Answer:
400 volt
Explanation:
[tex]v \: = \: \frac{kq}{r} [/tex]
since the charge is constant
V1*r1 = V2*r2
V2 = V1r1/r2
= (200×4) ÷ 2
= 800/2
= 400 Volt
At a distance of 2 m, the potential will be "400 Volt".
Electric potential:The proportion of energy required to transport a unit charge from either a reference point just to a particular place throughout the presence of such an electrical field is considered as Electric potential.
According to the question,
Distance, r₁ = 4 m
r₂ = 2 m
Charge, V₁ = 200 V
V₂ = ?
We know the relation,
→ V₁ × r₁ = V₂ × r₂
or,
V₂ = [tex]\frac{V_1\times r_1}{r_2}[/tex]
By substituting the values, we get
= [tex]\frac{200\times 4}{2}[/tex]
= [tex]\frac{800}{2}[/tex]
= 400 Volt
Thus the answer above is correct.
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11. A seesaw sits in static equilibrium. A child with a mass of 30 kg sits 1 m away from a pivot point. Another child sits 0.75 m away from the pivot point on the opposite side. The second child's mass is _____ kg.
Answer:
40 kgExplanation:
Find the diagram relating to the question for proper explanation of the question below.
Using the principle of moment
Sum of clockwise moments = Sum of anticlockwise moments
Moment = Force * perpendicular distance
For anti-clockwise moment:
Since the 30 kg moves in the anticlockwise direction according to the diagram
ACW moment = 30 * 1 = 30 kgm
For clockwise moment
If another child sits 0.75 m away from the pivot point on the opposite side, moment of the child in clockwise direction = M * 0.75 = 0.75M (M is the mass of the unknown child).
Equating both moments we have;
0.75M = 30
M = 30/0.75
M = 40 kg
The second child's mass is 40 kg
a) If a proton moved from a location with a 5.0 V potential to a location with 7.5 V potential, would its potential energy increase or decrease? Explain.b) If an electron moved from a location with 7.5 V potential to a location with a 5.0 V potential, would its potential energy increase or decrease? Explain
Answer:
A. The potential energy of the positively charge proton from 5v to 7.5v will increase because in an electric field work is done on the proton to move it to a point of higher potential
B. when an electron is moved from 7.5v to 5v it loses potential energy due to the charges lost so as to bring it to a point of lower potential
Explanation:
If a proton moved from a location with a 5.0 V potential to a location with 7.5 V potential, the potential will increase since the potential difference is positive. Electric work will therefore be done on the proton to move it to a high potential.
In contrast, If a proton moved from a location with a 7.5 V potential to a location with 5.0 V potential, the potential will decrease since the potential difference is negative. Electric work will therefore be done on the proton to move it to a low potential.
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What is the sound intensity level in decibels? Use the usual reference level of I0 = 1.0×10−12 W/m2.
Answer:
L = 130 decibels
Explanation:
The computation of the sound intensity level in decibels is shown below:
According to the question, data provided is as follows
I = sound intensity = 10 W/m^2
I0 = reference level = [tex]1 \times 10-12 W/m^2[/tex]
Now
Intensity level ( or Loudness)is
[tex]L = log10 \frac{I}{10}[/tex]
[tex]L = log10 \frac{10}{1\times 10^{-12}}[/tex]
[tex]L = log10 \times 1013[/tex]
[tex]= 13 \times 1 ( log10(10) = 1)[/tex]
Therefore
L = 13 bel
And as we know that
1 bel = 10 decibels
So,
The Sound intensity level is
L = 130 decibels
Point X is midway between the charges. In what section of the line will there be a point where the resultant electric field is zero?
Answer:
I believe the answer is in fact section (VW) on the line where the electric field result will be zero.
Explanation:
The direction of the electric field due to a positive charge is away from it and the direction of the electric field due to a negative one is towards it.
Two carts are connected by a loaded spring on a horizontal, frictionless surface. The spring is released and the carts push away from each other. Cart 1 has mass M and Cart 2 has mass M/3.
a) Is the momentum of Cart 1 conserved?
Yes
No
It depends on M
b) Is the momentum of Cart 2 conserved?
