Complete question:
a metal of work function 1.8eV is illuminated by light of wavelength 3.0x*10-7 m,calculate the maximum kinetic energy of the emitted photons.
Answer:
the maximum kinetic energy of the emitted photons is 3.742 x 10⁻¹⁹ J
Explanation:
Given;
work function, Ф = 1.8 eV
wavelength of the light, λ = 3 x 10⁻⁷ m
The maximum kinetic energy of the emitted photons is calculated from photoelectric equation.
[tex]E = K.E_{max} + \phi\\\\KE_{max} = E- \phi\\\\where;\\\\E \ is \ the \ energy \ of \ the \ incident \ light\\\\E = hf = h \frac{c}{\lambda} \\\\where;\\\\c \ is \ speed \ of \ light = 3 \times 10^8 \ m/s\\\\h \ is \ Planck's \ constant = 6.626 \times 10^{-34} \ Js\\\\E = \frac{ (6.626 \times 10^{-34})\times ( 3 \times 10^8)}{3\times 10^{-7}} \\\\E = 6.626 \times 10^{-19} \ J[/tex]
[tex]K.E_{max} = E - \phi\\\\K.E_{max} = 6.626\times 10^{-19} \ J \ - \ (1.8 \times 1.602 \times 10^{-19} \ J)\\\\K.E_{max} = 6.626\times 10^{-19} \ J \ - \ 2.884 \times 10^{-19} \ J\\\\K.E_{max} =3.742 \times 10^{-19} \ J[/tex]
Therefore, the maximum kinetic energy of the emitted photons is 3.742 x 10⁻¹⁹ J
you are stowing items and come across an aerosol bottle of hairspray.what should you do?
Answer:
below
Explanation:
The time-average power carried by a UPEMW propagating in vacuum is 0.05 W/m2. i) What is the amplitude value of the electric field and the amplitude value of the magnetic field in the wave
Answer:
The correct solution is "11.51 mA".
Explanation:
Given:
Time average power,
[tex]P_{avg}=0.05 \ W/m^2[/tex]
n = 377
As we now,
⇒ [tex]P_{avg}=\frac{E_0^2}{n}[/tex]
or,
⇒ [tex]E_0^2=0.05\times 377[/tex]
⇒ [tex]=4.341 \ V[/tex]
hence,
⇒ [tex]H_0=\frac{E_0}{n}[/tex]
By putting the values, we get
[tex]=\frac{4.341}{377}[/tex]
[tex]=11.51 \ mA[/tex]
Una pelota de basket es soltada desde 2.5 m de altura y rebota con una velocidad igual a 3/4 partes de la velocidad que llego. ¿ a qué altura alcanza la bola en el rebote ? ¿ cuánto tiempo transcurre desde que rebota ?
Answer:
Tenemos dos problemas a resolver acá:
Primero, debemos encontrar la velocidad con la que la pelota impacta el suelo.
Acá podemos usar la conservación de la energía.
E = U + K
U = energía potencial = m*g*H
m = masa
g = aceleración gravitatoria = 9.8m/s^2
H = altura
K = energía cinética = (m/2)*V^2
donde V es la velocidad.
Inicialmente, cuando la pelota es soltada, su velocidad es cero, entonces solo tenemos energía potencial:
Ei = U = m*(9.8m/s^2)*2.5m
Al final, cuando la pelota esta por impactar el suelo, la altura tiende a cero, entonces ya no hay energía potencial, solo hay energía cinética:
Ef = (m/2)*V^2
Y como la energía se conserva, la energía final es igual a la inicial, entonces:
m*(9.8m/s^2)*2.5m = (m/2)*V^2
Podemos resolver esto para V, y asi obtener la velocidad con la que la pelota impacta el suelo.
V = √(2*(9.8m/s^2)*2.5m) = 7m/s
Ahora respondamos la segunda parte.
