Answer:
a) The velocity when it passes the equilibrium point is [tex]\pm 2.451[/tex] meters per second.
b) The velocity when it is 0.10 meters from equilibrium is [tex]\pm 1.567[/tex] meters per second.
c) The total energy of the system is 1.802 joules.
d) The equation describing the motion of the mass, assuming that initial position is a maximum is [tex]x(t) = 0.13\cdot \sin (18.850\cdot t +0.5\pi)[/tex].
Explanation:
a) If all non-conservative forces can be neglected and spring has no mass, then the mass-spring system exhibits a simple harmonic motion (SHM). The kinematic formula for the position of the system ([tex]x(t)[/tex]), measured in meters, is:
[tex]x(t) = A\cdot \sin(\omega \cdot t +\phi)[/tex] (1)
Where:
[tex]A[/tex] - Amplitude, measured in meters.
[tex]\omega[/tex] - Angular frequency, measured in radians per second.
[tex]t[/tex] - Time, measured in seconds.
[tex]\phi[/tex] - Phase, measured in radians.
The kinematic equation for the velocity formula of the system ([tex]v(t)[/tex]), measured in meters per second, is derived from (1) by deriving it in time:
[tex]v(t) = \omega\cdot A\cdot \cos (\omega\cdot t+\phi)[/tex] (2)
The velocity when it passes the equilibrium point occurs when the cosine function is equal to 1 or -1. Then, that velocity is determined by following formula:
[tex]v = \pm \omega\cdot A[/tex] (3)
The angular frequency is calculated by this expression:
[tex]\omega = 2\pi\cdot f[/tex] (4)
Where [tex]f[/tex] is the frequency, measured in hertz.
If we know that [tex]f = 3\,hz[/tex] and [tex]A = 0.13\,m[/tex], then the velocity when it passes the equilibrium point, which is the maximum and minimum velocities of the mass:
[tex]\omega = 2\pi\cdot (3\,hz)[/tex]
[tex]\omega \approx 18.850\,\frac{rad}{s}[/tex]
[tex]v = \pm \left(18.850\,\frac{rad}{s} \right)\cdot (0.13\,m)[/tex]
[tex]v = \pm 2.451\,\frac{m}{s}[/tex]
The velocity when it passes the equilibrium point is [tex]\pm 2.451[/tex] meters per second.
b) First, we need to determine the spring constant of the system ([tex]k[/tex]), measured in newtons per meter, in terms of the angular frequency ([tex]\omega[/tex]), measured in radians per second, and mass ([tex]m[/tex]), measured in kilograms. That is:
[tex]k = \omega^{2}\cdot m[/tex] (5)
If we know that [tex]\omega \approx 18.850\,\frac{rad}{s}[/tex] and [tex]m = 0.60\,kg[/tex], then the spring constant is:
[tex]k = \left(18.850\,\frac{rad}{s} \right)^{2}\cdot (0.60\,kg)[/tex]
[tex]k = 213.194\,\frac{N}{m}[/tex]
Lastly, we determine the velocity when the mass is 0.10 meters from equilibrium by the Principle of Energy Conservation:
[tex]U_{k} + K = K_{max}[/tex] (6)
[tex]\frac{1}{2}\cdot k\cdot x^{2} + \frac{1}{2}\cdot m\cdot v^{2} = \frac{1}{2}\cdot m\cdot v_{max}^{2}[/tex] (7)
Where:
[tex]U_{k}[/tex] - Current elastic potential energy, measured in joules.
[tex]K[/tex] - Current translational kinetic energy, measured in joules.
[tex]K_{max}[/tex] - Maximum translational kinetic energy, measured in joules.
[tex]v[/tex] - Current velocity of the system, measured in meters per second.
[tex]m[/tex] - Mass, measured in kilograms.
[tex]v_{max}[/tex] - Maximum velocity of the system, measured in meters per second.
