Explanation:
bro I have no idea fam......
Rachel has good distant vision but has a touch of presbyopia. Her near point is 0.60 m. Part A When she wears 2.0 D reading glasses, what is her near point
Answer:
The right answer is "0.273 m".
Explanation:
Given:
Power (P),
[tex]\frac{1}{f} = 2D[/tex]
Near point,
u = 0.6 m
As we know,
⇒ [tex]\frac{1}{v} -\frac{1}{u}=\frac{1}{f} = 2[/tex]
By substituting the values, we get
⇒ [tex]\frac{1}{v} -\frac{1}{0.6} =2[/tex]
[tex]\frac{1}{v}=2+\frac{1}{0.6}[/tex]
[tex]\frac{1}{v} =\frac{1.2+1}{0.6}[/tex]
[tex]\frac{1}{v}=\frac{2.2}{0.6}[/tex]
By applying cross-multiplication, we get
[tex]0.6=2.2 \ v[/tex]
[tex]v = \frac{0.6}{2.2}[/tex]
[tex]S_{near} = 0.273 \ m[/tex]
Give reason why think before you use a simple cell ?
The disadvantages of simple cell are: It is not rechargeable. The battery needs to be disposed of after all the power has been used up. It can't produce electricity anymore. That is why, why think before you use a simple cell.
What are the benefits and drawbacks of simple cell?A battery designed to be used only once is called a simple cell. Small gadgets used in the house are frequently powered by simple cells.
The benefits of a simple cell include:
A simple cell can be used to power small electronic devices because of its modest size. (Games, lightsabers, radios on the go, cameras, hearing aids)Simple cell electrolyte is not very detrimental to the environment.Simple cells are reasonably priced.Among the drawbacks of a simple cell are:
The biggest drawback of a simple cell is that once it runs out of electricity, it cannot be replenished.Learn more about cell here:
https://brainly.com/question/30046049
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A simple pendulum consists of a ball of mass 3 kg hanging from a uniform string of mass 0.05 kg and length L. If the period of oscillation of the pendulum is 2 s, determine the speed of a transverse wave in the string when the pendulum hangs vertically.
Answer:
v = 3.12 m/s
Explanation:
First, we will find the length of the string by using the formula of the time period:
[tex]T = 2\pi \sqrt{\frac{l}{g}}\\\\l = \frac{T^2g}{4\pi^2}\\\\[/tex]
where,
l = length of string = ?
T = time period = 2 s
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]l = \frac{(2\ s)^2(9.81\ m/s^2)}{4\pi^2}\\\\l = 0.99\ m[/tex]
Now, we will find tension in the string in the vertical position through the weight of the ball:
T = W = mg = (3 kg)(9.81 m/s²)
T = 29.43 N
Now, the speed of the transverse wave is given as follows:
[tex]v=\sqrt{\frac{Tl}{m}}\\\\v=\sqrt{\frac{(29.43\ N)(0.99\ m)}{3\ kg}}\\\\[/tex]
v = 3.12 m/s
A 1.5kg block slides along a frictionless surface at 1.3m/s . A second block, sliding at a faster 4.3m/s , collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.0m/s . What was the mass of the second block?
Answer:
The mass of the second block=0.457 kg
Explanation:
We are given that
m1=1.5 kg
v1=1.3m/s
v2=4.3 m/s
V=2.0 m/s
We have to find the mass of the second block.
[tex]m_1v_1+m_2v_2=(m_1+m_2)V[/tex]
Let m2=m
Substitute the values
[tex]1.5(1.3)+m(4.3)=(1.5+m)(2)[/tex]
[tex]1.95+4.3m=3+2m[/tex]
[tex]4.3m-2m=3-1.95[/tex]
[tex]2.3m=1.05[/tex]
[tex]m=\frac{1.05}{2.3}[/tex]
[tex]m=0.457 kg[/tex]
Hence, the mass of the second block=0.457 kg
A string has its 4th harmonic at 31.5 Hz. What is the frequency of its third harmonic?
Answer:
The answer would be 7.5 Hz.
