A circular ring in the x y plane (radius r, centered at the origin) carries a uniform line charge λ. find the first three terms (n) is 2π λ ( 1 - (a/2r)² ( 3cos²0 -1).
Calculation:-
The first three of uniform charge carrying loop is.
V (r,theta) = q/r + 1/2r³ ( Q sin²0 cos²0 + q sin²0 sin²0 + Q cos²0)
= q/r - qa²/4r³ (3 cos²0 -1 )
= 2π λ ( 1 - (a/2r)² ( 3cos²0 -1)
A radius is a line section with one endpoint at the middle of the circle and the other endpoint at the circle. Radius = Diameter of a Circle: A line segment passing through the center of a circle, and having its endpoints at the circle, is referred to as the diameter of the circle. Diameter = 2 × radius.
underneath the Uniform rate techniques, we rate depreciation at the constant belongings which can be uniformly productive on a uniform basis every year. for this reason, the quantity of depreciation stays the same each 12 months.
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To practice Problem-Solving Strategy 24.1 Magnetic field problems.
The element niobium, which is a metal, is a superconductor at temperatures below 9 K . This means that at such low temperatures a niobium wire has no electrical resistance. The superconductivity of the wire, however, is destroyed if the magnetic field at the surface of the wire exceeds 0.10 T . What is the maximum current in a straightsuperconducting niobium wire of radius 1.5 mm ? You may assume that all the current flows in the center of the wire.
Prepare
Create a visual representation of the problem. First choose the best orientation for the wire, then draw the magnetic field created by the current. You can then determine your known variables and identify those that you must find.
Recall that you are to treat the current as if it all runs through the center of the wire, and you are interested in the value of the magnetic field at the surface of the wire. Sort the following variables as known or unknown: the magnitude of the magnetic field B, the current I, and the distance r between the current and the point at which the magnetic field has a value of B.
u0= 4pi*10^-7
r=1.5mm= 1.5*10^-3m
substitute these values
The answer is 750 A
Long straight wire :B =µ01/2πr = Icrit==(2π)*(r=D/2)(Bcrit))/µ0
a 6.0 v ideal battery is being used to charge a 5.0 x 10-6 f capacitor through a resistor r. the capacitor attains a potential difference of 3.6 v in 1.0 s after the charging begins. the value of the resistance r is
The value of the resistance R in the above case is 120000 Ohms.
In this case
Charge is given by
Q=CV
=6*5*10^-6
=30*10^-6 C
Then the current through the resistor, R is
I=Q/t
=(30*10^-6)/1
=30*10^-6 A
From Ohm's law
R=V/I
=3.6/(30*10^-6)
=120000 Ohms
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you shoot a 51 g pebble straight up with a catapult whose spring constant is 320 n/m. the catapult is initially stretched by 0.20 m. how high above the starting point does the pebble fly? ignore air resistance
Applying the Conservation of Energy, the point above the starting point of the pebble will fly is 12.8 m
Given Parameters:
Mass of the pebble (m) = 51 g
Spring Constant (k) = 320 N/m
Extension (x) = 0.20 m
Law Of Conservation Of Energy:
According to the rule of conservation of energy, energy cannot be created or destroyed. Energy doesn't evaporate when it is used by people.
Within a specific frame of reference, the total energy of an isolated system either remains constant or is conserved.
According to the law of conservation of energy
Kinetic Energy is equal to potential Energy,
[tex](1/2)kx^2 = mgh\\\\h = (kx^2) / (2mg)= [(320 N/m)(0.20m)^2] / [2(0.051kg)(9.8 m/s^2)]= 12.8 m[/tex]
Thus, The point above the starting point of the pebble will fly is 12.8 m.
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explain the relationship between diffraction angle and wavelength using the two dimensional diffraction relationship n
Diffraction is described as the bending of waves around the corners of an obstruction or through an opening into the region of the barrier.
