Answer:
Fluid Flow is a part of fluid mechanics and deals with fluid dynamics. Fluids such as gases and liquids in motion are called fluid flow. It involves the motion of a fluid subjected to unbalanced forces. This motion continues as long as unbalanced forces are applied.
Difference:
Whereas in real fluids velocity varies across the section. But when the size and shape of cross section are constant along the length of channels under consideration, the flow is said to be uniform. A non-uniform flow is one in which velocity is not constant at a given instant.
Zoning laws establish _______.
Answer:
Zoning ordinances detail whether specific geographic zones are acceptable for residential or commercial purposes. Zoning ordinances may also regulate: - size
- placement
- density
- height of structures
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Answer:
B
Explanation:
B on edg.
Parallel rays of monochromatic light with wavelength 582 nm illuminate two identical slits and produce an interference pattern on a screen that is 75.0 cm from the slits. The centers of the slits are 0.640 mm apart and the width of each slit is 0.434 mm. If the intensity at the center of the central maximum is 4.40×10^−4 W/m^2.
Required:
What is the intensity at a point on the screen that is 0.710 mm from the center of the central maximum?
Answer:
What is the intensity is 1.3349 × 10⁻⁷ w/m²
Explanation:
Given that;
λ = 582 nm = 582 × 10⁻⁹
R = 75.0 cm = 0.75 m
d = 0.640 mm = 0.000640 m
a = 0.434 mm = 0.000434 m
I₀ = 4.40×10⁻⁴ W/m²
y = 0.710 mm = 0.00071 m
Now to get our tanФ we say
tanФ = y/R = 0.00071 / 0.75 = 0.0009466
Ф is so small
∴ tanФ ≈ sinФ
So
∅ = 2πdsinФ / λ
we substitute
∅ = ( 2π × 0.000640 × 0.0009466 ) / 582 × 10⁻⁹
= 6.54 rad
Now
β = 2πasinФ / λ
we substitute
β = ( 2π × 0.000434 × 0.0009466 ) / 582 × 10⁻⁹
β = 4.435 rad
I = I₀ cos²(∅/2) [(sin(β/2))/(β/2)]²
we substitute
I = 4.40×10⁻⁴ cos(3.27)² [ (sin(2.2175)) / (2.2175) ]²
= 4.40×10⁻⁴ × 0.9967 × 0.0003044
= 1.3349 × 10⁻⁷ w/m²
Which statement best describes the liquid state of matter?
ОА.
It has definite shape but indefinite volume.
OB.
It has definite shape and definite volume.
Ос.
It has indefinite shape and indefinite volume.
OD.
It has indefinite shape but definite volume.
Answer:
OB.It has definite shape and definite volume
Make a graph of the data. You may use a graphing program. Think about what data should be on the y-axis and the x-axis. Be sure to label each axis and note the units used in the measurements. Be sure to draw a smooth curve through the points. Do not just connect the dots. Upload your data and graph in the essay box below and answer the following questions. Did the car travel at a constant speed? What was the average speed of the car? What are some practical applications for determining the motion of an object?
Answer:
yes it was a constant speed and the car traveled 10 meters in 20 seconds.
Explanation:
Answer:
It's a constant speed and the car traveled 10 meters in 20 seconds.
hope this helps!
Explanation:
Consider two different isotopes of the same neutral element. Which statements about these isotopes are true?
a. Both isotopes contain the same number of protons.
b. Both isotopes contain the same number of nucleons.
c. isotopes contain the same number of neutrons.
d. Both isotopes contain the same number of orbital electrons.
d. The sum of the protons and neutrons is the same for both isotopes.
Answer:
a. d.
Explanation:
isotopes have a diff number of neutrons
A speeding motorist traveling down a straight highway at 100 km/h passes a parked police car. It takes the police constable 1.0 s to take a radar reading and to start up his car. The police vehicle accelerates from rest at 2 m/s2 and finally catches up with the speeder. a) How much time has elapsed when the two cars meet?
