Answer:
The angle between the electric field and the axis of the filter is 54⁰
Explanation:
Apply the equation for intensity of light through a polarizer.
[tex]I = I_oCos^2 \theta[/tex]
where;
I is the intensity of the transmitted light
I₀ is the intensity of the incident light
θ is the incident angle
If only 35 % of the intensity of a polarized light wave passes through a polarizing filter, then the ratio of the intensity of the transmitted light to that of the intensity of the incident light is given by;
[tex]\frac{I}{I_o} = Cos^2 \theta\\\\\frac{35}{100} = Cos^2 \theta\\\\Cos^2 \theta = 0.35\\\\Cos\theta = \sqrt{0.35} \\\\Cos\theta = 0.5916\\\\\theta = Cos^{-1}(0.5916)\\\\\theta = 54 ^0[/tex]
Therefore, the angle between the electric field and the axis of the filter is 54⁰
The angle between the electric field and the axis of the filter for the polarized light wave which passes through a polarizing filter is 54°.
What is electric field?The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge.
From the Malus's law, the intensity of the polarized beam can be calculated with the following formula.
[tex]I=I_o\cos^2\theta[/tex]
Here, (I₀) is the intensity of the polarized beam incident on the observer θ is the angle of incident.
It is given that only 35 % of the intensity of a polarized light wave passes through a polarizing filter.
The angle between the electric field and the axis of the filter can be found out using the above formula as,
[tex]35\%=(100\%)\cos^2\theta\\\theta=\cos^{-1}\sqrt{\left(\dfrac{35}{100}\right)}\\\theta=54^o[/tex]
Hence, the angle between the electric field and the axis of the filter for the polarized light wave which passes through a polarizing filter is 54°.
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What net force is necessary to give a 2 kg mass that is initially at rest an acceleration of 5 m/s2?
Answer:
10 NExplanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 2 × 5
We have the final answer as
10 NHope this helps you
The legs includes which anatomical features? Select all that apply. CD occipital region patellar region plantar region crural region cranial region lumbar region DONE.f
Answer:
Patellar region
Plantar region
Crural region
Answer:
B, C, D
Patellar, Plantar, Crural regions
Explanation:
Legs are the lower limbs of the body, consisting of all these parts. Just did assignment too.
Help!!
A 30-N force is applied to a 4-kg object to move it with a constant
velocity of 2 m/s across a level surface. The coefficient of friction
between the object and the surface is approximately (Use the
approximation: g - 10 m/s/s.)
A 0.20
B O 0.50
C 0.55
D 0.75
Answer:
[tex]\mu=0.75[/tex]
Explanation:
Given that,
Force acting on an object, F = 30 N
Mass of the object, m = 4 kg
It is moving with a constant velocity of 2 m/s across a level surface.
We need to find the coefficient of friction between the object and the surface. Let it is μ. Force in terms of coefficient of friction is given by :
F = μ N, Where N is normal force, N = mg
[tex]\mu=\dfrac{F}{mg}\\\\\mu=\dfrac{30}{4\times 10}\\\\\mu=0.75[/tex]
So, the coefficient of friction between the object and the surface is 0.75.
As shown in the diagram below, a rope attached to a 500.-kilogram crate is used to exert a force of 45 newtons at an angle of 65 degrees above the horizontal 45 N at an angle of 65 degrees above the horizontal.
500 kg
The horizontal component of the force acting on the crate is?
Answer:
19.01 N
Explanation:
F = Force being applied to the crate = 45 N
[tex]\theta[/tex] = Angle at which the force is being applied = [tex]65^{\circ}[/tex]
Horizontal component of force is given by
[tex]F_x=F\cos\theta\\\Rightarrow F_x=45\times \cos65^{\circ}\\\Rightarrow F_x=19.01\ \text{N}[/tex]
The horizontal component of the force acting on the crate is 19.01 N.
An object is moving along a straight line, and the uncertainty in its position is 1.90 m.
Required:
Find the minimum uncertainty in the momentum of the object. Find the minimum uncertainty in the object's velocity, assuming that the object is (b) a golf ball (mass=0.045 kg) and (c) an electron.