Una vez la pelota rebota, su aceleración va a estar dada solamente por la aceleración gravitatoria, entonces tenemos:
A(t) = -9.8m/s^2
Para obtener su velocidad integramos:
V(t) = (-9.8m/s^2)*t + V0
donde V0 es la velocidad con la que la pelota reboto, que sabemos que es 3/4 de 7m/s
V0 = (3/4)*7m/s = (21/4) m/s
Así, la ecuación de la velocidad es:
V(t) = (-9.8m/s^2)*t + (21/4) m/s
Sabemos que la altura máxima se da cuando la velocidad es igual a cero, entonces primero calculemos el valor de t tal que esto ocurra:
V(t) = 0 = (-9.8m/s^2)*t + (21/4) m/s
t = (21/4) m/s/9.8m/s^2 = 0.54 s
Ahora debemos encontrar la ecuación de la posición y evaluarlo en este tiempo.
Para ello integramos de vuelta:
P(t) = (1/2)(-9.8m/s^2)*t^2 + (21/4 m/s)*t + P0
donde P0 es la posición inicial, como la pelota rebota en el suelo, la posición inicial es el suelo, el cual representamos con 0, entonces la ecuación de la posición es:
P(t) = (1/2)(-9.8m/s^2)*t^2 + (21/4 m/s)*t
La altura máxima estará dada por esta ecuación evaluada en t = 0.54 s
P(0.54s) = (1/2)(-9.8m/s^2)*(0.54s)^2 + (21/4 m/s)*0.54s = 1.81 m
La altura máxima es 1.81 metros.
Y entre que rebota y llega a esta altura máxima, transcurren 0.54 segundos.
When it comes to the movement of air, friction
A. increases with altitude.
B. is greater near the ground surface.
C. diminishes turbulence.
D. is responsible for weaker winds aloft.
Answer: When it comes to the movement of air, friction is greater near the ground surface.
Explanation:
A resistance in motion observed by an object while on another object is called friction.
For example, a vehicle moving on road will have friction between its tires and the road.
Friction is more near the ground surface rather than away from the ground surface.
Thus, we can conclude that when it comes to the movement of air, friction is greater near the ground surface.
what is measured by the ammeter
Answer:
amperes
Ammeter, instrument for measuring either direct or alternating electric current, in amperes. An ammeter can measure a wide range of current values because at high values only a small portion of the current is directed through the meter mechanism; a shunt in parallel with the meter carries the major portion.
Explanation:
hope it helps
Find the final velocity if the initial velocity of 8 m/s with an acceleration of 7 m/s2 over a 3 second interval?
I don't know about it your answer will give another people
Answer: Let the final velocity be v.
Given,
Initial velocity(u)=8m/s
Acceleration(a)=7m/s2
Time(t)=3 sec
Then,
v=u+at
=8+7*3 m/s
=29m/s
Therefore, the final velocity is 29m/s.
Kulsum’s TV uses 45 W. How much does it cost her to watch TV for one month (30 days). She watches TV for 4 hours/day during mid-peak time (10.4 cents/kWh).
Answer:
Total cost = 56.16 cents
Explanation:
Given the following data;
Power = 45 Watts
Time = 4 hours
Number of days = 30 days
Cost = 10.4 cents
To find how much does it cost her to watch TV for one month;
First of all, we would determine the energy consumption of the TV;
Energy = power * time
Energy = 45 * 4
Energy = 180 Watt-hour = 180/1000 = 0.18 Kwh (1 Kilowatts is equal to 1000 watts).
Energy consumption = 0.18 Kwh
Next, we find the total cost;
Total cost = energy * number of days * cost
Total cost = 0.18 * 30 * 10.4
Total cost = 56.16 cents
Please, describe low-frequency vs. high-frequency waves.
Answer: High-frequency sound waves are perceived as high-pitched sounds, while low-frequency sound waves are perceived as low-pitched sounds. The audible range of sound frequencies is between 20 and 20000 Hz, with greatest sensitivity to those frequencies that fall in the middle of this range.
Explanation: Obviously explained in the answer
A 2000-kg truck traveling at a speed of 6.0 m/s slows down to 4.0 m/s along a straight road. What
is the magnitude of the impulse?
The magnitude of the impulse of the truck is equal to 4000 Kg.m/s.
What is impulse?Impulse can be described as the integral of a force over the time interval for which it acts. Impulse is also a vector quantity since force is a vector quantity. Impulse can be applied to an object that generates an equivalent vector change in its linear momentum.
