Answer:
21.5 °C.
Explanation:
Given:
α = −5.00 × 10−4 °C−1
To = 4°C
Ro = 217 Ω
Rt = 215.1 Ω
Rt/Ro = 1 + α(T - To)
215.1/217 = 1 + (-5 × 10^−4) × (T - 4)
-0.00876 = -5 × 10^−4 × (T - 4)
17.5 = (T - 4)
T = 21.5 °C.
The temperature on a spring day can be calculated by using the relationship between temperature and resistance. Given the initial resistance, the change in resistance and the temperature coefficient of resistivity for carbon, the spring day temperature can be deduced to be 21.51 °C.
Explanation:The change in temperature can be determined by using the equation ΔR = R0αΔT, where ΔR is the change in resistance, R0 is the initial resistance, α is the temperature coefficient of resistivity and ΔT is the change in temperature. The initial resistance of the carbon resistor is given as 217.0 Ω, the resistance on a spring day was 215.1 Ω, so the change in resistance ΔR is 215.1 - 217.0 = -1.9 Ω. The temperature coefficient for carbon is given as -5.00×10-4 C-1, and α is -5.00×10-4 C-1. We can rearrange the above equation to solve for ΔT: ΔT = ΔR / (R0 * α) = -1.9 Ω / (217.0 Ω * -5.00×10-4 C-1) = 17.51 C.
Thus, the temperature on the spring day would be the winter day temperature plus the calculated change in temperature: T_spring = T_winter + ΔT = 4.00 °C + 17.51 °C = 21.51 °C.
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What is the speed of a point on the earth's surface located at 3/43/4 of the length of the arc between the equator and the pole, measured from equator
Answer:
[tex]v=177.95m/s[/tex]
Explanation:
First, determine circle's radius between Earth's pole and the location. This can be calculated as:
[tex]r=R_e_a_r_t_hCos(90\frac{3}{4})\\R_e_a_r_t_h=6.37\times10^6m\\r=6.37\times10^6\times Cos67.5\textdegree\\r=2,437,693.46m\\[/tex]
The angular speed of earth is constant and is :
[tex]w=\frac{2\pi}{T}=\frac{2\pi}{24\times 3600}\\=7.3\times10^{-5}rad/s[/tex]
Velocity is:
[tex]v=wr\\=7.3\times10^{-5}\times 2,437,693.46\\v=177.95m/s[/tex]
. The current flowing through a tungsten-filament light bulb is determined to follow i(t) = 114 sin(100πt) A. (a) Over the interval defined by t = 0 and t = 2 s, how many times does the current equal zero amperes? (b) How much charge is transported through the light bulb in the first second?
Answer:
a) 201
b) 0
Explanation:
note:
solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment
For the time interval from 0 to 2, sine wave reaches to zero for 201 times. Hence the current equals to zero for 201 times.
No charge transported through the light in the first second.
Given that, the current flowing through the bulb is [tex]I(t)=114 sin (100\pi t) \;\rm A[/tex].
The general equation of the current is,
[tex]I(t) =Asin(2\pi ft)[/tex]
So the frequency can be calculated as,
[tex]2\times \pi\times f\times t = 100\times \pi[/tex]
[tex]f =50 \;\rm Hz[/tex]
Hence the frequency of the sine wave is 50 Hz.
For the time interval from 0 to 2, the number of zero for the sine wave is,
[tex]No.\;of\; zero = (2\times50\times2 )+1[/tex]
[tex]No.\;of\;zero=201[/tex]
So, For the time interval from 0 to 2, sine wave reaches to zero for 201 times. Hence the current equals to zero for 201 times.
The charge can be calculated by the formula given below.
[tex]Q(t) = \int\limits^{t_1}_{t_2} {I(t)} \ dt[/tex]
[tex]Q(t)=\int\limits^1_0 {114sin(100\pi t)} \ dt[/tex]
[tex]Q(t) = \dfrac {-114cos(100\pi t)}{100\pi}[/tex]
[tex]Q(t) = \dfrac {-114cos(100\pi \times 1)}{100\pi}-\dfrac {-114cos(100\pi \times0)}{100\pi}[/tex]
[tex]Q(t)=0[/tex]
Hence, no charge transported through the light in the first second.
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https://brainly.com/question/1345174.
A bumper car with mass m1 = 109 kg is moving to the right with a velocity of v1 = 4.9 m/s. A second bumper car with mass m2-83 kg is moving to the left with a velocity of v2 =-3.6 m/s. The two cars have an elastic collision. Assume the surface is frictionless.
1) What is the velocity of the center of mass of the system?
2) What is the initial velocity of car 1 in the center-of-mass reference frame?
3) What is the final velocity of car 1 in the center-of-mass reference frame?
4) What is the final velocity of car 1 in the ground (original) reference frame? -
5) What is the final velocity of car 2 in the ground (original) reference frame?
Answer:
(1)
The velocity of the center of mas of the system is= 0.801 m/s
(2)
Initial velocity of car 1 in the center of mass reference frame is =4.099 m/s
(3)
The final velocity of car 1 in the center of mass reference frame is
- 4.099 m/s
(4)
The final velocity of car 1 in the ground (original ) reference frame is = -3.298 m/s
(5)
The final velocity of car 2 in the ground (original) reference frame is = 7.166 m/s
Explanation:
Given
m₁ = 109 kg
v₁= 4.9 m/s
m₂= 83 kg
v₂= -3.6 m/s
The two cars have an elastic collision.
(1)
The velocity of the center of mas of the system is
[tex]V_{cm}=\frac{m_1v_1+m_2v_2}{m_1+m_2}[/tex]
[tex]= \frac{109.4.9+83(-3.4)}{109+83}[/tex] m/s
= 0.801 m/s
(2)
Initial velocity of car 1 in the center of mass reference frame is
[tex]V_{1,i}[/tex] = initial velocity - [tex]V_{cm}[/tex]
= (4.9 - 0.801) m/s
=4.099 m/s
(3)
Since the collision is elastic, the car 1 will bounce of opposite direction.