Answer:
[tex]2.78\times 10^{-35}\ \text{kg m/s}[/tex]
[tex]6.178\times 10^{-34}\ \text{m/s}[/tex]
[tex]0.31\times 10^{-4}\ \text{m/s}[/tex]
Explanation:
[tex]\Delta x[/tex] = Uncertainty in position = 1.9 m
[tex]\Delta p[/tex] = Uncertainty in momentum
h = Planck's constant = [tex]6.626\times 10^{-34}\ \text{Js}[/tex]
m = Mass of object
From Heisenberg's uncertainty principle we know
[tex]\Delta x\Delta p\geq \dfrac{h}{4\pi}\\\Rightarrow \Delta p\geq \dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p\geq \dfrac{6.626\times 10^{-34}}{4\pi\times 1.9}\\\Rightarrow \Delta p\geq 2.78\times 10^{-35}\ \text{kg m/s}[/tex]
The minimum uncertainty in the momentum of the object is [tex]2.78\times 10^{-35}\ \text{kg m/s}[/tex]
Golf ball minimum uncertainty in the momentum of the object
[tex]m=0.045\ \text{kg}[/tex]
Uncertainty in velocity is given by
[tex]\Delta p\geq m\Delta v\geq 2.78\times 10^{-35}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{0.045}\\\Rightarrow \Delta v\geq 6.178\times 10^{-34}\ \text{m/s}[/tex]
The minimum uncertainty in the object's velocity is [tex]6.178\times 10^{-34}\ \text{m/s}[/tex]
Electron
[tex]m=9.11\times 10^{-31}\ \text{kg}[/tex]
[tex]\Delta v\geq \dfrac{\Delta p}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{9.11\times 10^{-31}}\\\Rightarrow \Delta v\geq 0.31\times 10^{-4}\ \text{m/s}[/tex]
The minimum uncertainty in the object's velocity is [tex]0.31\times 10^{-4}\ \text{m/s}[/tex].
Starting at 1.3 m/s, a runner accelerates at a constant 0.22 m/s2 for 6.0 s. What is the runner’s displacement during this time interval?
Answer:
answer is 11.76 meter
Explanation:
use 2nd equation of motion
S=ut+1/2at^2
5)
Using only the information available in the periodic table, consider the elements calcium and chlorine. From their
location on the periodic table, identify the oxidation state and number of valence electrons for calcium and chlorine.
Which statement most accurately describes the compound formed by calcium and chlorine?
C
A)
B)
Calcium, a nonmetal with an oxidation number of +2 will form a covalent
bond with chlorine, a halogen (nonmetal) with an oxidation number of -1
called calcium chloride (CaCl2).
Calcium, an alkaline earth metal with an oxidation number of +2 will form
covalent bond with chlorine, also a metal with an oxidation number of -1
called calcium dichloride (CaCla)
Calcium, an alkaline earth metal with an oxidation number of +2 will form
an ionic bond with chlorine, a halogen in group VILA with an oxidation
number of -1 called calcium chloride (CaCl2).
Calcium, an alkaline earth metal with an oxidation number of 2 will share
electrons to form an lonic bond with chlorine, a nonmetal with an
xidation number of -1 called calcium dichloride (CaCl).
D)
Answer:C,(Calcium,an alkaline earth metal with an oxidation number of +2 will form an ionic bond with chlorine,a halogen in group VllA with an oxidation number of -1 called calcium chloride (CaCl2)
Explanation:
on USAtestprep !!
Calcium, an alkaline earth metal with an oxidation number of +2 will form
an ionic bond with chlorine, a halogen in group VIIA with an oxidation
number of -1 called calcium chloride (CaCl₂). This is correct statement.
What is oxidation number?Simply said, the number assigned to each element in a chemical combination is the definition of an oxidation number. The total number of electrons that an atom in a molecule can share, lose, or gain while forming a chemical bond with an atom of a different element is known as the oxidation number.
Also known as oxidation state, oxidation number is a numerical value. But depending on whether we take into account the atoms' electronegativity or not, these phrases might occasionally have a different meaning. In coordination chemistry, the term "oxidation number" is often used.