If we know that [tex]k = 213.194\,\frac{N}{m}[/tex], [tex]x = 0.10\,m[/tex], [tex]m = 0.60\,kg[/tex] and [tex]v_{max} = \pm 2.451\,\frac{m}{s}[/tex], then the velocity of the mass-spring system is:
[tex]\frac{k}{m} \cdot x^{2} + v^{2} = v_{max}^{2}[/tex]
[tex]v^{2} = v_{max}^{2}-\frac{k}{m}\cdot x^{2}[/tex]
[tex]v = \sqrt{v_{max}^{2}-\frac{k\cdot x^{2}}{m} }[/tex] (8)
[tex]v = \sqrt{\left(\pm 2.451\,\frac{m}{s} \right)^{2}-\frac{\left(213.194\,\frac{N}{m} \right)\cdot (0.10\,m)^{2}}{0.60\,kg} }[/tex]
[tex]v \approx \pm 1.567\,\frac{m}{s}[/tex]
The velocity when it is 0.10 meters from equilibrium is [tex]\pm 1.567[/tex] meters per second.
c) The total energy of the system ([tex]E[/tex]), measured in joules, can be determined by the following expression derived from the Principle of Energy Conservation:
[tex]E = \frac{1}{2}\cdot m\cdot v_{max}^{2}[/tex] (9)
If we know that [tex]m = 0.60\,kg[/tex] and [tex]v_{max} = \pm 2.451\,\frac{m}{s}[/tex], then the total energy of the system is:
[tex]E = \frac{1}{2}\cdot (0.60\,kg)\cdot \left(\pm 2.451\,\frac{m}{s}\right)^{2}[/tex]
[tex]E = 1.802\,J[/tex]
The total energy of the system is 1.802 joules.
d) Given that initial position of the mass-spring system is a maximum, then we conclude that the equation of motion has the following parameters: ([tex]A = 0.13\,m[/tex], [tex]\omega \approx 18.850\,\frac{rad}{s}[/tex] and [tex]\phi = 0.5\pi\,rad[/tex])
From (1) we obtain the resulting formula:
[tex]x(t) = 0.13\cdot \sin (18.850\cdot t +0.5\pi)[/tex] (10)
The equation describing the motion of the mass, assuming that initial position is a maximum is [tex]x(t) = 0.13\cdot \sin (18.850\cdot t +0.5\pi)[/tex].
What are the effects of hot rolling on the metal grain structure?
Answer:Mark me as the brainliset
Explanation:The rollers impart compression stresses, while squeezing the metal. These results in changes of grains, grain fragmentation, and lattice distortion. In hot rolling the coarse grains are converted into smaller grains. Care has been taken that the correct range of temperature is maintained during the process.
Water can form large dewdrops in nature how would droplets made of vegetable oil instead of water be different
Answer:d
Explanation: oil would form droplets but only tiny ones because it’s surface tension is lower than that of water
Both water and vegetable oil would made droplets but Oil would form only very tiny droplets because its surface tension is lower than that of water.
What is surface tension?Surface tension is the propensity for liquid surfaces that are at rest to condense into the smallest surface area. Razor blades and insects (like water striders), which have a higher density than water, can float on the surface of the water without even becoming partially buried because to surface tension.
Surface tension at liquid-air contacts originates from the liquid molecules' stronger attraction to one another due to cohesion than to the air molecules (due to adhesion).
As the surface tension of the oil is lower than that of water, it forms droplets will smaller size.
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an effort force 40n is applied to a screwdriver to pry the lid off of a can of pain. the screwdriver applies 100n of force to the lid. what is the MA of the screwdriver?
Answer:
MA = 2.5
Explanation:
Given that,
Effort force applied to a screwdriver to pry the lid off of a can= 40 N
The screwdriver applies 100n of force to the lid = 100 N
We need to find the mechanical advantage of the screwdriver. Mechanical advantage of a machine is given by the ratio of load to the effort.
m = load/effort
= 100/40
= 2.5
Hence, the MA of the screwdriver is 2.5
A hammer strikes a nail with a 10N force for .01 seconds. Calculate the impulse of the hammer
Answer:
Impulse = 0.1 Kgm/s
Explanation:
Given the following data;
Force, F = 10N
Time, t = 0.01 seconds
To find impulse
An impulse can be defined as the net force acting an object for a very short period of time.
