field on the left side of the dipole compare to the right side for the same distance
Answer:
The magnitude of the electric field is same while the direction at the left and at the right is opposite to each other.
Explanation:
The direction of the electric field due to the dipole on the axial line is same as the direction of dipole moment.
The magnitude of the electric field due to an electric dipole on its axial line is
[tex]E=\frac{2kp}{r^3}[/tex]
where, k is the constant, p is the electric dipole moment and r is the distance from the center of dipole.
The magnitude of the electric field is same while the direction at the left and at the right is opposite to each other.
A charged particle is injected into a uniform magnetic field such that its velocity vector is perpendicular to the magnetic field lines. Ignoring the particle's weight, the particle will
Answer:
The charged particle will follow a circular path.
Explanation:
Formula for the magnetic force is;
F = qvb sin θ
Where;
where;
q = the charge
v = the velocity
B = the magnetic field
θ = the angle between the velocity and magnetic field
We are told that velocity vector is perpendicular to the magnetic field lines. Thus, angle is 90.
So sin θ = sin 90 = 1
Thus,
F = qvB
Now, since the velocity vector is perpendicular to the magnetic field line,it also means from flemmings right hand rule, that the magnetic force is as well perpendicular to both of them.
Therefore, we have:
- a force that is always perpendicular to the velocity and as well constant in magnitude since magnitude of velocity or magnetic field does not change.
What this statement implies is that the force is acting as a centripetal force, and therefore, the charged particle will be kept in a uniform circular motion.
A 1.0 ball moving at 2.0 / perpendicular to a wall rebounds from the wall at 1.5 /. If the ball was in contact with the wall for 0.1 , what force did the wall impart onto the ball?
Answer:
5N
Explanation:
We have a simple problem of momentum here.
ΔMomentum= mΔv= FΔt
Solve for F
mΔv/Δt=F
Plug in givens
1*(2-1.5)/0.1=F
F=5N
The amount of force that the wall imparts on the ball is 5.0N
According to Newton's second law, the formula for calculating the force applied is expressed as:
[tex]F=ma[/tex]
m is the mass of the object
a is the acceleration of the object
Since acceleration is the change in velocity of an object, hence [tex]a=\frac{\triangle v}{t}[/tex]
The applied force formula becomes [tex]F=\frac{m\triangle v}{t}[/tex]
Given the following parameters
m = 1.0kg
[tex]\triangle v=2.0-1.5\\\triangle v=0.5m/s[/tex]
t = 0.1sec
Substitute the given parameter into the formula
[tex]F=\frac{1.0\times 0.5}{0.1}\\F=\frac{0.5}{0.1}\\F=5N[/tex]
Hence the amount of force that the wall imparts on the ball is 5.0N
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For waves moving through the atmosphere at a constant velocity, higher frequency waves must have proportionally longer wavelengths.
a) true
b) false
Answer:
false.
Explanation:
We know that for a wave that moves with velocity V, with a wavelength λ, and a frequency f, we have the relation:
V = λ*f
So, if the velocity is constant and we increase the frequency to:
f' > f
we will have a new wavelength λ'
Such that:
V = f'*λ'
And V = f*λ
Then we have:
f'*λ' = f*λ
Solvinf for λ', we get:
λ' =(f/f')*λ
And because:
f' > f
then:
(f/f') < 1
Then:
λ' =(f/f')*λ < λ
So, if we increase the frequency, we need to decrease the wavelength.
So, for higher frequency waves, we must have proportionally shorter wavelengths.
Then we can conclude that the given statement:
"or waves moving through the atmosphere at a constant velocity, higher frequency waves must have proportionally longer wavelengths"
is false.
A student measure the length of a laboratory bench with a meter ruler. Which of the following values is the most approbriate way to record the result ? a.4.022m b.4.02m c.4.0m d.4m
Answer:
Well a meter stick has increments of a centimeter, and since 1 cm=0.01m he should record it as 4.02m(b)
Explanation:
Wood is an example of
A. Metalloid
B. Insulator
C. Nonmetal
D. Conductor
Receptor elétrico 5 pontos Dispositivo que converte energia elétrica em outra forma de energia, não exclusivamente térmica. Exemplos: motores elétricos, ventiladores, liquidificadores, geladeiras, aparelhos de sons, vídeos, celulares, computadores?
Answer:
Electromechanical transducer and Electrical receiver.
Explanation:
Electromechanical transducer is the part of a communication system which converted electrical waves or electrical energy into sound waves. The most common example loudspeaker while on the other hand, Electrical receiver is a device that converts electrical energy into another form of energy, except thermal. Examples are cell phones, computers and television.
