Answer:
The correct option is D.
Explanation:
The wavelength of the photon can be calculated with the following equation:
[tex] E = h\frac{c}{\lambda} [/tex]
Where:
E: is the energy of the photon = 2.90 eV
h: is the Planck's constant = 6.62x10⁻³⁴ J.s
c: is the speed of light = 3x10⁸ m/s
λ: is the wavelength
Hence, the photon's wavelength is:
[tex] \lambda = \frac{hc}{E} = \frac{6.62 \cdot 10^{-34} J*s*3.0 \cdot 10^{8} m/s}{2. 90 eV*\frac{1.6 \cdot 10^{-19} J}{1 eV}} = 428 nm [/tex]
Therefore, the correct option is D.
I hope it helps you!
Trong máy phát điện xoay chiều ba pha khi tổng điện áp tức thời của cuộn 1,2 là e1+e2=120V thì điện áp tức thời của cuộn 3 là
Answer:
I just noticd i dont speak this launguage
Explanation:
Radiation exerts pressure on surfaces on which it lalls (radintion pressure). Will this pressure be greater on a shiny surface or a dark surface
Answer:
Shiny surface.
Explanation:
We know that radiation pressure is the pressure over a surface exposed to electromagnetic radiation.
Where if the radiation is absorbed by the material (like in the case of a dark surface), the pressure is the energy density flux divided by the speed of light, while if the radiation is totally reflected (idealized case, but we can suppose that this happens for a shiny surface) the pressure is twice pressure for the absorbed case.
This is a simplification for the radiation pressure but is enough to conclude that the radiation pressure is always greater on reflective surfaces, then for this case, the pressure will be greater on a shiny surface than in a dark surface,
Which of the following accurately describes circuits
Answer:
In a series circuits, the current can flow through only one path from start to finish - accurately describes circuits.
Answer:
It is a series circuit, the current can flow through only one path from start to finish - accurately describes circuits.
Explanation:
I don't know the option but it is also the correct one.
You send a traveling wave along a particular string by oscillating one end. If you increase the frequency of oscillations, does the speed of the wave increase, decrease, or remain the same?
Answer:
The speed of the wave remains the same
Explanation:
Since the speed of the wave v = √(T/μ) where T is the tension in the string and μ is the linear density of the string.
We observed that the speed, v is independent of the frequency of the wave in the string. So, increasing the frequency of the wave has no effect on the speed of the wave in the string, since the speed of the wave in the string is only dependent on the properties of the string.
So, If you increase the frequency of oscillations, the speed of the wave remains the same.
2. The given graph shows that the object is
(a) in non-uniform motion
(b) in uniform motion
(c) at rest
(d) in an oscillatory motion.
distance
time
Answer:
(c) at rest
Explanation:
Given
See attachment for the distance time graph
Required
What does the graph illustrate?
From the graph, we can see that the line of distance is a horizontal line.
This suggests that a time increases, the distance remains unchanged
When distance remains unchanged over time, then it means the object is at rest.
Hence, (c) is correct
PLEASE HELP MEE THIS IS DUE IN 45 MINS
Answer:
The distance travelled does not depend on the mass of the vehicle. Therefore, [tex]s = d[/tex]
Explanation:
This deceleration situation can be analyzed by means of Work-Energy Theorem, where change in translational kinetic energy is equal to the work done by friction:
[tex]\frac{1}{2}\cdot m\cdot v^{2}-\mu\cdot m\cdot g \cdot s = 0[/tex] (1)
Where:
[tex]m[/tex] - Mass of the car, in kilogram.
[tex]v[/tex] - Initial velocity, in meters per second.
[tex]\mu[/tex] - Coefficient of friction, no unit.
[tex]s[/tex] - Travelled distance, in meters.
Then we derive an expression for the distance travelled by the vehicle:
[tex]\frac{1}{2}\cdot v^{2} = \mu \cdot g \cdot s[/tex]
[tex]s = \frac{v^{2}}{\mu\cdot g}[/tex]
As we notice, the distance travelled does not depend on the mass of the vehicle. Therefore, [tex]s = d[/tex]
Define Standard Unit
answer:
standard units are the units we usually use to measure the weight, length or capacity of objects.
explanation:
international system of units or SI units are standard units these units used for measuring quantities are accepted worldwide for example: kilogram is standard unit for mass, second is standard unit for measuring timeJacie made a model to show the water cycle. The model she made is shown
below.
Which process in the model represents condensation?
A. As water vapor transfers heat to ice cubes, it forms clear droplets outside the
plastic wrap.
B. As water vapor gains heat from ice cubes, it forms clear droplets outside the
plastic wrap.