The angle of diffraction is directly proportional to the size of the wavelength.
How to. Define relationship between angle of diffraction and wavelength using 2-D diffraction?
Angular displacement ∅ = wt = 2 × 10 = 20 rad
Here w = constant
Thus, angular acceleration will be zero.
Here,
∅=wt=4×20= =80 rad
w= constant
∅=wt=3×8=24 rad
w= constant.
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what is the speed of visible light waves when traveling through a vacuum? about 3 cross 10^8 m/s about 350 m/s about 1 cross 10^8 m/s light waves cannot travel through a vacuum. about 2 cross 10^8 m/s
The speed of visible light waves when travelling through a vacuum is about 3 X 10⁸ m/s.
The speed of electromagnetic waves, which includes visible light, radio light and x-rays is 3 X 10⁸ m/s. This is same as that of the speed of visible light in the presence of air.
The speed of light is depicted using the letter 'c', and it a universal constant value. It is a physical constant and important in many areas of physics.
The speed of light in water is 2.25 X 10⁸ m/s and that in glass is 2 X 10⁸ m/s. The speed varies with respect to the medium.
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after 10 seconds the car body is 1/3 foot above its equilibrium position and at the high point in its cycle. what were the initial conditions ?
force is required to stretch it a certain length than you would need to stretch a less stiff spring the same length.
How do you find the amplitude of oscillation?The angular frequency ω, period T, and frequency f of a simple harmonic oscillator are given by ω=√km, T = 2π√mk, and f = 12π√km, where m is the mass of the system and k is the force constant.if one increases the mass of a simple harmonic oscillator (and keeps everything else the same) the frequency of oscillation must decrease. This means that the oscillator will move more slowly. On the other hand if one increases the spring constant K the frequency will increase.So, recapping, the period of a mass on a spring does not depend on the amplitude. You can change the amplitude, but it will not affect how long it takes this mass to go through a whole cycle.x(t) = A cos(ωt + φ). A is the amplitude of the oscillation, i.e. the maximum displacement of the object from equilibrium, either in the positive or negative x-direction.0.43622 seconds
0.9158 foot
-4.43 ffot/sec
we first find the period of oscillation
= 2π√w/gk
= 2π√4000/32x2600
= 2π√0.00481
= 2π0.0694
= 0.43622
b. we find the angular velocity
2π/T
= 2π/0.43622
= 14.41 rad/sec
we find displacement
rom the calculation in the attachment
Ф = -144.1
initial condition
1*cos(-144.1 rad)
= 0.9158 foot
initial velocity of the car
= (-1)(14.41)sin(ω(0)-144.1)
= -4.44 foot/sec
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What is the centripetal acceleration acting on a 19 kg ball if it is being spun around by a tension force of 125 N?
Answer:
Explanation:
The centripetal acceleration:
aₙ = F /m = 125 /19 ≈ 6.6 m/s²
suppose two cards, one twice as massive as the other, fly apart when compressed, spring that joins them is released. how fast does the heavier car roll compared with the lighter card
Answer: The heavier car rolls with half the acceleration as the lighter car.
Which of the following statements are true of free-
falling objects of different mass? Include all that apply.
a. The rate of acceleration is the same despite their
different masses.
b. They experience the same force of gravity.
c. They have the same g value but a different
acceleration.
d. They have the same acceleration but a different g
value.
e. nonsense! None of these statements are true of
such objects.
Statement "a" is true.
it's the only one of the bunch.
The true statement for free-falling objects of different masses is the rate of acceleration is the same despite their different masses. The correct option is a.
What is acceleration?Acceleration is the rate at which the speed and direction of a moving object vary over time. A point or object going straight ahead is accelerated when it accelerates or decelerates. Even if the speed is constant, motion on a circle accelerates because the direction is always shifting.
If things of different masses are dropped from the same height, they will still impact the ground at the same time because Earth provides everything with the exact same acceleration.
