Answer:
A) 443 Hz
B) She has to loosen the string
Explanation:
A) Given;
Beat frequency;f_beat = 3 Hz
Frequency of electronically generated tone; f_e = 440 Hz
We know that formula for beat frequency is given by;
f_beat = |f1 - f2|
Now, applying it to this question, we have;
f_beat = f_v - f_e
Where f_v is frequency of the note played by the violinist
Thus, plugging in the relevant values;
3 = f_v - 440
f_v = 3 + 440
f_v = 443 Hz
B) In the concept of wave travelling in a string, the frequency is directly proportional to the square root of the force acting on the string.
Now, for the violinist to get her violin perfectly tuned to concert A from what it was when she heard the 3-Hz beats, the beat frequency will have to be zero. Which means it has to decrease by 3 Hz. For it to decrease, it means that the force applied has to decrease as we have seen that frequency is directly proportional to the square root of the force acting on the string.
Thus, she would have to loosen the string.
Your friend asks you for a glass of water and you bring her 5 millilitersof water. Is this more or less than what she was probably expecting?Explain your reasoning
x rays with wavelength of 2.0nm scatter from a nacl crystal with plane spacing 0.281 nm find the scattering
Explanation:
It is given that,
Wavelength of x-rays = 2 nm
Plane spacing, d = 0.281 nm
It is assumed to find the scattering angle for second order maxima.
For 2nd order, Bragg's law is given by :
[tex]2d\sin\theta=n\lambda[/tex]
For second order, n = 2
[tex]\sin\theta=\dfrac{n\lambda}{2d}\\\\\sin\theta=\dfrac{2\times 2\ nm}{2\times 0.28\ nm}\\\\\theta=\sin^{-1}(7.14)[/tex]
Here, θ is not defined. Also, the wavelength of x-rays is more than the plane spacing. It means that it cannot produce any diffraction maximum.
A bowling ball moves with constant velocity from an initial posi-tion of 1.6 m to a final position of 7.8 m in 3.1 s.
(a) What is the position time equation for the bowling ball?
(b) At what time is the ball at the position 8.6 m?
Answer:
calculate 1.6*7.8*3.1
Explanation:
Suppose a flagellum is rotating at a constant rps (revolution per second). What is the angular displacement of the flagellum in seconds?
Answer:
1794.47 radian.
Explanation:
Suppose a flagellum is rotating at a constant 952 rps (revolution per second). What is the angular displacement of the flagellum in 0.3seconds? (report your answer in SI units)
Let us suppose that a flagellum is rotating at a constant 952 revolution per second.
Firstly, we will convert 952 revolution per second to radian per second as follows :
952 revolution per second = 2π × 952 rad/s
=5981.592 rad/s
The angular displacement is given by the expression as follows :
[tex]\theta=\omega t\\\\\theta=5981.592\times 0.3\\\\\theta=1794.47\ \text{rad}[/tex]
So, the angular displacement of the flagellum in 0.3 seconds is 1794.47 radian.
If you are driving at a constant velocity of 5 m/s what is your
acceleration?
"Acceleration" means the rate at which velocity is changing.
If velocity is constant, then it isn't changing.
If velocity isn't changing, then acceleration is zero.
A ball starts from rest and undergoes uniform acceleration of 2.50m/s^2. What is the velocity of the ball 4s later?
Explanation:
Given:
v₀ = 0 m/s
a = 2.50 m/s²
t = 4 s
Find: v
v = at + v₀
v = (2.50 m/s²) (4 s) + 0 m/s
v = 10 m/s
The speed of the ball after 4s is 10 m/s.
There's not enough information to find its velocity. (We would need to know what direction it's moving.)
A round coin has a diameter of 19.55 mm, a thickness of 1.55 mm, and a mass of 2.50 g. What is its density?