Yes
No
It depends on M
c) Is the total momentum of Carts 1 and 2 conserved?
Yes
No
It depends on M
d) Which cart ends up moving faster?
Cart 1
Cart 2
They move at the same speed
e) If M = 6 kg and Cart 1 moves with a speed of 16 m/s, what is the speed of Cart 2?
0 m/s
4.0 m/s
5.3 m/s
16 m/s
48 m/s
64 m/s
Answer:
a) yes
b) no
c) yes
d)Cart 2 with mass [tex]\frac{M}{3}[/tex] is expected to be more faster
e) u₂ = 48 m/s
Explanation:
a) the all out linear momentum of an arrangement of particles of Cart 1 not followed up on by external forces is constant.
b) the linear momentum of Cart 2 will be acted upon by external force by Cart 1 with mass M, thereby it's variable and the momentum is not conserved
c) yes, the momentum is conserved because no external force acted upon it and both Carts share the same velocity after the reaction
note: m₁u₁ + m₂u₂ = (m₁ + m₂)v
d) Cart 2 with mass [tex]\frac{M}{3}[/tex] will be faster than Cart 1 because Cart 2 is three times lighter than Cart 1.
e) Given
m₁= M
u₁ = 16m/s
m₂ =[tex]\frac{M}{3}[/tex]
u₂ = ?
from law of conservation of momentum
m₁u₁= m₂u₂
M× 16 = [tex]\frac{M}{3}[/tex] × u₂(multiply both sides by 3)
therefore, u₂ = [tex]\frac{3(M .16)}{M}[/tex] ("." means multiplication)
∴u₂ = 3×16 = 48 m/s
A generator producing a varying current from 0 to 10 A was allowed to flow in a coil of magnetic field. After a time interval the current was observed to be 4 A. Describe how back EMF was enduced in self induction
Answer:
the emf generated by a magnetic coil of the generator, because it consists of a coil turning in a magnetic field; it opposes the voltage powering the motor, so at first the current reaches 10A but subsequently due to the back emf generated in the coil this emf opposes the self induced emf thereby reducing the current to 4A
Explanation:
Rafael models a longitudinal wave using a coiled-spring toy. If he wants to model a longitudinal wave that carries
more energy, how should he move the end of the spring?
Answer:
Rafeal should compress or pull the end of the spring to increase its energy of vibration.
Explanation:
Waves are phenomenon that causes a disturbance in a material medium without a permanent effect on the particles of the the medium. It had majorly two types; transverse and longitudinal.
Transverse waves are waves in which the direction of propagation of the wave generated is perpendicular to the direction of vibration of the particles of the medium. Examples are; water waves, light waves etc. But in longitudinal waves, the direction of propagation of the waves is the same as that of the particles of the medium. Examples are; waves in a slinky, waves in a spring etc.
Rafeal should compress the spring at the end or pull it at the end to increase its energy of vibration.
Describe the relationship between the density of electric field lines and the strength of the electric field?
Answer:
The greater the density of the electric field lines the stronger the electric field and vice versa
Explanation:
Electric field can be defined as the region where an electric force is experienced by a charged body. A charged body experiences a force whenever it is positioned close to another charged body.
An electric field may be described in terms of lines of force which represent the direction of a small positive charge placed at that point assuming that the charge is so small that it does not change appreciably in the presence of another charge. Arrows on the lines of force indicate the direction of the electric field.
The lines of force are indicated in such a way that the strength of the electric field is shown by the number or density of electric field lines crossing a unit area perpendicular to the lines. Hence, the greater the density of the electric field lines the stronger the the electric field and vice versa
An ultrasound machine uses 1.64 × 105 watts of power. If it draws 12.0 amps of current, what is the resistance?
Answer:
Explanation:
Power of electrical circuit = I² R where I is current and R is resistance
Putting the given data
1.64 x 10⁵ = 12² x R
R = 1.139 x 10³ ohm
= 1139 ohm .
Answer:
r=1140ohms
Explanation: plato family
Flower bed is filled with five types of flowers. Which placement of the flowers represents the highest entropy?
Answer:
B
Explanation:
ANSEWER :B IN ROWS ONLY
If, instead, the ball is revolved so that its speed is 3.7 m/s, what angle does the cord make with the vertical?