Answer:
t = 7.5 s
Explanation:
The distance traveled by the car at the time of meeting of the two cars must be the same. First, we calculate the distance traveled by the police car. For that we use 2nd equation of motion. Here, we take the time when police car starts to be reference. So,
s₁ = Vi t + (0.5)gt²
where,
s₁ = distance traveled by police car
Vi = Initial Velocity = 0 m/s
t = time taken
Therefore,
s₁ = (0 m/s)(t) + (0.5)(9.8 m/s²)t²
s₁ = 4.9 t²
Now, we calculate the distance traveled by the car. For constant speed and time to be 1 second more than the police car time, due to car starting time, we get:
s₂ = Vt = V(t + 1)
where,
s₂ = distance traveled by car
V = Velocity of car = (100 km/h)(1000 m/1 km)(1 h/ 3600 s) = 27.78 m/s
Therefore,
s₂ = 27.78 t + 27.78
Now, we know that at the time of meeting:
s₁ = s₂
4.9 t² = 27.78 t + 27.78
4.9 t² - 270.78 t - 27.78 = 0
solving the equation and choosing the positive root:
t = 6.5 s
since, we want to know the time from the moment car crossed police car. Therefore, we add 1 second of starting time in this.
t = 6.5 s + 1 s
t = 7.5 s
The 2-kg collar is attached to a spring that has an un-stretched length of 3.0 m. If the collar is drawn to point B and releases from rest, what is the speed when it arrives at point A. Note that k = 3.0 N/m and neglect friction.
Complete Question
The image for this question is shown on the first uploaded image
Answer:
[tex]v = 3.4 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the collar is [tex]m = 2 \ kg[/tex]
The original length is [tex]L = 3.0 \ m[/tex]
The spring constant is [tex]k = 3.0 \ N/m[/tex]
Generally the extension of the spring is mathematically evaluated as
[tex]e = 4 -3 = 1 \ m[/tex]
Now with Pythagoras theorem we can obtain the length from A to B as
[tex]AB = \sqrt{5 ^2 + 4^2}[/tex]
[tex]AB = 6.4 \ m[/tex]
The extension of the spring at B is
[tex]e_b = 6.4 - 3 = 3.4 \ m[/tex]
According to the law of energy conservation
The energy stored in the spring at point A + the kinetic energy of the spring = The energy stored on the spring at B
So
[tex]\frac{1}{2} * k * e + \frac{1}{2} * m* v^2 = \frac{1}{2} * k * e_b[/tex]
substituting values
[tex]\frac{1}{2} * 3 * 1^2 + \frac{1}{2} * 2* v^2 = \frac{1}{2} * 3 * 3.4^2[/tex]
=> [tex]v = 3.4 \ m/s[/tex]
A source of emf is connected by wires to a resistor, and electrons flow in the circuit. The wire diameter is the same throughout the circuit. Compared to the potential energy of an electron before entering the source of emf, the potential energy of an electron after leaving the source of emf is
Answer
The potential energy is less
Explanation:
From the question we are told that
The source of the emf is by wires to a resistor.
Now the potential energy of electron before leaving the source emf will be greater than the potential energy of an electron after leaving the source of emf because the resistor connected to the source emf will reduced the potential energy as it will convert some of the energy to heat
The potential energy of an electron after leaving the source of emf is lesser.
What is electro motive force?Electro motive force is the voltage or potential difference of an electrical energy device such as battery.
The source of the emf is by wires to a resistor.Now, the potential energy of an electron before leaving the source emf will be greater than the potential energy of an electron.
After leaving the source of emf because the resistor connected to the source of emf will reduce the potential energy as it will convert some of the energy to heat.
Thus the potential energy of an electron after leaving the source of emf is lesser.
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pls help me solve dis
Answer:
9 Ω
Explanation:
The following data were obtained from the question:
Resistor 1 (R1) = 3 Ω
Resistor 2 (R2) = 3 Ω
Resistor 3 (R3) = 3 Ω
Resistor 4 (R4) = 3 Ω
Resistor 5 (R5) = 3 Ω
Resistor 6 (R6) = 3 Ω
Resistor 7 (R7) = 3 Ω
Resistor 8 (R8) = 3 Ω
Resistor 9 (R9) = 3 Ω
Resistor 10 (R10) = 3 Ω
Resistor 11 (R11) = 3 Ω
Resistor 12 (R12) = 3 Ω
Equivalent Resistance (R) =.?