The final velocity of car 1 in the center of mass reference frame is
[tex]V_{1,f}[/tex] = - 4.099 m/s
(4)
The final velocity of car 1 in the ground (original ) reference frame
[tex]V'_{1,f}[/tex] = [tex]V_{cm}+V_{1,f}[/tex]
=(0.801- 4.099) m/s
= - 3.298 m/s
(5)
The momentum is conserved,
[tex]m_1v_1+m_2v_2=m_1v'_1+m_2v'_2[/tex]
[tex]\Rightarrow v'_2=\frac{m_1}{m_2}(v_1-v'_1) +v_2[/tex]
Here [tex]v'_1= V'_{1,f}[/tex] = - 3.298 m/s
[tex]\Rightarrow v'_2=\frac{109}{83}[4.9-(-3.298)]+(-3.6)[/tex]
=7.166 m/s
The final velocity of car 2 in the ground (original) reference frame is = 7.166 m/s
Final answer:
The velocity of the center of mass of the bumper car system is 1.225 m/s. Car 1 has an initial velocity of 3.675 m/s in the center-of-mass reference frame. After the elastic collision, car 1's final velocity is -2.45 m/s, and car 2's final velocity is 4.825 m/s in the ground reference frame.
Explanation:
Velocity of the Center of Mass, Velocities in Reference Frames, and Final Velocities After an Elastic Collision
The velocity of the center of mass (V cm) of a system in one dimension is given by the formula:
V cm = (m1 * v1 + m2 * v2) / (m1 + m2)
In this case, we have m1 = 109 kg moving at v1 = 4.9 m/s and m2 = 83 kg moving at v2 = -3.6 m/s. The velocity of the center of mass is therefore:
V cm = (109 kg * 4.9 m/s + 83 kg * (-3.6 m/s)) / (109 kg + 83 kg)
Calculating this we get:
V cm = (534.1 kg*m/s - 298.8 kg*m/s) / 192 kg = 235.3 kg*m/s / 192 kg = 1.225 m/s (rounded to three significant figures)
The initial velocity of car 1 in the center-of-mass reference frame is given by:
u1 = v1 - V cm
Substituting the known values:
u1 = 4.9 m/s - 1.225 m/s = 3.675 m/s
In an elastic collision, velocities in the center-of-mass reference frame are mirrored. Thus, the final velocity of car 1 in the center-of-mass reference frame remains the same but in the opposite direction:
u1' = -u1 = -3.675 m/s
To find the final velocity of car 1 in the ground reference frame, we add the velocity of the center of mass:
v1' = u1' + V cm = -3.675 m/s + 1.225 m/s = -2.45 m/s
For car 2, the same principle applies. The final velocity in the center-of-mass reference frame will be the opposite of the initial, and thus:
v2' = -v2 + V cm = 3.6 m/s + 1.225 m/s = 4.825 m/s
You are given a copper bar of dimensions 3 cm × 5 cm × 8 cm and asked to attach leads to it in order to make a resistor. If you want to achieve the smallest possible resistance, you should attach the leads to the opposite faces that measure.
A) 3 cm × 5 cm.
B) 3 cm × 8 cm.
C) 5 cm × 8 cm.
D) Any pair of faces produces the same resistance.
Answer:
[tex]c. 5cm \times 8cm[/tex]
Explanation:
The dimensions [tex]5cm\times 8cm[/tex] have the highest cross-sectional area combination of [tex]40cm^2[/tex].
-Resistance reduces with an increase in cross sectional area.
-[tex]Reason-[/tex]Electrons have alarger area to flow through.
Suppose a child drives a bumper car head on into the side rail, which exerts a force of 3400 N on the car for 0.400 s. (Assume the initial velocity is in the positive direction.)
(a)What impulse (in kg · m/s) is imparted by this force? (Indicate the direction with the sign of your answer.)
(b)Find the final velocity (in m/s) of the bumper car if its initial velocity was 3.30 m/s and the car plus driver have a mass of 200 kg. You may neglect friction between the car and floor. (Indicate the direction with the sign of your answer.)
Q2: Compare the kinetic energy of a 22,000 kg truck moving at 120 km/h with that of an 82.0 kg astronaut in orbit moving at 28,000 km/h.
Answer a) impulse is 1360 kg.m/s
Explanation: impulse is impact force times the time of impact
I = 3400 * 0.4 = 1360kg.m/s
Answer b) final velocity is 10.1m/s
Explanation: impulse is the change of momentum
I = m(v-u)
V and u are final and initial velocities respectively.
1360 = 200 (v - 3.3)
1360 = 200v - 660
V = (1360+660)÷200
V = 10.1m/s
Q2 answer is: astronaut has KE approximately 208 times that of the truck
Explanation :
KE = 0.5mv^2
For truck
KE = 11000 * 33.3^2 = 12197790J
For astronaut
KE = 42 * 7777.7^2 = 2540689926J
Comparing
2540689926/12197790 = 208.2
NB: speed has been converted to m/s by multiplying with 0.28 I.e 1000/3600
3/122 The collar has a mass of 2 kg and is attached to the light spring, which has a stiffness of 30 N/m and an unstretched length of 1.5 m. The collar is released from rest at A and slides up the smooth rod under the action of the constant 50‐N force. Calculate the velocity v of the collar as it passes position B.
Explanation:
According to the law of conservation of energy, work done by the force is as follows.
[tex]W_{F} = F Cos (30^{o}) \times 1.5[/tex]
= 64.95 J
Now, gain in potential energy is as follows.
P.E = mgh
= [tex]2 \times 9.8 \times 1.5[/tex]
= 29.4 J
Gain in potential energy will be as follows.
= [tex]\frac{1}{2}kx^{2}_{2} - \frac{1}{2}kx^{2}_{1}[/tex]
= [tex]\frac{1}{2} \times 30 N/m \times [(2.5 - 1.5)^{2} - (2 - 1.5)^{2}][/tex]
= 11.25
As,
[tex]W_{f} = u_{1} + u_{2} + \frac{1}{2}mv^{2}[/tex]
[tex]\frac{1}{2}mv^{2} = W_{f} - u_{1} - u_{2}[/tex]
= 64.95 J - 29.4 - 11.25
= 24.3
[tex]v^{2} = \frac{24.3 \times 2}{2}[/tex]
v = 4.92 m/s
Therefore, we can conclude that relative velocity at point B is 4.92 m/s.