C. As water vapor transfers heat to ice cubes, it forms colored droplets inside the
plastic wrap.
D. As water vapor gains heat from ice cubes, it forms colored droplets inside the
plastic wrap
Answer:
option C
Explanation:
as water vapor transfer heat, colored drops are seen inside the wrap.
A body is dropped from a height H. In how much time will it reach the ground?
Answer:
[tex]s = ut + \frac{1}{2} g {t}^{2} \\ t(u + \frac{1}{2} gt) = H \\ u + \frac{1}{2} gt = H \\ t = 2(H - u) \div g \\ t = \frac{(H - u)}{5} \\ u \: is \: speed \: or \: velocity[/tex]
With neat circuit diagrams where possible show any two (2) ways of direct current motor
excitations,
The tires of a car make 60 revolutions as the car reduces its speed uniformly from 95.0 km/h to 60.0 km/h. The tires have a diameter of 0.88 m. If the car continues to decelerate at this rate, how far does it go
Answer:
-2.869 rad/s2
Explanation:
Data given:
speed, vi at 95.0 km/h = 95 X (1 hour /3600 seconds) X (1000m / 1km)
Note that, for every 1 hour, there will be 60sec X 60sec = 3600 seconds
And for every 1km, there will be 1000m.
So, speed of 95.0 km/h = 26.389 m/s
speed, vi = r ω (radius X angular velocity)
angular velocity, ωi = v/r
ωi = 26.389 m/s ÷ half of 0.88 m diameter
= 59.975 rad/s
decelerating to speed, vf at 60.0 km/h = 60 X X (1 hour /3600 seconds) X (1000m / 1km)
= 16.667m/s
The angular velocity for this speed = 16.667m/s ÷ half of 0.88 m diameter
= 37.879rad/s
How far the car goes is equivalent to the angular acceleration which equals to (ωf^2 - ωi^2) ÷ 2θ
= (37.879rad/s)^2 - (59.975 rad/s)^2 ÷ 2 (60 rev X 2π rad/rev)
= -2.869 rad/s2
The launching catapult of the aircraft carrier gives the jet fighter a constant acceleration of 59 m/s2 from rest relative to the flight deck and launches the aircraft in a distance of 97 m measured along the angled takeoff ramp. If the carrier is moving at a steady 26 knots (1 knot = 1.852 km/h), determine the magnitude v of the actual velocity of the fighter when it is launched.
Answer:
Explanation:
We shall find the final velocity of aircraft with respect to aircraft carrier using the following relation.
v² = u² + 2as
v² = 0² + 2 x 59 x 97
v² = 11446
v = 107 m /s
velocity of aircraft carrier = 1.852 x 26 = 48.152 km/h
= 48.152 x 1000 / (60 x 60) m/s
= 13.37 m /s
This velocity of aircraft carrier will be added to the velocity of aircraft .
So absolute velocity of aircraft = 107 m /s + 13.37 m/s
= 120.37 m/s
what are the symptoms of hepatitis 'b'
A 56 kg pole vaulter falls from rest from a height of 5.1 m onto a foam rubber pad. The pole vaulter comes to rest 0.29 s after landing on the pad.
Required:
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
a. The athlete's velocity just before reaching the pad is [tex]35.21m/s[/tex]
b. The constant force exerted on the pole vaulter is 6799.52 N
a. We use Newton's equation of motion,
[tex]v=u+at\\\\S=ut+\frac{1}{2}at^{2}[/tex]
Where u is initial velocity, v is final velocity, a is acceleration , t is time and S represent distance.
Given that, s = 5.1 m , t = 0.29s, u = 0
Substitute in above equation.
[tex]5.1=\frac{1}{2}*a*(0.29)^{2} \\\\a=\frac{5.1*2}{0.084}=121.42m/s^{2}[/tex]
the athlete's velocity, [tex]v=0+121.42*(0.29)=35.21m/s[/tex]
b. The constant force exerted on the pole vaulter due to the collision is given as, [tex]Force=mass*acceleration[/tex]
[tex]Force=56*121.42=6799.52N[/tex]
Learn more:
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What is Velocity
Answer:
noun
the speed of something in a given direction.
"the velocities of the emitted particles"
(in general use) speed.
"the tank shot backwards at an incredible velocity"
Similar:
speed
pace
rate
tempo
momentum
impetus
swiftness
swift/fast pace
fastness
quickness
speediness
rapidity
briskness
expeditiousness
expedition
dispatch
acceleration
clip
fair old rate
fair lick
steam
nippiness
fleetness
celerity
ECONOMICS
the rate at which money changes hands within an economy.
noun: velocity of circulation; plural noun: velocities of circulation
Answer:
i hope this helps you
Explanation:
hii
the velocity of a ship in the unit of m/s moving initially along west is given by V(t) = 5-2t. It’s acceleration at t=1 s is given by:
1. 0 m/s^2
2. 2m/s^2 along west
3. 2m/s^2 along east
4. None
Whis one is correct?