According to Periodic table: calcium is a alkaline earth metal with an oxidation number of +2 whereas chlorine is a halogen in group VIIA with an oxidation number of -1. When they reacts chemically, they form an ionic compound named calcium chloride having chemical formula CaCl₂.
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A hunter aims directly at a target (on the same level) 120 m away. If the bullet leaves the gun at a speed of 250 m/s, by how much will it miss the target
Answer:
The distance the bullet will miss the target is 1.13 m.
Explanation:
Given;
Initial velocity of the bullet = 250 m/s
Distance of the target = 120 m
Time of motion;
t = 120 / 250
t = 0.48 s
During this time the bullet is under the gravitational pull and the distance it will miss the target is given by;
Y = V₀y + ¹/₂gt²
where;
V₀y is the initial vertical velocity = 0
Y = 0+ ¹/₂gt²
Y = ¹/₂(9.8)(0.48)²
Y = 1.13 m
Therefore, the distance the bullet will miss the target is 1.13 m.
How much energy is required to move a 1000 kg object from the Earth's surface to an altitude twice the Earth's radius?
An energy of [tex]4.171\times 10^{10}[/tex] joules to move a 1000-kg object from the Earth's surface to an altitude twice the Earth's radius.
Since the object must be moved away to a distance greater than the radius of the Earth, then change in gravitational potential energy must be based on Newton's Law of Gravitation.
By the Work-Energy Theorem, the work ([tex]W[/tex]), in joules, done on the object is equal to the change in gravitational potential energy ([tex]U_{g}[/tex]), in joules:
[tex]W = U_{g}[/tex] (1)
[tex]W = -G\cdot m\cdot M\cdot \left(\frac{1}{r_{f}}-\frac{1}{r_{o}} \right)[/tex] (1b)
Where:
[tex]G[/tex] - Gravitational constant, in cubic meters per kilogram-square second.[tex]m[/tex] - Mass of the object, in kilograms.[tex]M[/tex] - Mass of the Earth, in kilograms.[tex]r_{o}[/tex] - Initial distance, in meters.[tex]r_{f}[/tex] - Final distance, in meters.If we know that [tex]G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex], [tex]m = 1000\,kg[/tex], [tex]M = 5.972\times 10^{24}\,kg[/tex], [tex]r_{o} = 6.371\times 10^{6}\,m[/tex] and [tex]r_{f} = 19.113\times 10^{6}\,m[/tex], then the energy required to move the object from the Earth's surface is:
[tex]W = -\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (1000\,kg)\cdot (5.972\times 10^{24}\,kg)\cdot \left[\frac{1}{19.113\times 10^{6}\,m} - \frac{1}{6.371\times 10^{6}\,m} \right][/tex][tex]W = 4.171\times 10^{10}\,J[/tex]
An energy of [tex]4.171\times 10^{10}[/tex] joules to move a 1000-kg object from the Earth's surface to an altitude twice the Earth's radius.
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Helicopters rotor blades, could spin at high speed of 510 rpm. Find the angular displacement in radian for 3 hour(s) operation
Answer:
The angular displacement of the blade is 576,871.2 radians
Explanation:
Given;
angular speed of the Helicopters rotor blades, ω = 510 rpm (revolution per minute)
time of motion, t = 3 hours
The angular speed of the Helicopters rotor blades in radian per second is given as;
[tex]\omega = \frac{510 \ rev}{mins} *\frac{2 \pi \ rad}{1 \ rev} *\frac{1 \ min}{60 \ s}\\\\\omega = 53.414 \ rad/s[/tex]
The angular displacement in radian is given as;
θ = ωt
where;
t is time in seconds
θ = (53.414)(3 x 60 x 60)\\
θ = 576,871.2 radians
Therefore, the angular displacement of the blade is 576,871.2 radians
please help i’ll mark u branliest
Answer:
62
Explanation:
it doesn't need explanation
One object has a temperature twice as large as another object. If the objects have the same surface area, how much more power does the hotter object radiate than the cooler object
Answer:
The hotter object radiate more power than the cooler object 15 times i.e 15σeA[tex]T^{4}[/tex]
Explanation:
From Stefan's law, an object would radiate power with respect to its temperature.
i.e Radiative power, Q = σeA[tex]T^{4}[/tex]
where Q is the radiative power, σ is the constant, e is the emissivity of the object, A is the area of the object and T is the temperature.