Complete Question:
A 0.50-kg ball that is tied to the end of a 1.5-m light cord is revolved in a horizontal plane, with the cord making a 30° angle with the vertical.
(a) Determine the ball’s speed. (b) If, instead, the ball is revolved so that its
speed is 3.7 m/s, what angle does the cord make with the vertical?
(Check attached image for the diagram.)
Answer:
(a) The ball’s speed, v = 2.06 m/s
(b) The angle the cord makes with the vertical is 50.40⁰
Explanation:
If the ball is revolved in a horizontal plane, it will form a circular trajectory,
the radius of the circle, R = Lsinθ
where;
L is length of the string
The force acting on the ball is given as;
F = mgtanθ
This above is also equal to centripetal force;
[tex]mgTan \theta = \frac{mv^2}{R} \\\\Recall, R = Lsin \theta\\\\mgTan \theta = \frac{mv^2}{Lsin \theta}\\\\v^2 = glTan \theta sin \theta\\\\v = \sqrt{glTan \theta sin \theta} \\\\v = \sqrt{(9.8)(1.5)(Tan30)(sin30)} \\\\v = 2.06 \ m/s[/tex]
(b) when the speed is 3.7 m/s
[tex]v = \sqrt{glTan \theta sin \theta} \ \ \ ;square \ both \ sides\\\\v^2 = glTan \theta sin \theta\\\\v^2 = gl(\frac{sin \theta}{cos \theta}) sin \theta\\\\v^2 = \frac{gl*sin^2 \theta}{cos \theta} \\\\v^2 = \frac{gl*(1- cos^2 \theta)}{cos \theta}\\\\gl*(1- cos^2 \theta) = v^2cos \theta\\\\(9.8*1.5)(1- cos^2 \theta) = (3.7^2)cos \theta\\\\14.7 - 14.7cos^2 \theta = 13.69cos \theta\\\\14.7cos^2 \theta + 13.69cos \theta - 14.7 = 0 \ \ \ ; this \ is \ quadratic \ equation\\\\[/tex]
[tex]Cos\theta = \frac{13.69\sqrt{13.69^2 -(-4*14.7*14.7)} }{14.7} \\\\Cos \theta = 0.6374\\\\\theta = Cos^{-1}(0.6374)\\\\\theta = 50.40 ^o[/tex]
Therefore, the angle the cord makes with the vertical is 50.40⁰
Newton’s first law says that if motion changes, then a force is exerted. Describe a collision in terms of the forces exerted on both objects.
Answer:
In collision between equal-mass objects, each object experiences the same acceleration, because of equal force exerted on both objects.
Explanation:
In a collision two objects, there is a force exerted on both objects that causes an acceleration of both objects. These forces that act on both objects are equal in magnitude and opposite in direction.
Thus, in collision between equal-mass objects, each object experiences the same acceleration, because of equal force exerted on both objects.
What is the relation between the direction of electric field and equipotential line at the same point
Answer:
The direction of electric field and equipotential line at the same point are always PERPENDICULAR TO THE ELECTRIC FIELD.
Explanation:
Equipotential surface is a three dimensional part of equipotential lines.
Equipotential lines are a type of contour lines that is use to trace lines that have the same altitude on the map and the altitude is the electric potential.
Equipotential lines are always perpendicular to electric potential because the lines creates three dimension equipotential surface.
The direction of the electric field and equipotential line at the same point is always perpendicular.
Equipotential lines:The electric field and the electric potential both are dependent on distance. As we move farther from the source generating the electric field, the electric field strength, as well as the electric potential strength, decreases.
The electric field is inversely proportional to the square of the distance from the source, while the electric potential is inversely proportional to the distance itself.
The equipotential surface is a surface on which every point is at the same distance from the source so that each point is at the same potential.
Now, we can make an equipotential line joining these points. Since it is fixed at a surface, we can not move farther or closer to the source, because there will be a change in the distance and the potential will change.