The mass per unit length of the rope is 0.0500 kg/m. Find the tension. Express your answer in newtons.
Complete question:
A transverse wave on a rope is given by [tex]y \ (x, \ t) = (0.75 \ cm) \ cos \ \pi[(0.400 \ cm^{-1}) x + (250 \ s^{-1})t][/tex]. The mass per unit length of the rope is 0.0500 kg/m. Find the tension. Express your answer in newtons.
Answer:
The tension on the rope is 1.95 N
Explanation:
The general equation of a progressive wave is given as;
[tex]y \ (x,t) = A \ cos(kx \ + \omega t)[/tex]
Compare the given equation with the general equation of wave, the following parameters will be deduced.
A = 0.75 cm
k = 0.400π cm⁻¹
ω = 250π s⁻¹
The frequency of the wave is calculated as;
ω = 2πf
2πf = 250π
2f = 250
f = 250/2
f = 125 Hz
The wavelength of the wave is calculated as;
[tex]\lambda = \frac{2\pi}{k} \\\\\lambda = \frac{2\pi }{0.4 \pi} = 5 \ cm = 0.05 \ m[/tex]
The velocity of the wave is calculated as;
v = fλ
v = 125 x 0.05
v = 6.25 m/s
The tension on the rope is calculated as;
[tex]v = \sqrt{\frac{T}{\mu}} \\\\where;\\\\T \ is \ the \ tension \ of \ the \ rope\\\\\mu \ is \ the \ mass \ per \ unit \ length = 0.05 \ kg/m\\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu\\\\T = (6.25)^2\times (0.05)\\\\T = 1.95 \ N[/tex]
Therefore, the tension on the rope is 1.95 N
65. The weight of a body when totally immersed in a liquid is 4.2N if he weight of the liquid displaced is 2.5N. Find the weight of the body in air.
Answer:
Given, Apparent weight(W₂)=4.2N
Weight of liquid displaced (u)=2.5N
Let weight of body in air = W₁
Solution,
U=W₁-W₂
W₁=4.2=2.5=6.7N
∴Weight of body in air is 6.7N
Question 2 of 10
Which of the following statements is true of an isolated system?
A.The system has energy but no matter.
B.Energy separates the matter in the system from outside matter.
C.The matter within the system does not interact with matter outside the system
D.The matter within the system does not interact with other matter in the system
Answer:
D
Explanation:
because it is system removed from system and it dosent interact with them
The density of blood is 1055 kg/m3 . If the blood at the very top of your head exerts a minimum gauge pressure of 45 mm Hg (6000 Pa), estimate the gauge pressure at your heart in pascals.
Answer:
P = 10135.6 Pa
Explanation:
For this exercise we use that the pressure varies with the height
P = P₀ + ρ g h
where h is the height from the head to the heart, which is approximately
h = 40 cm = 0.40m and P₀ is the head pressure P₀ = 6000 Pa
P = 6000 + 1055 9.8 0.40
P = 6000 + 4135.6
P = 10135.6 Pa
Why does a compass give unreliable readings when used near electrical appliances
Answer:
Explanation:
Since the compass uses a magnetic field, if anything else magnetic is near it, the compass will start acting up. Making it unreliable so keep magnets away!
A 55-kg block, starting from rest, is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 140 N. Fp is parallel to the displacement of the block. The final speed of the block is 2.35 m/s.
a) How much work was converted to thermal energy? What work did friction do on the box?
b) What is the coefficient of friction?
Answer:
The answer is "151.25 J and -547.64 J".
Explanation:
[tex]u = 0\\\\v = 2.35\ \frac{m}{sec}\\\\d = 5.0 \ m\\\\[/tex]
Using formula:
[tex]v^2 = u^2 + 2 \times a \times d\\\\2.35^2 = 0^2 + 2 \times a \times 5\\\\a = \frac{2.35^2}{10} \\\\[/tex]
[tex]= 0.55 \ \frac{m}{sec^2}\\\\[/tex]
[tex]F_{net} = m \times a\\\\F_{net} = 55 \times 0.55 = 30.25\ N\\\\[/tex]
Calculating the Work by net force
[tex]W = F_{net}\times d\\\\W = 30.25 \times 5 = 151.25 \ J\\\\[/tex]
The above work is converted into thermal energy.
