[tex]\textsf{When an electron is hit by a }[/tex] [tex]\textsf{photon of light, it absorbs the quanta}[/tex] [tex]\textsf{of energy the photon was carrying}[/tex] [tex]\textsf{and moves to a higher energya}[/tex] [tex]\textsf{ state. Electrons therefore have to }[/tex] [tex]\textsf{jump around within the atom as }[/tex] [tex]\textsf{they either gain or lose energy. }[/tex]
When an electron is hit by a photon of light, it absorbs the quanta of energy the photon was carrying and moves to a higher energy state. Electrons therefore have to jump around within the atom as they either gain or lose energy.
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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 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].
Calculate the elastic energy stored up in a wire originally 5 meter
long and 10^-3 m in diameter which has been stretched by 3×10^-4 m due to a load of 10 kg.
Answer:
The elastic energy is 245 J.
Explanation:
Length, L = 5 m
Diameter, D = 10^-3 m
Stretch, l = 3 x 10^-4 m
Load, F = 10 x 9.8 = 98 N
Let the elastic energy is U.
[tex]U = \frac{1}{2}\times stress\times strain\times volume\\\\U = 0.5\times \frac{Force}{area}\times \frac{l}{L}\times Area\times L\\\\U = 0.5 \times F\times l\\\\U = 0.5\times 98\times 5\\\\U = 245 J[/tex]
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
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.
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
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
Light falling on a metal surface causes electrons to be emitted from the metal by the photoelectric effect.
As we decrease the frequency of this light, but do not vary anything else (there may be more than one correct answer),
A: the number of electrons emitted from the metal increases.
B: the maximum speed of the emitted electrons decreases.
C: the maximum speed of the emitted electrons does not change.
D: the work function of the metal increases.
(B)
Explanation:
The speed of the ejected electrons depends on the frequency of the incident radiation. The closer the energy of the incident photons to the work function of the metal, the slower is the speed of the ejected electrons. Intensity of the incident radiation has no effect on the speed of the ejected electrons, only its frequency.
Light falling on a metal surface causes electrons to be emitted from the metal by the photoelectric effect. As we decrease the frequency of this light, but do not vary anything else B: the maximum speed of the emitted electrons decreases.
What is photoelectric effect?
Photoelectric effect is the phenomenon in which electrically charged particles are released from or within a material when it absorbs electromagnetic radiation. The effect is often defined as the ejection of electrons from a metal plate when light falls on it.
According to Photoelectric effect the kinetic energy of the photo electrons emitted depend on the frequency of incident light , the more is the frequency the more is the kinetic energy of emitted electron and hence high will be the velocity of the emitted electrons and vise versa
since , in question the frequency has been decreased hence , the kinetic energy must be decreased therefore velocity will also get decreased
hence , correct option will be B: the maximum speed of the emitted electrons decreases.
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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
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
write down the unit of mass ,temperature ,power and density
Explanation:
mass=kilogram,temperature=Klevin,power=watt,density=kilogram per cubic metre
Explanation:
the unit of mass is kg , temperature is kelvin ,power is watt and density is kilogram per cubic meter.
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
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.
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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.
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.
Which of the following has a negative acceleration?
A. A car increases its speed moving forward.
B. A car sits at rest at a stop sign.
C. A car is slowing down as it approaches a traffic light.
D. A car is in cruise control at a constant speed.
Answer:
B. A car sits at rest at a stop sign.
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
Two horizontal pipes have the same diameter, but pipe B is twice as long as pipe A. Water undergoes viscous flow in both pipes, subject to the same pressure difference across the lengths of the pipes. If the flow rate in pipe B is Q=ΔV/Δt what is the flow rate in pipe A? Viscosity: Two horizontal pipes have the same diameter, but pipe B is twice as long as pipe A. Water undergoes viscous flow in both pipes, subject to the same pressure difference across the lengths of the pipes. If the flow rate in pipe B is what is the flow rate in pipe A?
a) Q√2
b) 16Q
c) 2Q
d) 4Q
e) 8Q
Answer:
c) 2Q
Explanation:
From the given information:
The pressure inside a pipe can be expressed by using the formula:
[tex]\Delta P = \dfrac{128 \mu L Q}{\pi D^4}[/tex]
Since the diameter in both pipes is the same, we can say:
[tex]D = D_A = D_B[/tex]
where;
length of the first pipe A [tex]L_A = L[/tex] and the length of the second pipe B [tex]L_B = 2L[/tex]
Since the difference in pressure is equivalent in both pipes:
Then:
[tex]\dfrac{128 \mu L_1Q_1}{\pi D_1^4} = \dfrac{128 \mu L_2Q_2}{\pi D_2^4}[/tex]
[tex]\dfrac{ L_1Q_1}{D_1^4} = \dfrac{ L_2Q_2}{D_2^4}[/tex]
[tex]\dfrac{ LQ_1}{D^4} = \dfrac{ 2LQ}{D^4}[/tex]
[tex]\mathbf{Q_1 = 2Q}[/tex]
The flow rate in pipe B is 2Q of the flow rate of the pipe A
What is flow rate?
