Answer:
The reason for the increase or decrease of the moon is due to the angular perception of the moon.
Explanation:
Also called lunar illusion, this phenomenon is due to the position in which the moon is, it can be at the zenith or on the horizon, both distances are different from each other with respect to the position of the person.
The zenith is the highest part of the sky and the horizon the lowest.
When there are landmarks such as trees, buildings or mountains on the horizon, the illusion of closeness is given and the illusion of distance is misinterpreted.
But when looking up at the sky as there is no reference point there will be a failure in the perception of size.
A 23 cm tall object is placed in front of a concave mirror with a radius of 37 cm. The distance of the object to the mirror is 86 cm. Calculate the focal length of the mirror.
Answer:
18.5 cm
Explanation:
From;
1/u + 1/v = 1/f
Where;
u= object distance = 86cm
image height = 23 cm
Radius of curvature = 37 cm
The radius of curvature (r) is the radius of the sphere of which the mirror forms a part.
Focal length (f) = radius of curvature (r)/2 = 37cm/2 = 18.5 cm
Therefore, the focal length of the mirror is 18.5 cm
Which of the following regions of the electromagnetic spectrum have longer wavelengths than visible light? 1. infrared radiation 2. ultraviolet radiation 3. microwave radiation
Answer:infrared radiation
Explanation:
Infrared radiation and microwave radiation of the electromagnetic spectrum have longer wavelengths than visible light.
What is electromagnetic wave?EM waves are another name for electromagnetic waves. When an electric field interacts with a magnetic field, electromagnetic waves are created. These electromagnetic waves make up electromagnetic radiations. It is also possible to say that electromagnetic waves are made up of magnetic and electric fields that are oscillating. The basic equations of electrodynamics, Maxwell's equations, have an answer in electromagnetic waves.
If we arrange electromagnetic wave with decrease in wavelength, we get:
Radio waves > microwave > Infrared > Visible light > Ultraviolet > X-rays > Gamma radiation.
Hence, Infrared radiation and microwave radiation of the electromagnetic spectrum have longer wavelengths than visible light.
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Consider two isolated spherical conductors each having net charge Q. The spheres have radii a and b, where b > a.
Which sphere has the higher potential?
1. the sphere of radius a
2. the sphere of radius b
3. They have the same potential
Answer:
1. the sphere of the radius a
Explanation:
Because the charge distribution for each case is spherically symmetric, we can choose a spher- ical Gaussian surface of radius r , concentric with the sphere in question.
So E = k (Q /r 2) (for r ≥ R ) , where R is the radius of the sphere being considered, either a or b .
With the choice of potential at r = ∞ being zero, the electric potential at any distance r from the center of the sphere can be expressed as V = - integraldisplay r ∞ E dr = k /Q r
(for r ≥ R ) .
On the spheres of radii a and b , we have V a = k (Q/ a)and V b = k (Q/ b), respectively.
So Since b > a , the sphere of radius a will have the higher potential.
Also recall Because E = 0 inside a conductor, the potential
Do you believe in ghost
Answer:
well its about our thinking but i do believe in ghost a little
Thomas and Lilian are walking down the street to get to the corner store. They walk 5 blocks up the street and turn right by the stop sign. Once they turn at the stop sign they continue walking for 8 more blocks. They make a left, walk 2 blocks and cross the street to arrive at the corner store. While there they purchase a few snacks, sit at the curb, and then walk back home where they originally started. Thomas and Lilian are discussing their walk in reference to their overall displacement and distance. They seem to be in disagreement about their journey. Thomas says their overall displacement and distance are both zero, because they are back where they started. Lilian thinks their total distance and displacement are greater than zero.
Which person do you most agree with?
You are not expected to actually calculate in order to solve this problem.
Answer:
Thomas is correct that the zero displacements
Lilian is right that the distance is greater than zero.
Explanation:
In this problem we have to be clear about the difference between displacement and distance.
The displacement is a vector, that is, it has a modulation and direction, in this case we can draw a vector for the outward trip and another vector for the return trip, both will have the same magnitude, but their directions are opposite, so the resulting vector is zero.
The distance is a scalar and its value coincides with the modulus of the distance vector, in our case the distance is d for the outward journey and d for the return journey, so the total distance is 2d, which is different from zero.