To calculate the velocity of the collar as it passes position B, we need to consider the forces acting on the collar and use Newton's second law of motion. The collar is attached to a light spring, so the force exerted by the spring can be calculated using Hooke's law. The net force acting on the collar is the sum of the force exerted by the spring and the constant 50-N force.
Explanation:To calculate the velocity of the collar as it passes position B, we need to consider the forces acting on the collar and use Newton's second law of motion. The collar is attached to a light spring, so the force exerted by the spring can be calculated using Hooke's law. The net force acting on the collar is the sum of the force exerted by the spring and the constant 50-N force. We can equate this net force to the mass of the collar times its acceleration. Solving for the acceleration gives us the magnitude of the velocity as it passes position B.
Using Hooke's law, the force exerted by the spring can be calculated as:
F = k * x
where F is the force, k is the stiffness of the spring, and x is the displacement of the collar from its equilibrium position. Since the collar is attached to a light spring, we can assume that the displacement of the spring is negligible. Therefore, the force exerted by the spring is zero.
The net force is then equal to the constant 50-N force:
Net force = 50 N
Applying Newton's second law, we have:
Net force = mass * acceleration
Solving for acceleration gives us:
Acceleration = Net force / mass = 50 N / 2 kg = 25 m/s^2
Since velocity is the derivative of displacement, we can integrate the acceleration with respect to time to find the velocity:
v = a * t
where v is the velocity, a is the acceleration, and t is the time.
Since the collar is released from rest at position A, the time taken to reach position B can be determined using the equation of motion:
s = u*t + (1/2)*a*t^2
where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.
Since the collar starts from rest, the initial velocity is zero. Solving for t gives us:
t = sqrt((2*s) / a) = sqrt((2*1.5 m) / 25 m/s^2) = 0.7746 s
Finally, substituting the values of acceleration and time into the equation for velocity gives us:
v = a * t = 25 m/s^2 * 0.7746 s = 19.365 m/s
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A solid metal sphere with radius 0.430 m carries a net charge of 0.270 nC . Part A Find the magnitude of the electric field at a point 0.106 m outside the surface of the sphere. Express your answer using three significant figures. E
Answer:
8.46 N/C
Explanation:
Using Gauss law
[tex]E=\frac {kQ}{r^{2}}[/tex]
Gauss's Law states that the electric flux through a surface is proportional to the net charge in the surface, and that the electric field E of a point charge Q at a distance r from the charge
Here, K is Coulomb's constant whose value is [tex]9\times 10^{9} Nm^{2}/C^{2}[/tex]
r = 0.43 + 0.106 = 0.536 m
[tex]E=\frac {9\times 10^{9}\times 0.270\times 10^{-9}}{0.536^{2}}=8.4581755402094007\approx 8.46 N/C[/tex]
The magnitude of the electric field at a point 0.106 m outside a solid metal sphere with a radius of 0.430 m and a net charge of 0.270 nC is approximately 892 N/C, to three significant figures.
Explanation:The subject of your question is Physics, specifically focussing on electrostatics and the calculation of the electric field outside a charged sphere. The formula for the electric field (E) due to a point charge is given by Coulomb's Law, which is E = kQ/r^2, where 'Q' is the charge, 'r' is the distance from the charge, and 'k' is Coulomb's constant, approximately 8.99 x 10^9 N.m^2/C^2. Since the electric field due to a uniformly distributed spherical charge behaves as if all the charge is concentrated at the center, you can use this formula for the magnitude of the electric field outside the sphere.
Substituting the provided values (converted to appropriate units), we find E = (8.99 x 10^9 N.m^2/C^2 x 0.270 x 10^-9 C)/(0.536 m)^2, which gives E approximately equal to 892 N/C to three significant figures.
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A speeder tries to explain to the police that the yellow warning lights she was approaching on the side of the road looked green to her because of the Doppler shift. How fast would she have been traveling if yellow light of wavelength 575.9 nm had been shifted to green with a wavelength of 564.2 nm
Answer:
The speed of the speeder is [tex]8.348x10^6m/s[/tex].
Explanation:
Spectral lines will be shifted to the blue part of the spectrum if the source of the observed light is moving toward the observer, or to the red part of the spectrum when is moving away from the observer (that is known as the Doppler effect).
That shift can be used to find the velocity of the object (in this case the speeder) by means of the Doppler velocity.
[tex]v = c\frac{\Delta \lambda}{\lambda_{0}}[/tex] (1)
Where [tex]\Delta \lambda[/tex] is the wavelength shift, [tex]\lambda_{0}[/tex] is the wavelength at rest, v is the velocity of the source and c is the speed of light.
[tex]v = c(\frac{\lambda_{0}-\lambda_{measured}}{\lambda_{0}})[/tex]
For this case [tex]\lambda_{measured}[/tex] is equal to 564.2 nm and [tex]\lambda_{0}[/tex] is equal to 575.9 nm.
[tex]v = (3x10^8m/s)(\frac{575.9 nm - 564.2 nm}{564.2 nm)})[/tex]
[tex]v = 8.348x10^6m/s[/tex]
Hence, the speed of the speeder is [tex]8.348x10^6m/s[/tex].
The question involves calculating the speed of a car based on the Doppler shift of light from yellow to green. The formula for Doppler shift of light is utilized to determine the relative velocity, but the practicality of such an effect for a moving car is negligible.
Explanation:The question relates to the concept of the Doppler Effect for light, where the observed wavelength of light emitted by a source changes due to relative motion between the source and the observer. In this case, to calculate how fast the speeder would have to be traveling for the Doppler shift to change the color of a yellow light (575.9 nm) to green (564.2 nm), we can use the Doppler shift formula for light:
f' = f (c + v) / (c - v), where:
f' is the observed frequency,f is the emitted frequency,c is the speed of light,v is the velocity of the source relative to the observer.Since frequency is inversely proportional to wavelength (f = c / λ), the formula can be rearranged in terms of wavelength. Solving for v when the source is moving towards the observer gives us the expression:
v = c (λ0 - λ) / λ, where λ0 is the original wavelength and λ is the observed wavelength.