3. Some guitarists like the feel of a set of strings that all have the same tension. For such a guitar, the G string (196 Hz) has a mass density of 0.31 g/m. What is the mass density of the A string (110 Hz)
Answer:
0.98 g/m
Explanation:
Note: Since Tension and frequency are constant,
Applying,
F₁²M₁ = F₂²M₂............... Equation 1
Where F₁ = Frequency of the G string, F₂ = Frequency of the A string, M₁ = mass density of the G string, M₂ = mass density of the A string.
make M₂ the subject of the equation
M₂ = F₁²M₁/F₂²............... Equation 2
From the question,
Given: F₁ = 196 Hz, M₁ = 0.31 g/m, F₂ = 110 Hz
Substitute these values into equation 2
M₂ = 196²(0.31)/110²
M₂ = 0.98 g/m
Express 6revolutions to radians
Answer:
About 37.70 radians.
Explanation:
1 revolution = 2[tex]\pi[/tex] radians
∴ 6 revolutions = (6)(2[tex]\pi[/tex] radians)
6 revolutions = 37.6991 or ≈ 37.70 radians
just me or does brainly just want people to watch ads or pay for answers that are sometimes wrong dont get it
Answer:
i think bro is becuse they want to get money also paying to this app if i put paying right im latin but that is my opinion and also becuse they want we learn somethings
Explanation:
A strontium vapor laser beam is reflected from the surface of a CD onto a wall. The brightest spot is the reflected beam at an angle equal to the angle of incidence. However, fringes are also observed. If the wall is 1.2 m from the CD, and the second bright fringe is 0.803 m from the central maximum, what is the spacing (in m) of grooves on the CD
Answer:
[tex]d=1.29*10^{-6}m[/tex]
Explanation:
From the question we are told that:
Distance of wall from CD [tex]D=1.4[/tex]
Second bright fringe [tex]y_2= 0.803 m[/tex]
Let
Strontium vapor laser has a wavelength \lambda= 431 nm=>431 *10^{-9}m
Generally the equation for Interference is mathematically given by
[tex]y=frac{n*\lambda*D}{d}[/tex]
Where
[tex]d=\frac{n*\lambda*D}{y}[/tex]
[tex]d=\frac{2*431 *10^{-9}m*1.4}{0.803}[/tex]
[tex]d=1.29*10^{-6}m[/tex]
01.04 Law of Conservation of Energy
science question
Answer:
law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time.
What is hydroelectric power ?
Answer quickly..!
Answer:
It's electricity produced from hydropower. It's also a form of energy that controls the power of water motion.
Explanation:
One pro about hydroelectric power is that it's renewable energy. But one con about hydroelectric power is that it can impact the environment in a negative way.
A wire, 0.60 m in length, is carrying a current of 2.0 A and is placed at a certain angle with respect to the magnetic field of strength 0.30 T. If the wire experiences a force of 0.18 N, what angle does the wire make with respect to the magnetic field
Answer:
[tex]\theta=30 \textdegree[/tex]
Explanation:
From the question we are told that:
Current [tex]I=2.0A[/tex]
Length [tex]L=0.60m[/tex]
Magnetic field [tex]B=0.30T[/tex]
Force [tex]F=0.18N[/tex]
Generally the equation for Force is mathematically given by
[tex]F = BIL sin\theta[/tex]
[tex]sin\theta=\frac{F}{BIL}[/tex]
[tex]\theta=sin^{-1}\frac{0.18}{0.3*2*0.6}[/tex]
[tex]\theta=30 \textdegree[/tex]
Calculate the potential energy stored in a metal ball of a mass of 80 kg kept at a height of 15m from the earth surface.What will be the potential energy when the metal ball is kept on the earth surface.