Therefore, the correct option is a, The rate of acceleration is the same despite their different masses.
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the magnitude of the magnetic field 50 cm from a long, thin, straight wire is 8.0 μt. what is the current through the long wire?
The current through the long wire is 20A.
What is a magnetic field in simple terms?
The magnetic field is the area around a magnet in which the effect of magnetism is felt. We use the magnetic field as a tool to describe how the magnetic force is distributed in the space around and within something magnetic in nature.
What causes a magnetic field?
Magnetism is caused by the motion of electric charges. Every substance is made up of tiny units called atoms. Each atom has electrons, particles that carry electric charges. Spinning like tops, the electrons circle the nucleus, or core, of an atom.
How do you find current in a wire in a magnetic field?
The strength of a magnetic field, , some distance away from a straight wire carrying a current, , can be found using the equation = 2 , where is a constant known as “the permeability of free space”
What happens when a wire with a current is placed in a magnetic field?
The wire will bend away from the poles of the magnet.
[tex]i=\frac{2\pi RB}{m_{0}}[/tex]
Given : [tex]R=0.50m, B=8*10^{-6}[/tex]
[tex]i=\frac{2\pi *0.50*8*10^{-6}}{4\pi *10^{-7}}[/tex]
[tex]i=20A[/tex]
Thus, the current flowing in the wire is 20A.
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suppose that an object is dropped, not from the earth's surface, but from a height rearth above the earth's surface. with what acceleration will it start to fall? give your answers as a fraction of gearth
The acceleration of the object when it will start falling from the height above the earth is ( 1/9 ) gₓ.
Acceleration due to gravity from the height = gₙ
Acceleration due to gravity on earth = gₓ
Distance between the core of earth and the point at a height above earth = h
Radius of the earth = rₓ
Now, for h = rₓ
= Using the gravitational formula
= gₙ = gₓ ( rₓ / (rₓ + rₓ))²
= gₙ = ( 1/4 ) gₓ
Similarly for h = 2rₓ,
= gₙ = gₓ ( rₓ / (rₓ + rₓ))²
= gₙ = ( 1/9 ) gₓ
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a machine works with an efficiency of 18%. the energy input is 500 joules/second. how much useful energy will the machine generate in 15 minutes?(1 point)
The useful energy the machine will generate in 15minutes is 81000J
What is efficiency of a machine?The efficiency of a machine indicates how well its input energy is converted to useful output energy or work. It is a major factor in the usefulness of a machine and is the fraction or percentage of the output divided by the input.
Therefore the efficiency of a machine is given as Energy output/ energy input ×100 or power output/ power input ×100
let the power output be x
therefore 18=( x/500)×100
100x = 18×500
x= 18×500/100= 90J/s
Energy = power ×time
in 15minutes = 15×60s, the machine will generate energy of 90×15×60= 81000J
Therefore the machine will generate a useful energy of 81000J
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In recent years, astronomers have found planets orbiting nearby stars that are quite different from planets in our solar system. Kepler-12b, has a diameter that is 1.7 times that of Jupiter (RJupiter = 6.99 x 107 m), but a mass that is only 0.43 that of Jupiter (MJupiter = 1.90 x 1027 kg). Part A What is the value of g on this large, but low-density, world? Express your answer with the appropriate units. НА ? g= Value Units Submit Request Answer Provide Feedback
3.86 m/s^2 is the value of gravity on this large, but low-density, world.
given :
Kepler-12b
diameter= 1.7 times of Jupiter (R_Jupiter = 6.99 × 10^7 m),
mass = 0.43 Jupiter (M_Jupiter = 1.90 × 10^27 kg ).
g = GM/r^2
g = 6.67×10^-11 × 0.43×1.9×10^27/( 1.7×6.99×10^7)^2
g = 3.859 ~ 3.86 m/s^2
Gravity, also referred to as gravitation, is the unchanging force of attraction that binds all matter together in mechanics. It is by far the weakest known force in nature, so it has no effect on determining the internal properties of common matter.