The speed of the car can thus be calculated accordingly. However, the effect of Doppler shift at such speeds is so small that it would not account for the perceptual change from yellow to green in a real-world scenario.
Consider a cylindric container with the radius of circle 50 cm and the length of 3 m. We fill this container with one mole of the oxygen gas O2 at the room temperature 20◦C. Assume it is an ideal gas. a) What is the density of the gas in the tank?
Explanation:
Below is an attachment containing the solution .
While sliding a couch across a floor, Andrea and Jennifer exert forces FA and F on the couch. Andrea's force is due north with a magnitude of 140.0 N and Jennifer's force is 260 east of north with a magnitude of 220.0 N (a) Find the net force (in N) in component form. net : (b) Find the magnitude (in N) and direction (in degrees counterclockwise from the east axis) of the net force magnitude direction X o counterclockwise from the east axis (c) If Andrea and Jennifer's housemates, David and Stephanie, disagree with the move and want to prevent its relocation, with what combined force Fos (in N) should they push so that the couch does not move? (Express your answer in vector form.) Fos
Answer:
See the answers and the explanation below.
Explanation:
To solve this problem we must make a free body diagram with the forces applied as well as the direction. In the attached images we can see the nomenclature of the direction of the forces and the free body diagram.
a)
Sum of forces in y-axis
Fy = 140 - (220*sin(10))
Fy = 101.8 [N]
Sum of forces in x-axis
Fx = - (220*cos(10))
Fx = - 216.65 [N]
b)
For the above result, for there to be balance we realize that we need one equal to the resulting y-axis but in the opposite direction and another opposite force in direction but equal in magnitude on the x-axis.
Fy = - 101.8 [N]
Fx = 216.65 [N]
c )
Now we need to use the Pythagorean theorem to find the result of these forces.
[tex]F = \sqrt{(101.8)^{2} +(216.65)^{2} } \\F=239.4[N][/tex]
And the direction will be as follows.
α = tan^(-1) (101.8 / 216.65)
α = 25.16° (south to the east)
F = 216.65 i - 101.8 j [N]
A concave mirror with a radius of curvature of 10 cm is used in a flashlight to produce a beam of near-parallel light rays.
The distance between the light bulb and the mirror is most nearly _____.
Answer: 5cm
Explanation: Since the radius of curvature is 10cm, the focal length of the mirror (f) is
f = r/2
Where r is the radius of curvature.
Since r = 10cm, f = 10/2 = 5cm.
To produce a parallel light rays of a flash light, it means the the image will be at infinity this making the image distance to be infinite.
From the mirror formulae
1/u + 1/v = 1/f
Where u = object distance =?, v = image distance = infinity and f = focal length = 5cm.
Let us substitute the parameters, we have that
1/u + 1/∞ = 1/5
1/∞ = 0
Hence 1/u = 1/5
u = 5cm.
Hence the object needs to be placed nearly 5cm to the mirror
Final answer:
The light bulb should be placed at the focal length of the concave mirror, which is 5 cm, to produce near-parallel light rays in a flashlight.
Explanation:
The distance between the light bulb and a concave mirror to produce a beam of near-parallel light rays in a flashlight is most nearly the focal length of the mirror. Since the mirror's radius of curvature (R) is 10 cm, using the formula R = 2f, we can calculate the focal length (f). Dividing the radius of curvature by 2, we get f = R/2 = 10 cm / 2 = 5 cm. Therefore, the light bulb should be placed approximately 5 cm from the mirror to achieve near-parallel rays.
A volley ball is hit directly toward the ceiling in a gymnasium with a ceiling height of L 0 m. If the initial vertical velocity is 13 m/s and the release height is 1.8 m will the ball hit the ceiling?
Complete Question:
A volley ball is hit directly toward the ceiling in a gymnasium with a ceiling height of 10 m. If the initial vertical velocity is 13 m/s and the release height is 1.8 m will the ball hit the ceiling?
Answer:
The ball will hit the ceiling
Explanation:
Given;
Initial vertical Velocity U = 13 m/s
Height of the ceiling = 10 m
Released height of the volley ball = 1.8 m
Height traveled by the volley ball, is calculated as follows;
[tex]V^2 =U^2 -2gH[/tex]
where;
V is final vertical velocity
[tex]2gH =U^2\\\\H = \frac{U^2}{2g} = \frac{(13)^2}{2(9.8)} = 8.62 m[/tex]
Remember this ball was released from 1.8 m height and it traveled 8.62 m.
Total distance traveled = 1.8 + 8.62 = 10.42 m
Therefore, the ball will hit the ceiling
. Determine the horizontal and vertical components of reaction at the hinge A and the horizontal reaction at the smooth surface B caused by the water pressure. The plate has a width of 4 ft. Start your solution by presenting the appropriate FBD. rhow = 1.94 slug/ft3 .
Answer:
Please find attached file for complete answer solution and explanation of same question.
Explanation:
A horizontal spring with stiffness 0.5 N/m has a relaxed length of 19 cm (0.19 m). A mass of 22 grams (0.022 kg) is attached and you stretch the spring to a total length of 26 cm (0.26 m). The mass is then released from rest. What is the speed of the mass at the moment when the spring returns to its relaxed length of 19 cm (0.19 m)?