The S.I. unit of impulse is N⋅s and the dimensionally equivalent unit of momentum is kg⋅m/s. A resultant force gives acceleration and changes the velocity of an object for as long as it acts.
Given the mass of the truck, m= 2000 Kg
The initial speed of the truck, u = 6 m/s
The final speed of the truck, v = 4 m/s
The change in the linear momentum is equal to the impulse.
I = ΔP = mv - mu
I = 2000 ×4 - 2000 × 6
I = 8000 - 12000
I = - 4000 Kg.m/s²
Therefore, the magnitude of the impulse is 4000 Kg.m/s².
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A pulley has a mechanical advantage of 1. What does this tell you about the size and direction of the input and output forces?
Answer:
The number of input force is the same as output. ... If it equals once, then both numbers are equal making it the same.Explanation:
Does this helps
Answer:
The number of input force is the same as output. Formula for MA (Mechanical Advantage) is Input Force/Output Force. If it equals once, then both numbers are equal making it the same. In order to raise MA, you must lower efficiency, something you learn around grade 8. Good luck!
P.S. Direction is the same for both, meaning if you pull something, the object you pull will come towards you.
A 55 kg pole vaulter falls from rest from a height of 5.4 m onto a foam rubber pad. The pole vaulter comes to rest 0.24 s after landing on the pad.
a. Calculate the athlete's velocity just before reaching the pad
b. Calculate the constant force exerted on the pole vaulter due to the collision.
Answer:
a) 10.3 m/s
b) 566 N
Explanation:
[tex]v {}^{2} = {u}^{2} + 2as \\ v {}^{2} = 0 {}^{2} + 2(9.81)(5.4) \\ v = 10.3 \: ms {}^{ - 1} [/tex]
[tex]force \: = \frac{d(mv)}{dt} \\ = 55(10.293) \\ = 566 \: newtons[/tex]
The athelete velocity will be 10.3 and constant force 566 N.
What is velocity?The displacement that an object or particle experiences with respect to time is expressed vectorially as velocity. The meter per second (m/s) is the accepted unit of velocity magnitude (also known as speed).
Alternately, the magnitude of velocity can be expressed in centimeters per second (cm/s). Depending on how many dimensions are included, there are numerous ways to indicate the direction of a velocity vector.
The car's velocity in relation to your body is zero when you are driving. The speed of the car in relation to you if you were to stand by the side of the road is 20 m/s northward.
Therefore, The athelete velocity will be 10.3 and constant force 566 N.
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A 97.0 kg ice hockey player hits a 0.150 kg puck, giving the puck a velocity of 48.0 m/s. If both are initially at rest and if the ice is frictionless, how far does the player recoil in the time it takes the puck to reach the goal 14.5 m away
Answer:
s₁ = 0.022 m
Explanation:
From the law of conservation of momentum:
[tex]m_1u_1 + m_2u_2 = m_1v_1+m_2v_2[/tex]
where,
m₁ = mass of hockey player = 97 kg
m₂ = mass of puck = 0.15 kg
u₁ = u₂ = initial velocities of puck and player = 0 m/s
v₁ = velocity of player after collision = ?
v₂ = velocity of puck after hitting = 48 m/s
Therefore,
[tex](97\ kg)(0\ m/s)+(0.15\ kg)(0\ m/s)=(97\ kg)(v_1)+(0.15\ kg)(48\ m/s)\\\\v_1 = -\frac{(0.15\ kg)(48\ m/s)}{97\ kg} \\v_1 = - 0.074 m/s[/tex]
negative sign here shows the opposite direction.
Now, we calculate the time taken by puck to move 14.5 m:
[tex]s_2 =v_2t\\\\t = \frac{s_2}{v_2} = \frac{14.5\ m}{48\ m/s} \\\\t = 0.3\ s[/tex]
Now, the distance covered by the player in this time will be:
[tex]s_1 = v_1t\\s_1 = (0.074\ m/s)(0.3\ s)[/tex]
s₁ = 0.022 m
What is the efficiency of a machine that uses 102 kJ of energy to do 98 kJ of work?