From the above diagram,
Resistor 1, 2, 3, 4, 5 and 6 are in series connection and in parralle connections with Resistor 7, 8, 9, 10, 11 and 12 which are also in series connection.
Thus we shall determine the equivalent resistance of Resistor 1, 2, 3, 4, 5 and 6
This is illustrated below:
Resistance Ra = R1 + R2 + R3 + R4 + R5 + R6
Ra = 3 + 3 + 3 + 3 + 3 + 3
Ra = 18 Ω
Next, we shall determine the equivalent resistance of Resistor 7, 8, 9, 10, 11 and 12.
This is illustrated below:
Resistance Rb = R7 + R8 + R9 + R10 + R11 + R12
Rb = 3 + 3 + 3 + 3 + 3 + 3
Rb = 18 Ω
Thus, Ra and Rb are in parallel connections. The equivalent resistance between A and B can be obtained as shown below :
Ra = 18 Ω
Rb = 18 Ω
Equivalent resistance R =?
1/R = 1/Ra + 1/Rb
1/R = 1/18 + 1/18
1/R = 2/18
1/R = 1/9
Invert
R = 9 Ω
Therefore, the equivalent resistance between A and B is 9 Ω.
50 g of pieces of brass are heated to 200 ° C and then placed in the aluminum container of a 50 g calorimeter containing 160 g of water. What is the equilibrium temperature, if the temperature of the container and the water is initially 20 ° C?
Answer:
24.7°C
Explanation:
Heat lost by brass = heat gained by aluminum and water
-q = q
-mCΔT = mCΔT + mCΔT
-(50 g) (0.380 J/g/°C) (T − 200°C) = (50 g) (0.900 J/g/°C) (T − 20°C) + (160 g) (4.186 J/g/°C) (T − 20°C)
-(19 J/°C) (T − 200°C) = (45 J/°C) (T − 20°C) + (670 J/°C) (T − 20°C)
-(19 J/°C) (T − 200°C) = (715 J/°C) (T − 20°C)
-19 (T − 200°C) = 715 (T − 20°C)
-19T + 3800°C = 715T − 14300°C
18100°C = 734T
T = 24.7°C
A Carnot heat engine and an irreversible heat engine both operate between the same high temperature and low temperature reservoirs. They absorb the same energy from the high temperature reservoir as heat. The irreversible engine:__________. A. does more work. B. has the same efficiency as the reversible engine. C. has the greater efficiency. D. transfers more energy to the low temperature reservoir as heat. E. cannot absorb the same energy from the high temperature reservoir as heat without violating the second law of thermodynamics.
Answer:
the answers d
A Carnot heat engine and an irreversible heat engine both operate between the same high-temperature and low-temperature reservoirs. They absorb the same energy from the high-temperature reservoir as heat. The irreversible engine cannot absorb the same energy from the high-temperature reservoir as heat without violating the second law of thermodynamics, therefore the correct option is E.
What is a Carnot engine?It is a theoretical engine assumed to have the maximum efficiency among all engines out there in the universe.
No other engine can have greater efficiency than the Carnot engine.
The Carnot engine work on the principle of the Carnot cycle which assume that the heat and work tranfer processes are reversible in nature. there is no loss due to the friction of the piston.
As we know from the second law of thermodynamics that no other engine can have greater efficiency than the Carnot engine.
Thus, the irreversible engine cannot absorb the same energy from the high-temperature reservoir as heat without violating the second law of thermodynamics, therefore the correct option is E.