So these lines are always perpendicular to the lines representing the electric field, which travel towards or away from the source
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The Law of Biot-Savart shows that the magnetic field of an infinitesimal current element decreases as 1/r2. Is there anyway you could put together a complete circuit (any closed path of current-carrying wire) whose field exhibits this same 1/r^2 decrease in magnetic field strength? Explain your reasoning.
Answer and Explanation:
There is no probability of obtaining such a circuit of closed track current carrying wire whose field of magnitude displays i.e. [tex]B \alpha \frac{1}{r^2}[/tex]
The magnetic field is a volume of vectors
And [tex]\phi\ bds = 0[/tex]. This ensures isolated magnetic poles or magnetic charges would not exit
Therefore for a closed path, we never received magnetic field that followed the [tex]B \alpha \frac{1}{r^2}[/tex] it is only for the simple current-carrying wire for both finite or infinite length.
The average speed of a nitrogen molecule in air is about 6.70×102 m/s, and its mass is 4.68×10-26 kg.
(a) If it takes 3.00×10-13 s for a nitrogen molecule to hit a wall and rebound with the same speed but
moving in the opposite direction, what is the average acceleration of the molecule during this time
interval?
(b) What average force does the molecule exert on the wall?
(c) What is the total force exerted
on the wall if it is struck by 1015 such molecules all at once?
Answer:
a) a = 3.06 10¹⁵ m / s , b) F= 1.43 10⁻¹⁰ N, c) F_total = 14.32 10⁻²⁶ N
Explanation:
This exercise will average solve using the moment relationship.
a ) let's use the relationship between momentum and momentum
I = ∫ F dt = Δp
F t = m [tex]v_{f}[/tex] - m v₀
F = m (v_{f} -v₀o) / t
in the exercise indicates that the speed module is the same, but in the opposite direction
F = m (-2v) / t
if we use Newton's second law
F = m a
we substitute
- 2 mv / t = m a
a = - 2 v / t
let's calculate
a = - 2 4.59 10²/3 10⁻¹³
a = 3.06 10¹⁵ m / s
b) F= m a
F= 4.68 10⁻²⁶ 3.06 10¹⁵
F= 1.43 10⁻¹⁰ N
c) if we hit the wall for 1015 each exerts a force F
F_total = n F
F_total = n m a
F_total = 10¹⁵ 4.68 10⁻²⁶ 3.06 10¹⁵
F_total = 14.32 10⁻²⁶ N
Suppose the frequency of a note on an organ is 18 Hz. What is the shortest organ pipe with both ends open that will resonate at this frequency
Answer:
9.53 m
Explanation:
The computation of shortest organ pipe with both ends open that will resonate at this frequency is shown below:-
[tex]\lambda = \frac{velocity}{frequency}[/tex]
[tex]= \frac{343}{18}[/tex]
= 19.06 m
Now the
Shortest organ pipe with both ends open is
= [tex]\frac{\lambda}{2}[/tex]
[tex]= \frac{19.06}{2}[/tex]
= 9.53 m
Basically we applied the above formulas so that first we easily determined the shortest organ pipe for both ends at this frequency
can I get help please?
Which of the following cities has both the highest water vapor content AND lowest relative humidity?City A: Temperature = 90 °F, Dew Point Temperature = 75 °FCity B: Temperature = 70 °F, Dew Point Temperature = 65 °FCity C: Temperature = 20 °F, Dew Point Temperature = 10 °F
Answer:
CITY A has the highest water vapour and the least relative humidity
Explanation:
The dew point temperature is the temperature above which the water vapour will convert into liquid water.
The relative humidity is the ratio is the current water vapor content to the maximum amount of water that can be present in the air
Therefore although the temperature may be high yet the relative humidity can be considerably low
Hence, the correct option must be CITY A
A 1.0-kg ball is attached to the end of a 2.5-m string to form a pendulum. This pendulum is released from rest with the string horizontal. At the lowest point in its swing when it is moving horizontally, the ball collides elastically with a 2.0-kg block initially at rest on a horizontal frictionless surface. What is the speed of the block just after the collision
Answer:
[tex]v_{2}=3.5 m/s[/tex]
Explanation:
Using the conservation of energy we have:
[tex]\frac{1}{2}mv^{2}=mgh[/tex]
Let's solve it for v:
[tex]v=\sqrt{2gh}[/tex]
So the speed at the lowest point is [tex]v=7 m/s[/tex]
Now, using the conservation of momentum we have:
[tex]m_{1}v_{1}=m_{2}v_{2}[/tex]
[tex]v_{2}=\frac{1*7}{2}[/tex]
Therefore the speed of the block after the collision is [tex]v_{2}=3.5 m/s[/tex]
I hope it helps you!