On Earth, everything has weight, or a gravitational pull that is imposed by the planet's mass and proportional to the object's mass. A measure of the force of gravity is the acceleration that freely falling objects experience. At the surface of the Earth, gravity accelerates at a rate of about 9.8 meters per second. As a result, an object's speed increases during free fall by about 9.8 meters per second. At the Moon's surface, a freely falling body accelerates to about 1.6 m/s2.
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a bicyclist is riding north on level ground and has zero net force. a single force pointing south then acts on the bicycle. which effect does the new force have on the motion of the bicycle? responses
If the single force pointing south acts on the bicycle then it will slow down or even stop, depending on the size of the force applied.
The cyclist is travelling north at a constant speed because the question says there is no net force acting on him.
The new force is against the direction of travel so will decelerate (slow down) the cyclist because,
Force, mass and acceleration are linked by the formula
= Force = mass x acceleration
= F = m X a
Moreover, it is given that the net force acting initially on the bicycle is zero.
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if the power delivered were 59.32 kw, evaluate the outer surface temperature, in k, and the rate of entropy production, in kw/k, for the gearbox as the system, keeping input power, h, and a from example 2.4 the same.
In the case of the gearbox as the system, the outer surface temperature is 297 K, and the rate of entropy formation is 2.3*10⁻³ kW. A surface's temperature is its ambient temperature.
More specifically, it can be talking about: Surface temperature, the temperature of the air at the earth's Surface temperature. A state of disorder, randomness, or uncertainty are frequently associated with the scientific notion of entropy, which is also a quantifiable physical attribute. Since work is produced by ordered molecular motion, entropy also serves as a proxy for a system's molecular disorder or unpredictability.
Given,
Power = 59.32
Q = ha(Ta-Tb) (Ta-Tb)
Q = (0.171)((1.0)(347k) (347k)
Q = 2.3*10⁻³ kW
Therefore the rate entropy of the gearbox is 2.3*10⁻³ kW.
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(a) a certain lens has a focal length of 79.0 cm in air and a refractive index of 1.55. find its focal length in water.
The focal length of lens in water is 23.75cm if focal length of same lens in air is 79cm and refractive index is 1.55
Delineation of the frequency and refraction points. The refractive file decides how much the way of light is bowed, or refracted, while entering a material. This is depicted by Snell's law of refraction, n1 sin θ₁ = n2 sin θ₂, where θ1 and θ2 are the point of frequency and point of refraction, separately, of a beam crossing the connection point between two media with refractive records n1 and n2.
The refractive records likewise decide how much light that is reflected while arriving at the point of interaction, as well as the basic plot for all out inward reflection, their power (Fresnel's conditions) and Brewster's angle.
According to lens maker's formula, we know that
1/f=(μ-1)(1/R₁-1/R₂)
where f is the focal length of lens and μ is refractive index ,and R₁,R₂ are the radius of given lens.
For focal length in air
=>1/f₁=(1.55-1)(1/R₁-1/R₂)-----eq1
Now, for focal length in water
=>1/f₂=(1.55/1.33-1)(1/R₁-1/R₂)------eq2
On dividing eq2 by eq1,we get
=>f₁/f₂ = [(1.55-1)×1.33] / (1.55-1.33)]
=>79 / f₂ =[(0.55)×1.33] / (0.22)]
=> f₂ =79 / [(0.7315/0.22)]
=> f₂ = 79/3.325
=>f₂ =23.75cm
Hence, focal length in water is 23.75cm.
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in the experiment, a meter is hooked up to a speaker to monitor the amplitude of the received sound. suppose the background signal level is 13 mv, the signal is 91 mv with no attenuator and is 25 mv with an attenuator in place. calculate pt/pi including the background correction. round your answer to 3 digits after the decimal point.
15.38 % is pt/pi including the background correction.