Let the temperature of the cooler object be represented by T.
So that its radiative power = σeA[tex]T^{4}[/tex]
Given that the temperature of the hotter object is twice as large as that of the cooler object.
Temperature of hotter object = 2T
So that its radiative power = σeA[tex](2T)^{4}[/tex]
= 16σeA[tex]T^{4}[/tex]
Radiative power difference between the two objects = 16σeA[tex]T^{4}[/tex] - σeA[tex]T^{4}[/tex]
= 15σeA[tex]T^{4}[/tex]
The hotter object radiate more power than the cooler object 15 times.
The hotter object radiates 15 times more power than the power of cooler object.
Absolute Temperature of one object = [tex]T_1[/tex]
Absolute Temperature of second object =[tex]T_2[/tex] = [tex]2T_1[/tex]
The Power emitted by the an object is given by the equation (1)
[tex]P= A\epsilon \sigma\;T^4[/tex].......(1)
Equation (1) is called as Stephan Boltzmann Law
Where
P = Power emitted by the object in Joule
A = Surface area of the object
[tex]\epsilon[/tex] = Emissivity of the object
T = Absolute Temperature
Let us consider emissivities are equal
[tex]So \; P_1/P_2 = T_1^4/T_2^4\\[/tex] ( Areas of both objects are equal)
[tex]P_1/P_2= T_1^4/16T1^4= 1/16[/tex]
[tex]P_2= 16 P_1[/tex]
Difference in Power = [tex]16P_1- P_1[/tex] = [tex]15P_1[/tex]
So the hotter object radiate 15 times more power than the power of cooler object.
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Artificial gravity is a must for any space station if humans are to live there for any extended length of time. Without artificial gravity, human growth is stunted and biological functions break down.
An effective way to create artificial gravity is through the use of a rotating enclosed cylinder, as shown in the figure. Humans walk on the inside of the outer edge of the cylinder, which has a diameter of =3335 m that is large enough such that its curvature is not readily noticeable to the inhabitants. (The space station in the figure is not drawn to scale.)
Once the space station is rotating at the necessary angular speed to create an artificial gravity of 1 , how many minutes would it take the space station to make one revolution?
I did [tex]2\pi \sqrt{\frac{1667.6}{9.8} } =81.96[/tex] then I converted 81.96 to minutes which was 1.4 but it still got marked wrong. 81.96 was wrong as well. Am I using the wrong equation for it? I'm not sure what to do.
The period of the enclosed cylinder is approximately 115.866 seconds.
The rotating enclosed cylinder is rotating at constant angular speed ([tex]\omega[/tex]), in radians per second, which means that experiments a constant radial angular acceleration ([tex]\alpha[/tex]), in radians per square second. Then, we derive an expression for the period of the cylinder, this is, the time needed by the cylinder to make one revolution:
[tex]g = \omega^{2}\cdot R[/tex] (1)
Where:
[tex]g[/tex] - Gravitational acceleration, in meters per square second. [tex]R[/tex] - Radius of the enclose cylinder, in meters.[tex]g = \frac{4\pi^{2}\cdot R}{T^{2}}[/tex]
[tex]T = 2\pi\cdot \sqrt{\frac{R}{g} }[/tex] (2)
Where [tex]T[/tex] is the period, in seconds.
If we know that [tex]R = 3335\,m[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the period of the enclosed cylinder is:
[tex]T = 2\pi\cdot \sqrt{\frac{3335\,m}{9.807\,\frac{m}{s^{2}} } }[/tex]
[tex]T \approx 115.866\,s\,(1.931\,min)[/tex]
The period of the enclosed cylinder is approximately 115.866 seconds.
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Using a launch speed of 40.0 m/s and any angle between 0 and 90 degrees, what would be the largest possible range for a projectile?
its 45
Answer:
The largest possible range of the projectile is 163.27 m.