Answer:
39200 joules
the potential energy will be zero
Explanation:
we know that potential energy is found by multiplying mass, acceleration due to gravity and height from the Earth's surface
so it will be
potential energy= mgh
80x9.8x15
= 39200 joules
the potential energy of the mental ball will be zero when kept on the Earth's surface because the height from the Earth's surface will be zero and zero multiplied to any number is zero only
I have a doubt with the second one, this is what I think it is. Consult your teacher if you think my answer for the second one is wrong
Answer:
392000 joules
Explanation:
hope it helpsss
Please help, I really need this. Thanks
Answer
Delta Q = change in thermal energy = c M * change in temperature
change in temperature = Q / (c * M)
change in temperature = -12 J / (390 J / Kg*deg * .012 kg
change in temp = -12 / (390 * .012) = - 2.56 deg C
how to make an uncharged particle positively charged
Answer:
If a neutral atom gains electrons, then it will become negatively charged. If a neutral atom loses electrons, then it become positively charged.
If 56.5 m3 of a gas are collected at a pressure of 455 mm Hg, what volume will the gas occupy if the pressure is changed to 632 mm Hg? *
Assuming ideal conditions, Boyle's law says that
P₁ V₁ = P₂ V₂
where P₁ and V₁ are the initial pressure and temperature, respectively, and P₂ and V₂ are the final pressure and temperature.
So you have
(455 mm Hg) (56.5 m³) = (632 mm Hg) V₂
==> V₂ = (455 mm Hg) (56.5 m³) / (632 mm Hg) ≈ 40.7 m³
Si un resorte de constante elástica 1300 n/m se comprime 12 cm ¿Cuanta energía almacena? Y si estira 12cm ¿Cuanta energía almacena?
La energía que almacena el resorte cuando se comprime y estira 12 cm es 9,4 J.
La energía potencial elástica del resorte se puede calcular con la siguiente ecuación:
[tex] E_{p} = \frac{1}{2}kx^{2} [/tex]
En donde:
k: es la constante del resorte = 1300 N/m
x: es la distancia de compresión o de elongación = 12 cm = 0,12 m
Dado que la energía es proporcional al cuadrado de la distancia recorrida por el resorte (x), la energía almacenada por el resorte durante la compresión será la misma que la energía almacenada por la elongación.
Por lo tanto, la energía almacenada es:
[tex]E_{p} = \frac{1}{2}kx^{2} = \frac{1}{2}1300 N/m*(0,12 m)^{2} = 9,4 J[/tex]
Entonces, la energía del resorte cuando se comprime y cuando se estira es la misma, a saber 9,4 J.
Para saber más sobre energía potencial visita este link: https://brainly.com/question/156316?referrer=searchResults
Espero que te sea de utilidad!
Answer:
Al comprimirse o estirarse 12 centímetros desde su posición sin deformar, el resorte almacena 9,360 joules.
Explanation:
La Energía Potencial Elástica almacenada por el resorte ([tex]U_{e}[/tex]), en joules, se calcula a partir de la Ley de Hooke, la definición de Trabajo y el Teorema del Trabajo y la Energía, cuya expresión se presenta abajo:
[tex]U_{e} = \frac{1}{2}\cdot k\cdot (x_{f}^{2}-x_{o}^{2})[/tex] (1)
Donde:
[tex]k[/tex] - Constante elástica del resorte, en newtons por metro.
[tex]x_{o}[/tex] - Posición inicial del resorte, en metros.
[tex]x_{f}[/tex] - Posición final del resorte, en metros.
Nótese que el resorte sin deformar tiene una posición de cero, la tensión tiene un valor positivo y la compresión, negativo.
Asumiendo que en ambos casos el resorte se encuentra inicialmente sin deformar, se reduce (1) a una forma de función par, es decir, una función que cumple con la propiedad de que [tex]f(x) = f(-x)[/tex], se encuentra que al comprimirse o estirarse en la misma medida almacena la misma cantidad de energía.
La cantidad de energía a almacenar es:
[tex]U_{e} = \frac{1}{2}\cdot \left(1300\,\frac{N}{m} \right)\cdot (0,12\,m)^{2}[/tex]
[tex]U_{e} = 9,360\,J[/tex]
Al comprimirse o estirarse 12 centímetros desde su posición sin deformar, el resorte almacena 9,360 joules.