Answer:
v = 0.0147 m / s
Explanation:
For this exercise let's use energy conservation
Starting point. Fully stretched spring
Em₀ = Ke = ½ k (x-x₀)²
Final point. Unstretched position
Emf = K = ½ m v²
Emo = Emf
½ k (x- x₀)² = ½ m v²
v = √m/k (x-x₀)
Let's calculate
v = √(0.022 / 0.5) (0.26-0.19)
v = 0.0147 m / s
The speed of the mass at the mean position is 0.333 m/s
Conservation of energy:The potential energy stored in a fully stretched spring
PE = ½ kx²
where x is the stretch of the spring = 26 -19 = 7 cm = 0.07 m
At the mean position, where x = 0, the PE stored in sprig is zero,
So according to the law of conservation of energy total energy must remain conserved so all the energy is converted into kinetic energy KE of the mass
KE = ½ mv²
where m is the mass and v is the velocity
½ kx² = ½ mv²
where k is the spring constant = 0.5 N/m
and m is the mass = 0.022 kg
[tex]v=\sqrt{\frac{k}{m} } x[/tex]
[tex]v=\sqrt{\frac{0.5}{0.022} } 0.07[/tex]
v = 0.333 m/s
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A 100 kg box as showsn above is being pulled along the x axis by a student. the box slides across a rough surface, and its position x varies with time t according to the equation x=.5t^3 +2t where x is in meters and t is in seconds.
(a) Determine the speed of the box at time t=0
(b) determine the following as functions of time t.
Answer:
a) 2 m/s
b) i) [tex]K.E = 50 (1.5t^2 + 2) ^2\\[/tex]
ii) [tex]F = 3tm[/tex]
Explanation:
The function for distance is [tex]x = 0.5t ^3 + 2t[/tex]
We know that:
Velocity = [tex]v= \frac{d}{dt} x[/tex]
Acceleration = [tex]a= \frac{d}{dt}v[/tex]
To find speed at time t = 0, we derivate the distance function:
[tex]x = 0.5 t^3 + 2t\\v= x' = 1.5t^2 + 2[/tex]
Substitute t = 0 in velocity function:
[tex]v = 1.5t^2 + 2\\v(0) = 1.5 (0) + 2\\v(0) = 2[/tex]
Velocity at t = 0 will be 2 m/s.
To find the function for Kinetic Energy of the box at any time, t.
[tex]Kinetic \ Energy = \frac{1}{2} mv^2\\\\K.E = \frac{1}{2} \times 100 \times (1.5t^2 + 2) ^2\\\\K.E = 50 (1.5t^2 + 2) ^2\\[/tex]
We know that [tex]Force = mass \times acceleration[/tex]
[tex]a = v'(t) = 1.5t^2 + 2\\a = 3t[/tex]
[tex]F = m \times a\\F= m \times 3t\\F = 3tm[/tex]
The speed of the box at time t=0 is 0 m/s. The acceleration, displacement, and velocity functions of the box as a function of time are a(t) = 3t, x(t) = .5t^3 + 2t and v(t) = 1.5t^2 + 2 respectively.
Explanation:For part (a), the speed of the box at time t=0 can be found by taking the derivative of the position function x(t), which gives us the velocity function v(t). Therefore, v(t) = 1.5t^2 + 2 and v(0) = 0. Thus, the speed of the box at t=0 is 0 m/s.
For part (b), the box's acceleration at any time t can be found by taking the derivative of the velocity function v(t), which gives us the acceleration function a(t). Therefore, a(t) = 3t, the displacement function as a function of time is the original function x(t) = .5t^3 + 2t and the velocity function is as mentioned above, v(t) = 1.5t^2 + 2.
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When she rides her bike, she gets to her first classroom building 36 minutes faster than when she walks. Of her average walking speed is 3 mph and her average biking speed is 12 mph, how far is it from her apartment to the classroom building
Answer:
[tex]d=2.4\ miles[/tex]
Explanation:
Given:
average walking speed, [tex]v_w=3\ mph[/tex]average biking speed, [tex]v_b=12\ mph[/tex]According to given condition:
[tex]t_w=t_b+\frac{36}{60}[/tex]
where:
[tex]t_w=[/tex] time taken to reach the building by walking
[tex]t_b=[/tex] time taken to reach the building by biking
We know that,
[tex]\rm time=\frac{distance}{speed}[/tex]
so,
[tex]\frac{d}{v_w} =\frac{d}{v_b} +\frac{36}{60}[/tex]
[tex]\frac{d}{3}=\frac{d}{12} +\frac{3}{5}[/tex]
[tex]d=2.4\ miles[/tex]
Answer:
The distance from her apartment to the classroom building is 2.4 miles.
Explanation:
Given that,
Time = 36 min
Walking average speed of her = 3 m/h
biking average speed of her = 12 m/h
If she takes n minutes to ride, then if the distance is d miles,
We need to calculate the distance
Using formula of time
[tex]t=t_{1}+t_{2}[/tex]
[tex]t=\dfrac{d}{v_{1}}+\dfrac{d}{v_{2}}[/tex]
Put the value into the formula
[tex]\dfrac{36}{60}=\dfrac{d}{3}-\dfrac{d}{12}[/tex]
[tex]d=\dfrac{36\times12}{60\times3}[/tex]
[tex]d=2.4\ miles[/tex]
Hence, The distance from her apartment to the classroom building is 2.4 miles.
. A proton, which moves perpendicular to a magnetic field of 1.2 T in a circular path of radius 0.080 m, has what speed? (qp = 1.6 · 10-19 C and mp = 1.67 · 10-27 kg)
Answer:
[tex]9.198\times 10^6 m/s[/tex]
Explanation:
We are given that
Magnetic field, B=1.2 T
Radius of circular path, r=0.080 m
[tex]q_p=1.6\times 10^{-19} C[/tex]
[tex]m_p=1.67\times 10^{-27} kg[/tex]
[tex]\theta=90^{\circ}[/tex]
We have to find the speed of proton.
We know that
Magnetic force, F=[tex]qvBsin\theta[/tex]
According to question
Magnetic force=Centripetal force
[tex]q_pvBsin90^{\circ}=\frac{m_pv^2}{r}[/tex]
[tex]1.6\times 10^{-19}\times 1.2=\frac{1.67\times 10^{-27}v}{0.08}[/tex]
[tex]v=\frac{1.6\times 10^{-19}\times 1.2\times 0.08}{1.67\times 10^{-27}}[/tex]
[tex]v=9.198\times 10^6 m/s[/tex]
You have a battery marked " 6.00 V 6.00 V ." When you draw a current of 0.207 A 0.207 A from it, the potential difference between its terminals is 5.03 V 5.03 V . What is the potential difference when you draw 0.523 A 0.523 A ?