Explanation:
Given;
launch speed, u = 40 m/s
angle of projection, θ; between 0⁰ and 90⁰
The range of a projection is given as;
[tex]R = \frac{u^2sin(2\theta )}{g}[/tex]
The largest possible range will occur at 45 degrees angle of projection;
[tex]R = \frac{u^2sin(2\theta )}{g} \\\\R = \frac{(40)^2sin(2\ \times \ 45^0 )}{9.8}\\\\R = \frac{(40)^2sin( 90^0 )}{9.8}\\\\R = \frac{(40)^2( 1 )}{9.8} \\\\R = 163.27 \ m\\\\[/tex]
Therefore, the largest possible range of the projectile is 163.27 m.
A boy on a bicycle rides in a circle of radius ro at speed vo. If the boy now rides at a radius equal to half the initial radius ro, by what approximate factor must he change his speed in order to have the same radial acceleration
Answer:
The speed must change by [tex]\frac{1}{\sqrt{2}}[/tex] factor in order to have the same radial acceleration.
Explanation:
The radial or centripetal acceleration is given by:
[tex] a_{c} = \frac{v^{2}}{r} [/tex]
Where:
v: is the speed = v₀
r: is the radius = r₀
[tex] a_{c} = \frac{v_{0}^{2}}{r_{0}} [/tex] (1)
If the radius is now equal to half the initial radius the speed must be:
[tex]a_{c} = \frac{v^{2}}{r_{0}/2}[/tex] (2)
By equating equation (1) and (2):
[tex] \frac{v_{0}^{2}}{r_{0}} = \frac{v^{2}}{r_{0}/2} [/tex]
[tex]v^{2} = \frac{v_{0}^{2}}{2}[/tex]
[tex] v = \frac{v_{0}}{\sqrt{2}} [/tex]
Therefore, the speed must change by [tex]\frac{1}{\sqrt{2}}[/tex] factor in order to have the same radial acceleration.
I hope it helps you!
A rolling ball has 8 joules of kinetic energy and is rolling 4m/s. Find it’s mass
Answer:
m = 1
Explanation:
K.E = 8J
v = 4m/s
m = ?
Now,
K.E = 1÷2mv^2
8 = 1÷2 × m × (4)^2
8 = 1÷2 × m × 16
8 = m × 8
m × 8 = 8
m = 8 ÷ 8 = 1
m = 1
VERIFICATION:K.E. = 1÷2mv^2
K.E = 1÷2 × 1 × 4^2
K.E. = 8J
-TheUnknownScientist
Jerry runs 60 meters east and then 20 meters west in 10
seconds. His average velocity is
m/s.
Answer: 8 meters per second
Explanation: If you add 60 to 20 you get 80 meters and since he ran those 80 meters in 10 seconds you divide 80 by ten and get 8 and then you get 8m/s
Which is one of Edwin Hubble’s findings that supports the big bang theory?
Answer:
Edwin Hubble found that galaxies are constantly moving away from us. According to his observations with the Hubble Space Telescope, galaxies are moving at different speeds. This shows that the universe is expanding. The farther away a galaxy is, the faster it is moving away. Found this on google hope this helps.
Answer:
A) the universe started at a central point
Explanation:
taking the quiz on eg. :))
The asteroid 234 Ida has a mass of about 4 × 1016 kg and an average radius of about 16 km. What is the acceleration due to gravity on 234 Ida? Assume that the asteroid is spherical; use G = 6.67 × 10–11 Nm2/kg2.
A. 1 cm/s2
B. 2 cm/s2
C. 5 cm/s2
D. 6 cm/s2
Answer:
1 cm/s²
Explanation:
I just took the quiz
The asteroid 234 Ida has a mass of about 4×1016 kg and an average radius of about 16 km. The acceleration due to gravity will be 1.04 cm/s². Hence, option A is correct.
What is the acceleration due to gravity?The acceleration an object experiences as a result of gravitational force is known as acceleration due to gravity. M/s² is its SI unit. Its vector nature—which includes both magnitude and direction—makes it a quantity. The unit g stands for gravitational acceleration. At sea level, the standard value of g on the earth's surface is 9.8 m/s².
The formula for the acceleration due to gravity is g=GM/r².