Answer:
Density, d = 5.37 g/cm³
Explanation:
Given that,
Diameter of coin is 19.55 mm
Thickness of the cone is 1.55 mm
Mass of the coin is 2.5 g
We need to find its density. Density is equal to mass upon volume. So,
[tex]d=\dfrac{m}{V}[/tex]
V is volume, V = A d, d is thickness
Radius = d/2 = (1.955/2) cm=0.9775 cm
Thickness, d = 1.55 mm = 0.155 cm
[tex]d=\dfrac{m}{\pi r^2 \times d}\\\\d=\dfrac{2.5}{\pi (0.9775\ cm)^2\times 0.155\ cm}\\\\d=5.37\ g/cm^3[/tex]
So, the density of the coin is 5.37 g/cm³.
The radioactive hydrogen isotope 3H is called tritium. It decays by beta-minus decay with a half-life of 12.3 years.
A. What is the daughter nucleus of tritium?
B. A watch uses the decay of tritium to energize its glowing dial. What fraction of the tritium remains 20 years after the watch was created?
Answer:
a) A Helium nucleus with two protons and one neutron
b) 0.324 of the original amount
Explanation:
a) When Tritium decays, the nucleus emits an electron and an antineutrino, which then changes it from a Triton with one proton and two neutrons, to a Helium nucleus with two protons and one neutron.
The decay constant for the decay of Tritium is gotten from
[tex]t_{1/2}[/tex] = 0.693/k
where
[tex]t_{1/2}[/tex] is the half life = 12.3 yrs
k is the decay constant = ?
substituting, we have
12.3 = 0.693/k
k = 0.693/12.3 = 0.0563
The fraction that will remain after 20 years can be calculated from
[tex]N[/tex] = [tex]N_{0} e^{-kt}[/tex]
where
[tex]N[/tex] is the final remaining amount of Tritium after 20 years
[tex]N_{0}[/tex] is the original amount of Tritium
k is the decay constant = 0.0563
t is the time = 20 years
Equation can be re-written as
[tex]N/N_{0}[/tex] = [tex]e^{-kt}[/tex]
where
[tex]N/N_{0}[/tex] is the fraction of the tritium that will be remaining after 20 years.
substituting values, we have
[tex]N/N_{0}[/tex] = [tex]e^{-0.0563 * 20}[/tex]
[tex]N/N_{0}[/tex] = [tex]e^{-1.126}[/tex]
[tex]N/N_{0}[/tex] = 0.324 of the original amount
Which statement is true? A Displacement can never be greater than distance. B Displacement can never be less than distance. C Displacement can never be equal to distance. D Displacement is always equal to distance.
Answer:
__________________
I think the answer is A.
__________________
A person stands on a scale in an elevator. His apparent weight will be the greatest when the elevator(a) is standing still.(b) is moving upward at constant velocity.(c) is accelerating upward.(d) is moving downward at constant velocity.(e) is accelerating downward.
Answer:
(c) is accelerating upward
Explanation:
When a person stand in an elevator moving upwards, he feels heavier because the elevator's floor presses harder on his feet, and the scale will show a higher reading than when the elevator is at rest.
According to Newton's second law
R = mg + ma
where;
R is the reading of the scale = apparent weight of the person
mg is the normal weight of the person
ma is the upward force acting on the person
Therefore, his apparent weight will be the greatest when the elevator is accelerating upward
(c) is accelerating upward
An observer sits in a boat watching wave fronts move past the boat. The distance between successive wave crests is 0.80 m, and they are moving at 2.6 m/s.What is the wavelength of these waves? What is the frequency of these waves? What is the period of these waves?
Answer:
The wavelength is [tex]\lambda = 0.80 \ m[/tex]
The frequency is [tex]f = 3.25 \ Hz[/tex]
Explanation:
From the question we are told that
The distance between two successive crest is 0.80 m
The velocity is v = 2.6 m/s
Generally wavelength is the distance between two successive crest
So
[tex]\lambda = 0.80 \ m[/tex]
Now the frequency is mathematically represented as
[tex]f = \frac{v}{\lambda }[/tex]
[tex]f = \frac{2.6}{0.80 }[/tex]
[tex]f = 3.25 \ Hz[/tex]
A 0.140-kg baseball is dropped and reaches a speed of 1.20 m/s just before it hits the ground. It rebounds with a speed of 1.00 m/s. What is the change of the ball's momentum
Answer:
The change in momentum is Δp= 0.028 kg m/sExplanation:
An impulse describes a change in momentum. The change in momentum of an object is its mass times the change in its velocity.