What is the definition of an amplitude simple?The greatest deviation from equilibrium of a point on either a vibrating body or wave in terms of displacement or distance traveled. It is similar to the half-length of the vibration path.
Briefing :
Background [tex]=13mv[/tex]
corrected signal [tex]= 91mv - 13 mv = 78mv[/tex]
with attenuator[tex]=25mv - 13mv = 12 mv[/tex]
[tex]\frac{P_{t} }{P_{i} } = \frac{12}{78} . 100[/tex]
[tex]= 15.38 %[/tex] %
Therefore , the pt/pi is 15.38 %
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A 15-kilogram cart is at rest on a horizontal surface. A 5-kilogram box is placed in the cart. Compared to the mass and inertia of the cart, the cart-box system has.
Compared to the mass and inertia of the cart, the cart-box system has more mass and more inertia.
Inertia refers to the position in which a body repels a change in its position of motion or rest. Moreover, it also includes a change in the direction of an object.A 15-kilogram cart at rest and a 5-kilogram box would make up a 20-kilogram cart and box that is at rest on a horizontal surface.The mass changed into something more, of course, as a result of combining the two object into one and by combining the two objects' mass, the inertia that it previously possessed as a cart by itself was increased when the inertia of the box was also combined to the cart.To know more about Inertia visit:
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Answer:
More mass and more inertia
Explanation:
No explanation
for a counter flow double pipe heat exchanger insulated from its surroundings, the mean outlet temperature of the initially cold fluid can exceed the outlet temperature of the initially hot fluid.
Yes, In case of the counter-flow warmth exchangers the hot and bloodless fluids enter the warmth exchanger from the alternative ends and the outlet temperature of the bloodless fluid may exceed the hole temperature of the new fluid.
Observe that in a counter-flow warmth exchanger the outlet temperature of the cold fluid can exceed the opening temperature of the hot fluid but this can not take place in a parallel glide device. right here, ΔTk refers to temperature difference among the recent and cold fluids at factor okay (okay = 1 or 2).
This distributes the warmth extra calmly across the warmth exchanger and permits for maximum efficiency. In principle, the bloodless fluid can exit the heat exchanger at a better temperature than the temperature of the recent fluid outlet, despite the fact that in reality that is very tough to achieve.
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a 0.030-kg bullet is fired vertically at 200 m/s into a 0.15-kg baseball that is initially at rest. how high does the combined bullet and baseball rise after the collision, assuming the bullet embeds itself in the ball?
If a 0.030-kg bullet is shot vertically at 200 m/s into a 0.15-kg baseball, the combined height of the bullet and baseball following the collision is at most 56.6 m, provided the bullet embeds itself in the ball.
This phase will begin immediately prior to the collision and conclude immediately following the impact (when the two objects are on the verge to move up as a single unit). Note that we are assuming the ball will initially be moving at a speed of 200 m/s when the bullet then collision with it. At this point, we define the system as the ball plus the bullet. We are ignoring any outside factors, such as air resistance or even gravitational force, that can affect this system. As a result, the final velocity of the system immediately following the impact can be determined using the conservation of momentum principle, p=0. The gravitational potential energy is being ignored because we assume that the ball and the bullet were at the same level just before (and just after) colliding.
Yf = v1^2*2*9.8
Yf = 33.3^2*2*9.8
Yf = 56.6 metres.
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which planet(s) will experience an increase in gravitational force and an increase in orbital speed from this added mass? explain your reasoning.
Saturn, Uranus, and Neptune
The reason is that more mass is added interior to those planets' orbits. With a greater interior mass, Saturn, Uranus, and Neptune will experience a larger gravitational force and an increased orbital speed.
A planet's orbital speed changes, depending on how far it is from the Sun. The closer a planet is to the Sun, the stronger the Sun's gravitational pull on it, and the faster the planet moves. The farther from the Sun, the weaker the Sun's gravitational pull, and the slower it moves in its orbit.