The flow rate is defined as the flow of the fluid across the cross section in per unit time.
From the given information:
The pressure inside a pipe can be expressed by using the formula:
[tex]\Delta p=\dfrac{128\mu LQ}{\pi D^4}[/tex]
Since the diameter in both pipes is the same, we can say:
[tex]D=D_A=D_B[/tex]
where;
length of the first pipe A [tex]L_A=L[/tex] and the length of the second pipe B
[tex]L_B=2L[/tex]
Since the difference in pressure is equivalent in both pipes:
Then:
[tex]\dfrac{128\mu L_1Q_1}{\pi D_1^4}=\dfrac{128\mu L_2Q_2}{\pi D_2^4}[/tex]
[tex]\dfrac{L_1Q_1}{D_1^4}=\dfrac{L_2Q_2}{D_2^4}[/tex]
[tex]\dfrac{LQ_1}{D_1^4}=\dfrac{2LQ}{D_2^4}[/tex]
[tex]Q_1=2Q[/tex]
Hence the flow rate in pipe B is 2Q of the flow rate of the pipe A
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Particle A has less mass than particle B. Both are pushed forward across a frictionless surface by equal forces for 1 s. Both start from rest. Which is true? A. A has more momentum. B. B has more momentum. C. A and B have the same momentum D. Not enough information.
Answer:
Both will have the same momentum.
P = M v momentum
v = a t for uniform acceleration
P = M a t
But a = F / M
P = M (F / M) t = F t so both have the same momentum
Which of these rotational quantities is analogous to mass in a linear system?
a.
Angle in radians
b.
Angular acceleration
c.
Torque
d.
Rotational inertia
Answer:
d
Explanation:
i think it is rotational inertia
because analogue of mass in rotational motion is moment of inertia. It plays the same role as mass plays in transnational motion.
hope it's right & helps !!!!!!!!!
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
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]
An AM radio transmitter broadcasts 50.0 kW of power uniformly in all directions. I live 10 km from this station. What is the maximum strength of Electric Field in my house
Answer:
[tex]E_0=0.173N/C[/tex]
Explanation:
From the question we are told that:
Power [tex]P=50kw=>50*10^3w[/tex]
Distance [tex]d=10km=10000m[/tex]
Generally the equation for Intensity is mathematically given by
[tex]I=\frac{P}{4\pi d^2} w/m^2[/tex]
[tex]I=\frac{50*10^3}{4 \pi 10000^2} w/m^2[/tex]
[tex]I=3.98*10^{-5}w/m^2[/tex]
Generally Intensity is also
[tex]I=\frac{1}{2}cE_0^2e[/tex]
Where
[tex]e=8.854*10^{-12}Nm^2/c^2[/tex]
Therefore
[tex]E_0=\sqrt{\frac{2I}{c *e}}[/tex]
[tex]E_0=\sqrt{\frac{2*3.98*10^{-5}}{3*10^8 *8.854*10^{-12}}}[/tex]
[tex]E_0=0.173N/C[/tex]
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
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
A standard bathroom scale is placed on an elevator. A 28 kg boy enters the elevator on the first floor and steps on the scale. What will the scale read (in newtons) when the elevator begins to accelerate upward at 0.5 m/s2
Answer:
Explanation:
Newton's Second Law is pretty much the standard for all motion that involves a force. It applies to gravitational force and torque and friction and weight on an elevator. The main formula for force is
F = ma. We have to adjust that to take into account that when the elevator is moving up, that "surge" of acceleration weighs down a bit on the scale, causing it to read higher than the actual weight until the acceleration evens out and there is no acceleration at all (no acceleration simply means that the velocity is constant; acceleration by definition is a change in velocity, and if there is no change in velocity, there is 0 acceleration). The force equation then becomes
[tex]F_n-w=ma[/tex] where [tex]F_n[/tex] is normal force. This is what the scale will read, which is what we are looking for in this problem (our unknown). Since we are looking for [tex]F_n[/tex], that is what we will solve this literal equation for:
[tex]F_n=ma+w[/tex] . m is the mass of the boy, a is the acceleration of the elevator (which is going up so we will call that acceleration positive), and w is weight. We have everything but the unknown and the weight of the boy. We find the weight:
w = mg so
w = 28(9.8) and
w = 274.4 N BUT rounding to the correct number of significance we have that the weight is actually
w = 270 N.
Filling in the elevator equation:
[tex]F_n=28(.50)+270[/tex] and according to the rules of significant digits, we have to multiply the 28(.50) {notice that I did add a 0 there for greater significance; if not that added 0 we are only looking at 1 significant digit which is pretty much useless}, round that to 2 sig fig's, and then add to 170:
[tex]F_n=14+270[/tex] and adding, by the rules, requires that we round to the tens place to get, finally:
[tex]F_n=280N[/tex] So you see that the surge in acceleration did in fact add a tiny bit to the weight read by the scale; conversely, if he were to have moved down at that same rate, the scale would have read a bit less than his actual weight). Isn't physics like the coolest thing ever!?
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.
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
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]
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]