Answer:
V= 3.55 V
Explanation:
As the potential difference between the battery terminals, is less than the rated value of the battery, this means that there is some loss in the internal resistance of the battery. We can calculate this loss, applying Ohm's law to the internal resistance, as follows:[tex]V_{rint} = I* r_{int}[/tex]
The value of the potential difference between the terminals of the battery, is just the voltage of the battery, minus the loss in the internal resistance, as follows:[tex]V = V_{b} - V_{rint} = 6.00 V - 0.207A* r_{int}[/tex]
We can solve for rint, as follows:[tex]r_{int} = \frac{V_{b} - V}{I} = \frac{6.00 V - 5.03V}{0.207A} = 4.7 \Omega[/tex]
When the circuit draws from battery a current I of 0.523A, we can find the potential difference between the terminals of the battery, as follows:[tex]V = V_{b} - V_{rint} = 6.00 V - 0.523A* 4.7 \Omega = 3.55 V[/tex]
As the current draw is larger, the loss in the internal resistance will be larger too, so the potential difference between the terminals of the battery will be lower.A straight wire carries a current of 238 mA from right to left. What is the magnetic field at a point 10.0 cm directly below the wire? Give the magnitude here, but make sure you can find the direction too
Answer:
B = (4.76 × 10⁻⁷) T
Explanation:
From Biot Savart's law, the magnetic field formula is given as
B = (μ₀I)/(2πr)
B = magnetic field = ?
I = current = 238 mA = 0.238 A
μ₀ = magnetic constant = (4π × 10⁻⁷) H/m
r = 10 cm = 0.1 m
B = [4π × 10⁻⁷ × 0.238)/(2π×0.1)]
B = (4.76 × 10⁻⁷) T
The direction of the magnetic field is in the clockwise direction wrapped around the current-carrying wire.
Hope this Helps!!!
The place you get your hair cut has two nearly parallel mirrors 6.50 m apart. As you sit in the chair, your head is 3.00 m from the nearer mirror. Looking toward this mirror, you first see your face and then, farther away, the back of your head. (The mirrors need to be slightly nonparallel for you to be able to see the back of your head, but you can treat them as parallel in this problem.) How far away does the back of your head appear to be?
Answer:
[tex]k=13\ m[/tex]
Explanation:
Given:
distance between two nearly parallel mirrors, [tex]d=6.5\ m[/tex]distance between the face and the nearer mirror, [tex]x=3\ m[/tex]So, the distance between the back-head and the mirror, [tex]y=6.5-3=3.5\ m[/tex]From the given information by the laws of reflection we can deduce the distance of the first reflection of the back of the head of person in the rear mirror.
Distance of the first reflection of the back of the head in the rear mirror from the object head:
[tex]y'=2\times y[/tex]
[tex]y'=7\ m[/tex] is the distance of the image from the object back head.
Now the total distance of this image from the front mirror:
[tex]z=y'+x[/tex]
[tex]z=7+3[/tex]
[tex]z=10\ m[/tex]
Now the second reflection of this image will be 10 meters inside in the front mirror.So, the total distance of the image of the back of the head in the front mirror from the person will be:
[tex]k=x+z[/tex]
[tex]k=3+10[/tex]
[tex]k=13\ m[/tex]
Final answer:
The back of the head appears to be 16.00 meters away due to multiple reflections between two parallel mirrors 6.50 meters apart, when your head is positioned 3.00 meters from the nearer mirror.
Explanation:
The question revolves around a flat mirror problem in physics where multiple reflections between two parallel mirrors are considered. When a person looks into a flat mirror, the image of any object (like the back of the head) appears to be the same distance behind the mirror as the object is in front of it. So if your head is 3.00 meters away from the nearer mirror, the first image of the back of your head will appear 3.00 meters behind that mirror.
However, since there are two mirrors, this first image will then act as an object for the second, farther mirror, creating a new image. The additional distance to this second image will be twice the distance between the two mirrors. So, the total apparent distance will be the distance to the first image plus 6.50 meters times 2, which is 16.00 meters (3.00+6.50+6.50 = 16.00 meters). Therefore, the back of your head appears to be 16.00 meters away.
Two insulated current-carrying straight wires of equal length are arranged in the lab so that Wire A carries a current northward and Wire B carries a current eastward, the wires crossing at their midpoints separated only by their insulation. Which of the statements below is true?
a. There are no forces in this situation.
b. The net force on Wire B is southward.
c. There are forces, but the net force on each wire is zero.
d. The net force on Wire A is westward.
Since there is a meeting of the cables at their midpoints, it is therefore understood that the force on them is the same but in the opposite direction, this to maintain the static balance between the two.
This can also be corroborated by applying the right hand rule for the force, at which depends of the magnetic field. The net force is zero because the cable segment to the left of the vertical cable feels an opposite force in the direction of the cable segment to the right.
Then the forces cancel.
Therefore the correct answer is C. Therefore the net force on each wire is zero
An electromagnetic wave is propagating towards the west in free space. At a certain moment the direction of the magnetic field vector associated with this wave points vertically upward.
What is the direction of the electric field vector?
A. vertical and pointing down.
B. vertical and pointing up.
C. horizontal and pointing north.
D. horizontal and pointing south.
E. horizontal and pointing east.
Answer:
The direction of the electric field vector is horizontal and pointing north.
Option (C) is correct option.
Explanation:
Given :
The direction of wave propagation is toward the west.
The direction of magnetic field vector is vertically upward.
According to the theory of electromagnetic wave propagation, the electric field and magnetic field is perpendicular to each other and the direction of propagation is also perpendicular to both electric and magnetic field vector.
⇒ [tex]\vec{E} + \vec{B} = \vec {k}[/tex]
From right hand rule, the fingers goes towards horizontal and pointing north and curl the finger goes towards vertically upward and thumb will give you the direction of wave propagation toward west.
Hence, the direction of electric field vector is horizontal and pointing north.
The correct direction of the electric field vector in an electromagnetic wave propagating to the west with an upward magnetic field is horizontal and pointing north, following the right-hand rule for perpendicularity and direction of electromagnetic waves.