The two students have some reason, but neither complete,
The displacement is zero because it is a vector and
the distance is different from zero (2d) because it is a scalar
Thomas is correct that the zero displacements
Lilian is right that the distance is greater than zero.
Therefore I agree with both, because each one has a 50% of the reason
What is utilization of energy
Explanation:
Energy utilization focuses on technologies that can lead to new and potentially more efficient ways of using electricity in residential, commercial and industrial settings—as well as in the transportation sector
00
Use base units to check whether the following equations are balance
(a) pressure = depth x density gravitational field strength,
(b) energy = mass x (speed of light).
dod to molt solid
Answer:
In a column of fluid, pressure increases with depth as a result of the weight of the overlying fluid. Thus a column of fluid, or an object submerged in the fluid, experiences greater pressure at the bottom of the column than at the top. This difference in pressure results in a net force that tends to accelerate an object upwards.
The pressure at a depth in a fluid of constant density is equal to the pressure of the atmosphere plus the pressure due to the weight of the fluid, or p = p 0 + ρ h g , p = p 0 + ρ h g , 14.4
Granite: 2.70 × 10 32.70 × 10 3
Lead: 1.13 × 10 41.13 × 10 4
Iron: 7.86 × 10 37.86 × 10 3
Oak: 7.10 × 10 27.10 × 10 2
2. A parallel-plate capacitor has a capacitance of C. If the area of the plates is doubled and
the distance between the plates is doubled, what is the new capacitance?
A) C/4
B) C/2
C)C
D) 4C
(C)
Explanation:
The capacitance C of a parallel plate capacitor is given by
[tex]C = \epsilon_0 \dfrac{A}{d}[/tex]
Let C' be the new capacitance where the area and the plate separation distance are doubled. This gives us
[tex]C' = \epsilon_0\dfrac{A'}{d'} = \epsilon_0\left(\dfrac{2A}{2d}\right) = \epsilon_0 \dfrac{A}{d} = C[/tex]
what are the applications of pascal's principle
Explanation:
The applications are, hydraulic lift- to transmit equal pressure throughout a fluid.Hydraulic jack- used in the braking system of cars.use of a straw- to suck fluids, which goes because of air pressure.The question simply asks, where pressure can be applied. There are many others, such as lift pump.2.) Is it possible to have negative velocity but positive acceleration? If so, what would
this mean?
Answer:
Yes, yes it would
Explanation:
Fish is cold blooded animal
A 12-V battery is connected across a device with variable resisstance. As the resistance of the device increases, determine whether the following quantities increase, decrease, or remain unchanged. Indicate your answers with I, D, or U respectively.
a. The current through the device
b. The voltage across the device
c. The power consumed by the device
Given the solution [tex]y_{1}(x)[/tex] from EDO below, develop a second solution.
[tex]x\frac{d^{2}y }{dx^{2} } +3\frac{dy}{dx} -y=0,\\y_{1} (x)=1+\frac{x}{3} +\frac{x^{2} }{24} +\frac{x^{3} }{360} + ...[/tex]
We're given
[tex]\displaystyle y_1(x) = 1 + \frac x3 + \frac{x^2}{24} + \frac{x^3}{360} + \cdots = \sum_{n=0}^\infty a_nx^n[/tex]
so let's see if we can find a closed form for the n-th term's coefficient.
Notice that
[tex]\displaystyle a_0 = 1 \\\\ a_1 = \frac13 = \frac1{1\times3} \\\\ a_2 = \frac1{24} = \frac1{(1\times3) \times (2\times4)} \\\\ a_3 = \frac1{360} = \frac1{(1\times3) \times (2\times4) \times (3\times5)}[/tex]
If the pattern continues, the next few terms are likely
[tex]\displaystyle a_4 = \frac1{8640} = \frac1{(1\times3) \times (2\times4) \times (3\times5) \times (4\times6)} \\\\ a_5 = \frac1{302400} = \frac1{(1\times3) \times (2\times4) \times (3\times5) \times (4\times6) \times (5\times7)} \\\\ a_6 = \frac1{14515200} = \frac1{(1\times3) \times (2\times4) \times (3\times5) \times (4\times6) \times (5\times7) \times (6\times8)}[/tex]
which leads up to the n-th term,
[tex]\displaystyle a_n = \frac1{(1\times3) \times (2\times4) \times \cdots \times (n\times(n+2))} = \frac2{n!(n+2)!}[/tex]
where the numerator is multiplied by 2 in order to "complete" the factorial pattern in (n + 2)!.