The change in moment is given by the expression below
Δp=m⋅(Δv)=m⋅(vf−vi) .
Given data
mass m= 0.140-kg
initial velocity vi= 120 m/s
final velocity vf= 1 m/s
substituting we have
Δp=m⋅(Δv)=0.14⋅(1−1.2)
Δp=m⋅(Δv)=0.14⋅(-0.2)
Δp= 0.028 kg m/s
The change in momentum was found to be Δp= 0.028 kg m/s
At time t=0 , a cart is at x=10 m and has a velocity of 3 m/s in the −x -direction. The cart has a constant acceleration in the +x -direction with magnitude 3 m/s^2 < a < 6 m/s^2 . Which of the following gives the possible range of the position of the cart at t=1 s ?
Answer:
Explanation:
The minimum magnitude of acceleration = 3 m /s²
displacement at t = 1
s = ut + 1 /2 at²
= -3 x 1 + .5 x 3 x 1²
= - 3 + 1.5
= - 1.5 m
position at t = 1 s
= 10 - 1.5
= 8.5 m
The maximum magnitude of acceleration = 6 m /s²
displacement at t = 1
s = ut + 1 /2 at²
= -3 x 1 + .5 x 6 x 1²
= - 3 + 3
= 0
position at t = 1 s
= 10 +0
= 10 m
So range of position is 8.5 m to 10 m .
We want to find the range of possible positions for the cart at t = 1s.
The range is:
14.5m ≤ p ≤ 16m
Finding the cart's motion equations.
First, we know that the acceleration of the cart is a, a constant (but we don't know the exact value yet) so we write the acceleration as:
a(t) = a
To get the velocity equation we integrate over time, and we know that the velocity at t = 0s is 3m/s, so that will be the constant of integration.
v(t) = a*t + 3m/s
To get the position we integrate again, the initial position is x = 10m, so that will be the constant of integration.
p(t) = (a/2)*t^2 + (3m/s)*t + 10m
Now, to get the range of possible positions for t = 1s we need to use the minimum and maximum accelerations and evaluate the position equation in t = 1s.
The minimum acceleration is a = 3m/s^2, so we have the minimum position:
p(1s) = (3m/s^2/2)*(1s)^2 + (3m/s)*1s + 10m = 14.5m
The maximum acceleration is 6m/s^2, so the maximum position is:
P(1s) = (6m/s^2/2)*(1s)^2 + (3m/s)*1s + 10m = 16m
So the range of the position at t = 1s is:
14.5m ≤ p ≤ 16m
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Which question cannot be answered through making measurements?
A. Should wolves be reintroduced into national parks?
B. What is the birthrate for wolves in their natural habitats?
O c. How would reintroducing wolves into a national park affect the
deer population there?
D. Which species in a national park do wolves hunt?
"Should wolves be reintroduced into national parks" is a question that cannot be answered through making measurements, therefore the correct answer is option C.
What is a unit of measurement?A unit of measurement is a specified magnitude of a quantity that is established and used as a standard for measuring other quantities of the same kind. It is determined by convention or regulation. Any additional quantity of that type can be stated as a multiple of the measurement unit.
A length, for instance, is a physical quantity. A defined, predetermined length is represented by the length unit known as the meter.
A qualitative type of question cannot be answered while making a quantitative measurement.
Thus, the question of whether wolves should be reintroduced into national parks cannot be answered by taking measurements, so option C is the appropriate response.
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What If? Fluoride ions (which have the same charge as an electron) are initially moving with the same speed as the electrons from part (a) through a different uniform electric field. The ions come to a stop in the same distance d. Let the mass of an ion be M and the mass of an electron be m. Find the ratio of the magnitude of electric field the ions travel through to the magnitude of the electric field found in part (a). (Use the following as necessary: d, K, m, M, and e for the charge of the electron.)
Answer:
E₁ / E₂ = M / m
Explanation:
Let the electric field be E₁ and E₂ for ions and electrons respectively .
Force on ions = E₁ e where e is charge on ions .