There is given an ideal capacitor with two plates at a distance of 3 mm. The capacitor is connected to a voltage source with 12 V until it is loaded completely. Then the capacitor is disconnected from the voltage source. After this the two plates of the capacitor are driven apart until their distance is 5 mm. Now a positive test charge of 1 nC is brought from the positively charged plate to the negatively charges plate. How large is the kinetic energy of the test charge? The test charge of 1 nC can be regarded to be so small that it does not influence the electric field between the two plates of the capacitor.
Answer:
K = 2 10⁻⁸ J
Explanation:
Let's solve this exercise in parts, we start by finding the charge on each plate of the capacitor
C = Q / ΔV
C = ε₀ A / d
ε₀ A / d = Q / ΔV
Q = ε₀ A ΔV / d (1)
indicate the potential difference ΔV₁ = 12 V, the distance between the plates d₁ = 3 mm = 0.003 m,
as the power supply is disconnected and the capacitor is ideal the charge remains constant
in the second part we separate the plates at d₂ = 5 mm = 0.005 m, using equation 1
ΔV₂ = [tex]\frac{Q d_2}{ \epsilon_o A}[/tex]
we substitute the equation for Q
ΔV₂ = [tex]\frac{d_2}{\epsilon_o A} \ \frac{\epsilon_o A \Delta V }{d_1}[/tex]
ΔV₂ = [tex]\frac{d_2}{d_1} \ \Delta V_1[/tex]
in the third part we use the concepts of energy
starting point. Test charge near positive plate
Em₀ = U = q ΔV₂
final point. Test charge near negative plate
Em_f = K
energy is conserved
Em₀ = Em_f
q ΔV₂ = K
K = q ΔV₁ [tex]\frac{d_2}{d_1}[/tex]
we calculate
K = 1 10⁻⁹ 12 0.005/0.003
K = 2 10⁻⁸ J
The position of a particle is given by ~r(t) = (3.0 t2 ˆi + 5.0 ˆj j 6.0 t kˆ) m
Answer:
[tex]v=(6ti+6k)\ m/s[/tex]
Explanation:
Given that,
The position of a particle is given by :
[tex]r(t) = (3.0 t^2 i + 5.0j+ 6.0 tk) m[/tex]
Let us assume we need to find its velocity.
We know that,
[tex]v=\dfrac{dr}{dt}\\\\=\dfrac{d}{dt}(3.0 t^2 i + 5.0j+ 6.0 tk) \\\\=(6ti+6k)\ m/s[/tex]
So, the velocity of the particle is [tex](6ti+6k)\ m/s[/tex].
When you are standing on Earth, orbiting the Sun, and looking at a broken cell phone on the ground, there are gravitational pulls on the cell phone from you, the Earth, and the Sun. Rank the gravitational forces on the phone from largest to smallest. Assume the Sun is roughly 109 times further away from the phone than you are, and 1028 times more massive than you. Rank the following choices in order from largest gravitational pull on the phone to smallest. To rank items as equivalent, overlap them.
a. Pull phone from you
b. Pull on phone from earth
c. Pull on phone from sun
Answer:
The answer is "Option b, c, and a".
Explanation:
Here that the earth pulls on the phone, as it will accelerate towards Earth when we drop it.
We now understand the effects of gravity:
[tex]F \propto M\\\\F\propto \frac{1}{r^2}\\\\or\\\\F \propto \frac{M}{r^2}\\\\Sun (\frac{M}{r^2}) = \frac{10^{28}}{(10^9)^2} = 10^{10}[/tex]
The force of the sun is, therefore, [tex]10^{10}[/tex] times greater and the proper sequence, therefore, option steps are:
b. Pull-on phone from earth
c. Pull-on phone from sun
a. Pull phone from you
Two train 75 km apart approach each other on parallel tracks, each moving at
15km/h. A bird flies back and forth between the trains at 20km/h until the trains pass
each other. How far does the bird fly?
Answer:
The correct solution is "37.5 km".