A 4-kW resistance heater in a water heater runs for 3 hours to raise the water temperature to the desired level. Determine the amount of electric energy used in both kWh and kJ.
Answer:
12kWhr
Explanation:
Energy = Power * Time
Power = 4kW
Time = 3hrs
Substitute into the formula
Energy used up = 4kW * 3hrs
Energy used up = 12kWhr
The graph below shows the distance traveled by the skateboarder on each of the different road conditions. Using the graph, determine which of the roads was dry, wet, or muddy. Explain your answer using complete sentences.
Answer:
Road A- dry
Road B- mud
Road C- wet
Explanation:
Surface conditions do affect the ease and speed with which a skateboarder can move, on a muddy surface, the tyres of the skate boards finds it difficult to establish adequate fictional force between the skates trees and the traveling surface. Hence, the muddy surface presents a very slippery travel ground for the skate, hence leading the to skateboarder needing to apply caution.
The speed on a wet surfave is height as the amount of firece that will be applied in other to accelerate is very small. The surface is wet and hence serves as a lubricant between the contact surface.
The dry road also has a high speed but lower than a wet surface, frictional force is high here and this tend to slow the skateboarder down except in sloppy terrains.
In an experiment to measure the temperature of a Bunsen burner flame, a 250 g piece of iron is held in the flame for several minutes until it reaches the same temperature as the flame . The hot metal is then quickly transferred to 285 g of water contained in a 40.0 g copper calorimeter at 15.0 oC. The final temperature of the copper and water is 80.0 oC.
Using your answer from determine the temperature of the Bunsen flame.
Answer:
wait
Explanation:
A pilot drops a package from a plane flying horizontally at a constant speed. Neglecting air resistance, when the package hits the ground the horizontal location of the plane will Group of answer choices be behind the package. be over the package. be in front of the package depend of the speed of the plane when the package was released.
Answer:
The location of helicopter is behind the packet.
Explanation:
As the packet also have same horizontal velocity as same as the helicopter, and also it has some vertical velocity as it hits the ground.
The horizontal velocity remains same as there is no force in the horizontal direction. The vertical velocity goes on increasing as acceleration due to gravity acts.
So, the helicopter is behind the packet.
An airplane, starting from rest, moves down the runway at constant acceleration for 23 s and then takes off at a speed of 66 m/s. What is the average acceleration of the plane (in m/s2)?
Answer:
46
Explanation:
A camera lens with focal length f = 50 mm and maximum aperture f>2
forms an image of an object 9.0 m away. (a) If the resolution is limited
by diffraction, what is the minimum distance between two points on the
object that are barely resolved? What is the corresponding distance
between image points? (b) How does the situation change if the lens is
“stopped down” to f>16? Use λ= 500 nm in both cases
Answer:
The minimum distance between two points on the object that are barely resolved is 0.26 mm
The corresponding distance between the image points = 0.0015 m
Explanation:
Given
focal length f = 50 mm and maximum aperture f>2
s = 9.0 m
aperture = 25 mm = 25 *10^-3 m
Sin a = 1.22 *wavelength /D
Substituting the given values, we get –
Sin a = 1.22 *600 *10^-9 m /25 *10^-3 m
Sin a = 2.93 * 10 ^-5 rad
Now
Y/9.0 m = 2.93 * 10 ^-5
Y = 2.64 *10^-4 m = 0.26 mm
Y’/50 *10^-3 = 2.93 * 10 ^-5
Y’ = 0.0015 m
An object that sinks in water has a mass in air of 0.0675 kg. Its apparent mass when submerged in water is 0.0424 kg. What is the specific gravity SG of the object? What material is the object probably made?
Answer:
1. SG
true
=2.689
2. The object is probably some sort of minerals and rocks such as Feldspar, Corals, Beryl, etc.
Explanation:
Given:
mass in the air= 0.0675 kg
mass in water= 0.0424 kg
The specific gravity of the object will be 2.6892. It is the ratio of the density of the given fluid and the standard fluid.
What is density?Density is specified as the mass divided by the volume. It is represented by the unit of measurement as kg/m³.