So we have
[tex]\displaystyle y_1(x) = \sum_{n=0}^\infty \frac2{n!(n+2)!} x^n[/tex]
Now we use reduction of order to find a linearly independent solution of the form [tex]y_2(x) = v(x)y_1(x)[/tex], with derivatives
[tex]\displaystyle \frac{\mathrm dy_2}{\mathrm dx} = v(x) \frac{\mathrm dy_1}{\mathrm dx} + y_1(x) \frac{\mathrm dv}{\mathrm dx} \\\\ \frac{\mathrm d^2y_2}{\mathrm dx^2} = v(x) \frac{\mathrm d^2y_1}{\mathrm dx} + 2 \frac{\mathrm dv}{\mathrm dx} \frac{\mathrm dy_1}{\mathrm dx} + y_1(x) \frac{\mathrm d^2v}{\mathrm dx^2}[/tex]
Substitute [tex]y_2[/tex] and its derivatives into the DE, and simplify the resulting expression to get a DE in terms of v(x) :
[tex]\displaystyle x y_1 \frac{\mathrm d^2v}{\mathrm dx^2} + \left(2x\frac{\mathrm dy_1}{\mathrm dx}+3y_1\right)\frac{\mathrm dv}{\mathrm dx} + \left(x\frac{\mathrm d^2y_1}{\mathrm dx^2}+3\frac{\mathrm dy_1}{\mathrm dx}-y_1\right)v = 0[/tex]
but since we know [tex]y_1(x)[/tex] satisfies the original DE, the last term vanishes and we're left with
[tex]\displaystyle x y_1 \frac{\mathrm d^2v}{\mathrm dx^2} + \left(2x\frac{\mathrm dy_1}{\mathrm dx}+3y_1\right)\frac{\mathrm dv}{\mathrm dx} = 0[/tex]
Reduce the order by substituting [tex]w(x)=\dfrac{\mathrm dv}{\mathrm dx}[/tex] to get yet another DE in w(x) :
[tex]\displaystyle x y_1 \frac{\mathrm dw}{\mathrm dx} + \left(2x\frac{\mathrm dy_1}{\mathrm dx}+3y_1\right)w = 0[/tex]
This equation is separable:
[tex]\displaystyle \frac{\mathrm dw}w = - \frac{2x\frac{\mathrm dy_1}{\mathrm dx}+3y_1}{xy_1}\,\mathrm dx \\\\ \frac{\mathrm dw}w = -\left(\frac2{y_1}\frac{\mathrm dy_1}{\mathrm dx} + \frac3x\right)\,\mathrm dx[/tex]
From here you would integrate to solve for w(x), then integrate again to solve for v(x), and finally for [tex]y_2(x)[/tex] by multiplying [tex]y_1(x)[/tex] by v(x). Using the fundamental theorem of calculus, you would find
[tex]\displaystyle \ln|w| = -2 \int_1^x \frac{{y_1}'(\xi)}{y_1(\xi)}\,\mathrm d\xi - 3\ln|x| + C_1 \\\\ w = \frac{C_1}{x^3} \exp\left(-2 \int_1^x \frac{{y_1}'(\xi)}{y_1(\xi)} \,\mathrm d\xi\right)\right) \\\\ v = C_1 \int_1^x \frac1{\omega^3} \exp\left(-2 \int_1^\omega \frac{{y_1}'(\xi)}{y_1(\xi)}\,\mathrm d\xi\right) \,\mathrm d\omega + C_2[/tex]
so that you end up with
[tex]\displaystyle y_2(x) = C_1 y_1(x) \int_1^x \frac1{\omega^3} \exp\left(-2 \int_1^\omega \frac{{y_1}'(\xi)}{y_1(\xi)}\,\mathrm d\xi\right) \,\mathrm d\omega + C_2y_1(x)[/tex]
But the second term is already accounted for by [tex]y_1(x)[/tex] itself, so the second solution is
[tex]\displaystyle y_2(x) = \boxed{y_1(x) \int_1^x \frac1{\omega^3} \exp\left(-2 \int_1^\omega \frac{{y_1}'(\xi)}{y_1(\xi)}\,\mathrm d\xi\right) \,\mathrm d\omega}[/tex]
You could go the extra mile and try to find a power series expression for this solution, but that's a lot of work for little payoff IMO.