Now,
[tex]F_{net} = F_p - F_f\\\\F_p = 140 \ N\\\\F_f = u_k\times m \times g = u_k \times 55 \times 9.81\\\\F_f = 539.55 \times u_k\\\\30.25 = 140 - u_k \times 55 \times 9.81\\\\u_k = \frac{(140 - 30.25)}{(55\times 9.81)}\\\\uk = 0.203 = \text{Coefficient of friction}\\\\W_f = -F_f \times d\\\\W_f = -0.203 \times 55 \times 9.81 \times 5\\\\Work\ done\ by\ friction = -547.64 \ J[/tex]
Why is the force of attraction between the Earth and ourselves so huge compared to the attraction between two apples?
Answer:
Answer in explanation
Explanation:
The force of attraction between two bodies is governed by Newton's Law of Gravitation:
[tex]F = \frac{Gm_1m_2}{r^2}[/tex]
where,
G = Universal Gravitational Constant
m₁ = mass of the first body
m₂ = mass of the second body
r = distance between the two bodies
F = Force
Hence, it is clear from the formula that the magnitude of the force is directly proportional to the product of the masses of the objects. So in the case of the earth and ourselves, the mass of the earth is very large in order of 10²⁴ kg. Due to this huge mass, the attraction between the earth and ourselves is so huge as compared to the attraction between two apples. Because the masses of the apple are very small in grams.
Two electrons are passing 20.0 mm apart. What is the electric repulsive force that they exert on each other
Answer:
0.5766422350752*10^-24 N
Explanation:
Couloumb's law states that states that there is an electrical force acting on 2 static charges. The magnitude is directly proportional to the product of the 2 charges.
Strength of electrons : q1 = q2 = 1.602 x 10-19. C
Substitute and solve:
F = (9*10^9)(1.602 x 10-19)(1.602 x 10-19) / (0.02)^2
Done.
What is life like in a cave camp? Do you think you would like to experience this? Why or why not?
Answer:
There's no risk of animals or bad weather interfering with your campsite, either. You don't even really need a tent. A sleeping pad, sleeping bag and a mindful eye to pick up everything you brought in is all you really need to enjoy overnight caving. Do your research
Explanation:
Hi, so i have to find T1, can some1 help?
30.1 N
Explanation:
Given:
[tex]W_1 = 16\:\text{N}[/tex]
[tex]W_2 = 8\:\text{N}[/tex]
Let's write the components of the net forces at the intersections. Note that the system is equilibrium so all the net forces are zero.
Forces involving W1:
[tex]x:\:\:\:-T_1 + T_3\cos \alpha = 0\:\: \\ \text{or}\:\:T_2 = T_3\cos \alpha\:\:\:\:\:(1)[/tex]
[tex]y:\:\:\:T_3\sin \alpha - W_1 = 0\:\:\: \\ \text{or}\:\:\:T_3\sin \alpha = W_1\:\:\:\:\:\:(2)[/tex]
Forces involving W2:
[tex]x:\:\:\:T_1\sin 53 - T_3\sin \alpha = W_2\:\:\:\:\:\:\:(3)[/tex]
[tex]y:\:\:\:T_4 - T_1\cos 53 - T_3\cos \alpha = 0\:\:\:\;(4)[/tex]
Substitute (2) into (3) and we get
[tex]T_1\sin 53 - W_1 = W_2[/tex]
Solving for [tex]T_1[/tex],
[tex]T_1 = \dfrac{W_1 + W_2}{\sin 53} = 30.1\:\text{N}[/tex]
The drawings show (in cross section) two solid spheres and two spherical shells. Each object is made from copper and has a net charge, as the plus and minus signs indicate. Which drawing correctly shows where the charges reside when they are in equilibrium?
a) shows a lot of negative signs in the interior of circle
b) shows a lot of positive signs in the interior of circle
c) shows a hollowed out "hole" in the interior of the circle, with negative signs surrounding the opening.
d) shows a hollowed out "hole" in the interior of the circle, with positive signs surrounding the exterior edge
Answer:
d
Explanation:
The minimum energy configuration in electrostatics states that Charges always reside on the surface of a conductor. If anyhow they were inside, an electric field would exist inside and would act to move them to the surface,
Therefore, the drawing that shows where the charges reside when they are in equilibrium is a hollowed-out "hole" in the interior of the circle, with positive signs surrounding the exterior edge. This means that the d part is the correct answer.
NEED HELP ASAP- Please show work
The angular position of an object is given by θ = 4t3 +10t −40 , where θ is in radians and t is in seconds what is:
(a) (5 points) The angular velocity at t = 2 s?
(b) (5 points) The angular acceleration at t = 2 s?