According to the question, the given values are :
Mass, M = 4 × 1016 kg or
M = 4 × 10¹⁶.
Radius, r = 16 km or,
r = 16000 meter.
G = 6.67 × 10⁻¹¹ Nm²/kg²
g = (6.67 × 10⁻¹¹ ) (4 × 10¹⁶) / 16000²
g = 0.0104 m/s² or,
g = 1.04 cm/s².
Hence, the acceleration due to gravity will be 1.04 m/s²
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The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4. Determine the magnitude of force at point and determine if the ladder will slip.
This question is incomplete, the complete question;
The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.
Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb
Answer:
- the magnitude of force at point A is 79.1033 lb
- since FA < FA_max; Ladder WILL NOT slip
Explanation:
Given that;
∑'MA = 0
⇒ NB [Lsin∅] - W[L/2.cos∅] = 0
NB = W / 2tan∅ -------let this be equation 1
∑Fx = 0
⇒ FA - NB = 0
FA = NB
therefore from equation 1
FA = NB = W / 2tan∅
we substitute in our values
FA = NB = 76 / 2tan(60°) = 21.9393 lb
Now ∑Fy = 0
NA - W = 0
NA = W = 76 lb
Net force at A will be
FA' = √( NA² + FA²)
= √( (W)² + (W / 2tan∅)²)
we substitute in our values
FA' = √( (76)² + (21.9393)²)
= √( 5776 + 481.3328)
= √ 6257.3328
FA' = 79.1033 lb
Therefore the magnitude of force at point A is 79.1033 lb
Now maximum possible frictional force at A
FA_max = μ × NA
so, FA_max = 0.4 × 76
FA_max = 30.4 lb
So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.
Therefore since FA < FA_max; Ladder WILL NOT slip
A 150 kg boy and his bike are traveling 12 m/s when he slams on his breaks and stop at his friend’s house. How much impulse is required to produce this change in momentum?
Please someone help me with this I’ll give brainliest
Answer:
J = 1800 kg-m/s
Explanation:
Given that,
Mass of a boy, m = 150 kg
Initial velocity of a boy, u = 12 m/s
Finally, it stops, v = 0
We need to find the impulse is required to produce this change in momentum. We know that impulse is equal to the change in momentum. So,
[tex]J=m(v-u)\\\\=150\times (0-12)\\\\=-1800\ kg-m/s\\\\|J|=1800\ kg-m/s[/tex]
So, the impulse is equal to 1800 kg-m/s
Two tectonic plates moving toward one another are at a
ANSWER CHOICES
convergent boundary.
divergent boundary.
subduction boundary.
transform boundary.
Answer:
A. cause i just took the test
Explanation:
Answer:
its A
Explanation:
no explanation is needed, just trust me.
please answer this question
You compress a spring by x, and then release it. Next you compress the spring by 2x. How much more work did you do the second time than the first
Answer:
Work done is [tex]\frac{3}{2}[/tex]k[tex]x^{2}[/tex].
Explanation:
The work done by the spring is the same as the potential energy stored in the spring.
So that,
work done = potential energy = [tex]\frac{1}{2}[/tex] k[tex]x^{2}[/tex]
where k is the spring constant of the material of the spring, and x is the compression.
When the spring is compressed by x;
work done = [tex]\frac{1}{2}[/tex] k[tex]x^{2}[/tex]
When the spring is compressed by 2x;
work done = [tex]\frac{1}{2}[/tex] k[tex](2x)^{2}[/tex]
= [tex]\frac{1}{2}[/tex] k(4[tex]x^{2}[/tex])
= 2k[tex]x^{2}[/tex]
Therefore,
The work done the second time more than the first = 2k[tex]x^{2}[/tex] - [tex]\frac{1}{2}[/tex] k[tex]x^{2}[/tex]
= [tex]\frac{3}{2}[/tex]k[tex]x^{2}[/tex]
The work done the second time more than the first is [tex]\frac{3}{2}[/tex]k[tex]x^{2}[/tex].
Spring work is equivalent to the effort done to extend the spring, Work did you do the second time than the first time will be [tex]\rm \frac{3}{2} Kx^2[/tex].