Air bags greatly reduces the chance og injury in a car accident.explain how they do si in terms of energy transfer
Answer:
in an accident, when the body collides with the air bags, the collision time of impact between the two bodies will increase due to the presence of air bags in the car. Larger is the impact time smaller is the transformation of energy between the body and air bag. That is why air bags greatly reduce the chance of injury in a car accident.
a car moves for 10 minutes and travels 5,280 meters .What is the average speed of the car?
Answer:use the formular distance over time i.e distance/time. Make sure to convert the distance from metres to kilometers and time from minutes to hours .
Explanation:
The average speed of the car is 31,680 meters per hour.
To calculate the average speed of the car, you need to divide the total distance traveled by the time it took to travel that distance.
Given:
Time taken (t) = 10 minutes = 10 minutes × (1 hour / 60 minutes) = 10/60 hours = 1/6 hours
Distance traveled (d) = 5,280 meters
Average Speed (v) = Distance (d) / Time (t)
Average Speed (v) = 5280 meters / (1/6) hours
To simplify, when you divide by a fraction, it's equivalent to multiplying by its reciprocal:
Average Speed (v) = 5280 meters × (6/1) hours
Average Speed (v) = 31,680 meters per hour
Hence, the average speed of the car is 31,680 meters per hour.
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Take an electric field sensor and move it in a straight line, crossing the equipotential lines. Describe the relationship between the distance between the equipotential lines and the strength of the electric field.
Answer:
E = - dV / dx
Explanation:
The equipotential lines are lines or surfaces that have the same power, therefore we can move in them without carrying out work between equipotential lines, work must be carried out, therefore the electric field changes.
The electric field and the potential are related by
E = - dV / dx
therefore when the change is faster, that is, the equipotential lines are closer, the greater the electric field must be.
A transformer consists of a 500-turn primary coil and a 2000-turn secondary coil. If the current in the secondary is 3.0 A, what is the current in the primary
Answer:
12AExplanation:
Formula for calculating the relationship between the electromotive force (emf), current and number of turns of a coil in a transformer is expressed as shown:
[tex]\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex] where;
Vs and Vp are the emf in the secondary and primary coil respectively
Ns and Np are the number if turns in the secondary and primary coil respectively
Ip and Is are the currents in the secondary and primary coil respectively
Since the are all equal to each other, then we can equate any teo of the expression as shown;
[tex]\dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]
Given parameters
Np = 500-turns
Ns = 2000-turns
Is = 3.0Amp
Required
Current in the primary coil (Ip)
Using the relationship [tex]\dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]
[tex]I_p = \dfrac{N_sI_s}{N_p}[/tex]
[tex]I_p = \dfrac{2000*3}{500} \\\\I_p = \frac{6000}{500}\\ \\I_p = 12A\\[/tex]
Hence the current in the primary coil is 12Amp
A simple pendulum is 3.00 m long. (a) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating upward at 3.00 m/s2? s (b) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating downward at 3.00 m/s2? s (c) What is the period of this pendulum if it is placed in a truck that is accelerating horizontally at 3.00 m/s2? s
Answer:
a,)3.042s
b)4.173s
c)3.281s
Explanation:
For a some pendulum the period in seconds T can be calculated using below formula
T=2π√(L/G)
Where L = length of pendulum in meters
G = gravitational acceleration = 9.8 m/s²
Then we are told to calculate
(a) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating upward at 3.00 m/s2?
Since oscillations for this pendulum is located in the elevator that is accelerating upward at 3.00 then
use G = 9.8 + 3.0 = 12.8 m/s²
Period T=2π√(L/G)
T= 2π√(3/12.8)
T=3.042s
b) (b) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating downward at 3.00 m/s2?
