Electronic transitions that end up in the state with quantum number n = 2 are what the Balmer series for the hydrogen atom refers to. The longest wavelength is 656.3x10-9 m, and the energy is 3.03X10^33J
Given quantum number (n) = 2
This will be transition 3-2 photon energy (E)= hc/λ since we must take into account the photon with the longest wavelength that completes the electron transition in the state with the quantum number n=2
photon energy (E)= hc/λ
(a) Then the energy (El) of longest wavelength= 6.63×10 −34 x 3x10^8/656.3x10-9
E = 3.03x10^33J = 1.89eV
(b) We know that 1/λl = R(1/n1^2 - 1/n2^2) where R = Rydberg constant
1/λl = 1.097×107m−1× (1/2^2 - 1/3^2)= 656.3x10-9m
(c) From given diagram we say thee shortest transition is from level 6 to level 2
n1 = 2 and n2 = 6
Its photon energy (Es) = 6.63×10 −34 x 3x10^8/411x10-9 = 3.023eV
(d) Wavelength of shortest transition series 1/λ = 1.097×107x(1/2^2 -1/6^2) = 411nm
(e)Shortest wavelength is emitted in Balmer series if the transition of electron takes place from n2=∞ to n1=2.
Shortest wavelength in Balmer series 1/λ = R(1/2^2 -1/∞)=364.6nm
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Energy never disappears, but it does change forms! Write a brief paragraph to describe three examples of energy changing forms around you. Think about electricity,
Some of the forms which energy can have, are:
- Thermal energy
- Kinetic energy
- Potential energy
- Electric potential energy
- Radiation
Some examples of transformations between different forms of energy, are:
- Potential energy to kinetic energy during the course of a roller coaster.
- Kinetic energy to thermal energy when an object decelerates due to friction while sliding on a surface.
-Kinetic energy to thermal energy when a meteor melts upon entering the atmosphere due to friction.
- Kinetic energy to electric potential energy in a generator.
- Electric potential energy to radiation in a digital screen.
- Electric potential energy to thermal energy in a light bulb.
- Thermal to kinetic energy in a heat engine.
Use these examples to identify energy changing forms around you and write a brief paragraph.
Gerry is looking at salt under a powerful microscope and notices a crystalline structure. What can be known about the salt sample that Gerry is looking at?
1. The atoms have spread out from each other.
2.The atoms are sliding past each other.
3.The atoms have no particular pattern
4.The atoms are vibrating in place
Answer:
4.The atoms are vibrating in place
Explanation:
The answer is; The atoms are vibrating in place
We know that, molecules in the crystal have a definite position in the crystal and are bonded to each other by electrostatic forces. However, since the molecules have some energy, they vibrate in their positions. Their energy, however, is not high enough to cause them to overcome the strong bonding (unless the crystal is heated or the atoms are irradiated).
A coin is placed 17.0 cm from the axis of a rotating turntable of variable speed. When the speed of the turntable is slowly increased, the coin remains fixed on the turntable until a rate of 26.0 rpm (revolutions per minute) is reached, at which point the coin slides off. What is the coefficient of static friction between the coin and the turntable?
Answer: The coefficient of static friction between the coin and the turntable is 0.13.
Explanation:
As we know that,
Centripetal force = static frictional force
\(\frac{mv^{2}}{r} = F_{s}\)
or, \(\frac{mv^{2}}{r} = \mu_{s} \times m \times g\)
v = \(\sqrt{\mu_{s} \times r \times g}\)
or, \(\mu_{s} = \frac{v^{2}}{rg}\) ......... (1)
Here, it is given that
r = 17 cm, \(\omega\) = 26 rpm,
and v = \(r \omega\) ..........(2)
Putting equation (2) in equation (1) we get the following.
\(\mu_{s} = \frac{r^{2}\omega^{2}}{rg}\)
= \(\frac{17 \times 10^{-2} \times (26 \times [\frac{2 \times \pi}{60}]^{2})}{9.8}\)
= 0.128
= 0.13 (approx)
Thus, we can conclude that the coefficient of static friction between the coin and the turntable is 0.13.
A girl weighing 45kg is standing on the floor, exerting a downward force of 200N on the floor. The force exerted on her by the floor is ..............