What is spring work?Spring work is equivalent to the effort done to extend the spring, which is dependent on both the spring constant k and the distance stretched.
The potential energy stored as a result of the deformation of an elastic item, such as spring stretching, is referred to as elastic potential energy.
Work done by spring = potential energy
[tex](PE)_{spring }= \frac{1}{2} Kx^2[/tex]
Case 1
spring is compressed by x
[tex](PE)_1{spring }= \frac{1}{2} Kx^2[/tex]
Case 2
spring is compressed by 2x
[tex]\rm (PE)_2{spring }= \frac{1}{2} K(2x)^2\\\\\rm (PE)_2{spring }= \frac{1}{2}\times 4K(x)^2\\\\\rm (PE)_2{spring }= 2K(x)^2[/tex]
The difference in the potential energy is found by;
[tex](PE)_2-(PE)_1=2Kx^2-\frac{1}{2} Kx^2=\frac{3}{2} Kx^2[/tex]
Hence spring work did you do the second time than the first time will be [tex]\rm \frac{3}{2} Kx^2[/tex].
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A jeweler's grinding wheel slows down at a constant rate from 185 rad/s to 105 rad/s while it rotates through 16.0 revolutions. How much time does this take?
Answer:
t = 0.6933 s
Explanation:
This is a rotational kinematics exercise, let's find the angular acceleration of the wheel
w² = w₀² + 2 α θ
α = (w² - w₀²) / 2 θ
Let's reduce the angles to the SI system
θ = 16 rev (2π rad / 1 rev) = 32π rad
let's calculate
α = (105² - 185²) / (2 32π)
α = -115.39 rad / s²
now let's use the relation
w = w₀ + α t
t = (w- w₀) /α
t = (105 - 185) / (- 115.39)
t = 0.6933 s
A transformer consists of a 500 turn primary coil and a 2000-turnsecondary coil. If the current in the secondary is 3.0A, what isthe current in the primary?and WHy?
Answer:
The correct solution will be "12.0 A".
Explanation:
The given values are:
[tex]N_p= 500 \ turn[/tex]
[tex]N_s= 200 \ turn[/tex]
[tex]I_s= 3.0 \ A[/tex]
By using the transformer formula, we get
⇒ [tex]\frac{N_p}{N_s} =\frac{I_s}{I_p}[/tex]
⇒ [tex]I_p = I_s\times \frac{N_s}{N_p}[/tex]
On substituting the given values, we get
⇒ [tex]=3.0 \ A\times \frac{2000}{500}[/tex]
⇒ [tex]=12.0 \ A[/tex]
To remove a stain using a solvent the stain has to become dissolved in the solvent
True
False
Answer:
True
Explanation:
have a good day:)
Answer: This statement is True
Determine the focal length of a plano-concave lens (refractive index n =1.5) with 24 cm radius of curvature on its curve surface
1)-96 cm
2)-24 cm
3)-48 cm
4)-72 cm
Answer:
Option 3: -48 cm
Explanation:
We are given:
refractive index; n = 1.5
radius of curvature; r2 = 24 cm
Formula for the focal length is given as;
1/f = (n - 1) × [(1/r1) - (1/r2)]
As r1 tends to infinity, 1/r1 = 0
Thus,we now have;
1/f = (n - 1) × (-1/r2)
Plugging in the relevant values;
1/f = (1.5 - 1) × (-1/24)
1/f = -0.02083333333
f = -1/0.02083333333
f = -48 cm
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Calculate the rotational inertia of a meter stick, with mass 0.499 kg, about an axis perpendicular to the stick and located at the 35.2 cm mark. (Treat the stick as a thin rod.)
Answer:
The rotational inertia of the meter stick is 0.0618 kgm².
Explanation:
Given;
mass of the meter stick, m = 0.499 kg
perpendicular distance to the rotational axis, r = 35.2 cm = 0.352 m
The rotational inertia or moment of inertia for a point mass is given by;
I = mr²
where;
m is the point mass
r is the perpendicular distance
Substitute the givens and solve for moment of inertia.
I = (0.499)(0.352)²
I = 0.0618 kgm²
Therefore, the rotational inertia of the meter stick is 0.0618 kgm².