Explanation:
Given:
Distance between the trains,
d = 75 km
Speed of each train,
= 15 km/h
The relative speed will be:
= [tex]15 + (-15)[/tex]
= [tex]30 \ km/h[/tex]
The speed of the bird,
V = 15 km/h
Now,
The time taken to meet will be:
[tex]t=\frac{Distance}{Relative \ speed}[/tex]
[tex]=\frac{75}{30}[/tex]
[tex]=2.5 \ h[/tex]
hence,
The distance travelled by the bird in 2.5 h will be:
⇒ [tex]D = V t[/tex]
[tex]=15\times 2.5[/tex]
[tex]=37.5 \ km[/tex]
An airplane propeller is 2.16 m in length (from tip to tip) with mass 100 kg and is rotating at 2900 rpm (rev/min) about an axis through its center. You can model the propeller as a slender rod. What is its rotational kinetic energy? Suppose that, due to weight constraints, you had to reduce the propeller's mass to 75.0% of its original mass, but you still needed to keep the same size and kinetic energy. What would its angular speed have to be, in rpm?
a) The rotational kinetic energy of the airplane propeller is 1792152.287 joules.
b) The angular speed of the airplane propeller is approximately 3348.631 revolutions per minute.
How to determine the angular speed of a airplane propeller
Let consider the airplane propeller a rigid body, the rotational kinetic energy of the propeller (K), in joules, is described by the following formula:
K = 0.5 · I · ω² (1)
Where:
I - Moment of inertia of the airplane propeller, in kilogram-square meters.ω - Angular speed, in radians per secondIn addition, the moment of inertia of a slender rod rotating around its center is:
I = 0.0833 · M · L² (2)
Where:
M - Mass of the propeller, in kilogramsL - Length of the propeller, in metersa) If we know that M = 100 kg, L = 2.16 m and ω = 303.687 rad/s, then the rotational kinetic energy of the propeller is:
K = 0.5 · [0.0833 · (100 kg) · (2.16 m)²] · (303.687 rad/s)²
K = 1792152.287 J
The rotational kinetic energy of the airplane propeller is 1792152.287 joules. [tex]\blacksquare[/tex]
b) By (1) and (2) we know that the mass of the propeller is inversely proportional to the square of the angular speed. Therefore, we have the following relationship:
[tex]M_{o}\cdot \omega_{o}^{2} = M_{f}\cdot \omega_{f}^{2}[/tex]
[tex]\omega_{f} = \sqrt{\frac{M_{o}}{M_{f}} }\cdot \omega_{o}[/tex] (3)
If we know that [tex]\omega_{o} = 2900\,\frac{rev}{min}[/tex], [tex]M_{o} = 100\,kg[/tex] and [tex]M_{f} = 75\,kg[/tex], then the angular speed of the airplane propeller is:
[tex]\omega_{f} = \left(2900\,\frac{rev}{min} \right)\cdot \sqrt{\frac{100\,kg}{75\,kg} }[/tex]
[tex]\omega_{f} \approx 3348.631\, \frac{rev}{min}[/tex]
The angular speed of the airplane propeller is approximately 3348.631 revolutions per minute. [tex]\blacksquare[/tex]
To learn more on rotational kinetic energy, we kindly invite to check this verified question: https://brainly.com/question/20261989
A charged particle having mass 6.64 x 10-27 kg (that of a helium atom) moving at 8.70 x 105 m/s perpendicular to a 1.30-T magnetic field travels in a circular path of radius 18.0 mm. What is the charge of the particle
Answer:
the charge of the particle is 2.47 x 10⁻¹⁹ C
Explanation:
Given;
mass of the particle, m = 6.64 x 10⁻²⁷ kg
velocity of the particle, v = 8.7 x 10⁵ m/s
strength of the magnetic field, B = 1.3 T
radius of the circle, r = 18 mm = 1.8 x 10⁻³ m
The magnetic force experienced by the charge is calculated as;
F = ma = qvB
where;
q is the charge of the particle
a is the acceleration of the charge in the circular path
[tex]a = \frac{v^2}{r} \\\\ma = qvB\\\\q = \frac{ma}{vB} \\\\q = \frac{mv^2}{rvB} = \frac{mv}{rB} \\\\q = \frac{(6.64\times 10^{-27} ) \times (8.7\times 10^5)}{(1.8\times 10^{-2}) \times (1.3)} \\\\q = 2.47 \ \times 10^{-19} \ C[/tex]
Therefore, the charge of the particle is 2.47 x 10⁻¹⁹ C
An ink-jet printer steers charged ink drops vertically. Each drop of ink has a mass of 10-11 kg, and a charge due to 500,000 extra electrons. It goes through two electrodes that gives a vertical acceleration of 104 m/s2. The deflecting electric field is _____ MV/m.