G = 9.8 – 3.0 = 6.8 m/s²
T= 2π√(3/6.8)
T=4.173s
C)(c) What is the period of this pendulum if it is placed in a truck that is accelerating horizontally at 3.00 m/s2?
Net acceleration is
g'= √(g² + a²)
=√(9² + 3²)
Then period is
T=2π√(3/11)
T=3.281s
A hydraulic system is being used to lift a 1500-kg car. If the large piston under the car has a diameter of 50 cm, the small piston has a diameter of 4.0 cm, and the car is lifted a distance of 1.3 m, how much work is done on the car
Answer:
W = 122.3 J
Explanation:
First, we need to find out the force applied to the smaller piston. We know that the pressure applied to smaller piston must be equally transmitted to the larger piston. Therefore,
P₁ = P₂
F₁/A₁ = F₂/A₂
F₂ = F₁(A₂/A₁)
where,
F₁ = Force of Larger Piston = Weight of car = mg = (1500 kg)(9.8 m/s²)
F₁ = 14700 N
F₂ = Force applied to smaller piston = ?
A₁ = Area of larger piston = πd₁²/4
A₂ = Area of smaller piston = πd₂²/4
Therefore,
F₂ = (14700 N)[(πd₂²/4)/(πd₁²/4)]
F₂ = (14700 N)(d₂²/d₁²)
where,
d₁ = diameter of large piston = 50 cm
d₂ = diameter of small piston = 4 cm
Therefore,
F₂ = (14700 N)[(4 cm)²/(50 cm)²]
F₂ = 94.08 N
Now, for the work done on the car:
Work Done = W = F₂ d
where,
d = displacement of car = 1.3 m
Therefore,
W = (94.08 N)(1.3 m)
W = 122.3 J
A 2100 kg truck traveling north at 38 km/h turns east and accelerates to 55 km/h. (a) What is the change in the truck's kinetic energy
Answer:
Change in kinetic energy (ΔKE) = 12.8 × 10⁴ J
Explanation:
Given:
Mass of truck(m) = 2,100 kg
Initial speed(v1) = 38 km/h = 38,000 / 3600 = 10.56 m/s
Final speed(v2) = 55 km/h = 55,000 / 3600 = 15.28 m/s
Find:
Change in kinetic energy (ΔKE)
Computation:
Change in kinetic energy (ΔKE) = 1/2(m)[v2² - v1²]
Change in kinetic energy (ΔKE) = 1/2(2100)[15.28² - 10.56²]
Change in kinetic energy (ΔKE) = 1,050[233.4784 - 111.5136]
Change in kinetic energy (ΔKE) = 1,050[121.9648]
Change in kinetic energy (ΔKE) = 128063.04
Change in kinetic energy (ΔKE) = 12.8 × 10⁴ J
during the course of songraphic exam, you notice lateral splaying of echoes in the far field. what can you do to improve the image
Answer:
lateral splaying of echoes in the far field can be improved by Increasing the maximum number of transmit focal zones and optimize their location.
A straight wire that is 0.56 m long is carrying a current of 2.6 A. It is placed in a uniform magnetic field, where it experiences a force of 0.24 N. The wire makes an angle of 900 with the magnetic field. What is the magnitude of the magnetic field
Answer:
0.165TeslaExplanation:
The Force experienced by the wire in the uniform magnetic field is expressed as F = BILsin∝ where;
B is the magnetic field (in Tesla)
I is the current (in amperes)
L is the length of the wire (in meters)
∝ is the angle that the conductor makes with the magnetic field.