As seen in the equation v = SQRT(G * Mcentral / R), the mass of the central body (earth) and the radius of the orbit affect orbital speed.
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in the two cases shown the mass and the spring are identical but the amplitude of the simple harmonic motion is twice as big in case 2 as in case 1. 1)how are the maximum velocities in the two cases related?
The maximum velocities in the two cases related are Vmax,2 = 2 Vmax,1
Simple harmonic motion :
In physics, simple harmonic motion is the repeated back-and-forth motion through an equilibrium, or center, position so that the maximum displacement on one side of this position is equal to the maximum displacement on the other. Each whole vibration occurs at the same time interval.
Complete question:
In the two cases shown the mass and the spring are identical but the amplitude of the simple harmonic motion is twice as big in Case 2 as in Case 1.
1)How are the maximum velocities in the two cases related?
Vmax,2 = Vmax,1
Vmax,2 = 2 Vmax,1
Vmax,2 = 4 Vmax,1
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using newton law of gravity calculate the lenghth of the year. consider the orbit of earth is a circule
Using Newton's Law of gravity, the length of the year is 364.24 days.
According to Newton's law, there exist a force of gravitation between the sun and the earth, The gravitational force is responsible for the revolution of earth around the sun,
So, we can write,
MW²/R = GMm/R²
Where,
M is the mass of earth,
m is the mass of the sun,
R is the distance between the sun and the earth,
G is the universal Gravitational constant.
Solving the relation,
W² = Gm/R
W = √(Gm/R)
The time period of revolution is given by,
T = 2π/W
So, we can write,
T = 2π√(R/Gm)
Bow, putting all the values,
T = 2π(147 x 10⁶/2 x 10³⁰ x 6 x 10⁻¹¹)
T = 364.24 Days.
So, the length of the year is 364.24 years.
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what is the size of an image of a 1.00 mm object, such as lettering inside a ring, held at 8.00 cm? take the lens-to-retina distance to be 2.00 cm.
Given details :
a) For this exercise we use the equation for thin lenses
1 / f = 1 / p + 1 / q
where f is the focal length, p the distance to the object and q the distance to the image
In this case q is the distance from the lens to the retina q = 2.00 cmThe distance p is the distance from the object to the lens p = 8.00 cm1 / f = 1 / 8.00 + 1/2
1 / f = 0.625 cm⁻¹
power is the inverse of the focal length in meters
P = 1 / f [m]
P = 0.625
P = 62.5 diopter
b) the size of the image
m = - p / q
m = - 2 / 8.00
m = -0.25 X
if we assume that the object is right
m = h´ / h
h´ = m h
h´ = -0.25 x 2
h´ = -0.5 mm
if the height of the object h is straight the image and inverted
c) if the object were at p = 25.0 cm
m = - 2/25
m = -0.08
h´ = -0.08 x 2
h´ = -0.16 mm
the image would be much smaller
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NASA's Skylab, the largest spacecraft ever to fall back to the Earth, reentered the Earth's atmosphere on July 11,1979, and broke into a myriad of pieces. One of the largest fragments was a 1770-kg, lead-lined film vault, which landed with an estimated speed of 120 m/s.
If its speed at an altitude of 146 km was 7950 m/s, how much nonconservative work was done on the film vault during reentry?
When the movie vault landed, it had a kinetic energy of 12744000J. The energy an object has as a result of motion is known as kinetic energy in physics.
It is described as the effort required to move a mass-determined body from rest to the indicated velocity. The body holds onto the kinetic energy it acquired during its acceleration until its speed changes. Where m is the mass and v is the velocity, the definition of kinetic energy is:
Equation 1 can be used to calculate the kinetic energy of the movie vault when it landed.
As a result, 12744000J of kinetic energy was present when the movie vault landed.
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An electron is accelerated from rest through a potential difference V. Its final speed is proportional to???