Answer:
E = 1.25 MV / m
Explanation:
For this exercise let's use Newton's second law
F = m a
where the force is electric
F = q E
we substitute
q E = m a
E = m a / q
indicate there are 500,000 excess electrons
q = 500000 e
q = 500000 1.6 10⁻¹⁹
q = 8 10⁻¹⁴ C
the mass is m = 10⁻¹¹ kg and the acceleration a = 10⁴ m / s²
let's calculate
E = 10⁻¹¹ 10⁴ / 8 10⁻¹⁴
E = 0.125 10⁷ V / m = 1.25 10⁶ V / m
E = 1.25 MV / m
A large metal sphere has three times the diameter of a smaller sphere and carries three times the charge. Both spheres are isolated, so their surface charge densities are uniform. Compare (a) the potentials (relative to infinity) and (b) the electric field strengths at their surfaces.
Answer:
A. Equals to that of the smaller sphere
B. 3 times less than that of the smaller sphere
Explanation:
(a) Equals to that of the smaller sphere
The potential of an isolated metal sphere, with charge Q and radius R, is kQ=R, so a sphere with charge 3Q and radius 3R has the same potential
b) 3 times less than that of the smaller sphere
However, the electric field at the surface of the smaller sphere is ?=? 0 = kQ=R2 , so tripling Q and R reduces the surface field by a factor of 1/3
The instrument includes a light source, which is passed through a Choose... , which isolates a single wavelength to pass through an aperture to reach the Choose... . Then, the light travels to the Choose... , which measures the intensity of light reaching it.
Answer:
Following are the response to the given question:
Explanation:
It's being used to measure the amount of light absorbed after traveling through a test tube (the amount of solar radiation received). For several quantitative estimations, this technique is widely employed. Spectrometer and Spectrometer were two devices that are used together to light intensity and light intensity.
It creates and diffuses phosphorescent light into the selected frequency, while the Spectrometer measures the strength of attenuation by the sample solution.
Diffraction beams or prisms are being used to convert polychromatic illumination into monochrome light.
Afterward, the sunlight has a certain hue. Once it reaches the specimen cuvette, it begins absorption. It falls on a sensor that transforms its intensity into such an electronic current.
Here are some ways to fill in such gaps:
In order to reach the specimen cuvette, the light from the light source must be routed via an aperture in order to be isolated by either a diffraction pattern. Light travels to the detector, which detects its intensity.
A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. How far horizontally will it travel in 2 seconds?
A. 30 m
B. 90 m
C. 45 m
D. 60 m
Answer:
It will travel Vx * t = 30 m/s * 2 s = 60 m
A 0.160 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.710 m/s. It has a head-on collision with a 0.296 kg glider that is moving to the left with a speed of 2.23 m/s. Suppose the collision is elastic.
Required:
a. Find the magnitude of the final velocity of the 0.157kg glider.
b. Find the magnitude of the final velocity of the 0.306kg glider.
The masses of the gliders provided in the question differ from the masses mentioned in the "Required" section. I'll use the first masses throughout.