Two spaceships are observed from earth to be approaching each other along a straight line. Ship A moves at 0.40c relative to the earth observer, while ship B moves at 0.60c relative to the same observer. What speed does the captain of ship A report for the speed of ship B
Answer:
0.80 c
Explanation:
The computation of speed is shown below:-
Here, The speed of the captain ship A report for speed of the ship B which is
[tex]S = \frac{S_A + S_B}{1 + \frac{(S_AS_B)}{c^2} }[/tex]
where
[tex]S_A[/tex] indicates the speed of the ship A
[tex]S_B[/tex] indicates the speed of the ship B
and
C indicates the velocity of life
now we will Substitute 0.40c for A and 0.60 for B in the equation which is
[tex]S = \frac{0.40c + 0.60c}{1 + \frac{(0.40c)(0.60c)}{c^2} }[/tex]
after solving the above equation we will get
0.80 c
So, The correct answer is 0.80c
A skier goes down a slope and detaches from the ground moving in the horizontal direction with a speed of 25m / s. The slope has an inclination of 35 °
a) At what point does the skier make contact again with the ground?
Answer:
107 m down the incline
Explanation:
Given:
v₀ₓ = 25 m/s
v₀ᵧ = 0 m/s
aₓ = 0 m/s²
aᵧ = -10 m/s²
-Δy/Δx = tan 35°
Find: d
First, find Δy and Δx in terms of t.
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (0 m/s) t + ½ (-10 m/s²) t²
Δy = -5t²
Δx = v₀ₓ t + ½ aₓ t²
Δx = (25 m/s) t + ½ (0 m/s²) t²
Δx = 25t
Substitute:
-(-5t²) / (25t) = tan 35°
t/5 = tan 35°
t = 5 tan 35°
t ≈ 3.50 s
Now find Δy and Δx.
Δy ≈ -61.3 m
Δx ≈ 87.5 m
Therefore, the distance down the incline is:
d = √(x² + y²)
d ≈ 107 m
A small motor is mounted on the axis of a space probe with its rotor (the rotating part of the motor) parallel to the axis of the probe. Its function is to control the rotational position of the probe about the axis. The moment of inertia of the probe is [07] times that of the rotor. Initially, the probe and rotor are at rest. The motor is turned on and after some period of time, the probe is seen to have rotated by positive 32.6 degrees. Through how many revolutions has the rotor turned
Answer:
0.634 rev
Explanation:
It is stated that,
The moment of inertia of the probe = 7 times the moment of inertia of the rotor
After running for some time, the probe is seen to have rotated through positive 32.6°.
According to the laws of conservation of angular momentum, the angular momentum of the probe must be equal to the angular momentum of the rotor.
==> Irωr = Ipωp ....... equ 1
integrating the angular speed ω with respect to time t leaves us with
Ir∅r = Ip∅p ...... equ 2
where,
Ir = moment of inertia of the rotor
ωr = angular speed of the rotor
Ip = moment of inertia of the rotor
ωp = angular speed of the probe
∅r = angular position of the rotor
∅p = angular position of the probe
equ 2 can be rewritten as
Ip/Ir = ∅r/∅p
from the statement, Ip/Ir = 7
therefore,
7 = ∅r/∅p = ∅r/32.6
∅r = 7 x 32,6 = 228.2°
converting to rev = [tex]\frac{228.2}{360 }[/tex] = 0.634 rev
A 8.00-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of 1.70 s. Find the force constant of the spring.
Answer:
109.32 N/m
Explanation:
Given that
Mass of the hung object, m = 8 kg
Period of oscillation of object, T = 1.7 s
Force constant, k = ?