Answer:
Look at work
Explanation:
Θ= 4t^3+10t-40
a) In order to find ω, we need to find displacement so plug in t=2 to find Θ.
Θ= 4*8+20-40=12
use ω=Θ/t
Plug in values
ω=6 rad/s
b) In order to find α we use ω/t.
Plug in values
α=6/2= 3 rad/s^2
The total resistance of a parallel circuit is 25 ohms. If the total current is 100mA, how much current is through a 220 ohm resistor that makes up part of the parallel circuit?
Answer:
The current across the resistance is 0.011 A.
Explanation:
Total resistance, R = 25 ohms
Total current, I = 100 mA = 0.1 A
Let the voltage is V.
By the Ohm's law
V = I R
V = 0.1 x 25 = 2.5 V
Now the resistance is R' = 220 ohm
As they are in parallel so the voltage is same. Let the current is I'.
V = I' x R'
2.5 = I' x 220
I' = 0.011 A
How do the magnitudes of the currents through the full circuits compare for Parts I-III of this exercise, in which resistors are combined in series, in parallel, and in combination
Answer: hello tables and data related to your question is missing attached below are the missing data
answer:
a) I = I₁ = I₂ = I₃ = 0.484 mA
b) I₁ = 0.016 amps
I₂ = 0.0016 amps
I₃ = 7.27 * 10^-4 amps
c) I₁ = 1.43 * 10^-3 amp
I₂ = 0.65 * 10^-3 amps
Explanation:
A) magnitude of current for Part 1
Resistors are connected in series
Req = r1 + r2 + r3
= 3300 Ω ( value gotten from table 1 ) ,
V = 1.6 V ( value gotten from table )
hence I ( current ) = V / Req = 1.6 / 3300 = 0.484 mA
The magnitude of current is the same in the circuit
Vi = I * Ri
B) magnitude of current for part 2
Resistors are connected in parallel
V = 1.6 volts
Req = [ ( R1 * R2 / R1 + R2 ) * R3 / ( R1 * R2 / R1 + R2 ) + R3 ]
= [ ( 100 * 1000 / 100 + 1000) * 2200 / ( 100 * 1000 / 100 + 1000 ) + 2200]
= 87.30 Ω
For a parallel circuit the current flow through each resistor is different
hence the magnitude of the currents are
I₁ = V / R1 = 1.6 / 100 = 0.016 amps
I₂ = V / R2 = 1.6 / 1000 = 0.0016 amps
I₃ = V / R3 = 1.6 / 2200 = 7.27 * 10^-4 amps
C) magnitude of current for part 3
Resistors are connected in combination
V = 1.6 volts
Req = R1 + ( R2 * R3 / R2 + R3 )
= 766.66 Ω
Total current ( I ) = V / Req = 1.6 / 766.66 = 2.08 * 10^-3 amps
magnitude of currents
I₁ = ( I * R3 ) / ( R2 + R3 ) = 1.43 * 10^-3 amps
I₂ = ( I * R2 ) / ( R2 + R3 ) = 0.65 * 10^-3 amps
A dandelion seed floats to the ground in a mild wind with a resultant velocity of 26.0 cm/s. If the horizontal component velocity due to the wind is 10.0 cm/s, what is the vertical component velocity? Show all work.
Answer:
24 cm/s
Explanation:
Applying
Pythagoras theorem,
a² = b²+c²............. Equation 1
Where a = resultant, b = vertical component, c = horizontal component
From the question,
Given: a = 26 cm/s, c = 10 cm/s
Substitute these values into equation 1
26² = b²+10²
676 = b²+100
b² = 676-100
b² = 576
b = √576
b = 24 cm/s
what are the dynamic properties of a nucleus
What is the total number of moles of products involved in the following reaction?
CaCO3 (s) + 2HCl (aq) - CaCl2 (aq) + CO2 (g) + H20 (g)
O 6
2.
3
5
Answer:
3
Explanation:
You must first make sure the equation is balanced. This one is. Then, you simply add up the coefficients of each compound on the products side of the equation. When the coefficient is not specified, you can assume it is 1 mole. So, in this equation, there is 1 mole of CaCl₂, 1 mole of CO₂, and 1 mole of H₂O = 3 moles.
The reactant side of the equation also has three moles:
1 mole of CaCO₃ and 2 moles of HCl.
boat carrying people more than its capacity is attributes of sinking why
Answer:
Upthrust on boat becomes lesser than Weight of boat
Explanation:
When there are more people than the capacity, The weight of the boat acting downwards increases. However, the upthrust acting on the submerged part of the boat is constant. Since Weight > Upthrust, there is a net force downwards, leading to sinking.