Momentum is conserved, so the total momentum of the system is the same before and after the collision:
m₁ v₁ + m₂ v₂ = m₁ v₁' + m₂ v₂'
==>
(0.160 kg) (0.710 m/s) + (0.296 kg) (-2.23 m/s) = (0.160 kg) v₁' + (0.296 kg) v₂'
==>
-0.546 kg•m/s ≈ (0.160 kg) v₁' + (0.296 kg) v₂'
where v₁' and v₂' are the gliders' respective final velocities. Notice that we take rightward to be positive and leftward to be negative.
Kinetic energy is also conserved, so that
1/2 m₁ v₁² + 1/2 m₂ v₂² = 1/2 m₁ (v₁' )² + 1/2 m₂ (v₂' )²
or
m₁ v₁² + m₂ v₂² = m₁ (v₁' )² + m₂ (v₂' )²
==>
(0.160 kg) (0.710 m/s)² + (0.296 kg) (-2.23 m/s)² = (0.160 kg) (v₁' )² + (0.296 kg) (v₂' )²
==>
1.55 kg•m²/s² ≈ (0.160 kg) (v₁' )² + (0.296 kg) (v₂' )²
Solve for v₁' and v₂'. Using a calculator, you would find two solutions, one of which we throw out because it corresponds exactly to the initial velocities. The desired solution is
v₁' ≈ -3.11 m/s
v₂' ≈ -0.167 m/s
and take the absolute values to get the magnitudes.
If you want to instead use the masses from the "Required" section, you would end up with
v₁' ≈ -3.18 m/s
v₂' ≈ -0.236 m/s
A point charge of +35 nC is above a point charge of –35 nC on a vertical line. The distance between the charges is 4.0 mm. What are the magnitude and direction of the dipole moment ?
Answer:
Magnitude = 140 x 10⁻¹² Cm
Direction = upwards
Explanation:
A pair of two equal and opposite point charges forms an electric dipole.
The magnitude of the moment of such dipole is the product of the magnitude of any of the charges (since the charges are the same in magnitude) and the distance of separation between them. i.e
p = q x d ----------(i)
Where;
p = dipole moment
q = magnitude of any of the charges
d = distance between the charges.
The direction of the dipole moment is from the negative charge to the positive charge.
(a) From the question, the charges are +35 nC and -35 nC, and the distance between them is 4.00mm.
This implies that;
q = 35 nC = 35 x 10⁻⁹C
d = 4.00mm = 4.0 x 10⁻³ m
Substitute the values of q and d into equation (i) to give;
p = 35 x 10⁻⁹C x 4.00 x 10⁻³ m
p = (35 x 4.0) x (10⁻⁹ x 10⁻³) C m
p = 140 x 10⁻¹² Cm
The magnitude of the dipole moment is 140 x 10⁻¹² Cm
(b) From the question, the +35nC charge is above the -35nC charge on a vertical line as shown below;
o +35nC
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|
|
o -35nC
Since the direction should point from the negative charge to the positive charge, this means that the direction of the dipole moment of the two charges is upwards (due North).
o +35nC
↑
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o -35nC
A lead ball is dropped into a lake from a diving board 5.20 m above the water. After entering the water, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the water. The ball reaches the bottom 4.50 s after it is released. How deep is the lake?
Answer:
35.047 m
Explanation:
The time it takes the lead ball to reach the surface of the water is
s = ut+gt²/2............. Equation 1
Where t = time it takes the lead ball to reach the surface of water, u = initial velocity of the lead ball, g = acceleration due to gravity, s = heigth.
From the question,
Given: s = 5.20 m, u = 0 m/s (dropped from a height)
Constant: g = 9.8 m/s²
5.2 = 0+9.8t²/2
t² = (5.2×2)/9.8
t² = 10.4/9.8
t² = 1.06
t = √(1.06)
t = 1.03 s
Hence, time taken for the lead ball to reach the bottom of the lake is
t' = 4.5-1.03
t' = 3.47 seconds
v² = u²+2gs............... Equation 2
Where v = final velocity of the lead ball
Substitute into equation 2
v² = 0+2(9.8)(5.2)
v² = 101.92
v = √(101.92)
v = 10.1 m/s
Therefore, depth of the lake is
D = vt'
D = 10.1(3.47)
D = 35.047 m