Select one:
a.
No force exerted
b.
Less than 2000N
c.
Equal to 200 N
d.
Greater than 200 N
Answer:
c.
Equal to 200 N..........
How a Motor Works
Figure 5
A motor is made of several basic parts, each of which
is described below. Study the information about each part. Then,
write the number of each description in the appropriate circle on
the image.
Answer:
hi \( \) \( \)
In the following circuit (Fig.3), calculate the intensity I through the resistance 3 using the principle of superposition.
Answer:
time
Explanation:
Were is the computer located
Answer:
where u put it last time or retrace ur steps to where u last put it
A typical machine tests the tensile strength of a sheet of material cut into a standard size of 5.00 centimeters wide by 10.0 centimeters long. The machine consists of one clamp that holds the entire width (5.00 centimeters) so that it hangs vertically. A second clamp is placed on the lower end of the object, to which a variable downward force is applied. The force is slowly increased until the object ruptures, and the breaking force is recorded.
A strip of aluminum foil with a thickness of 15.0 micrometers and matching the size recommendations of the machine is placed in the machine and tested. The force needed to rupture the foil is found to be 233 newtons. What is the tensile strength of the aluminum foil sample?
Answer:
Explanation:
tensile strength is stress that is needed to break the wire made of the material .
Here force required to break the sheet of material = 233 N
cross sectional area of the foil = breadth x thickness
= 5 x 10⁻² x 15 x 10⁻⁶ m²
= 75 x 10⁻⁸ m²
breaking stress = force / cross sectional area
= 233 / 75 x 10⁻⁸
= 3.1 x 10⁸ Pa .
Tensile strength = 3.1 x 10⁸ Pa .
A box of mass 210 kg is pulled from rest with a string of tension 1300n inclined at 35° to the horizontal. if the box moved with a speed of 10m/s and frictional force between the box and surface is 100 n, calculate the distance covered.
If A box of mass 210 kg is pulled from rest with a string of tension 1300n inclined at 35° to the horizontal. if the box moved with a speed of 10m/s and the frictional force between the box and surface is 100 n, Then the distance covered by the box is 10.89 meters.
To calculate the distance covered by the box, we need to analyze the forces acting on it and apply the work-energy principle.
Given:
Mass of the box, m = 210 kg
Tension in the string, T = 1300 N
The angle of inclination, θ = 35°
Frictional force, f = 100 N
Initial speed, u = 0 m/s
Final speed, v = 10 m/s
First, let's resolve the tension force into components parallel and perpendicular to the incline. The parallel component of the tension force can be calculated as:
T_parallel = T * cos(θ)
Next, let's calculate the net force acting on the box along the incline. The net force is given by:
Net force = T_parallel - f
Now, using Newton's second law, we can calculate the acceleration (a) of the box:
Net force = m * a
From the given information, we have the final velocity (v), initial velocity (u), and acceleration (a). We can use the following kinematic equation to calculate the distance covered (s):
v^2 = u^2 + 2as
Rearranging the equation, we get:
s = (v^2 - u^2) / (2a)
Now, let's plug in the given values and calculate the distance covered:
T_parallel = 1300 N * cos(35°) ≈ 1067.35 N
Net force = 1067.35 N - 100 N = 967.35 N
a = (967.35 N) / (210 kg) ≈ 4.61 m/s^2
s = (10 m/s)^2 - (0 m/s)^2 / (2 * 4.61 m/s^2) ≈ 10.89 m
Therefore, the distance covered by the box is approximately 10.89 meters.
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Can anyone explain how to tackle this problem please:
Figure 15.5 shows a 50 kg lead cylindrical piston which floats on 0.37 mol of compressed ideal air at 30°C. How far does the piston move if the temperature is increased to 300°C?