Given parameters
L = 0.56 m
I = 2.6A
F = 0.24N
∝ = 90°
Required
magnitude of the magnetic field (B)
Substituting the given values into the formula given above we will have;
F = BILsin∝
0.24 = B * 2.6 * 0.56 sin90°
0.24 = B * 2.6 * 0.56 (1)
0.24 = 1.456B
1.456B = 0.24
Dividing both sides by 1.456 will give;
1.456B/1.456 = 0.24/1.456
B ≈ 0.165Tesla
Hence the magnitude of the magnetic field is approximately 0.165Tesla
lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A in a circular orbit of radius 5000 km and period 4.0 hours. What is the radius of Planet R47A
[tex]2.6×10^6\:\text{m}[/tex]
Explanation:
The acceleration due to gravity g is defined as
[tex]g = G\dfrac{M}{R^2}[/tex]
and solving for R, we find that
[tex]R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)[/tex]
We need the mass M of the planet first and we can do that by noting that the centripetal acceleration [tex]F_c[/tex] experienced by the satellite is equal to the gravitational force [tex]F_G[/tex] or
[tex]F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)[/tex]
The orbital velocity v is the velocity of the satellite around the planet defined as
[tex]v = \dfrac{2\pi r}{T}[/tex]
where r is the radius of the satellite's orbit in meters and T is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as
[tex]\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}[/tex]
Solving for M, we get
[tex]M = \dfrac{4\pi^2 r^3}{GT^2}[/tex]
Putting this expression back into Eqn(1), we get
[tex]R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}[/tex]
[tex]\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}[/tex]
[tex]\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}[/tex]
[tex]\:\:\:\:= 2.6×10^6\:\text{m}[/tex]
A metal cube with sides of length a=1cm is moving at velocity v0→=1m/sj^ across a uniform magnetic field B0→=5Tk^. The cube is oriented so that four of its edges are parallel to its direction of motion (i.e., the normal vectors of two faces are parallel to the direction of motion).Find E, the magnitude of the induced electric field inside the cube. Express your answer numerically, in newtons per coulomb.
Answer:
the magnitude of the electric field is 1.25 N/C
Explanation:
The induced emf in the cube ε = LB.v where B = magnitude of electric field = 5 T , L = length of side of cube = 1 cm = 0.01 m and v = velocity of cube = 1 m/s
ε = LB.v = 0.01 m × 5 T × 1 m/s = 0.05 V
Also, induced emf in the cube ε = ∫E.ds around the loop of the cube where E = electric field in metal cube
ε = ∫E.ds
ε = Eds since E is always parallel to the side of the cube
= E∫ds ∫ds = 4L since we have 4 sides
= E(4L)
= 4EL
So,4EL = 0.05 V
E = 0.05 V/4L
= 0.05 V/(4 × 0.01 m)
= 0.05 V/0.04 m
= 1.25 V/m
= 1.25 N/C
So, the magnitude of the electric field is 1.25 N/C
The magnitude of the electric field is 1.25 N/C
Calculation of the magnitude of the electric field:But before that the following calculations need to be done.
ε = LB.v = 0.01 m × 5 T × 1 m/s
= 0.05 V
Now
ε = ∫E.ds
here ε = Eds because E is always parallel to the side of the cube
So,
= E∫ds ∫ds
= 4L so we have 4 sides
Now
= E(4L)
= 4EL
So,4EL = 0.05 V
Now
E = 0.05 V/4L
= 0.05 V/(4 × 0.01 m)
= 0.05 V/0.04 m
= 1.25 V/m
= 1.25 N/C
hence, The magnitude of the electric field is 1.25 N/C
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explain how does a convex lens produce real image from virtual object
Answer:
Real images are formed where rays of light actually converge, whereas virtual images occur with they are perceived to converge. Real images can be produced by passing light through converging lenses or with a concave mirror of some sort.
Explanation:
hope that helped a little :)
Real images are formed when the object is located beyond the focal length from the lens. A virtual image is formed if the object is place between focal length and pole of the converging lens.
What is convex lens?The converging lens which forms the real images by the intersection of light rays.
Real images are produced by light passing through converging lens or a concave mirror.
The real meeting of the rays coming from the focal point of the rays after reflection. Real images are formed where rays of light actually converge, whereas virtual images occur with they are appear to converge. Real images are formed when the object is located beyond the focal length from the lens. A virtual image is formed if the object is place between focal length and pole of the converging lens.
Thus, a convex lens produce real image from virtual object.
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How much charge is stored on two parallel-plate capacitors by the 12V battery if one is filled with air and the other is filled with a dielectric (k=3.00)
The question is incomplete, the complete question is;
how much charge is stored on the parallel-plate capacitors by the 12.0 V battery? One is filled with air, and the other is filled with a dielectric for which k = 3.00; both capacitors have a plate area of 5.00×10 −3 m 2 and a plate separation of 2.00 mm.