Answer:
E = V q energy attained by charge q thru potential energy V
E = 1/2 m v^2 kinetic energy of charge q
1/2 m v^2 = V q equating energy attained
v^2 = 2 V q / m
v = (2 V q / m)^1/2
The final speed is proportional to the square roots of V, q, and inverse of square root of m.
A planet moves in an elliptical orbit around the sun. The mass of the sun is Ms . The minimum and maximum distances of the planet from the sun are R1 and R2, respectively. Using Kepler's 3rd law and Newton's law of universal gravitation, find the period of revolution of the planet as it moves around the sun. Assume that the mass of the planet is much smaller than the mass of the sun. Use for the gravitational constant. Express the period in terms of G, Ms,R1 , and R2.
Assume that the planet's mass is substantially lower than that of the sun. Use in relation to the gravitational constant. Describe the duration in terms of G, Ms,R1 , and R2 is π√(R1+R2)³2GM.
By applying Newton's law, we can state that the centripetal force experienced by the planet due to its elliptical orbit will equalize the gravitational force exerted by the sun on the planet.
mv₂/(R1+R2)/2=GMm/(R1+R2/2)²
Here, m is the mass of the planet, v is the orbital velocity of the planet and G=6.67×10⁻¹¹N⋅m²⋅kg⁻² is the universal gravitational constant.
The period of revolution is given by,
P=2π(R1+R2)/v
Substituting we get,
P=2π((R1+R2)/2)√2GM/R1+R2
=π√(R1+R2)³2GM.
Hence, the time period of revolution is π√(R1+R2)³2GM.
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What is the maximum speed at which a car can safely travel around a circular track of radius 55. 0 m if the coefficient of friction between the tire and road is 0. 350?.
The maximum speed at which a car can safely travel around a circular track of radius 55. 0 m is 10.84 m/s
What does it imply to speak of centripetal forces?Any object is drawn toward the center of a curved surface by a centripetal force. Imagine that you are a yo-yo being spun. The yo-yo is set in motion by your arm movement, and as you twirl it, the tension of the string keeps it spinning.
Given : r=30 m
μ=0.4
g=9.8m/s
2The maximum speed
v max =μgr= 0.4×9.8×30 =10.84 m/s
How would you describe centripetal force to a young person?A force known as centripetal force keeps things moving in a circular motion. The object will keep rotating in a circle as long as the centripetal force persists and all other forces are equally balanced. One object that moves in a circle as a result of centripetal force is a yo-yo.
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Nuclear decay occurs according to first-order kinetics. A nuclide decays in 23. 0 minutes from 12. 9 g to 2. 04 g. What is the rate constant for the nuclide?.
The rate constant for the nuclide is 0.08.
We know that rate of first order reaction is given by: [tex]kt = 2.303 * log(R'/R)[/tex]
where k = rate of reaction
t = time = 23 minutes
R' = initial mass = 12.9 grams
R = final mass = 2.04 grams
Putting these values in above equation we get k = 0.08.
i.e. the rate constant for the nuclide is 0.08.
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a 100 meter long copper wire has a resistance of 100 ω. another copper wire has the same mass but is 10 meters long. find the resistance of the second wire. group of answer choices 1 ω 10 ω 100 ω 1,000 ω
The resistance of the second wire will be 10 ω.
What is resistance?
Electrical resistance, or resistance to electricity, is a force that opposes the flow of current. It acts as a gauge for the difficulty of current flow in this way. Ohms (Ω) are used to express resistance values. Electricity would then flow from high to low when two terminals have an electron differential. That flow is countered by resistance. The current decreases as resistance increases. On the other hand, the current increases as the resistance decreases. Resistance could be calculated as a number using the circuit's voltage and current.
The resistance of a wire is inversely proportional to its length, so the second wire, which is 10 meters long, will have a resistance that is 10 times less than the first wire, which is 100 meters long. The resistance of the second wire will be 100 ω ÷ 10 = 10 ω.
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