A. 65 cm
B. 140 cm
C. 73 cm
D. 730 cm
Answer:
P V = N R T ideal gas equation
V2 / V1 = T2 / T1 since P, N, R are constant
V2 = 573 / 303 * V1 = 1.89 V1
V = π R^2 h volume of cylinder
V1 = 3.14 * .05^2 * h1 = .00785 h1
V2 = .0148 h1
A (h1 + h) = A * h2
h = h2 - h1 = (.0148 - .00785) h1 = .00695 h1 distance moved by piston
Use V1 = N R T1 / P1 = A h1 to calculate h1
h1 = N R T1 / (A * P1) A * F/A = F to simplify denominater
h1 = .37 * 8.31 * 303 / (50 * 9.8) = 1.90 m P = F / A
Δh = .00695 h1 = .0132 m = 1.32 cm
Math should be checked!
why is it more painful to walk on gravel with your shoes off then on (3 marks please)
Answer:
Because shoes protect our feet from some of the most harmful platforms
Gravel has some small pebbles on it sometimes (or other sharp objects)
Gravel is pretty hard.
is a pair of scissorrs a lever and wedge?
Answer:
Wedge
Explanation:
Scissors are a pair of wedges that are combined together.
a student throws a coin vertically downward frok the top of a building. the coin leaves the throwers hand with a speed of 15.0m/s. what is its speed after falling freely for 2.00s?
Answer:
Final speed after 2 seconds = 34.6 m/s
Explanation:
Given:
Initial speed of coin (u) = 15 m/s
Time taken = 2 seconds
Find:
Final speed after 2 seconds
Computation:
Gravitational acceleration of earth = 9.8 m/s²
Using first equation of motion;
v = u + at
or
v = u + gt
where,
v = final velocity
u = initial velocity
g = Gravitational acceleration
t = time taken
v = 15 + 9.8(2)
v = 15 + 19.6
Final speed after 2 seconds = 34.6 m/s
what is achmedis principle
Answer:
Archimedes principle states that when a body is partially/wholly/completely immersed in fluid (liquid or gas),it experiences an upthrust which is equal to the weight of the fluid displaced....
the student applies a net force of 4.5 N to the 1.5 kg textbook.
Calculate the magnitude of the textbook's acceleration.
Answer:
3 m/s²
Explanation:
acceleration = Force / mass; 4.5/1.5 = 3
When a student applies a net force of 4.5 Newtons to the 1.5-kilogram textbook, then the magnitude of the acceleration would be 3 meters/second².
What is acceleration?The rate of change of the velocity with respect to time is known as the acceleration of the object. Generally, the unit of acceleration is considered as meter/seconds².
As given in the problem, when a student applies a net force of 4.5 Newtons to the 1.5-kilogram textbook, then we have to find the magnitude of the acceleration,
the applied force on the textbook = 4.5 Newtons
the mass of the textbook is 1.5 kilograms
acceleration of the textbook = force / mass
=4.5 Newtons /1.5 kilograms
=3 meters/second²
Thus, the magnitude of the acceleration would be 3 meters/second².
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The light ray is traveling from Acrylite into air. The refractive index for air is 1.00. If the angle of incidence and the angle of refraction are known, how could you determine the refractive index of Acrylite?
Answer:
Refraction. A light ray traveling through some plastic has a frequency of 5.5 x 1014 Hz. It is ... making an angle of 32° with the normal to the interface; it passes through the ... [5 points] (a) What is the refractive index of the glass? ... What you do know is the wavelength of the light in glass, 321 nm, and you know the frequency.
Explanation:
The refractive index of Acrylate is determined as: ratio of sine of incident angle to sine of refractive angle.
What is Snell's law?The relationship between the angles of incidence and refraction for light or other waves flowing through a border between two different isotropic media, such as water, glass, or air, is described by Snell's law, a formula.
The amount a light ray bends as it travels through different media is measured by the refractive index.
Given that: The refractive index for air is 1.00.
Let the angle of incidence from medium air to Acrylate is = i
the angle of refraction from medium air to Acrylate is = r.
Let the refractive index for Acrylate = n.
Then according to Snell's law:
1×sini = n× sinr
n = sini/sinr.
Hence, the refractive index of Acrylate is ratio of sine of incident angle to sine of refractive angle.
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PLEASE HELPPP ASAP, I'LL GIVE BRAINLEST
Answer:
a) Joules
b) W (symbol for work) (J is symbol for Joules)
c) seconds
d) watts
A commuter backs her car out of her garage with a constant acceleration of 1.20 m/s2. (a) How long in seconds does it take her to reach a speed of 1.90 m/s
The time in seconds it will take her to reach a speed of 1.9 m/s is 1.58s.