Mathematically, impulse is given by the formula;
Impulse = force * time
Substituting into the equation, we have;
Impulse = 10 * 0.01
Impulse = 0.1 kgm/s
Therefore, the impulse of the hammer is 0.1 kilogram meter per seconds.
The impulse of the hammer is 0.1 Ns.
To solve the given problem we need to use the formula for calculating impulse.
Impulse: This can be defined as the product of force and time on a body.
The formula of impulse is given below.
I = Ft....................... Equation 1
Where :
I = Impulse of the harmerF = Force on the nailt = timeFrom the question,
Given:
F = 10 Nt = 0.01 sSubstitute these values into equation 1
I = 10(0.01)I = 0.1 NsHence, The impulse of the hammer is 0.1 Ns.
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A circular wire loop lies inside a region of space containing a magnetic field. The direction of the magnetic field is out of the screen and parallel to the central axis of the loop. The magnitude of the magnetic field increases as a function of time. A circular loop oriented parallel to the plane of the screen lies inside a region containing magnetic field B. The field is directed out of the screen and is increasing. What is the direction of the induced current in the loop
Answer:
clockwise
Explanation:
According to the law given by Lenz, known as the Lenz law, it is said that a current induced in the circuit which is due to the change in the magnetic field and is so directed so as to oppose the change in the flux and to apply a force in the opposite direction if the force.
Here, as the magnetic field is directed out of the screen, the current flows in the direction which is clockwise in the loop and it opposes the increasing magnetic field.
The clockwise induced current will produce magnetic field in to the screen.
True/False: The core, mantle, and crust explanation for the Earth's layers is outdated and now we use the 5 layer explanation. *
Answer:
False
Explanation:
We still use the old way. It's easier to understand the old way, and its NOT outdated.
How fast should a rocket ship move for its mass to be dilated to 167% of its rest
mass?
Answer:
The two forces acting on rockets at the moment of launch are the thrust upwards and the weight downwards. Weight is the force due to gravity and is calculated (at the Earth’s surface) by multiplying the mass (kilograms) by 9.8.The resultant force on each rocket is calculated using the equation resultant force = thrust – weight.
Hopefully, this answer helps you! :)
How can a particle's position determine the
amount of potential energy available to a
system?
Potential energy of a particle is directly proportional its position, thus, potential energy of particle increases with increase in the height of the particle and decreases with a decrease in the height of the particle.
The potential energy of a particle is the energy possessed by the particle by virtue of its position. This potential energy is form of mechanical energy and it is expressed mathematically as the product of the mass of the particle, acceleration due to gravity and the particles displacement or position.
P.E = mgh
where;
P.E is the potential energy of the particle
m is the mass of the particle
g is the acceleration due to gravity
h is the position of the particle.
Potential energy of a particle is directly proportional its position, thus, potential energy of particle increases with increase in the height of the particle and decreases with a decrease in the height of the particle.
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Which best describes the law of conservation of mass?
O The coefficients in front of the chemicals in the reactants should be based on the physical state of the products,
O Products in the form of gases are not considered a part of the total mass change from reactants to products
O When reactants contain both a solid and a liquid, the solid counts toward the overall mass and the liquid does not
O The mass of the reactants and products is equal and is not dependent on the physical state of the substances.
Explanation:
The law of conservation of mass states that mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations. According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants.
Which example provides a complete scientific description of an object in motion?
(A) The hiker followed the north trail a distance of two kilometers in thirty minutes
(B) The tiger had to run at 50 kilometers per hour across the field to catch the zebra
(C) The hockey player sent the puck flying toward the north goal to score the winning point
(D) The vibration of the jackhammer broke through the south-facing rock in under 10 seconds
Answer:
C
Explanation:
it explains how an object that was in motion continued motion after the force
5. The wire in consists of two segments of different diameters but made from the same metal. The current in segment 1 is I1. a. (1 pt) Compare the currents in the two segments. That is, is I2 greater than, less than, or equal to I1? Explain. b. (1 pt) Compare the current densities J1 and J2 in the two segments. c. (1 pt) Compare the electric field strengths E1 and E2 in the two segments. d. (1 pt) Compare the drift speeds vd1 and vd2 in the two segments.