Answer:
4. None
Explanation:
Applying,
a(t) = dV(t)/dt
Where a(t) = Acceleration of the ship at a given time.
From the question,
Given: V(t) = 5-2t
Therefore,
a(t) = dV(t)/dt = 5 m/s²
Hence it's acceleration is 5 m/s²
From the question,
The right option is 4. None
How long does take for a freely falling object to reach 4.0 m/s
Answer :
Considering initial velocity is 0,
It takes about 0.4 seconds.
Use the equation v-u =at
4-0 = 9.8×t
t = 4/9.8 = 0.4 seconds approximately.
A ball is thrown vertically upward at 24.0 ms can reach a height of 28.8m ( neglecting air resistance).The speed,in m/s,when it is halfway to its highest point is (using g= 10 ms ^2)
Answer:
The answer is "[tex]16.79\ \frac{m}{s}[/tex]"
Explanation:
In this question, the halfway indicates the height that is [tex]\frac{28.8}{2}=14.4 \ m[/tex]
Using formula:
[tex]v^2=u^2+2as\\\\v^2=24^2+2(-10)(14.4)\\\\[/tex]
[tex]v^2=576-288\\\\v^2=288\\\\v=\sqrt{288}\\\\v=16.97 \ \frac{m}{s}[/tex]
A circle has a radius of 20 cm and a central angle that measures 216° find the length of the arc defined by the central angle
Answer:
4320cm
Explanation:
(length of the arc)/(length of the radius)=central angle
lengthof the arc/20=216
lengthof the arc=216×20
length of the arc=4320cm
hope this helps, let me know if I'm right!
The length measured along the arc is called length of the arc. The length of the arc defined by the central angle is 4320cm.
What is central angle?Central angle is the angle subtended by the arc at the center of circle.
central angle = length of the arc / length of the radius
216 = length of the arc/20
length of the arc = 216×20
length of the arc = 4320cm
Thus, the length of the arc defined by the central angle is 4320cm.
Learn more about central angle.
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make a list of five agriculture based occupation
Answer:
Agricultural Engineer.
Agricultural Food Scientist.
Agricultural Inspector.
Agricultural Manager.
Agricultural Specialist.
They are;
Agricultural EngineerAgricultural specialistFarm ManagerAgricultural economistCommercial mixed farmingThank you hope it helps you. May god bless you❤What are 3 artificial and 2 natural sources of electromagnetic radiation?
Answer: its b bro
Explanation:
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sometimes balance point may not be obtained on the potentiometer wire why
A teacher performs a demonstration to show the properties of an unknown substance. The teacher cuts off a piece of gray shiny substance and drops it in water. The substance floats and reacts violently with the water. Based on this Information, what type of element the unknown substance?
A. metalloid
B. metal
C. nonmental
D. halogen
Answer:
nonmental
Explanation:
Option B is correct. The substance floats and reacts violently with the water. Based on this Information, metal is the unknown substance.
What is metal?A metal, a glossy substance that transmits electricity and heat reasonably efficiently when newly manufactured, polished, or broken.
Metals are either malleable or ductile. Metals can be chemical elements like iron.
Metal is substance floats and reacts violently with the water. When particles collide, energy is transferred. A change in temperature can be used to identify this.
The substance floats and reacts violently with the water. Based on this Information, metal is the unknown substance.
Hence, option B is correct.
To learn more about metal, refer to the link;
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In the Biomedical and Physical Sciences building at MSU there are 135 steps from the ground floor to the sixth floor. Each step is 16.6 cm tall. It takes 5 minutes and 30 seconds for a person with a mass of 73.5 kg to walk all the way up. How much work did the person do?
Answer:
W = 16.4 kJ
Explanation:
Given that,
There are 135 steps from the ground floor to the sixth floor.
Each step is 16.6 cm tall.
The mass of a person, m = 73.5 kg
We need to find the work done by the person. We know that,
Work done = Fd
Where
d is the displacement, d = 135 × 0.166 = 22.41
So,
W = 73.5 × 10 × 22.41
= 16471.35 J
or
W = 16.4 kJ
So, 16.4 kJ is the work done by the person.
You throw a glob of putty straight up toward the ceiling, which is 3.50 m above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 9.10 m/s.