Express the unit of force in terms of fundamental unit
Answer:
The fundamental unit of force is kg.m/s²
Explanation:
According to Newton's second law of motion, force is given as the product of mass and acceleration.
Mathematically, force can be expressed as; F = ma
where;
F is the force
M is mass of the object, unit of mass = kg
a is acceleration of the object, unit of acceleration = m/s²
Force = kg x m/s²
Force = kg.m/s² = Newton [N]
Therefore, the fundamental unit of force is kg.m/s²
TIME REMAINING
45:13
A framed picture hangs from two cords attached to the ceiling.
A picture of a picture frame hanging by two cables at the center of the frame at the same length and angle from the vertical.
Which shows the correct free body diagram of the hanging picture?
A free body diagram with two force vectors, the first pointing downward labeled F Subscript g Baseline, the second pointing upward labeled F Subscript N Baseline.
A free body diagram with three force vectors, the first pointing south labeled F Subscript p Baseline, the second pointing northeast labeled F Subscript T Baseline, and the third pointing northwest labeled F Subscript N.
A free body diagram with three force vectors, the first pointing south labeled F Subscript g Baseline, the second pointing northeast labeled F Subscript T Baseline and the third pointing northwest labeled F Subscript T.
A free body diagram with two force vectors, the first pointing downward labeled F Subscript p Baseline, the second pointing upward labeled F Subscript T Baseline.
Answer:The answer is C
Explanation:
recognizing forms of energy
Answer:
hi the question isn't obvious and need a photo I guess
A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is rotating at 10 rev/s; 60 revolutions later, its angular speed is 15 rev/s. Calculate
(a) the angular acceleration,
(b) the time required to complete the 60 revolutions,
(c) the time required to reach the 10 rev/s angular speed, and
(d) the number of revolutions from rest until the time the disk reaches the 10 rev/s angular speed.
Explanation:
Given:
[tex]\omega_0[/tex] = 10 rev/s = [tex]20\pi\:\text{rad/s}[/tex]
[tex]\omega[/tex] = 15 rev/s = [tex]30\pi\:\text{rad/s}[/tex]
[tex]\theta[/tex] = 60 rev = [tex]120\pi\:\text{rads}[/tex]
a) the angular acceleration [tex]\alpha[/tex] is given by
[tex]\alpha = \dfrac{\omega^2 - \omega_0^2}{2\theta}[/tex]
[tex]\:\:\:\:\:\:\:=\dfrac{(30\pi)^2 - (20\pi)^2}{240\pi} = 6.5\:\text{rad/s}^2[/tex]
b) [tex]t = \dfrac{\omega - \omega_0}{\alpha} = \dfrac{30\pi - 20\pi}{6.5} = 4.8\:\text{s}[/tex]
c) [tex]t = \dfrac{\omega - \omega_0}{\alpha}[/tex]
[tex]=\dfrac{20\pi - 0}{6.5} = 9.7\:\text{s}[/tex]
d)[tex]\theta = \frac{1}{2}\alpha t^2[/tex]
[tex]\:\:\:\:\:\:\:=\frac{1}{2}(6.5\:\text{rad/s}^2)(9.7\:\text{s})^2 = 305.8\:\text{rad}[/tex]
[tex]\:\:\:\:\:\:\:= 48.7\:\text{revs}[/tex]
Betelgeuse (in Orion) has a parallax of 0.00451 + 0.00080 arcsec,as measured by the Hipparcos satellite. What is the distance to Betelgeuse, and what is the uncertainty in that measurement?
We have that the distance to Betelgeuse, and the uncertainty in that measurement is
[tex]d=(221.7\pm39.33)pc[/tex]Uncertainty U = 0.00080
From the Question we are told that
Betelgeuse (in Orion) has a parallax of 0.00451 + 0.00080
Generally
[tex]Distance\ in\ parsecs =\frac{ 1}{(parallax\ measured\ in\ arcseconds}[/tex]
Where
Parallax [tex]P =0.00451[/tex]
Uncertainty [tex]U = 0.00080[/tex]
Generally the equation for the distance is mathematically given as
[tex]d=(\frac{1}{P}pc\pm(\frac{U}{P}*100\%))[/tex]
Therefore
[tex]d=(\frac{1}{0.00451}pc\pm(\frac{0.00080}{0.00451}*100\%))[/tex]
[tex]d=(221.7\pm39.33)pc[/tex]
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