What is time?:Time can be defined as the measurable period or duration, during which an action, process, or condition exists or continue.
The time taken for the commuter to reach a speed of 1.9 m/s, can be calculated using the formula below.
Formula:
a = (v-u)/t................. Equation 1Where:
a = acceleration of the commuterv = final velocity of the commuteru = initial velocity of the commutert = time taken for the commuter to reach the speed.make t the subject of equation 1
t = (v-u)/a................. Equation 2From the question,
Given:
v = 1.90 m/sa = 1.2 m/s²u = 0 m/sSubstitute these values into equation 1.
t = (1.9-0)/1.2t = 1.58 s.Hence, The time in seconds it will take her to reach a speed of 1.9 m/s is 1.58s
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forces applied to the collision
Answer: I think this is the answer, In a collision, there is a force on both objects that causes an acceleration of both objects; the forces are equal in magnitude and opposite in direction. For collisions between equal-mass objects, each object experiences the same acceleration.
Explanation: I had a question similar to this, Hope this helps!
In the diagram below, the batteries are 1.5 V and each lamp has a
resistance of 1.1 ohm. When the switch is closed, what is the current
through each lamp and what is the power dissipated in each lamp?
The current through each lamp is 0.545 A and the power dissipated in each lamp is 0.323 W.
In the diagram the batteries are 1.5V each and each lamp has a resistance of 1.1Ω when the switch is closed.
When S₁ is closed. The current is passing through lamps. The lamps are connected in the series connection the current passing through each is same.
Applying KVL through lamps
1.1 I + 1.1 I + 1.1 I + 1.1 I + 1.1 I - 1.5 - 1.5 = 0
5( 1.1 I) = 3
I = 3/5.5 = 0.545 A
Current through each lamp is 0.545 A
The resistor value of each lamp is same and current passing through each lamp is also same.
Therefore, power dissipation through each lamp is also same.
P = I²R = (0.545)²× 1.1 = 0.323 W
Hence, the current through each lamp is 0.545 A and the power dissipated in each lamp is 0.323 W.
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A venturi meter used to measure flow speed in the pipe. Derive an expression for the flow speed "H1" interns of the crossectional areas "A1" and "A2" and the difference in height "h" of the liquid levels in the two vertical tubes ?
Answer:
v₁ = √[ 2gh / ((A₁ / A₂)² − 1) ]
Explanation:
Use Bernoulli's equation:
P₁ + ½ ρ v₁² + ρgz₁ = P₂ + ½ ρ v₂² + ρgz₂
Since there's no elevation change between points 1 and 2, z₁ = z₂.
P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²
Assuming incompressible fluid, the volumetric flow rate is the same at points 1 and 2.
Q₁ = Q₂
v₁ A₁ = v₂ A₂
v₂ = v₁ A₁ / A₂
Substituting:
P₁ + ½ ρ v₁² = P₂ + ½ ρ (v₁ A₁ / A₂)²
P₁ + ½ ρ v₁² = P₂ + ½ ρ v₁² (A₁ / A₂)²
P₁ − P₂ = ½ ρ v₁² (A₁ / A₂)² − ½ ρ v₁²
P₁ − P₂ = ½ ρ v₁² ((A₁ / A₂)² − 1)
v₁² = 2 (P₁ − P₂) / (ρ ((A₁ / A₂)² − 1))
v₁² = 2 (ρgh) / (ρ ((A₁ / A₂)² − 1))
v₁² = 2gh / ((A₁ / A₂)² − 1)
v₁ = √[ 2gh / ((A₁ / A₂)² − 1) ]
Converting from the US customary system to the metric system, 16 fluid ounces is approximately equal to a) 1 gallon. O b) 160 milliliters. c) 480 milliliters. O d) 960 milliliters.
Answer:
c) 480 mL
Explanation:
16 fluid ounces is 473.176473 mL exactly. The closest answer choice is ...
480 mL
_____
1 gallon = 128 fl oz = 3.785411784 liters exactly
How does friction impact work? (that's the whole question)
What are some possible injuries that can occur as a result of anabolic steroid use?
How does gravitational force relate to distance?