Acceleration on ions a = E₁ e / M . Let initial velocity of both be u . Final velocity v = 0
v² = u² - 2as
0 = u² - 2 x E₁ e d / M
u² = 2 x E₁ e d / M
Similarly for electrons
u² = 2 x E₂ e d / m
Hence
2 x E₁ e d / M = 2 x E₂ e d / m
E₁ / E₂ = M / m
The period of a simple pendulum is 8 s. If the same pendulum experiment is repeated on a plant of g = 1/4 gE, the new period is
Answer:
The new period is 16 s
Explanation:
The period of a simple pendulum is given by
[tex]T=2\pi \sqrt{\frac{L}{g} }[/tex]
Where [tex]T[/tex] is the period
[tex]L[/tex] is the length of the string
[tex]g[/tex] is the acceleration due to gravity
2π is constant
For the first experiment,
[tex]T =8 s[/tex]
[tex]g = gE[/tex]
Then,
[tex]8 = 2\pi \sqrt{\frac{L}{gE} } \\[/tex]
[tex]8\sqrt{gE} = 2\pi \sqrt{L}[/tex] ....... (1)
For the second experiment
[tex]T = T_{2}[/tex]
[tex]g = \frac{1}{4} gE[/tex]
Hence,
[tex]T=2\pi \sqrt{\frac{L}{g} }[/tex] becomes
[tex]T_{2} =2\pi \sqrt{\frac{L}{\frac{1}{4}gE } }[/tex]
Then,
[tex]T_{2} \sqrt{\frac{1}{4}gE } = 2\pi\sqrt{L}[/tex] ....... (2)
Since 2π is constant and
The same pendulum is used for the second experiment, then [tex]L[/tex] is also constant.
∴ [tex]2\pi \sqrt{L}[/tex] is constant for both experiment.
Hence, we can equate equations (1) and (2) such that
[tex]8\sqrt{gE} = T_{2} \sqrt{\frac{1}{4}gE }[/tex]
Then,
[tex]\frac{8\sqrt{gE} }{\sqrt{\frac{1}{4}gE } } = T_{2} \\\frac{8\sqrt{gE} }{\frac{1}{2}\sqrt{gE} } } } = T_{2} \\ 16 = T_{2} \\T_{2} = 16 s[/tex]
Hence, the new period is 16 s
A simple pendulum swings for 8 seconds. The new period is determined by repeating the pendulum experiment on a plant with g = 1/4 gE.
Move to a position where the gravitational acceleration is greater.
- Reduce the pendulum's length.
Explanation: The frequency of a simple pendulum is given by where L is the length of the pendulum g is the gravitational acceleration
From the formula, we notice that - As the value of g increases, the frequency of the pendulum increases as well
- As the value of L increases, the frequency of the pendulum decreases
So, in order to increase the frequency of the pendulum, we can: - Move to a location where the acceleration due to gravity, g, is larger
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URGENT!!!!! If a 6 Ω resistor is in a circuit connected to a voltage source, and the current through the circuit is 2 amps, what is the amount of voltage across the circuit? 8 V 12 V 0.33 V 3 V
Answer:
Explanation:
according to ohms law we know that
v=IR
given current =2 amps
given resistance =6Ω
so voltage is
v=2*6 =12 V
The voltage across the circuit is equal to 12 V when 2 A current flows through the circuit with a 6 Ω resistor. Therefore, option (B) is correct.
What is Ohm's law?Ohm’s law can be described as the voltage difference across any conductor being directly proportional to the electric current flowing through that conductor while assuming temperature and all the other physical conditions remain constant.
Ohm’s law can apply only if the temperature is kept constant. Ohm’s S.I. unit is rho represented by the symbol Ω, and the current raises the temperature.
The mathematical expression of potential difference given by Ohm's law is:
V = IR
Where V is the voltage and R is the resistance and I the current.
Given, the resistance of the circuit, R = 6 ohm
The current flowing through the circuit, I = 2 A
The potential or voltage difference across the electric circuit will be:
V = I × R
V = 2 × 6
V = 12 Volt
Therefore, the potential difference across the electric circuit is equal to 12 V.