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The allowed energies of a simple atom are 0.0 eV, 4.0 eV, and 6.0 eV. Part A What wavelength(s) appear(s) in the atom's emission spectrum
Answer:
3.1 × 10^- 7 m and 2.1 × 10^-7 m
Explanation:
First we must convert each value of energy to Joules by multiplying its value by 1.6 ×10^-19. After that, we can now obtain the wavelength from E= hc/λ
Where;
h= planks constant
c= speed of light
λ= wavelength of light
For 6.0ev;
E= 6.0 × 1.6 ×10^-19
E= 9.6 × 10^-19 J
From
E= hc/λ
λ= hc/E
λ= 6.6 × 10^-34 × 3 × 10^8/9.6 × 10^-19
λ= 2.1 × 10^-7 m
For 4.0 eV
4.0 × 1.6 × 10^-19 = 6.4 × 10^-19 J
E= hc/λ
λ= hc/E
λ= 6.6 × 10^-34 × 3 × 10^8/6.4 × 10^-19
λ= 3.1 × 10^- 7 m
(a) The wavelength of the atom's emission spectrum when the energy is 4 eV is [tex]3.1 \times 10^{-7} \ m[/tex]
(b) The wavelength of the atom's emission spectrum when the energy is 6 eV is
[tex]2.1 \times 10^{-7} \ m[/tex]
The wavelength of the atom's emission spectrum is calculated as follows;
[tex]E = hf\\\\E = \frac{hc}{\lambda}[/tex]
where;
λ is the wavelengthh is Planck's constantFor 4 eV;
[tex]\lambda = \frac{hc}{E} \\\\\lambda = \frac{(6.626 \times 10^{-34}) \times 3\times 10^8}{4 \times 1.602 \times 10^{-19}} \\\\\lambda = 3.1 \times 10^{-7} \ m[/tex]
For 6 eV;
[tex]\lambda = \frac{hc}{E} \\\\\lambda = \frac{(6.626 \times 10^{-34}) \times 3\times 10^8}{6 \times 1.602 \times 10^{-19}} \\\\\lambda = 2.1 \times 10^{-7} \ m[/tex]
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The coefficient of static friction between a 3.00 kg crate and the 35.0o incline is 0.300. What minimum force F must be applied perpendicularly to the incline to prevent the crate from sliding down
Answer:
32.13 N
Explanation:
Given that
mass of the crate, m = 3 kg
angle of inclination, = 35°
coefficient of static friction, = 0.3
To solve this, we can assume that the minimum force is F Newton, then use the formula
mgsinA = coefficient of static friction * [F + mgcosA]
=>3 * 9.8 * sin35 = 0.3 * [F + 3 * 9.8 * cos35]
=> 29.4 * 0.5736 = 0.3 * [F + 29.4 * 0.8192]
=> 16.86 = 0.3 [F + 24.08]
=> 16.86 = 0.3F + 7.22
=> 16.86 - 7.22 = 0.3F
=> 0.3F = 9.64
=> F = 9.64/0.3
=> F = 32.13 N
Therefore, the Force that must be applied is 32.13 N
The net force that must be applied is 9.8 N.
The minimum force required is Fnet.
Fnet = -Ff + mgsinθ
But Ff = μN = μmgcosθ
Fnet = - μmgcosθ + mgsinθ
Where;
m = 3.00 kg
μ = 0.300
θ = 35.0o
Substituting values;
Fnet = mgsinθ - μmgcosθ
Fnet = (3 × 10 × sin 35.0o) - (3 × 0.300 × 10 × cos 35.0o)
= 17.2 - 7.4
Fnet = 9.8 N
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A liquid of density 1250 kg/m3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the flow speed is 9.93 m/s and the pipe diameter d1 is 11.1 cm. At Location 2, the pipe diameter d2 is 16.7 cm. At Location 1, the pipe is Δy=8.89 m higher than it is at Location 2. Ignoring viscosity, calculate the difference ΔP between the fluid pressure at Location 2 and the fluid pressure at Location 1.