Recall that the period of oscillation of a Simple Harmonic Motion is given as
T = 2π √(m/k), where
T = period of oscillation
m = mass of object and
k = force constant if the spring
Since we are looking for the force constant, if we make "k" the subject of the formula, we have
k = 4π²m / T², now we go ahead to substitute our given values from the question
k = (4 * π² * 8) / 1.7²
k = 315.91 / 2.89
k = 109.32 N/m
Therefore, the force constant of the spring is 109.32 N/m
A 2.3kg bicycle wheel has a diameter of 50cm. What torque must you apply to take the wheel from 0rpm to 120rpm in 5.5s?
Answer:
τ = 0.26 N.m
Explanation:
First we find the moment of inertia of the wheel, by using the following formula:
I= mr²
where,
I = Moment of Inertia = ?
m = mass of wheel = 2.3 kg
r = radius of wheel = 50 cm/2 = 25 cm = 0.25 m
Therefore,
I = (2.3 kg)(0.25 m)²
I = 0.115 kg.m²
Now, we find the angular acceleration of the wheel:
α = (ωf - ωi)/t
where,
α = angular acceleration = ?
ωf = final angular velocity = (120 rpm)(2π rad/1 rev)(1 m/60 s) = 12.56 rad/s
ωi = Initial Angular Velocity = 0 rad/s
t = time = 5.5 s
Therefore,
α = (12.56 rad/s - 0 rad/s)/(5.5 s)
α = 2.28 rad/s²
Now, the torque is given as:
Torque = τ = Iα
τ = (0.115 kg.m²)(2.28 rad/s²)
τ = 0.26 N.m
A player is positioned 35 m[40 degrees W of S] of the net. He shoot the puck 25 m [E] to a teammate. What second displacement does the puck have to travel in order to make it to the net?
Answer:
x=22.57 m
Explanation:
Given that
35 m in W of S
angle = 40 degrees
25 m in east
From the diagram
The angle
[tex]\theta=90-40=50^o[/tex]
From the triangle OAB
[tex]cos40^o=\frac{35^2+25^2-x^2}{2\times 35\times 25}[/tex]
[tex]1340.57=35^2+25^2-x^2[/tex]
x=22.57 m
Therefore the answer of the above problem will be 22.57 m
If an object on Earth weighs 100N what is its weight in pounds?
Answer:
10.2 lbs
Explanation:
m=F/g
m=100N/9.8
m=10.2040816 lbs
The weight of the object in pounds will be 220.46 pounds m/s².
We have an object on Earth whose weight is 100 N.
We have to determine its weight in pounds.
Pounds is used to represent which Physical quantity ?Pounds is used to represent the mass of the body.
According tot the question -
Weight on earth = 100 N = 100 Kg . m/s²
1 Kg = 2.2046 Pounds
Therefore, the weight on earth in pounds will be = 100 x 2.2046 pounds m/s² = 220.46 pounds m/s².
Hence, the weight of the object in pounds will be 220.46 pounds m/s².
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If, the limits of the visible spectrum are approximately 3000 A.U. and 5000 A.U. respectively. Determine the angular breadth of the first order visible spectrum produced by a plane diffraction grating having 12000 lines per inch when light is incident normally on the grating.
Answer:
θ₁ = 0.04º , θ₂ = 0.00118º
Explanation:
The equation that describes the diffraction pattern of a network is
d sin θ = m λ
where the diffraction order is, in this case they indicate that the order
m = 1
θ = sin⁻¹ (λ / d)
Trfuvsmod ls inrsd fr ll red s SI units
d = 12000 line / inc (1 inc / 2.54cm) = 4724 line / cm
the distance between two lines we can look for it with a direct proportions rule
If there are 4724 lines in a centimeter, the distance for two hundred is
d = 2 lines (1 cm / 4724 line) = 4.2337 10⁻⁴ cm
let's calculate the angles
λ = 300 10-9 m
θ₁ = sin⁻¹ (300 10-9 / 4,2337 10-4)
θ₁ = sin⁻¹ (7.08 10-4)
θ₁ = 0.04º
λ = 5000
θ₂ = sin-1 (500 10-9 / 4,2337 10-4)
θ₂ = 0.00118º
Monochromatic light falls on two very narrow slits 0.048 mm apart. Successive fringes on a screen 6.50 m away are 8.5 cm apart near the center of the pattern. Determine the wavelength and frequency of the light.