Answer:
hello your question is incomplete attached below is the complete question
answer :
a) I1 = I2
b) J1 > J2
c) E 1 > E2
d) ( vd1 ) > ( vd2 )
Explanation:
a) The currents in the two segments are the same i.e. I1 = I2 and this is because the segments are connected in series
b) Comparing the current densities J1 and J2 in the two segments
note : current density ∝ 1 / area
The area of the second segment is > the area of first segment therefore
J1 > J2
J1 ( current density of first segment )
J2 ( current density of second segment )
c) Comparing the electric field strengths E1 and E2
note : electric field strength ∝ current density
since current density of first segment is > current density of second segment and conductivity of the materials are the same hence
E 1 > E2
d) Comparing the drift speeds Vd1 and Vd2
( vd1 ) > ( vd2 )
this because ; vd ∝ current density
Currents in the two segments are equal. Current densities, electric field strengths and drift speeds of the first segment are greater than the second segment.
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.
The wire in consists of two segments of different diameters but made from the same metal.
a. Compare the currents in the two segments.The current in segment 1 is [tex]I_1[/tex] and the current in segment 2 is [tex]I_2[/tex]. As, the segments are connected in the series connection. Thus, the value of both the currents will be equal.
[tex]I=I_1=I_2[/tex]
b. Compare the current densities J1 and J2 in the two segments.The current density is total electric current which is flowing in a unit cross sectional area. Current density is inversely p[proportional to the cross sectional area.
The cross-sectional area of the second segment is greater than the first one as shown in the image attached below.
Thus, the current densities of first segment will be J1 greater than the current density of second segment J2.
[tex]J_1 > J_2[/tex]
c. Compare the electric field strengths E1 and E2 in the two segments.The current density of the first segment is greater than the second one. Thus, the electric field of the first segment will also be greater than the second one for the same material.
[tex]E_1 > E_2[/tex]
d. Compare the drift speeds vd1 and vd2 in the two segments.Current density is directly proportional to the drift velocity. As the current densities of first segment is greater than the current density of second segment.Therefore, there drift speed,
[tex]v_{d1} > v_{d2}[/tex]
Hence, the currents in the two segments are equal. Current densities, electric field strengths and drift speeds of the first segment are greater than the second segment.
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A race car starts from rest and accelerates down a track at a rate of 3.0 m/s2
How fast is the car moving after 10 seconds?
30 m/s
Explanation:We are given:
Initial Velocity (u) = 0 m/s
Acceleration of the Car (a) = 3 m/s²
Time Interval (t) = 10 seconds
Speed of the Car After 10 seconds:
From the First equation of motion:
v = u + at
replacing the given values
v = 0 + (3)(10)
v = 30 m/s
Hence, the car is moving at a velocity of 30 m/s after 10 seconds
A particle performing circular motion along the path of radius 2 metre with linear acceleration 5 metre per second square find its angular acceleration
Answer:
What is the question though?
Explanation:
Parent: My son spends three hours on homework every night. He has no time to see his friends. He has no time to relax. We are watching him turn into a homework machine instead of a human being.
Answer:
this is a homework helping website I don't know If I can help you with this problwm
A pail in a water well is hoisted by means of a frictionless winch, which consists of a spool and a hand crank. When Jill turns the winch at her fastest water-fetching rate, she can lift the pail the 28.0 m to the top in 11.0 s. Calculate the average power supplied by Jill's muscles during the upward ascent. Assume the pail of water when full has a mass of 7.30 kg.
Answer:
182.28 W
Explanation:
Here ,
m = 7.30 Kg
distance , d= 28.0 m
time , t = 11.0 s
average power supplied = change in potential energy/time
average power supplied = m×g×d/time
average power supplied = 7.30×9.81×28/11
average power supplied = 182.28 W
the average power supplied is 182.28 W
A 2800 kg speedboat starting from rest attains a speed of 16 m/s in 8.0 s as a combination of 1200 N of air resistance and water drag act on the boat. How much power is being expended by speedboat at the maximum speed?