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The Type K thermocouple has a sensitivity of about 41 micro-Volts/℃, i.e. for each degree difference in the junction temperature, the output changes by 41 micro-Volts. If you have a 32-bit ADC, what is the smallest temperature change you can detect if the ADC range is 10 V?
Answer:
5.68*10^-5 °C
Explanation:
ADC is an acronym that means analog to digital conversion.
The numbers of bit in any ADC is the level of Voltage it possess.
The question states 32 bits, this means that it's total voltage levels is 2^32
Also, the question gives us a range of 10 V for the ADC. Now we can evaluate the resolution of the voltage as:
Δv = 10 / (2^32)
Δv = 10 / 4.29^9
Δv = 2.33*10^-9
The thermocouple sensitivity is 41 micro volts/°C
Thermocouple sensitivity = Δv/ΔT, where ΔT is the temperature change
ΔT = Δv/thermocouple sensitivity
ΔT = 2.33*10^-9 / 41*10^-6
ΔT = 5.68*10^-5 °C
Thus, the smallest temperature change is 5.68*10^-5 °C
Why is an element considered a pure substance?
Answer:
Pure substance are made of only one kind of particles and fixed constant.
Explanation:
Pure substance are combined in fixed ratio into separate by chemical methods into a physical and chemical methods.
Pure substance are the classified by the compounds and elements, that pure substance fixed into melting and boiling points.Pure substance is used to the chemical reaction product, and there are homogeneous nature only one type.Pure substance include that the copper, oxygen and gold, and water or crystals as that used in pure substance.Pure substance are made the single element,and mainly used in uniform composition throughout.Pure substance element are contain the only one atom to perform the substance the physical and chemical.Pure substance elements into similar substance by heat and electricity reactions with other substance.Pure substance compound contain the two or more elements combined in a fixed proportion.Pure substance compound perform the separate by the physical methods, by that the electric chemical methods.Pure substance compound has fixed composition, and it has two elements hydrogen and oxygen in a ratio to combined.Answer:
A compound is a pure substance composed of two or more different atoms chemically bonded to one another. A compound can be destroyed by chemical means. It might be broken down into simpler compounds, into its elements or a combination of the two
Explanation:
Two tiny, spherical water drops, with identical charges of -6.19 × 10-16 C, have a center-to-center separation of 1.22 cm. (a) What is the magnitude of the electrostatic force acting between them? (b) How many excess electrons are on each drop, giving it its charge imbalance?
Answer:
a
[tex]F = 2.32*10^{-17} \ N[/tex]
b
[tex]n =3869 \ electrons[/tex]
Explanation:
From the question we are told that
The charge on each water drop is [tex]q_1=q_2=q = - 6.19*10^{-16} \ C[/tex]
The distance of separation is [tex]d = 1.22\ cm = 0.0122 \ m[/tex]
Generally the electrostatic force between the water drops is mathematically represented as
[tex]F = \frac{k * q_1 * q_2 }{ d^ 2}[/tex]
Here k is the coulombs constant with value [tex]k = 9*10^9 \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]
So
[tex]F = \frac{9*10^9 * -6.19 *10^{-16} * (-6.19*10^{-16}) }{ 0.0122^ 2}[/tex]
[tex]F = 2.32*10^{-17} \ N[/tex]
Generally the quantity of charge is mathematically represented as
[tex]q = n * e[/tex]
Here n is the number of electron present
and e is the charge on one electron with value [tex]e = 1.60*10^{-19} \ C[/tex]
So
[tex]n = \frac{6.19 *10^{-16}}{1.60*10^{-19}}[/tex]
[tex]n =3869 \ electrons[/tex]
GnuPG (GPG) uses __________ to create a random set of keys. keyrings hashing entropy steganography
Answer:
entropy
Explanation:
GPG is an acronym for GNU Privacy Guard, which is considered as free software, is a command-line tool for generating cryptographically secure keys.
These cryptographically secure keys can be created by using entropy, through reliable entropy source, like /dev/random, which in turn derives its entropy from physical devices connected to the machine, such as keyboards, mice, and disks.
Hence, the right answer is ENTROPY
In a compound microscope, the objective has a focal length of 1.0 cm, the eyepiece has a focal length of 2.0 cm, and the tube length is 25 cm. What is the magnitude of the overall magnification of the microscope?