Answer:
The wavelength is [tex]\lambda = 6.28 *10^{-7}=628 nm[/tex]
The frequency is [tex]f = 4.78 Hz[/tex]
Explanation:
From the question we are told that
The slit distance is [tex]d = 0.048 \ mm = 4.8 *0^{-5}\ m[/tex]
The distance from the screen is [tex]D = 6.50 \ m[/tex]
The distance between fringes is [tex]Y = 8.5 \ cm = 0.085 \ m[/tex]
Generally the distance between the fringes for a two slit interference is mathematically represented as
[tex]Y = \frac{\lambda * D}{d}[/tex]
=> [tex]\lambda = \frac{Y * d }{D}[/tex]
substituting values
[tex]\lambda = \frac{0.085 * 4.8*10^{-5} }{6.50 }[/tex]
[tex]\lambda = 6.27 *10^{-7}=628 nm[/tex]
Generally the frequency of the light is mathematically represented as
[tex]f = \frac{c}{\lambda }[/tex]
where c is the speed of light with values
[tex]c = 3.0 *10^{8} \ m/s[/tex]
substituting values
[tex]f = \frac{3.0*10^8}{6.28 *10^{-7}}[/tex]
[tex]f = 4.78 Hz[/tex]
What do energy advisors mean by the phrase "the greenest kilowatt is the one you never use?"
Answer:
"Energy deficiency, no coal-burning, no-cost mining pollution" is the correct answer.
Explanation:
“The greenest kilowatt-hour seems to be the one this really doesn't should use,” explained Joe Stepenovitch, co-owner as well as COO of something like the electricity IQ Group. Whether a kilowatt becomes generated is far less essential instead of not needing to do something with it.It, therefore, reduces operational costs, appeals to progressives and green-conscious consumers, prepares the business for impending emissions reductions policy caps, as well as coincides with you including an imminent future focused on renewable energy sources.At t = 0, a battery is connected to a series arrangement of a resistor and an inductor. If the inductive time constant is 36.0 ms, at what time is the rate at which energy is dissipated in the resistor equal to the rate at which energy is stored in the inductor's magnetic field?
Answer:
The rate at which energy is dissipated in the resistor is equal to the rate at which energy is stored in the inductor's magnetic field in 24.95 ms.
Explanation:
The energy stored in the inductor is given as
E₁ = ½LI²
The rate at which energy is stored in the inductor is
(dE₁/dt) = (d/dt) (½LI²)
Since L is a constant
(dE₁/dt) = ½L × 2I (dI/dt) = LI (dI/dt)
(dE₁/dt) = LI (dI/dt)
Rate of Energy dissipated in a resistor = Power = I²R
(dE₂/dt) = I²R
When the rate at which energy is dissipated in the resistor equal to the rate at which energy is stored in the inductor's magnetic field
(dE₁/dt) = (dE₂/dt)
OK (dI/dt) = I²R
L (dI/dt) = IR
Current in a this kind of series setup of inductor and resistor at any time, t, is given as
I = (V/R) (1 - e⁻ᵏᵗ)
k = (1/time constant) = (R/L)
(dI/dt) = (kV/R) e⁻ᵏᵗ = (RV/RL) e⁻ᵏᵗ = (V/L) e⁻ᵏᵗ
L (dI/dt) = IR
L [(V/L) e⁻ᵏᵗ] = R [(V/R) (1 - e⁻ᵏᵗ)
V e⁻ᵏᵗ = V (1 - e⁻ᵏᵗ)
e⁻ᵏᵗ = 1 - e⁻ᵏᵗ
2 e⁻ᵏᵗ = 1
e⁻ᵏᵗ = (1/2) = 0.5
e⁻ᵏᵗ = 0.5
In e⁻ᵏᵗ = In 0.5 = -0.69315
- kt = -0.69315
kt = 0.69315
k = (1/time constant)
Time constant = 36.0 ms = 0.036 s
k = (1/0.036) = 27.78
27.78t = 0.69315
t = (0.69315/27.78) = 0.02495 = 24.95 ms
Hope this Helps!!!