Answer:
The pressure difference is [tex]\Delta P = 1.46 *10^{5}\ Pa[/tex]
Explanation:
From the question we are told that
The density is [tex]\rho = 1250 \ kg/m^3[/tex]
The speed at location 1 is [tex]v_1 = 9.93 \ m/s[/tex]
The diameter at location 1 is [tex]d_1 = 11.1\ cm = 0.111 \ m[/tex]
The diameter at location 2 is [tex]d_1 = 16.7\ cm = 0.167 \ m[/tex]
The height at location 1 is [tex]h_1 = 8.89 \ m[/tex]
The height at location 2 is [tex]h_2 = 1 \ m[/tex]
Generally the cross- sectional area at location 1 is mathematically represented as
[tex]A_1 = \pi * \frac{d^2}{4}[/tex]
=> [tex]A_1 = 3.142 * \frac{ 0.111^2}{4}[/tex]
=> [tex]A_1 = 0.0097 \ m^2[/tex]
Generally the cross- sectional area at location 2 is mathematically represented as
[tex]A_2 = \pi * \frac{d_1^2}{4}[/tex]
=> [tex]A_2= 3.142 * \frac{ 0.167^2}{4}[/tex]
=> [tex]A_2 =0.0219 \ m^2[/tex]
From continuity formula
[tex]v_1 * A_1 = v_2 * A_2[/tex]
=> [tex]v_2 = \frac{A_1 * v_1}{A_2 }[/tex]
=> [tex]v_2 = \frac{0.0097 * 9.93}{0.0219 }[/tex]
=> [tex]v_2 = 4.398 \ m/s[/tex]
Generally according to Bernoulli's theorem
[tex]P_1 + \rho * g * h_1 + \frac{1}{2} \rho * v_1^2 = P_2 + \rho * g * h_2 + \frac{1}{2} \rho * v_2^2[/tex]
=> [tex]P_2 - P_1 = \frac{1}{2} \rho (v_1 ^2 - v_2^2 ) + \rho* g (h_1 - h_2)[/tex]
=> [tex]\Delta P = \frac{1}{2}* 1250* (9.93 ^2 - 4.398^2 ) + 1250* 9.8 (8.89- 1)[/tex]
=> [tex]\Delta P = 1.46 *10^{5}\ Pa[/tex]
what happens to the speed of the
Skateboard/refrigerator when there is no longer a force being applied ?
Answer:
The speed stays constant after the force stops pushing.
Explanation:
Speed always stays constant when the force stops pushing it.
A tire swing hanging from a branch reaches nearly to the ground. How could you estimate the height of the branch using only a stopwatch?
Answer:
Explanation:
With the help of expression of time period of pendulum we can calculate the height of the branch . The swinging tire can be considered equivalent to swinging bob of a pendulum . Here length of pendulum will be equal to height of branch .
Let it be h . Let the time period of swing of tire be T then from the formula of time period of pendulum
[tex]T = 2\pi\sqrt{\frac{l}{g} }[/tex] where l is length of pendulum .
here l = h so
[tex]T = 2\pi\sqrt{\frac{h}{g} }[/tex]
[tex]h = \frac{T^2g}{4\pi^2}[/tex]
If we calculate the time period of swing of tire , we can calculate the height of branch .
The time period of swing of tire can be estimated with the help of a stop watch . Time period is time that the tire will take in going from one extreme point to the other end and then coming back . We can easily estimate it with the help of stop watch .
If a thermometer measured the temperature in an oven as 400oF five days in a row when the temperature was actually 397oF, this measuring instrument would be considered quite:
Answer:
It can be said to be reliable although it is not valid
Explanation:
This is because Reliability means an indicator of consistency, A measure should produce similar or the same results consistently if it measures the same quantity. So does the thermometer measures over 5days but it is not valid because it deviates from the real value
Estimate the distance (in cm) between the central bright region and the third dark fringe on a screen 5.00 m from two double slits 0.500 mm apart illuminated by 500-nm light.
Answer:
y = 1.75 cm
Explanation:
In the double-slit experiment the equation for destructive interference is
d sin tea = (m + ½)
λ
let's use trigonometry to find the angle
tan θ = y / L
as all the experiment does not occur at small angles
tan θ = sin θ / cos θ = sin θ = y / L
we substitute
y = (m + 1/2 ) λ L / d
we calculate
y = (3 + ½) 500 10⁻⁹ 5.00 / 0.5 10⁻³
y = 1.75 10⁻² m
y = 1.75 cm
Sodium has a work function of 2.46 eV.
(a) Find the cutoff wavelength and cutoff frequency for the photoelectric effect.
(b) What is the stopping potential if the incident light has a wavelength of 181 nm?