Answer:
m = 312.5
Explanation:
Given that,
The focal length of the objecive lens, [tex]f_o=1\ cm[/tex]
The focal length of eye piece, [tex]f_e=2\ cm[/tex]
length of the tube, L = 25 cm
We need to find the magnitude of the overall magnification of the microscope. It is given by the formula as follows :
[tex]m=\dfrac{L}{f_o}\times \dfrac{D}{f_e}[/tex]
D = 25 cm
So,
[tex]m=\dfrac{25}{1}\times \dfrac{25}{2}\\\\m=312.5[/tex]
So, the overall magnification of the microscope is 312.5.
Determine the weight in newtons of a woman whose weight in pounds is 157. Also, find her mass in slugs and in kilograms. Determine your own weight in newtons.
Answer:
Weight of the woman in Newton
[tex]x = 698.6 \ N[/tex]
Mass of the woman in slug
[tex]Mass = 4.86 \ slug[/tex]
Mass of the woman in kg
[tex]Mass = 16 \ kg[/tex]
My weight in Newton
[tex]W = 784 \ N[/tex]
Explanation:
From the question we are told that
The weight of the woman in pounds is [tex]W = 157 \ lb[/tex]
Converting to Newton
1 N = 0.22472 lb
x N = 157
=> [tex]x = \frac{157 * 1}{0.22472}[/tex]
=> [tex]x = 698.6 \ N[/tex]
Obtaining the mass in slug
[tex]Mass = \frac{W}{g}[/tex]
Here [tex]g = 32.2 ft/s^2[/tex]
So
[tex]Mass = \frac{157 }{32.2}[/tex]
[tex]Mass = 4.86 \ slug[/tex]
Obtaining the mass in kilogram
[tex]Mass = \frac{W}{g}[/tex]
Here [tex]g = 9.8 \ m/s[/tex]
So
[tex]Mass = \frac{157 }{9.8}[/tex]
[tex]Mass = 16 \ kg[/tex]
Generally weight is mathematically represented as
[tex]W = m * g[/tex]
Given that my mass is 80 kg then my weight is
[tex]W = 80 *9.8[/tex]
[tex]W = 784 \ N[/tex]
With the sun and the earth back in their regular positions,consider a space probe with mass mp = 125 kg launched from the earth towardthe sun. When the probe is exactly halfway between the earth andthe sun along the line connecting them, what is the direction ofthe net gravitational force acting on the probe? Ignore the effects of other massive objectsin the solar system, such as the moon and other planets.
A. The force is toward the sun.
B. The force is toward the earth.
C. There is no net force because neither the sunnor the earth attracts the probe gravitationally at themidpoint.
D. There is no net force because the gravitationalattractions on the probe due to the sun and the earth are equal insize but point in opposite directions, so they cancel each otherout.
Answer:
A
Explanation:
Gravitational force given by
F_ps =G(m_pM_s)/r^2
G= gravitational constant
m_p=mass of the probe =125 Kg
M_s= mass of sun
r= distance between probe and Sun.
Similarly, F_pe = G(m_pM_e)/r^2.
m_p = mass of probe
M_e = mass of earth
Clearly, M_s> M_e . Therefore, F_ps> F_pe.
Hence, A. The force is toward the sun.
A parallel beam of light containing orange (610 nm) and violet (410 nm) wavelengths goes from fused quartz to water, striking the surface between them at a 60.0°60.0° incident angle. What is the angle between the two colors in water?
Answer:
0.98°
Explanation:
First we find the refracting angle of orange in water
Which is given as
စr= sin^-1. ( 1.456 sin60°/1.33)
= 71.2°
Then that of violet in water
စv=sin^-1. ( 1.468sin60°/1.342)
= 72.3°
So angle between boths colours is the difference
71.2- 72.3= 0.98°
cle 2 of charge 4.00q are held at separation L 9.00 cm on an x axis. If particle 3 of charge q3 is to be located such that the three particles remain in place when released, what must be the (a) x and (b) y coordinates of particle 3, and (c) the ratio q3 /q
Answer:
x=L/2 y=0 and the charge q3 is ¼ of the charge q
Explanation:
For this exercise we will use Coulomb's law.