Explanation:The subject of this question is the direction of the electric field vector in an electromagnetic wave that is propagating towards the west, with a magnetic field vector that points vertically upward. According to the properties of electromagnetic waves, the electric field (E) and magnetic field (B) are perpendicular to each other and to the direction of wave propagation. Therefore, if the magnetic field is pointing upward and the wave is moving west, the electric field must be pointing horizontally. Since the right-hand rule dictates that E x B gives the direction of wave propagation, and the wave is moving west, the electric field cannot be pointing east or west as it would not satisfy the right-hand rule. Thus, the electric field is either pointing north or south. To determine the correct direction between north and south, we rely on the right-hand rule: the fingers of the right hand point in the direction of E, the curled fingers point towards B, and the thumb points in the direction of the wave propagation (west in this case). If the magnetic field points up, and propagation is to the west, the electric field must be directed to the north. Therefore, the correct answer is C. horizontal and pointing north.
Thin Layer Chromatography consists of three parts: The analyte, the stationary phase, and mobile phase. Match each of these terms to what it was in our experiment. Stationary Phase ____ a) The solvent
Mobile Phase ____ b) Silica
Analyte ____ c) One of the analgesiscs
Answer:
Analyte⇒ one of analgesics
stationery phase⇒ silica
mobile phase⇒ solvent
Explanation:
during the thin layer chromatography non volatile mixtures are separated.The technique is performed on the plastic or aluminum foil that is coated with a thin layer.
One wire possesses a solid core of copper, with a circular cross-section of radius 3.78 mm. The other wire is composed of 19 strands of thin copper wire bundled together. Each strand has a circular cross-section of radius 0.756 mm. The current density J in each wire is the same, J=2950 A/m².
1. How much current does each wire carry?
2. The resistivity of copper is 1.69 x 10⁻⁸ ohm m. What is the resistance of a 1.00 m length of each wire?
Answer:
a) Current in wire 1 = 0.132 A
Current in wire 2 = 0.101 A
b) Resistance of wire 1 = R₁ = 0.000376 Ω = (3.76 × 10⁻⁴) Ω = 0.376 mΩ
Resistance of wire 2 = R₂ = 0.000495 Ω = (4.95 × 10⁻⁴) Ω = 0.495 mΩ
Explanation:
Current density, J = (current) × (cross sectional area)
Current density for both wires = J = 2950 A/m²
For wire 1,
Cross sectional Area = πr² = π(0.00378²)
A₁ = 0.00004491 m²
For wire 2,
With the assumption that the strands are well banded together with no spaces in btw.
Cross sectional Area = 19 × πr² = π(0.000756)²
A₂ = 0.00003413 m²
Current in wire 1 = I₁ = J × A₁ = 2950 × 0.00004491 = 0.132 A
Current in wire 2 = I₂ = J × A₂ = 2950 × 0.00003413 = 0.101 A
b) Resistance = ρL/A
ρ = resistivity for both wires = (1.69 x 10⁻⁸) Ω.m
L = length of wire = 1.00 m for each of the two wires
A₁ = 0.00004491 m²
A₂ = 0.00003413 m²
R₁ = ρL/A₁ = (1.69 x 10⁻⁸ × 1)/0.00004491
R₁ = 0.000376 Ω = (3.76 × 10⁻⁴) Ω = 0.376 mΩ
R₂ = ρL/A₂ = (1.69 x 10⁻⁸ × 1)/0.00003413
R₂ = 0.000495 Ω = (4.95 × 10⁻⁴) Ω = 0.495 mΩ
Hope this helps!!
A supply plane needs to drop a package of food to scientists working on a glacier in Greenland. The plane flies 110 m above the glacier at a speed of 150 m/s . You may want to review (Page) . For help with math skills, you may want to review:
Answer:
The distance is 709.5 m.
Explanation:
Given that,
Speed = 150 m/s
Distance = 110 m
Suppose, How far short of the target should it drop the package?
We need to calculate the time
Using equation of motion
[tex]s=ut+\dfrac{1}{2}gt^2[/tex]
[tex]t^2=\dfrac{2s}{g}[/tex]
Where, g = acceleration due to gravity
t = time
Put the value into the formula
[tex]t=\sqrt{\dfrac{2\times110}{9.8}}[/tex]
[tex]t=4.73\ sec[/tex]
We need to calculate the distance
Using formula of distance
[tex]d= vt[/tex]
Put the value into the formula
[tex]d=150\times4.73[/tex]
[tex]d=709.5\ m[/tex]
Hence, The distance is 709.5 m.
In some recent studies it has been shown that women are men when competing in similar sports (most notably in soccer and basketball). Select the statement that explains why this disparity might exist. a. The cross-sectional area of the ACL is typically larger in men, and therefore experiences less strain for the sam tensile force and Young's modulus. b. The Young's modulus of women's ACLS is typically smaller than that of men's, resulting in more stress for the same amount of strain. c. The cross-sectional area of the ACL is typically smaller in women, and therefore experiences less stress for the same tensile force. d. The ACL of women is more elastic than the ACL of men.
Answer: The correct option is B (The Young's modulus of women's ACLS is typically smaller than that of men's, resulting in more stress for the same amount of strain)
Explanation:
Anterior cruciate ligament (ACL) is one of the important ligaments found at the knee joint which helps to stabilise the joint. It connects the femur to the tibia bone at the knee joint.
Anterior cruciate ligament tear is one of the common knee joint injury which is seen in individuals( especially females) involved in sports( example soccer and basketball which involves sudden change in direction causing the knee to rotate inwards)
ACL tear occurs through both contact and non contact mechanisms. The contact mechanism of ACL injury occurs when force is directly applied at the lateral part of the knee while in non contact mechanism,tear occurs when the tibia is externally rotated on the planted foot.
Research has proven that women are prone to have ACL tear than men when competing in similar sports. This disparity exists due to structural differences that pose as risk factors. These includes
- the female ACL size is smaller than the male.