Answer:
Explanation:
given, work function of Φ = 2.46 eV.
converting the eV to joule, we have
2.45 * 1.6*10^-19 J
The cutoff wavelength is the wavelength where the incoming light does not have enough energy to free an electron, i.e. all of
the photon’s energy will be channeled into trying overcoming the work function barrier.
It is mathematically given as
Φ = hf
f = Φ/h
f = (2.46 * 1.6*10^-19) / 6.63*10^-34
f = 3.936*10^-19 / 6.63*10^-34
f = 5.94*10^14 Hz as our cut off frequency
λf = c,
λ = c/f
λ = 3*10^8 / 5.94*10^14
λ = 5.05*10^-7
λ = 505 nm as our cut off wavelength
K(max) = hf - Φ
K(max) = hc/λ - Φ
K(max) = [(6.63*10^-34 * 3*10^8) / 181*10^-9] - 3.936*10^-19
K(max) = (1.989*10^-25/181*10^-9) - 3.936*10^-19
K(max) = 1.1*10^-18 - 3.936*10^-19
K(max) = 7.064*10^-19 J or 4.415 eV
V(s) = K(max) / e
V(s) = 4.612 V
An airplane is traveling at 400 mi/h. It touches down at an airport 2000 miles away. How long was the airplane airborne?
Answer:
5 hours
Explanation:
Given that
Speed of the airplane, v = 400 mile/hr
Distance of the airport, s = 2000 miles
This is quite a straightforward question that deals with one of the basic formulas in physics.
Speed.
speed is said to the the ratio of distance covered with respect to the time taken. This can be mathematically expressed as
Speed, v = distance covered, d / time taken, t
v = d / t
In the question above, we're looking for the time taken. So, so make t, subject of formula.
t = d / v, now we proceed to substituting the earlier given values into this equation.
t = 2000 / 400
t = 5 hrs,
therefore we can conclude that the airplane was airborne for 5 hours
Firecrackers A and B are 600 m apart. You are standing exactly halfway between them. Your lab partner is 300 m on the other side of firecracker A. You see two flashes of light, from the two explosions, at exactly the same instant of time.
Define event 1 to be "firecracker 1 explodes" and event 2 to be "firecracker 2 explodes." According to your lab partner, based on measurements he or she makes, does event 1 occur before, after, or at the same time as event 2? Explain.
Answer:
See the explanation
Explanation:
Given:
Distance of Firecrackers A and B = 600 m
Event 1 = firecracker 1 explodes
Event 2 = firecracker 2 explodes
Distance of lab partner from cracker A = 300 m
You observe the explosions at the same time
to find:
does event 1 occur before, after, or at the same time as event 2?
Solution:
Since the lab partner is at 300 m distance from the firecracker A and Firecrackers A and B are 600 m apart
So the distance of fire cracker B from the lab partner is:
600 m + 300 m = 900 m
It takes longer for the light from the more distant firecracker to reach so
Let T1 represents the time taken for light from firecracker A to reach lab partner
T1 = 300/c
It is 300 because lab partner is 300 m on other side of firecracker A
Let T2 represents the time taken for light from firecracker B to reach lab partner
T2 = 900/c
It is 900 because lab partner is 900 m on other side of firecracker B
T2 = T1
900 = 300
900 = 3(300)
T2 = 3(T1)
Hence lab partner observes the explosion of the firecracker A before the explosion of firecracker B.
Since event 1 = firecracker 1 explodes and event 2 = firecracker 2 explodes
So this concludes that lab partner sees event 1 occur first and lab partner is smart enough to correct for the travel time of light and conclude that the events occur at the same time.
An electron accelerates through a 12.5 V potential difference, starting from rest, and then collides with a hydrogen atom, exciting the atom to the highest energy level allowed. List all the possible quantum-jump transitions by which the excited atom could emit a photon. 4 rightarrow 3 4 rightarrow 2 4 rightarrow 1 3 rightarrow 2 3 rightarrow 1 2 rightarrow 1
Answer:
Initial state Final state
3 ⇒ 2
3 ⇒ 1
2 ⇒ 1
Explanation:
For this exercise we must use Bohr's atomic model
E = - 13.606 / n²
where is the value of 13.606 eV is the energy of the ground state and n is the integer.
The energy acquired by the electron in units of electron volt (eV)
E = e V
E = 12.5 eV
all this energy is used to transfer an electron from the ground state to an excited state
ΔE = 13.6060 (1 / n₀² - 1 / n²)
the ground state has n₀ = 1
ΔE = 13.606 (1 - 1/n²)
1 /n² = 1 - ΔE/13,606
1 / n² = 1 - 12.5 / 13.606
1 / n² = 0.08129
n = √(1 / 0.08129)
n = 3.5
since n is an integer, maximun is
n = 3
because it cannot give more energy than the electron has
From this level there can be transition to reach the base state.
Initial state Final state
3 ⇒ 2
3 ⇒ 1
2 ⇒ 1
The possible quantum-jump transitions by which the excited atom emits a photon are :
Initial state Final state
3 ---> 2
3 ----> 1
2 ----> 1
Given data :
Potential difference through which an electron accelerates = 12.5 V
Energy acquired by the the electrons = 12.5 eV ( e * 12.5 )
The Model we will use to determine the possible quantum jump transition is Bohr's atomic model
E = - 13.606 / n²
where ; n = integer
energy at ground state = 13.606 eV
The energy acquired by the electrons ( 12.5 eV ) is used to move the electron from its ground state to an excited state.
Therefore
ΔE = 13.606 * (1 / n₀² - 1 / n²) ---- ( 1 )
where n₀ = 1
Back to equation ( 1 )
ΔE = 13.606 (1 - 1/n²) -- ( 2 )
Resolving equation ( 2 )
1 / n² = 0.08129
n = 3.5 . Therefore the maximum integer = 3
Hence The collision between the electron and the hydrogen atom will undergo three ( 3 ) transition to reach the base state.
In Conclusion The possible quantum-jump transitions by which the excited atom emits a photon are :
Initial state Final state
3 ---> 2
3 ----> 1
2 ----> 1
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A particle with charge -5 C initially moves at v = (1.00 i^ + 7.00 j^ ) m/s. If it encounters a magnetic field B =80 Tkˆ, find the magnetic force vector on the particle.
Answer:
The magnetic force is [tex]\= F = 400\r j + 2800\r i[/tex]
Explanation:
From the question we are told that
The charge is [tex]q = -5C[/tex]
The velocity is [tex]v = (1.00\ \r i + 7.00 \ \r j )\ m/s[/tex]
The magnetic field is [tex]B = 80 \r k \ \ T[/tex]
Generally the magnetic force is mathematically represented as
[tex]\= F = q \= v \ \ X \ \ \= B[/tex]
=> [tex]\= F = -5 (1.0 \r i + 7.0 \r j ) \ \ X \ \ 80 \r k[/tex]
=> [tex]\= F = -5.0 \r i + 35\r j \ \ \ X \ \ 80\r k[/tex]
=> [tex]\= F = 400\r j + 2800\r i[/tex] N/B - Applied cross - product of unit vector
Two gratings A and B have slit separations dA and dB, respectively. They are used with the same light and the same observation screen. When grating A is replaced with grating B, it is observed that the first-order maximum of A is exactly replaced by the second-order maximum of B.
a.) I already found that the ratio of db/da is 2
b.) Find the next two principal maxima of grating A and the principal maxima of B that exactly replace them when the gratings are switched. Identify these maxima by their order numbers
Answer:
mA=2,mB=4
mA=2,mB=4 and
mA=3,mB=6
Explanation:
First of all we need to write the equation of the networks
sin θ = mA λ / dA
sin θ = mB λ / dB
Equating we have
mA λ/ dA = mB λ / dB
We are given the ratio as
dB / dA = 2
So
mA 2 = mB
Finally overlapping orders
We have
mA=2,mB=4
mA=2,mB=4
and mA=3,mB=6
What kind of wave is formed (transverse or longitudimal wave, pick one) is formed by ripples on a calm pond? With explanation! Please help, most detailed answer will get brainliest and many points.
Answer:
Transverse
Explanation:
It's tranverse because the water molecules are moving repeatedly up and down vertically when the waves move horizontally across the waters surface.
What is the tension in the cord after the system is released from rest? Both masses (A and B) are 10-kg.