F₁₂ = k q₁ q₂ / r₁₂²
From this expression we see that like charges repel and charges of different signs attract.
Let's apply this expression to our case, they indicate that the two charges are of equal magnitude and sign, therefore the force is repulsive, so that it is in equilibrium with a third charge (q₃) this must be of the opposite sign and be between the two charge (q)
let's apply Newton's second law to one of the charges, for example the one on the left
-F₁₂ + F₁₃ = 0
F₁₂ = F₁₃
k q₁ q₂ / r₁₂² = k q₁ q₃ / r₁₃²
q₂ / r₁₂² = q₃ / r₁₃²
q₃ = q₂ (r₁₃ / r₁₂)²
The problem indicates the charge q₁ = q₂ = 4 q and the distance between them is r₁₂ = L = 9 cm = 0.09 m, we substitute
q₃ = 4q (r₁₃ / L)²
Let's analyze the situation a bit that the charge 1 and 2 are in equilibrium with a single charge 3 this must be symmetrical between the two charge (the same force), therefore its position on the x axis must be r₁₃ = L/2 and how it is on the y axis = 0
let's substitute
q₃ = 4q (L / 2L)²
q₃ = 4q 1/4
q₃ = q
the charge q3 is ¼ of the charge q
what is an atomic nucleus
Answer:
The atomic nucleus is the small, dense region consisting of protons and neutrons at the center of an atom, discovered in 1911 by Ernest Rutherford based on the 1909 Geiger–Marsden gold foil experiment.
Explanation:
A resistor is connected in series with a power supply of 20.00 V. The current measure is 0.50 A. What is the resistance of the resistor
Answer:
40 Ω
Explanation:
The following data were obtained from the question:
Potential difference (V) = 20 V
Current (I) = 0.50 A
Resistance (R) =?
From ohm's law:
V = IR
Where:
V is the potential difference.
I is the current.
R is the resistance.
With the above formula, we can obtain the resistance of the resistor as follow:
Potential difference (V) = 20 V
Current (I) = 0.50 A
Resistance (R) =?
V = IR
20 = 0.5 × R
Divide both side by 0.5
R = 20/0.5
R = 40 Ω
Therefore, the resistance of the resistor is 40 Ω
26500 in scientific notation
Answer:
in scientific notation 26500 is 2.65×10⁴
Answer:
2.56*10^4
26500
(We have to move the decimal 4 places)
2.6500*10^4
So the answer is 2.6500*10^4 or 2.56*10^4
Hope this helps! :)
g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart
Answer:
The speed of q₂ is [tex]4\sqrt{10}\ m/s[/tex]
Explanation:
Given that,
Distance = 0.4 m apart
Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.
We need to calculate the speed of q₂
Using conservation of energy
[tex]E_{i}=E_{f}[/tex]
[tex]\dfrac{1}{2}mv_{i}^2+\dfrac{kq_{1}q_{2}}{r_{i}}=\dfrac{kq_{1}q_{2}}{r_{f}}+\dfrac{1}{2}mv_{f}^2[/tex]
[tex]\dfrac{1}{2}m(v_{i}^2-v_{f}^2)=kq_{1}q_{2}(\dfrac{1}{r_{f}}-\dfrac{1}{r_{i}})[/tex]
Put the value into the formula
[tex]\dfrac{1}{2}\times1.5\times10^{-3}(20^2-v_{f}^2)=9\times10^{9}\times-2\times10^{-6}\times-8\times10^{-6}(\dfrac{1}{(0.4)}-\dfrac{1}{(0.8)})[/tex]
[tex]0.00075(400-v_{f}^2)=0.18 [/tex]
[tex]400-v_{f}^2=\dfrac{0.18}{0.00075}[/tex]
[tex]-v_{f}^2=240-400[/tex]
[tex]v_{f}^2=160[/tex]
[tex]v_{f}=4\sqrt{10}\ m/s[/tex]
Hence, The speed of q₂ is [tex]4\sqrt{10}\ m/s[/tex]