Answer:
For me
Explanation:
first of all
find the distance=speed×time
16×8=128
power= workdone /time
power=force×distance/time
power=1200×128/8
power=153600/8
powere=19200
A ball weighing 60 N is swinging like a pendulum at the end of a 6.0 m rope. As the ball swings through its lowest point, its speed is measured at 5.0 m/s. What is the tension in the rope at this point?
A. 58.6 N
B. 85.4 N
C. 90.2 N
D. 97.3 N
Answer:
85.4 N
Explanation:
Weight of the ball, W = 60 N
W = mg, m is mass
m = W/g
m = 60/9.8 = 6.12 kg
Length of a rope, r = 6 m
Speed, v = 5 m/s
We need to find the tension in the rope at this point. Tension is equal to the centripetal force It is given by :
[tex]F=\dfrac{mv^2}{r}+mg\\\\F=\dfrac{6.12\times (5)^2}{6}+60\\\\=85.5\ N[/tex]
So, the correct option is (b) " 85.4 N".
in the figure if coefficient of friction between block and surface is 0.2 then find acceleration produced by the resultant force acting on the block (take g= 10 m/s²)
☃ First of all we have to calculate frictional force by using below equation :
→ F = μR
→ F = 0.2 × 5 × 10
→ F = 10 N
Hence,the frictional force (F) is 10 N.☃ But we know that frictional force is acting on opposite direction of the motion :
⇏Net force = 15 - 10
⇏Net force = 5 N
Hence,the net force is 5 N.☃ By using Newton's second law we get :
Let the required acceleration produced by the block be 'a'.➝ Force = mass × acceleration
➝ 5 = 5 × a
➝ a = 5 ÷ 5
➝ a = 1 m/s²
Hence,the acceleration produced by the resultant force acting on the block is 1 m/s².A high-resistance material is used as an insulator between the conductors of a length of coaxial cable. The resistance material, which forms a hollow tube, has an inner radius a and an outer radius b, and the insulator provides a resistance R between the conductors. If a second insulator, made of the same material and having the same length, is made with double both the inner radius and the outer radius of the first, what resistance would it provide between the conductors
This question is incomplete, the complete question is;
A high-resistance material is used as an insulator between the conductors of a length of coaxial cable. The resistance material, which forms a hollow tube, has an inner radius a and an outer radius b, and the insulator provides a resistance R between the conductors. If a second insulator, made of the same material and having the same length, is made with double both the inner radius and the outer radius of the first, what resistance would it provide between the conductors
a) (In2)R
b) 4R
c) R/(In2)
d) 2R
e) R
Answer: Option e) R is the correct answer.
Explanation:
Given that;
Inner radius = a
Outer radius = b
Conical Cylinder
∫dR = ∫(edr/2πrL)
R = e/2πL In e |ᵇₐ
R = e/2πL In(b/a) ------------- let this be equation 1
Taking a look at the second cone
a' = 2a
b' = 2a
R' = e/2πL In(2b/2a)
{L = L'}
R' = e/2πL In(b/a) -------let this be equation 2
now lets compare the two equation
R = e/2πL In(b/a)
R' = e/2πL In(b/a)
so R' = R
Therefore Option e) R is the correct answer.
Examples of ohm law and conductor resistance :
So I need help with these things, I did 2 examples but with these 3 I have problems.
I appreciate alot if anyone can handle this
1.Calculate the resistance of a Cu 2-wire line of total length for 1 line of 500m, if it has a cross section of 6mm2.
3.Calculate the voltage drop across the resistor R = 1.2 kΩ when a current I = 125 mA flows through it
2.Calculate the current flowing through the resistor R = 47 Ω if the source has a voltage U = 24 V.
I"m attaching also Microsoft Word file.
Thanks in advance!