1. What is the speed of the putty just before it strikes the ceiling? Express your answer with the appropriate units.
2. How much time from when it leaves your hand does it take the putty to reach the ceiling? Express your answer with the appropriate units.
Answer:
Explanation:
Given that:
the putty initial speed (u) = 9.10 m/s
distance (s) between hand and the ceiling = 3.50 m
the speed of the putty prior to the time it hits the ceiling can be determined by considering the second equation of motion.
v² - u² = 2as
Since the putty is moving in a vertical motion(i.e. in an upward direction)
v² - u² = -2gs
v² = u² - 2gs
[tex]v = \sqrt{u^2 - 2gs}[/tex]
[tex]v = \sqrt{(9.10)^2 -( 2* 9.8) (3.50 -0)}[/tex]
[tex]v = \sqrt{82.81 -19.6 (3.50)}[/tex]
[tex]v = \sqrt{82.81 -68.6}[/tex]
[tex]v = \sqrt{14.21}[/tex]
v = 3.77 m/s
2.
The time it takes to reach the ceiling from the moment it leaves your hand can be calculated by using the first equation of motion:
v = u + at
In an upward direction
v = u - gt
making time t the subject;
v - u = -gt
[tex]t = \dfrac{v-u}{-g}[/tex]
[tex]t = \dfrac{3.77 - 9.10}{-9.8}[/tex]
[tex]t = \dfrac{-5.33}{-9.8}[/tex]
t = 0.54s
A body of mass 2.0 kg makes an elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its original speed.
(a) What is the mass of the other body?
(b) What is the speed of the two-body center of mass if the initial speed of the 2.0 kg body was 4.0 m/s?
Answer:
(a) the mass of the second body is 1.2 kg
(b) the speed of the two-body center of mass 2.5 m/s
Explanation:
Given;
mass of the body, m₁ = 2 kg
let the mass of the second body = m₂
let the initial speed of the first body, = u₁ = 4 m/s
then, the final speed of the first body, v₁= ¹/₄u₁ = 0.25u₁
initial speed of the second body, u₂ = 0
let the final speed of the second body = v₂
Apply principle of conservation of linear momentum to determine the mass of the second body;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
2u₁ + 0(m₂) = 2(0.25u₁) + m₂v₂
2u₁ = 0.5u₁ + m₂v₂
1.5(4) = m₂v₂
6 = m₂v₂
Apply one-directional velocity
u₁ + v₁ = u₂ + v₂
u₁ + (0.25u₁) = 0 + v₂
1.25u₁ = v₂
1.25(4) = v₂
5 = v₂
Then, the mass of the second body is calculated as;
m₂v₂ = 6
5m₂ = 6
m₂ = 6/5
m₂ = 1.2 kg
(b) the speed of the two-body center of mass after collision;
[tex]V_c_m = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} \\\\V_c_m = \frac{2(0.25\times 4) \ + \ 1.2(5)}{2\ + \ 1.2} \\\\V_c_m = 2. 5 \ m/s[/tex]
Does the same battery always deliver the same amount of flow to any circuit? Mention two observations of any circuits in this lab that support your answer. Explain.
Answer:
Yes
Explanation:
Given that the battery is the same the PD ( potential difference ) in the circuit will also be the same likewise the flow of charge in the circuit,
Hence the same amount of charge flow is delivered to any circuit.
attached below are examples
A pressure sensor was used to measure the unsteady pressure in a cylinder. The sensor output was acquired for 15 seconds at a rate of 202 Hz and spectral analysis was performed using FFT. What is the maximum frequency that will be displayed on the power spectrum plot
Answer:
Maximum frequency on power spectrum plot = 101 Hz
Explanation:
Given:
Time taken for output = 15 seconds
Frequency rate = 202 Hz
Find:
Maximum frequency on power spectrum plot
Computation:
Maximum frequency = Given frequency rate / 2
Maximum frequency on power spectrum plot = Frequency rate / 2
Maximum frequency on power spectrum plot = 202 / 2
Maximum frequency on power spectrum plot = 101 Hz
79. If the takeoff velocity of an airplane on a runway is 300 km/hr with an acceleration of 2 m/s. What is the takeoff time
of the airplane.
Answer:
The take off time is equal to 41.66s.
Explanation:
Given that,
The initial speed of the airplane, u = 0
Acceleration, a = 2 m/s²
Let v is the take off time. Using first equation of motion,
v = u +at
Put v = 300 km/hr = 83.3 m/s
So,
[tex]t=\dfrac{v}{a}\\\\t=\dfrac{83.33}{2}\\\\t=41.66\ s[/tex]
So, the take off time is equal to 41.66s.
what is the suitable way of using social media
Answer:
not using it too much and getting addicted
Explanation: