Answer:
The smallest value is n= 2
Explanation:
The balmer equation is given below
1/λ = R(1/4 - 1/n₂²).
R= 1.0973731568508 × 10^7 m^-1
λ= 400*10^-9 m
(400*10^-9)= 1.0973731568508 × 10^7 (1/4-1/n²)
(400*10^-9)/1.0973731568508 × 10^7
= 1/4 - 1/n²
364.51 *10^-16= 1/4 - 1/n²
1/n²= 1/4 -364.51 *10^-16
1/n² = 0.25-3.6451*10^-14
1/0.25= n²
4= n²
√4= n
2= n
The smallest value is N= 2
Select the correct answer. Which of Newton's laws explains why your hands get red when you press them hard against a wall? A. Newton's law of gravity B. Newton's first law of motion C. Newton's second law of motion D. Newton's third law of motion
Answer:
D newton third law
Explanation:
good luck
A golfer hits a 42 g ball, which comes down on a tree root and bounces straight up with an initial speed of 15.6 m/s. Determine the height the ball will rise after the bounce. Show all your work.
Answer:
12.2 m
Explanation:
Given:
v₀ = 15.6 m/s
v = 0 m/s
a = -10 m/s²
Find: Δy
v² = v₀² + 2aΔy
(0 m/s)² = (15.6 m/s)² + 2 (-10 m/s²) Δy
Δy = 12.2 m
[tex] \LARGE{ \boxed{ \rm{ \green{Answer:}}}}[/tex]
Given,
The initial speed is 15.6 m/s The mass of the ball is 42g = 0.042kgFinding the initial kinetic energy,
[tex]\large{ \boxed{ \rm{K.E. = \frac{1}{2}m {v}^{2}}}}[/tex]
⇛ KE = (1/2)mv²
⇛ KE = (1/2)(0.042)(15.6)²
⇛ KE = 5.11 J
|| ⚡By conservation of energy, the potential energy at the highest point will also be 5.11 J, since there is no kinetic energy at the highest point because the ball is not moving (we neglect energy lost due to air resistance, heat, sound, etc.) ⚡||
So, we have:
[tex] \large{ \boxed{ \rm{P.E. = mgh}}}[/tex]
⇛ h = PE/(mg)
⇛ h = 5.11 J /(0.042 × 9.8)
⇛ h = 12.41 m
✏The ball will rise upto a height of 12.41 m
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an aluminum atom has an atomic number of 13 and a mass number of 27,how many
a)protons
b) electrons
pls write the formula too
Element is
[tex]\boxed{\sf {}^{27}Al_{13}}[/tex]
Atomic number=13Mass number=27[tex]\\ \sf\longmapsto No\:of\:Protons=Atomic \:Number=13[/tex]
And[tex]\\ \sf\longmapsto No\:of\:Neutrons=Mass\:number-Atomic\:Number[/tex]
[tex]\\ \sf\longmapsto No\:of\:Neutrons=27-13[/tex]
[tex]\\ \sf\longmapsto No\:of\:Neutrons=14[/tex]
And
[tex]\\ \sf\longmapsto No\:of\:electrons=No\:of\:Protons=13[/tex]
a baseball is given an initial velocity with magnitude v at the angle beta above the surface of an incline which in turn inclined at angle teta above horizontal calculate the distance measured along incline from the launch point to where the baseball strike the incline
Explanation:
The maximum height of an object, given the initial launch angle and initial velocity is found with:h=v2isin2θi2g h = v i 2 sin 2 θ i 2 g .
hat a 15 kg body is pulled along a horizontal fictional table by a force of 4N what is the acceleration of the body
Answer:
Acceleration of the body is:
[tex]a=0.27\,\,m/s^2[/tex]
Explanation:
Use Newton's second Law to solve for the acceleration:
[tex]F=m\,\,a\\a=\frac{F}{m} \\a=\frac{4\,N}{15\,\,kg} \\a=0.27\,\,m/s^2[/tex]
The temperature of the hot spots caused by the impact of transferred matter onto the surface of a pulsar can be 108 K. What is the peak wavelength in the blackbody spectrum of such a spot, and in what range of the electromagnetic spectrum does it occur
Given that,
Temperature = 10⁸ K
We need to calculate the peak wavelength in the blackbody spectrum
Using formula of peak wavelength
[tex]peak\ wavelength = \dfrac{2.898\times10^{-3}}{T}[/tex]
Where, T= temperature
Put the value into the formula
[tex]peak\ wavelength = \dfrac{2.898\times10^{-3}}{10^{8}}[/tex]
[tex]peak\ wavelength = 2.90\times10^{-11}\ m[/tex]
[tex]peak\ wavelength = 290\ nm[/tex]
This range of wavelength is ultraviolet.
Hence, The peak wavelength in the blackbody spectrum is 290 nm and the range of wavelength is ultraviolet electromagnetic spectrum .
Expectant mothers many times see their unborn child for the first time during an ultrasonic examination. In ultrasonic imaging, the blood flow and heartbeat of the child can be measured using an echolocation technique similar to that used by bats. For the purposes of these questions, please use 1500 m/s as the speed of sound in tissue. I need help with part B and C
To clearly see an image, the wavelength used must be at most 1/4 of the size of the object that is to be imaged. What frequency is needed to image a fetus at 8 weeks of gestation that is 1.6 cm long?
A. 380 kHz
B. 3.8 kHz
C. 85 kHz
D. 3.8 MHz
Answer:
380 kHz
Explanation:
The speed of sound is taken as 1500 m/s
The length of the fetus is 1.6 cm long
The condition is that the wavelength used must be at most 1/4 of the size of the object that is to be imaged.
For this 1.6 cm baby, the wavelength must not exceed
λ = [tex]\frac{1}{4}[/tex] of 1.6 cm = [tex]\frac{1}{4}[/tex] x 1.6 cm = 0.4 cm =
0.4 cm = 0.004 m this is the wavelength of the required ultrasonic sound.
we know that
v = λf
where v is the speed of a wave
λ is the wavelength of the wave
f is the frequency of the wave
f = v/λ
substituting values, we have
f = 1500/0.004 = 375000 Hz
==> 375000/1000 = 375 kHz ≅ 380 kHz
A vertical spring stretches 3.8 cm when a 13-g object is hung from it. The object is replaced with a block of mass 20 g that oscillates in simple harmonic motion. Calculate the period of motion.
Answer:
The period of motion is 0.5 second.
Explanation:
Given;
extension of the spring, x = 3.8 cm = 0.038 m
mass of the object, m = 13 g = 0.013 kg
Determine the force constant of the spring, k;
F = kx
k = F / x
k = mg / x
k = (0.013 x 9.8) / 0.038
k = 3.353 N/m
When the object is replaced with a block of mass 20 g, the period of motion is calculated as;
[tex]T = 2\pi\sqrt{\frac{m}{k} } \\\\T = 2\pi\sqrt{\frac{0.02}{3.353} } \\\\T = 0.5 \ second[/tex]
Therefore, the period of motion is 0.5 second.
Astronomers discover an exoplanet (a planet of a star other than the Sun) that has an orbital period of 3.75
Earth years in its circular orbit around its sun, which is a star with a measured mass of 3.23×1030kg
. Find the radius of the exoplanet's orbit.
Answer:
[tex]r=4.24\times 10^{11}\ m[/tex]
Explanation:
Given that,
Orbital time period, T = 3.75 earth years
Mass of star, [tex]m=3.23\times 10^{30}\ kg[/tex]
We need to find the radius of the exoplanet's orbit. It is a concept of Kepler's third law of motion i.e.
[tex]T^2=\dfrac{4\pi^2}{GM}r^3[/tex]
r is the radius of the exoplanet's orbit.
Solving for r we get :
[tex]r=(\dfrac{T^2GM}{4\pi^2})^{1/3}[/tex]
We know that, [tex]1\ \text{earth year}=3.154\times 10^7\ \text{s}[/tex]
So,
[tex]r=(\dfrac{(3.75\times 3.154\times 10^7)^2\times 6.67\times 10^{-11}\times 3.23\times 10^{30}}{4\pi^2})^{1/3}\\\\r=4.24\times 10^{11}\ m[/tex]
So, the radius of the exoplanet's orbit is [tex]4.24\times 10^{11}\ m[/tex].
A planar electromagnetic wave is propagating in the x direction. At a certain point P and at a given instant, the magnitude of the electric field of the wave is 0.082 V/m . What is the magnetic vector of the wave at point P at that instant?
A) (0.27 nT)k
B) (-0.27 nT)k
C) (0.27 nTİ
D) (6.8 nT)k
E) (-6.8 nT))
Answer:
b
Explanation:
You add 500 mL of water at 10°C to 100 mL of water at 70°C. What is the
most likely final temperature of the mixture?
O A. 80°C
OB. 10-C
OC. 20°C
O D. 60°C
Answer:
Option (c) : 20°C
Explanation:
[tex]t(final) = \frac{w1 \times t1 + w2 \times t2}{w1 + w2} [/tex]
T(final) = 500* 10 + 100*70/600 = 20°C
If the rods with diameters and lengths listed below are made of the same material, which will undergo the largest percentage length change given the same applied force along its length?a. d, 3L b. 3d, L c. 2d, 2L d. 4d, L
Answer:
The highest percentage of change corresponds to the thinnest rod, the correct answer is a
Explanation:
For this exercise we are asked to change the length of the bar by the action of a force applied along its length, in this case we focus on the expression of longitudinal elasticity
F / A = Y ΔL/L
where F / A is the force per unit length, ΔL / L is the fraction of the change in length, and Y is Young's modulus.
In this case the bars are made of the same material by which Young's modulus is the same for all
ΔL / L = (F / A) / Y
the area of the bar is the area of a circle
A = π r² = π d² / 4
A = π / 4 d²
we substitute
ΔL / L = (F / Y) 4 /πd²
changing length
ΔL = (F / Y 4 /π) L / d²
The amount between paracentesis are all constant in this exercise, let's look for the longitudinal change
a) values given d and 3L
ΔL = cte 3L / d²
ΔL = cte L /d² 3
To find the percentage, we must divide the change in magnitude by its value and multiply by 100.
ΔL/L % = [(F /Y 4/π 1/d²) 3L ] / 3L 100
ΔL/L % = cte 100%
b) 3d and L value, we repeat the same process as in part a
ΔL = cte L / 9d²
ΔL = cte L / d² 1/9
ΔL / L% = cte 100/9
ΔL / L% = cte 11%
c) 2d and 2L value
ΔL = (cte L / d ½ )/ 2L
ΔL/L% = cte 100/4
ΔL/L% = cte 25%
d) value 4d and L
ΔL = cte L / d² 1/16
ΔL/L % = cte 100/16
ΔL/L % = cte 6.25%
The highest percentage of change corresponds to the thinnest rod, the correct answer is a
A loop of wire in the shape of a rectangle rotates with a frequency of 143 rotation per minute in an applied magnetic field of magnitude 2 T. Assume the magnetic field is uniform. The area of the loop is A = 2 cm2 and the total resistance in the circuit is 7 Ω.
1. Find the maximum induced emf.
e m fmax =
2. Find the maximum current through the bulb.
Imax
Answer:
1. e m fmax = 0.00598 Volt
2. Imax = 0.000854 Amp
Explanation:
1. Find the maximum induced emf.
e m fmax =
Given that e m fmax = N*A*B*w
N = 1
A = 2 cm^2 = 0.0002 m^2
f = 143 rotation per minute = 143/min
f = (143/min) * (1 min/60 sec) = 2.38/sec
w = 2Πf = 2 * Π * 2.38 = 14.95 rad/sec
B = 2T
e m fmax = N*A*B*w
e m fmax = 1 * 0.0002 * 2 * 14.95
e m fmax = 0.00598 Volt.
2. Find the maximum current through the bulb.
Imax = e m fmax / R
Where R is the total resistance in the circuit is 7 Ω.
Imax = 0.00598/7 = 0.000854 Amp.
Imax = 0.000854 Amp
1) The maximum induced EMF in the loop of wire is; EMF_max = 9.52 × 10^(-4) V
2) The maximum current through the bulb is;
I_max = 1.36 × 10^(-4) A
We are given;
Number of turns; N = 1
Magnitude of magnetic field; B = 2 T
Area; A = 2 cm² = 0.0002 m²
Angular frequency; ω = 143 /min = 2.38 /s
Resistance; R = 7 Ω.
1) Formula for maximum induced EMF is;
EMF_max = NAωB
Plugging in the relevant values gives;
EMF_max = 1 × 0.0002 × 2.38 × 2
EMF_max = 9.52 × 10^(-4) V
2) Formula for maximum current through the bulb is given as;
I_max = EMF_max/R
Plugging in the relevant values;
I_max = (9.52 × 10^(-4))/7
I_max = 1.36 × 10^(-4) A
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A 4.00-Ω resistor, an 8.00-Ω resistor, and a 24.0-Ω resistor are connected together. (a) What is the maximum resistance that can be produced using all three resistors? (b) What is the minimum resistance that can be produced using all three resistors? (c) How would you connect these three resistors to obtain a resistance of 10.0 Ω? (d) How would you connect these three resistors to obtain a resistance of 8.00 Ω?
Answer:a) 4+8+24=36
B) 1/4+1/8+1/24=10
C) yu will connect them in parallel connection.
D) you will connect two in parallel then the remaining one in series to the ons connected in parallel.
Explanation:
(a)The maximum resistance that can be produced using all three resistors will be 36 ohms.
(b)The minimum resistance that can be produced using all three resistors will be 10 ohms.
(c)The three resistors to obtain a resistance of 10.0 Ω will be in the parallel connection.
(d) You connect these three resistors to obtain a resistance of 8.00 Ω will be in parallel. Two will be linked in parallel, and the last one will be connected in series to the two that are connected in parallel.
What is resistance?Resistance is a type of opposition force due to which the flow of current is reduced in the material or wire. Resistance is the enemy of the flow of current.
The maximum resistance that can be produced using all three resistors is obtained by adding all the given resistance;
[tex]\rm R_{max}=(4 +8+24 )\ ohms \\\\ R_{max}=36 \ ohms[/tex]
The minimum resistance that can be produced using all three resistors is obtained when connected in the parallel.
[tex]\rm R_{min}=\frac{1}{4} +\frac{1}{8} +\frac{1}{24} \\\\ R_{min}=10 \ ohm[/tex]
(c)The three resistors to obtain a resistance of 10.0 Ω will be in the parallel connection.
(d) You connect these three resistors to obtain a resistance of 8.00 Ω will be in parallel. Two will be linked in parallel, and the last one will be connected in series to the two that are connected in parallel.
Hence,the maximum resistance that can be produced using all three resistors will be 36 ohms.
To learn more about the resistance, refer to the link;
https://brainly.com/question/20708652
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A doctor counts 68 heartbeats in 1.0 minute. What are the corresponding period and frequency of the heart rhythm
Answer:
[tex]f=1.13s^{-1}=1.13Hz[/tex]
Explanation:
Hello,
In this case, a frequency stands for a rate in which some action is done per unit of time. In this case, for the heartbeat, since 68 actions (heartbeats) occur in 1.0, the frequency turns out:
[tex]f=\frac{68}{1.0min}=68min^{-1}[/tex]
Or as most commonly used in Hz ([tex]s^{-1}[/tex]):
[tex]f=68\frac{1}{min} *\frac{1min}{60s}=1.13s^{-1}=1.13Hz[/tex]
Best regards.
What is not one of the main uses of springs?
A. Car suspension
B. Bike suspension
C. The seasons
D. Clock making
Consider a bus traveling to the west (negative x direction) that begins to slow down as it approaches a traffic light. Which statement concerning its acceleration in the x direction is correct
Complete question is;
Consider a bus traveling to the west (-x direction) begins to slow down as it approaches a traffic light. Which statement concerning its acceleration in the x direction is correct
a) The bus is decelerating and its acceleration is positive.
b) The bus is decelerating, and its acceleration is negative.
c) The acceleration is zero.
d) A statement cannot be made using the information given.
Answer:
Option A - bus is decelerating and acceleration is positive.
Explanation:
We are told that the bus is travelling in (negative x direction) and begins to slow down. Since the bus is slowing down, it means that the bus is undergoing a negative acceleration which is called deceleration.
Thus, the bus is decelerating.
Since it is moving in the negative x-axis, it means acceleration is now; -(-a) which gives +a.
Thus, bus is decelerating and acceleration is positive.
A solid block is attached to a spring scale. When the block is suspended in air, the scale reads 21.2 N; when it is completely immersed in water, the scale reads 18. 2 N. What are the volume and density of the block?
Answer:
7066kg/m³
Explanation:
The forces in these cases (air and water) are: Fa =mg =ρbVg Fw =(ρb −ρw)Vg where ρw = 1000 kg/m3 is density of water and ρb is density of the block and V is its density. We can find it from this two equations:
Fa /Fw = ρb / (ρb −ρw) ρb = ρw (Fa /Fa −Fw) =1000·(1* 21.2 /21.2 − 18.2)
= 7066kg/m³
Explanation:
Answer:
The volume of the block is 306 cm³
The density of the block is 7.07 g/cm³
Explanation:
Given;
weight of block in air, [tex]W_a[/tex] = 21.2 N
Weight of block in water, [tex]W_w[/tex] = 18.2 N
Mass of the block in air;
[tex]W_a = mg[/tex]
21.2 = m x 9.8
m = 21.2 / 9.8
m = 2.163 kg
mass of the block in water;
[tex]W_w = mg[/tex]
18.2 = m x 9.8
m = 18.2 / 9.8
m = 1.857 kg
Apply Archimedes principle
Mass of object in air - mass of object in water = density of water x volume of object
2.163 kg - 1.857 kg = 1000 kg/m³ x Volume of block
0.306 kg = 1000 kg/m³ x Volume of block
Volume of the block = [tex]\frac{0.306 \ kg}{1000 \ kg/m^3}[/tex]
Volume of the block = 3.06 x 10⁻⁴ m³
Volume of the block = 306 cm³
Determine the density of the block
[tex]Density = \frac{mass}{volume} \\\\Density =\frac{2163 \ g}{306 \ cm^3} \\\\Density = 7.07 \ g/cm^3[/tex]
A wire is carrying current vertically downward. What is the direction of the force due to Earth's magnetic field on the wire?
Answer:
The direction of the force will be towards the east
Explanation:
From the question we are told that
The direction of the downward
Generally according to Fleming's right-hand rule(
Thumb - direction of force
Middle finger - direction of current
Index finger - direction of the magnetic field
) and the fact that the earth magnetic field acts from south to north with respect to the four cardinal points then the direction of the force will be toward the east with respect to the four cardinal point on the earth
Lamar has been running sprints to prepare for his next football game.He has found that he can maintain his maximum speed for 45 yards.He’s thinking of running in a 5km race in a few months,but doesn’t know if he can maintain his maximum speed for the entire 5 km.Can you help him determine how far he can?
Answer:
Kindly check explanation
Explanation:
Length of race = 5km
Maximum speed = 45 yards
Converting from yards to kilometer :
1km = 1093.613 yards
x = 45 yards
(1093.613 * x) = 45
x = 45 / 1093.613
x = 0.0411480 km
Where x = maximum length for which he can maintain his maximum speed expressed in kilometers.
Therefore, with the available information, it can be concluded that Lamar cannot maintain his maximum speed for the entire 5km race and will only be able maintain his maximum speed for 0.0411 kilometers.
Lamar cannot maintain his maximum speed for the entire 5km race and will only be able maintain his maximum speed for 0.0411 kilometers.
The calculation is as follows;
Length of race = 5km
Maximum speed = 45 yards
Converting from yards to kilometer :
1km = 1093.613 yards
x = 45 yards
[tex](1093.613 \times x) = 45[/tex]
[tex]x = 45 \div 1093.613[/tex]
x = 0.0411480 km
here x represent maximum length for which he can maintain his maximum speed expressed in kilometers.
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Copper Pot A copper pot with a mass of 2 kg is sitting at room temperature (20°C). If 200 g of boiling water (100°C) are put in the pot, after a few minutes the water and the pot come to the same temperature. What temperature is this in °C?
Answer:
The final temperature is 61.65 °C
Explanation:
mass of copper pot [tex]m_{c}[/tex] = 2 kg
temperature of copper pot [tex]T_{c}[/tex] = 20 °C (the pot will be in thermal equilibrium with the room)
specific heat capacity of copper [tex]C_{c}[/tex]= 385 J/kg-°C
The heat content of the copper pot = [tex]m_{c}[/tex][tex]C_{c}[/tex][tex]T_{c}[/tex] = 2 x 385 x 20 = 15400 J
mass of boiling water [tex]m_{w}[/tex] = 200 g = 0.2 kg
temperature of boiling water [tex]T_{w}[/tex] = 100 °C
specific heat capacity of water [tex]C_{w}[/tex] = 4182 J/kg-°C
The heat content of the water = [tex]m_{w}[/tex][tex]C_{w}[/tex][tex]T_{w}[/tex] = 0.2 x 4182 x 100 = 83640 J
The total heat content of the water and copper mix [tex]H_{T}[/tex] = 15400 + 83640 = 99040 J
This same heat is evenly distributed between the water and copper mass to achieve thermal equilibrium, therefore we use the equation
[tex]H_{T}[/tex] = [tex]m_{c}[/tex][tex]C_{c}[/tex][tex]T_{f}[/tex] + [tex]m_{w}[/tex][tex]C_{w}[/tex]
where [tex]T_{f}[/tex] is the final temperature of the water and the copper
substituting values, we have
99040 = (2 x 385 x [tex]T_{f}[/tex]) + (0.2 x 4182 x
99040 = 770[tex]T_{f}[/tex] + 836.4
99040 = 1606.4[tex]T_{f}[/tex]
[tex]T_{f}[/tex] = 99040/1606.4 = 61.65 °C
iven a 36.0 V battery and 14.0 Ω and 84.0 Ω resistors, find the current (in A) and power (in W) for each when connected in series.
Answer:
0.367A = Current of both resistors
For resistor 1: 1.89W; For resistor 2: 11.3W
Explanation:
When the resistors are connected in series, the equivalent resistance is the sum of both resistors, that is:
R = 14.0Ω + 84.0Ω = 98.0Ω
Using Ohm's law, we can find the current of the circuit (Is the same for both resistors):
V = RI
V / R = I
36.0V / 98.0Ω = I
0.367A = Current of both resistorsPower is defined as:
P = I²*R
For resistor 1:
P = 0.367A²*14.0Ω = 1.89W
For resistor 1:
P = 0.367A²*84.0Ω = 11.3W
Please help!
Much appreciated!
Answer:
Rp = 3.04×10² Ω.
Explanation:
From the question given:
1/Rp = 1/4.5×10² Ω + 1/ 9.4×10² Ω
Rp =?
We can obtain the value of Rp as follow:
1/Rp = 1/4.5×10² + 1/ 9.4×10²
Find the least common multiple (lcm) of 4.5×10² and 9.4×10².
The result is 4.5×10² × 9.4×10²
Next, divide the result of the lcm by each denominator and multiply the result obtained with the numerator as shown below:
1/Rp = (9.4×10² + 4.5×10²) /(4.5×10²) (9.4×10²)
1/Rp = 13.9×10²/4.23×10⁵
Cross multiply
Rp × 13.9×10² = 4.23×10⁵
Divide both side by 13.9×10²
Rp = 4.23×10⁵ / 13.9×10²
Rp = 3.04×10² Ω.
What is the maximum wavelength of incident light for which photoelectrons will be released from gallium
Answer:
292 nm
Explanation:
The work function of gallium ∅ = 94.25 eV = 6.81 x 10^-19 J
at maximum wavelength, the energy of the photons is equal to its work function
Energy of the electron = hf
but hf = hc/λ
where h is the planck's constant = 6.63 × 10-34 m^2 kg/s
c is the speed of light = 3 x 10^8 m/s
λ is the wavelength that this occurs, which is the maximum wavelength
Equating, we have
hc/λ = ∅
substituting, we have
(6.63 × 10-34 x 3 x 10^8)/λ = 6.81 x 10^-19
(1.989 x 10^-25)/(6.81 x 10^-19) = λ
λ = 292.07 x 10^-9 = 292 nm
A 10kg block with an initial velocity of 10 m/s slides 1o m across a horizontal surface and comes to rest. it takes the block 2 seconds to stop. The stopping force acting on the block is about
Answer:
-50N
Explanation:
F=ma=m(Vf-Vi)/t
m=10kgVf=0m/sVi=10m/st=2sF=(10)(-10)/(2)=-50N
So the force acting on the block is -50N, where the negative sign simply tells us that the force is opposite to the direction of movement.
A pair of narrow, parallel slits separated by 0.230 mm is illuminated by green light (λ = 546.1 nm). The interference pattern is observed on a screen 1.50 m away from the plane of the parallel slits.
A) Calculate the distance from the central maximum to the first bright region on either side of the central maximum.
B) Calculate the distance between the first and second dark bands in the interference pattern.
Answer:
A) y = 3.56 mm
B) y = 3.56 mm
Explanation:
A) The distance from the central maximum to the first bright region can be found using Young's double-slit equation:
[tex] y = \frac{m\lambda L}{d} [/tex]
Where:
λ: is the wavelength = 546.1 nm
m: is first bright region = 1
L: is the distance between the screen and the plane of the parallel slits = 1.50 m
d: is the separation between the slits = 0.230 mm
[tex] y = \frac{m\lambda L}{d} = \frac{1*546.1 \cdot 10^{-9} m*1.50 m}{0.230 \cdot 10^{-3} m} = 3.56 \cdot 10^{-3} m [/tex]
B) The distance between the first and second dark bands is:
[tex] \Delta y = \frac{\Delta m*\lambda L}{d} [/tex]
Where:
[tex] \Delta m = m_{2} - m_{1} = 2 - 1 = 1 [/tex]
[tex] \Delta y = \frac{1*546.1 \cdot 10^{-9} m*1.50 m}{0.230 \cdot 10^{-3} m} = 3.56 \cdot 10^{-3} m [/tex]
I hope it helps you!
What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?
Answer:
[tex]I=2.71\times 10^{-5}\ A[/tex]
Explanation:
A 6.0-cm-diameter parallel-plate capacitor has a 0.46 mm gap.
What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?
Let given is,
The diameter of a parallel plate capacitor is 6 cm or 0.06 m
Separation between plates, d = 0.046 mm
The potential difference across the capacitor is increasing at 500,000 V/s
We need to find the displacement current in the capacitor. Capacitance for parallel plate capacitor is given by :
[tex]C=\dfrac{A\epsilon_o}{d}\\\\C=\dfrac{\pi r^2\epsilon_o}{d}[/tex], r is radius
Let I is the displacement current. It is given by :
[tex]I=C\dfrac{dV}{dt}[/tex]
Here, [tex]\dfrac{dV}{dt}[/tex] is rate of increasing potential difference
So
[tex]I=\dfrac{\pi r^2\epsilon_o}{d}\times \dfrac{dV}{dt}\\\\I=\dfrac{\pi (0.03)^2\times 8.85\times 10^{-12}}{0.46\times 10^{-3}}\times 500000\\\\I=2.71\times 10^{-5}\ A[/tex]
So, the value of displacement current is [tex]2.71\times 10^{-5}\ A[/tex].
1.2miles=__________km
Answer:
1.931 kilometres is the answer of 1.2 miles
Answer and Explanation:
1 mile = 1.609 km
Set up a fraction to cancel the miles to get the kilometers.
[tex]\frac{1.2mi}{?km} *\frac{1.609}{1mi} = 1.9308km[/tex] <- This is the answer.
#teamtrees #PAW (Plant And Water)
How do you measure potential and kinetic energy?
Answer:
potential energy is a stored energy or energy of position (gravitational).
Kinetic energy is a energy of motion.
Explanation:
in the formula K is for the kinetic and the P stand for the potential.
Light of wavelength 476.1 nm falls on two slits spaced 0.29 mm apart. What is the required distance from the slits to the screen if the spacing between the first and second dark fringes is to be 4.2 mm?
Answer:
The distance is [tex]D = 2.6 \ m[/tex]
Explanation:
From the question we are told that
The wavelength of the light is [tex]\lambda = 476.1 \ nm = 476.1 *10^{-9} \ m[/tex]
The distance between the slit is [tex]d = 0.29 \ mm = 0.29 *10^{-3} \ m[/tex]
The between the first and second dark fringes is [tex]y = 4.2 \ mm = 4.2 *10^{-3} \ m[/tex]
Generally fringe width is mathematically represented as
[tex]y = \frac{\lambda * D }{d}[/tex]
Where D is the distance of the slit to the screen
Hence
[tex]D = \frac{y * d}{\lambda }[/tex]
substituting values
[tex]D = \frac{ 4.2 *10^{-3} * 0.29 *10^{-3}}{ 476.1 *10^{-9} }[/tex]
[tex]D = 2.6 \ m[/tex]