Answer:
98 N.
Explanation:
Given data: mass= 10 kg, gravity= 9.8 m/s2
required: tension in the cord= ?
solution:
formula of tension= mass x gravity
by putting values of mass and gravity, we get
tension= 10 x 9.8
tension= 98 N. Ans
If the mass of the object which is attached with both ends of cord is 10 kg, so the tension which is a opposite force of weight is 98 N.
A vertical scale on a spring balance reads from 0 to 220 N. The scale has a length of 15.0 cm from the 0 to 220 N reading. A fish hanging from the bottom of the spring oscillates vertically at a frequency of 2.70 Hz. Ignoring the mass of the spring, what is the mass m of the fish?
Mass of fish is 5.09Kg
Explanation:
First to find the spring constant K using
k = F/s
= 220/0.15 = 1466.7 N/m
So using the formula
T = 2π√(m/k)
f = 1/T = 1/2πx √(k/m)
f² x 4π²= k/m
So
m = k/(f² x π²)
m = 1466.7/(2.7² x 4π²)
m = 5.09 kg
If you wanted to find the area of the hot filament in a light bulb, you would have to know the temperature (determinable from the color of the light), the power input, the Stefan-Boltzmann constant, and what property of the filament
Answer:
To find out the area of the hot filament of a light bulb, you would need to know the temperature, the power input, the Stefan-Boltzmann constant and Emissivity of the Filament.
Explanation:
The emissive power of a light bulb can be given by the following formula:
E = σεAT⁴
where,
E = Power Input or Emissive Power
σ = Stefan-Boltzmann constant
ε = Emissivity
A = Area
T = Absolute Temperature
Therefore,
A = E/σεT⁴
So, to find out the area of the hot filament of a light bulb, you would need to know the temperature, the power input, the Stefan-Boltzmann constant and Emissivity of the Filament.
Represent a vector of 100 N in North-East direction
Answer:
please find the attachment to this question.
Explanation:
In this question, we represent the 100N in the North-East direction, but first, we define the vector representation:
It is generally represented through arrows, whose length and direction reflect the magnitude and direction of the arrow points. In this, both size and direction are necessary because the magnitude of a vector would be a number that can be compared to one vector.
Please find the attachment:
Please answer the following questions about uniform circular motion.?
Part (a) A planet orbits a star in a circular orbit at a constant orbital speed, which of the following statements is true?
All of these are correct.
The magnitude of the orbital velocity of the planet is unchanged, thus there is no acceleration and therefore no force action on the planet.
None of these are correct.
The planet experiences a centripetal force pulling towards its star.
The planet experiences no centripetal force.
The planet experiences a centripetal force pushing it away from its star.
Part (b) When a planet is orbiting a star, which force plays the role of the centripetal force?
The force resulting from the planets’ velocity around the star.
The force resulting from the centripetal acceleration.
The gravitational force
Part (c) Which of the following are true statements about uniform circular motion?
An object in uniform circular motion experiences a tangential force.
An object in uniform circular motion experiences a centripetal force, an equal and opposite centrifugal force, and a tangential force.
An object in uniform circular motion experiences a centripetal force and a tangential force.
An object in uniform circular motion experiences a centripetal force.
None of these choices are true.
An object in uniform circular motion experiences no forces.
Answer:
a) The planet experiences a centripetal force towards its star
b) The universal attractive force (Gravitational force)
c)None of these choices are true.
Explanation:
This problem raise several claims, let's review some aspects of circular motion
F = m a
the centripetal acceleration is
a = v² / r
where v is the speed (modulus of velocity) that is constant and r is the radius
The direction of the acceleration is perpendicular to the motion.
Let's review the different claims
Part a) the orbital velocity is constant
The correct statement is: The planet experiences a centripetal force towards its star
Part b) what is the centripetal force
The correct statement: The universal attractive force (Gravitational force)
Part c) which statement is true
1) False. There can be no tangential force
2) False. There is a centripetal force that creates the movement, but there is no centrifugal force because the system is accelerated and there is no tangential force because the movement is circular.
3) False. There is no tangential force
4) True none is true
5) False. There is a force because movement has acceleration