The mass of the object in air;
m=Vρ₀
m=0.0675 kg
Buoyant force on the object;
B= Vρₐg
For equilibrium;
N+B=m₀g
n=m₀g-Vρₓg
N/g=m₀-Vρₓ
N/g=0.0424 kg
[tex]\rm \frac{V\rho_0}{V\rho_x} =\frac{0.0675 }{m_0-0.0424 \ kg} \\\\ \frac{\rho_0}{\rho_x} =\frac{0.0675}{0.0675-0.0424} \\\\ \frac{\rho_0}{\rho_x} =2.6892[/tex]
Hence, the specific gravity of the object will be 2.6892.
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In addition to absorption of a photon, energy can be transferred to an atom by collision. Consider a hydrogen atom in its ground state. Incident on the atom are electrons having a kinetic energies of 10.5 eV. What is a possible result?
The question is incomplete, the complete question is;
In addition to absorption of a photon, energy can be transferred to an atom by collision. Consider a hydrogen atom in its ground state. Incident on the atom are electrons having a kinetic energies of 10.5 eV. What is a possible result?
A) The atom moves to a state of lower energy
B) The atom is ionized
C) One of the electrons leaves the atom
D) The atom can be excited to a higher energy state
Answer:
The atom can be excited to a higher energy state
Explanation:
According to the Bohr model of the atom, electrons in an atom can be excited from a lower to a higher energy level when energy is absorbed by the atom.
If electrons having an energy of 10.5ev are incident on a hydrogen atom, this energy is transferred to the atom by collision. Since the energy transferred is less than the ionization energy of hydrogen atom in its ground state(13.6ev), the atom is not ionized.
Rather, the atom is excited from ground state to a higher energy level.
A 2890-lb car is traveling with a speed of 58 mi/hr as it approaches point A. Beginning at A, it decelerates uniformly to a speed of 18 mi/hr as it passes point C of the horizontal and unbanked ramp. Determine the total horizontal force F exerted by the road on the car just after it passes point B.
Answer:
4592.57 lb
Explanation:
The missing diagram for this question is attached in the image below.
Given that:
the weight of the car = 2890 lb
At point A, the speed of the car [tex](V_A)[/tex] = 58 mi/hr
At point C, the speed of the car [tex](V_C)[/tex] = 18 mi/hr
To ft/s:
[tex](V_A)[/tex] = 58 mi/hr × 5280 ft/1 mi × 1 hr/3600 s
[tex](V_A)[/tex] = 85.07 ft/s
[tex](V_C)[/tex] = 18 mi/hr × 5280 ft/1 mi × 1 hr/3600 s
[tex](V_C)[/tex] = 26.4 ft/s
Between A to C, the total distance is;
[tex]S_{AC} = S_{AB}} + S_{BC} \\ \\ S_{AC} = 331 + \dfrac{\pi r}{2} \\ \\ S_{AC}= 331 + \dfrac{\pi \times 207}{2} \\ \\ S_{AC} = 656.154 \ ft[/tex]
Now, we need to determine the deceleration of the car using the formula:
[tex]V_C^2 = V_A^2 + 2 aS_{AC}[/tex]
[tex]26.4^2 = 85.07^2 + 2 a (654.154)[/tex]
[tex]696.96 = 7236.9049+ 2 a (654.154)[/tex]
[tex]696.96-7236.9049 = 2 a (654.154)[/tex]
[tex]-6539.9449 = 2 a (654.154)[/tex]
[tex]a= \dfrac{-6539.9449} {2(654.154)}[/tex]
a = -4.99 ft/s²
The velocity of the car as it passes via B
[tex]v_B^2 = v_A^2 + 2aS_{AB}[/tex]
[tex]v_B^2 = 85.07^2 + 2(-4.99 \times 331)[/tex]
[tex]v_B =\sqrt{ 85.07^2 + 2(-4.99 \times 331)}[/tex]
[tex]v_B =\sqrt{ 85.07^2 +3303.38}[/tex]
[tex]v_B =\sqrt{ 10540.2849}[/tex]
[tex]v_B =102.67 \ ft/s[/tex]
Along B, the car's acceleration is:
[tex]a_B = \sqrt{a^2 + (\dfrac{v_B^2}{r})^2}[/tex]
[tex]a_B = \sqrt{(-4.99)^2 + \dfrac{102.67^2}{207}^2 }[/tex]
[tex]a_B = 51.17 \ ft/s^2[/tex]
Finally, the total horizontal force F exerted = m[tex]a_B[/tex]
[tex]= (\dfrac{2890}{32.2}) \times 51.17[/tex]
= 4592.57 lb
A ball is thrown horizontally at a speed of 24 meters per second from the top of a cliff. If the ball hits the ground 6.0 seconds later, approximately how high is the cliff? ( EASY QUESTION.. PLZZ HELPPP MEEE I WILL MARK YOU THE BRAINLIEST PLZZ)
Answer:
144 meters
Explanation:
the ball is thrown with a speed of 24 meters per second right so if the ball reaches the ground in 6 seconds. the hight of the cliff must be S=v.t
S (height cliff)=24m/s×6s=144
Select the correct answer.