-Mathew
Answer:
see attached file
Explanation:
As an astronaunt travels from the surface of the earth to a postion that is four times
as far away from the center of the earth, the astronaut's
a) mass decreases
b)
mass remains the same
c) weight increases
d) weight remains the same
Answer: mass remains the same
Explanation:
My teacher gave me the answer
As an astronaut travels from the surface of the earth to a position that is four times as far away from the center of the earth, the astronaut's:
B. Mass remains the sameAccording to the given question, we can see that an astronaut is on a journey to a new location which is four times further from the center of the earth. We need to find out if there is a change to the mass of the astronaut.
As a result of this, we can see that the mass of the astronaut remains the same as he travels to a position which is four times further from the center of the earth, because the gravitational forces are the same.
Therefore, the correct answer is option B
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a boy pushing a 200kg object on the floor it begins from rest and reaches 20 m/s² in 10 minutes, what is it's acceleration?
Answer:
I hope the above pic helps you mate
I will always help you understanding your assingments
have a great day
#Captainpower:)
An object at rest is suddenly broken apart into two fragments by an explosion. The first fragment (with mass m1) acquires twice the kinetic energy of the second one (with mass m2).
What is the ratio of their masses?
Answer:
m1 ÷ m2 = 1 ÷ 2
Explanation:
The ratio of their masses is as follows:
As it is given that the mass m1 is twice of mass m2
So the equation could be written as
[tex]\frac{1}{2} m1 v1 v1 = 2 \times \frac{1}{2} m2 v2 v2\\\\\frac{m1}{m2} = \frac{2 v2 v2}{v1 v1} ..........(1)[/tex]
Here we used the conservation of momentum
Prior to the explosion, the object is in rest so the momentum would be zero but after the explosion the total momentum would be m1v1 + m2v2, that is also zero.
So,
[tex]m1 v1 + m2 v2 = 0\\\\m1 v1 = - m2 v2\\\\\frac{m1}{m2} = \frac{- v2}{v1}[/tex]
Now squaring to the both sides
m1 m1 ÷ (m2 m2) = v2 v2 ÷ (v1 v1) ........(2 )
Solved both the equations
After solving it, the ratio is
m1 ÷ m2 = 1 ÷ 2
A box of weighing 60N is placed on the ground. The bottom of the box measures 2m by 1m. What is the pressure on the ground? *
Answer:
pressure=force/area
p=60N/(2m×1m)
=35N/M
Isaac Newton discovered the three laws of motion. Which example best illustrates Newton's third law of motion, which describes action and reaction forces?
O a rocket takes off from the ground
O a book slides forward on a seat when the car stops
O a heavier ball falls at the same speed as a lighter ball
O a vehicle rounds a curve and changes its acceleration
The third law of motion defined the action and reaction force that the heavy ball should be fall at the same kind of the speed.
What is third law of motion?It defined that at the time when the two bodies should be interacted so here the forces should be applied to one another that should be equivalent to the magnitude also the direction should be opposite.
Also, we can called as the action law and the reaction.
Therefore, we can conclude that the third option is correct.
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What is the water cycle ?
Answer:
The water cycle shows the continuous movement of water within the Earth and atmosphere. ... Liquid water evaporates into water vapor, condenses to form clouds, and precipitates back to earth in the form of rain and snow. Water in different phases moves through the atmosphere (transportation).
Explanation:
It's the water cycle.
what are the importance of informal education?
Answer:
Informal education is important because it can help individual to learn how to react and control situations.
It help individual to improve on its existing knowledge, new skills or ideas. This kind of education can happen any where and it can add values to the learner.
Explanation:
Informal education is a type of education that is learned from different life experiences, happenings outside a structured curriculum.
Informal education is important because it can help individual to learn how to react and control situations.
It help individual to improve on its existing knowledge, new skills or ideas. This kind of education can happen any where and it can add values to the learner.