The capacitance of the capacitor is the quantity of charge stored by the capacitor.
Given that;
C1= εo k * A/d
εo = permittivity of free space
C1 = 8.85 x 10-12 farad per meter *1 * 5.00×10 −3 m 2/2 * 10^-3
= 2.21 * 10^-11 F
C2 = 8.85 x 10-12 * 3 * 5.00×10 −3 m 2/2 * 10^-3
= 6.63 * 10^-11 F
q1 = C1V1 = 2.21 * 10^-11 C * 12 V
= 2.65 * 10^-10 C
q2 = C2V2 = 6.63 * 10^-11 F * 12 V
= 7.96 * 10^-10 C
qtotal = 2.65 * 10^-10 C + 7.96 * 10^-10 C
qtotal = 1.061 * 10^-9C
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A 2-slit arrangement with 60.3 μm separation between the slits is illuminated with 482.0 nm light. Assuming that a viewing screen is located 2.14 m from the slits, find the distance from the first dark fringe on one side of the central maximum to the second dark fringe on the other side. A. 24.1 mm B. 34.2 mm C. 68.4 mm D. 51.3 mm
Answer:
The distance is [tex]y = 0.03425 \ m[/tex]
Explanation:
From the question we are told that
The distance of separation is [tex]d = 60.3 \mu m= 60.3 *10^{-6}\ m[/tex]
The wavelength is [tex]\lambda = 482.0 \ nm = 482.0 *10^{-9} \ m[/tex]
The distance of the screen is [tex]D = 2.14 \ m[/tex]
Generally the distance of a fringe from the central maxima is mathematically represented as
[tex]y = [m + \frac{1}{2} ] * \frac{\lambda * D}{d}[/tex]
For the first dark fringe m = 0
[tex]y_1 = [0 + \frac{1}{2} ] * \frac{482*10^{-9} * 2.14}{ 60.3*10^{-6}}[/tex]
[tex]y_1 = 0.00855 \ m[/tex]
For the second dark fringe m = 1
[tex]y_2 = [1 + \frac{1}{2} ] * \frac{482*10^{-9} * 2.14}{ 60.3*10^{-6}}[/tex]
[tex]y_2 = 0.0257 \ m[/tex]
So the distance from the first dark fringe on one side of the central maximum to the second dark fringe on the other side is
[tex]y = y_1 + y_2[/tex]
[tex]y = 0.00855 + 0.0257[/tex]
[tex]y = 0.03425 \ m[/tex]
Good morning 2 all ,What is mechanical advantage write its formula. Have a good day thank you ✌
The ratio of foort dustance to load distance in a simple machine is called mechanical advantage or MA.
MA= Effort Distance / Load Distance
Which statement belongs to Dalton’s atomic theory? Atoms have a massive, positively charged center. Atoms cannot be created or destroyed. Atoms can be broken down into smaller pieces. Electrons are located in energy levels outside of the nucleus.
Answer:
the correct statement is
* atoms cannot be created or destroyed
Explanation:
The Datlon atomic model was proposed in 1808 and represents atoms as the smallest indivisible particle of matter, they were the building blocks of matter and are represented by solid spheres.
Based on the previous descriptive, the correct statement is
* atoms cannot be created or destroyed
Answer:
the Answer is b hope it help
Explanation:
A battery establishes a voltage V on a parallel-plate capacitor. After the battery is disconnected, the distance between the plates is doubled without loss of charge. Accordingly:_____.
a. stay same
b. increases
c. decreases
d. the capacitance decreases and the voltage between the plates increases.
Answer:
d.
Explanation:
Since, the capacitance( decreases )
therefore voltage between the plates(increases ).
Hence, option d is correct.
C =εA/d.
d is doubled, therefore C decrease ( inverse relation).
D) The capacitance decreases and the voltage between the plates increases.
BatteryA battery establishes a voltage V on a parallel-plate capacitor. After the battery is disconnected, the distance between the plates is doubled without loss of charge. Accordingly, the capacitance decreases and the voltage between the plates increases.
The capacitance - (decreases)
The voltage between the plates- (increases ).
Thus, the correct answer is D.
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