Answer:
Gravitational force related to distance because when an object is closer to the center of the earth the stronger the gravitational force is the farther an objects it from the center of the earth the weaker the gravitational force is. For example if you are on the moon there is no gravity because the gravitational force of the moon to the center of the earth is father away then for example standing on earth.
90 V
R₁
60
R2
30
R3
30
Based on the circuit above, what would be the current through the R3 resistor?
In the parallel combination, the current through R3 is 3 A. In the series combination, the current through R3 is 0.75 A.
To determine the current through resistor R3 in both the parallel and series combinations, we need to apply Ohm's Law and the appropriate formulas for calculating total resistance and current in each configuration.
First, let's consider the parallel combination:
In a parallel combination, the voltage across each resistor is the same. Therefore, the voltage across R3 is also 90 V.
Using Ohm's Law (V = I × R), we can calculate the current flowing through R3 in the parallel combination:
I_parallel = V / R3
= 90 V / 30 Ω
= 3 A
So, in the parallel combination, the current through R3 is 3 A.
Now, let's consider the series combination:
In a series combination, the total resistance is the sum of the individual resistances:
R_total = R1 + R2 + R3
= 60 Ω + 30 Ω + 30 Ω
= 120 Ω
To find the current through the series combination, we can use Ohm's Law:
I_series = V / R_total
= 90 V / 120 Ω
= 0.75 A
Therefore, in the series combination, the current through R3 is 0.75 A.
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Note the complete questions is User
90 V R₁=60 R2= 30, R3 = 30
Based on the circuit above, what would be the current through the R3 resistor in parallel and in series combinantion.
If the fundamental frequency of a musical instrument is 42 Hz, what is the frequency of the second harmonic? 21 Hz 42 Hz 84 Hz 126 Hz
The frequency of the second harmonic is equal to 84 Hz. Therefore, option C is correct.
What is the frequency?The frequency of the wave can be described as the number of cycles of a wave in one second. The frequency has S.I. measurement units which can be represented as per second or Hz.
The wavelength can be explained as the distance between the two adjacent regions in phase. Two crests or two troughs are separated by a distance known as wavelength.
Given, the frequency of the wave: f = nν/2L
The n is the number of harmonics, n = 2
For the second harmonics:
f₂ = 2f₀
Given the fundamental frequency of a musical instrument = 42 Hz
f₂ = 2 × 42
f₂ = 84 Hz
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An electron in the first energy level of the electron cloud has an electron in the third energy level
Answer:
a lower energy than
Explanation:
sorry im a month late but is lower energy than
Question 4:
Two blocks are suspended motionless near the surface of
Earth by a cord passing over a frictionless pulley, as shown
to the right. Each block weighs 50 N. The force exerted on
each block by the cord is .... ?
(2 Marks)
If two blocks are suspended motionless near the surface of Earth by a cord passing over a frictionless pulley, the force exerted on each block by the cord is 50 N
Since each block weighs 50 N, the force exerted on each block by the cord is same.
T1 = T2 = T
W1 = W2 = 50 N
Since the blocks are suspended motionless,
∑ Fy = 0
- W1 - W2 + T1 + T2 = 0
- 50 - 50 + T + T = 0
2 T = 100
T = 50 N
Therefore, the force exerted on each block by the cord is 50 N.
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1. What does Louis de Broglie's principle propose about electrons?
They have energy and momentum.
They behave as a wave and a particle.
They contain high-energy and low-energy photons.
They interact with other electrons to form new substances.
Louis de Broglie's principle of electrons states that (b) They behave as a wave and a particle.
The de Broglie equation is one of the equations that is commonly used to define the wave properties of matter. It basically describes the wave nature of the electron.Electromagnetic radiation, exhibit dual nature of a particle (having a momentum) and wave (expressed in frequency, wavelength). Microscopic particle-like electrons also proved to possess this dual nature property.Louis de Broglie in his thesis suggested that any moving particle, whether microscopic or macroscopic will be associated with a wave character. It was called ‘Matter Waves’. He further proposed a relation between the velocity and momentum of a particle with the wavelength, if the particle had to behave as a wave.Particle and wave nature of matter, however, looked contradictory as it was not possible to prove the existence of both properties in any single experiment. This is because of the fact that every experiment is normally based on some principle and results related to the principle are only reflected in that experiment and not the other.To know more about electrons visit:
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