Consider a star that is a sphere with a radius of 6.32 108 m and an average surface temperature of 5350 K. Determine the amount by which the star's thermal radiation increases the entropy of the entire universe each second. Assume that the star is a perfect blackbody, and that the average temperature of the rest of the universe is 2.73 K. Do not consider the thermal radiation absorbed by the star from the rest of the universe. J/K
Answer:
The value is [tex]\Delta s = 8.537 *10^{25 } \ J/K[/tex]
Explanation:
From the we are told that
The radius of the sphere is [tex]r = 6.32 *10^{8} \ m[/tex]
The temperature is [tex]T_x = 5350 \ K[/tex]
The average temperature of the rest of the universe is [tex]T_r = 2.73 \ K[/tex]
Generally the change in entropy of the entire universe per second is mathematically represented as
[tex]\Delta s = s_r - s_x[/tex]
Here [tex]s_r[/tex] is the entropy of the rest of the universe which is mathematically represented as
[tex]s_r = \frac{Q}{T_r}[/tex]
Here Q is the quantity of heat radiated by the star which is mathematically represented as
[tex]Q = 4 \pi * r^2 * \sigma * T^4_x[/tex]
Here [tex]\sigma[/tex] is the Stefan-Boltzmann constant with value
[tex]\sigma = 5.67 * 10^{-8 }W\cdot m^{-2} \cdot K^{-4}.[/tex]
=> [tex]Q = 4 \pi * (6.32*10^{8})^2 * 5.67 * 10^{-8 } * 5350 ^4[/tex]
=> [tex]Q = 2.332 *10^{26} \ J[/tex]
So
[tex]s_r = \frac{2.332 *10^{26}}{2.73}[/tex]
=> [tex]s_r = 8.5415 *10^{25}\ J/K[/tex]
Here [tex]s_x[/tex] is the entropy of the rest of the universe which is mathematically represented as
[tex]s_x = \frac{Q}{T_x}[/tex]
=> [tex]s_x = \frac{2.332 *10^{26} }{5350}[/tex]
=> [tex]s_x = 4.359 *10^{22} \ J/K[/tex]
So
[tex]\Delta s = 8.5415 *10^{25} - 4.359 *10^{22}[/tex]
=> [tex]\Delta s = 8.537 *10^{25 } \ J/K[/tex]
This question involves the concepts of entropy and the thermal radiation
The entropy of the entire universe is increased by "8.41 x 10²⁵ J/k
".
The increase in entropy is given as follows:
[tex]\Delta s = s-s_T[/tex]
where,
Δs = increase in entropy = ?
σ = Stefan-Boltzman's constant = 5.67 x 10⁻⁸ W/m².k⁴
A = surface area = 4πr² = 4π(6.32 x 10⁸ m)² = 5.01 x 10¹⁸ m²
Tr = Absolute temperature of the star = 5350 K
T = absolute temperature of the rest of the universe = 2.73 k
Q = thermal radiation energy
Q = [tex]\sigma A T_r^4=(5.67\ x\ ^{-8}\ W/m^2.k^4)(5.01\ x\ ^{18}\ m^2)(5350\ k)^4=2.3\ x\ 10^{26}\ J[/tex]
s = entropy of the universe = [tex]\frac{Q}{T}=\frac{2.3\ x\ 10^{26}\ J}{2.73 k}=8.42\ x\ 10^{25}\ J/k[/tex]
[tex]s_T[/tex] = entropy of the star = [tex]\frac{Q}{T_r}=\frac{2.3\ x\ 10^{26}\ J}{5350\ k}=4.3\ x\ 10^{22}\ J/k[/tex]
Therefore,
Δs = 8.42 x 10²⁵ J/k - 4.3 x 10²² J/k
Δs = 8.41 x 10²⁵ J/k
Learn more about entropy here:
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A chimpanzee climbs a vine up and to the right for a total displacement of 50 m 50m50, start text, m, end text that makes a 60 ° 60°60, degree angle from the ground. What was the horizontal displacement of the chimpanzee in m mstart text, m, end text?
Answer:
25m
Explanation
Let the horizontal displacement formula be expressed as;
Dx = Dcos theta
theta is the angles subtended by the displacement
Given
D = 50m
theta = 60°
Required
horizontal displacement of the chimpanzee
Substitute into the formula
Recall that: Dx = Dcos theta
Dx = 50cos60°
Dx = 50(0.5)
Dx = 25m
Hence the horizontal displacement of the chimpanzee is 25m
From the anatomical position, the digits are ____ to the brachium.
A. Anterior
B. Inferior
C. Superior
D. Medical