What are the directions of an object's velocity and acceleration vectors when the object moves in a circular path with a constant speed?
OA. The question is meanimgless, since the acceleration is zero.
ов.
The vectors point in opposite directions.
Oc.
Both vectors point in the same direction.
OD
The vectors are perpendicular,
Answer:
A
Explanation:
If the object is moving at a constant speed, the object isn't accelerating as the velocity doesn't change.
Answer: C.
Explanation: plato users
A car accelerates uniformly from rest to a speed of 55.0 mi/h in 13.0 s. (a) Find the distance the car travels during this time. m (b) Find the constant acceleration of the car. m/s2
Answer:
(a) 159.84 m
(b) 1.89 m/s²
Explanation:
Applying,
(a)
s = (v+u)t/2.................. Equation 1
Where s = distance traveled by the car, u = initial velocity, v = final velocity, t = time.
From the question,
Given: u = 0 m/s ( from rest), v = 55 mi/h = (55/2.237) m/s = 24.59 m/s, t = 13 s
Substitute these values into equation 1
s = (24.59+0)13/2
s = 159.84 m
(b)
Also applying,
a = (v+u)/t................. Equation 2
Where a = acceleration of the car.
substituting into equation 2,
a = (24.59+0)/13
a = 1.89 m/s²
plz answer the question
Answer:
Ray A - incident ray
Ray B - reflected ray
On topographic maps, contour lines that are farther apart indicate what ?
Answer:
if I am correct, they indicate less steep terrain. think of it as the steeper the terrain the closer together the lines would be. hope that makes sense for you guys.
Answer:
gentle slopes
Explanation:
12. What type of circuit is the diagram below?
series circuit
parallel circuit
Answer:
parallel circuit
Explanation:
An electric circuit can be defined as an interconnection of electrical components which creates a path for the flow of electric charge (electrons) due to a driving voltage.
Generally, an electric circuit consists of electrical components such as resistors, capacitors, battery, transistors, switches, inductors, etc.
Basically, the components of an electric circuit can be connected or arranged in two forms and these includes;
I. Series circuit
II. Parallel circuit: it's an electrical circuit that has the same potential difference (voltage) across its terminals or ends. Thus, its components are connected within the same common points so that only a portion of current flows through each branch.
Hence, the type of circuit that the above diagram above represents is a parallel circuit.
Answer:
parallel circuit
Explanation:
I got it right on my exam
Which describes farsightedness? O Distant objects are blurry. O Concave lenses can correct it. O Objects appear larger when wearing corrective glasses. O Corrective glasses do not change apparent the size of objects.
Answer:
O Distant objects are blurry. describes farsightedness.
Explanation:
Farsightedness (hyperopia) is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry. The degree of your farsightedness influences your focusing ability.Farsightedness (hyperopia) is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry.
Ethyl alcohol is :
a. None of the above
b. Semi polar solvent
c. Polar solvent
d. Non-Polar solvant
Answer:
D. Non- polar solvant
Explanation:
l think that's it
Answer:
I think the answer is D polar solvent