- the ACL of female has a lower modulus if elasticity( that is, less stiff) than in males leading to greater joint mobility than in the male.. therefore the option, (The Young's modulus of women's ACLS is typically smaller than that of men's, resulting in more stress for the same amount of strain) is correct.
According to the Can Manufacturers Institute, the energy used to make an aluminum can from recycled aluminum is 5% of the energy used to make an aluminum can from virgin ore. In a typical year, 1.7 billion pounds of aluminum cans are recycled.
Part A
How much energy is thermally transferred to get this mass of aluminum from 20 ∘C to its melting point, 660 ∘C?
4.45 * 10¹⁴ J is transferred to get this mass of aluminum from 20°C to its melting point, 660⁰C.
The quantity of heat required to change the temperature of a substance is given by:
Q = mcΔT
Where Q is the heat, m is the mass of the substance, ΔT is the temperature change = final temperature - initial temperature. c is the specific heat capacity
m = 1.7 billion pounds = 77 * 10⁷ kg, ΔT = 660 - 20 = 640°C, c = 903 J/kg•K
Hence:
Q = 77 * 10⁷ kg * 903 J/kg•K * 640°C
Q = 4.45 * 10¹⁴ J
4.45 * 10¹⁴ J is transferred to get this mass of aluminum from 20°C to its melting point, 660⁰C.
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The energy required to heat 1.7 billion pounds of aluminum from 20 degrees Celsius to 660 degrees Celsius is approximately 4.398 × 10^17 Joules.
Explanation:The thermal energy transferred, or heat, to raise the temperature of a substance is given by the formula q=mcΔT where 'm' is the mass, 'c' is the specific heat capacity, and 'ΔT' is the change in temperature. For aluminum, the specific heat capacity is 0.897 Joules per gram per degree Celsius (J/g°C).
First, we need to convert 1.7 billion pounds of aluminum into grams since the specific heat value is in grams. There are about 453,592.37 grams in a pound, so this gives us about 7.711 × 10^14 grams of aluminum.
The change in temperature (ΔT) is the final temperature minus the initial temperature, or 660 degrees Celsius - 20 degrees Celsius, which equals 640 degrees Celsius.
So, to find the total energy required, we use the formula and substitute the known values: q=(7.711 × 10^14 g)*(0.897 J/g°C)*(640°C), which equals approximately 4.398 × 10^17 Joules.
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The Young’s modulus of nickel is Y = 2 × 1011 N/m2 . Its molar mass is Mmolar = 0.059 kg and its density is rho = 8900 kg/m3 . Given a bar of nickel of length 13 m, what time does it take for sound waves to propagate from one end to the other? Avogadro’s number is NA = 6.02 × 1023 atoms. Answer in units of s.
Answer:
Atomic Size and Mass:
convert given density to kg/m^3 = 8900kg/m^3 2) convert to moles/m^3 (kg/m^3 * mol/kg) = 150847 mol/m^3 (not rounding in my actual calculations) 3) convert to atoms/m^3 (6.022^23 atoms/mol) = 9.084e28 atoms/m^3 4) take the cube root to get the number of atoms per meter, = 4495309334 atoms/m 5) take the reciprocal to get the diameter of an atom, = 2.2245e-10 m/atom 6) find the mass of one atom (kg/mol * mol/atoms) = 9.7974e-26 kg/atom Young's Modulus: Y=(F/A)/(dL/L) 1) F=mg = (45kg)(9.8N/kg) = 441 N 2) A = (0.0018m)^2 = 3.5344e-6 m^2 3) dL = 0.0016m 4) L = 2.44m 5) Y = 1.834e11 N/m^2 Interatomic Spring Stiffness: Ks,i = dY 1) From above, diameter of one atom = 2.2245e-10 m 2) From above, Y = 1.834e11 N/m^2 3) Ks,i = 40.799 N/m (not rounding in my actual calculations) Speed of Sound: v = ωd 1) ω = √(Ks,i / m,a) 2) From above, Ks,i = 40.799 N/m 3) From above, m,a = 9.7974e-26 kg 4) ω=2.0406e13 N/m*kg 5) From above, d=2.2245e-10 m 6) v=ωd = 4539 m/s (not rounding in actual calculations) Time Elapsed: 1) length sound traveled = L+dL = 2.44166 m 2) From above, speed of sound = 4539 m/s 3) T = (L+dL)/v = 0.000537505 s
You are riding a ferris wheel while sitting on a scale. A ferris wheel with radius 9.7 m and a period 32s. Find the scale reading for a 60kg person at the bottom of the ferris wheel and the top of the ferris wheel, assuming it moves at a constant rate.
Answer:
The scale reading at the top = 565.8N
The scale reading at the bottom = 610.44N
Explanation:
We are given:
t = 32seconds
R = 9.7meters
mass (m) = 60Kg
We take g as gravitational field = 9.8
Therefore, to find the scale reading (N) at the top, let's use the formula:
[tex] N_t = m (g - w^2 R) [/tex]
Where w = 2π/t
w = 2π/32 = 0.197
Substituting the figures into the equation, we have
[tex] N = 60 (9.8 - 0.197^2 * 9.7) [/tex]
N = 60 (9.8 - 0.37)
N = 60 * 9.43
Na = 565.8N
To find scalar reading at the bottom, we use;
[tex] N_b = m(g + w^2 R) [/tex]
= 60 (9.8 + 0.37)
= 60 * 10.17
[tex] N_b = 610.44 [/tex]
Answer:
The scale reading at the top is [tex]z_{top} = 565.6\mu N[/tex]
The scale reading at the bottom is [tex]z_{bottom} = 610.356\ N[/tex]
Explanation:
From the question
The radius is [tex]r = 9.7 m[/tex]
The period is [tex]T = 32sec[/tex]
The mass is [tex]m =60kg[/tex]
Generally the mathematical representation for angular velocity of the wheel is
[tex]\omega = \frac{2 \pi}{T}[/tex]
[tex]= \frac{2*3.142}{32}[/tex]
[tex]= 0.196 \ rad/sec[/tex]
The velocity at which the point scale move can be obtained as
[tex]v = r\omega[/tex]
[tex]= 9.7* 0.196[/tex]
[tex]= 1.9 m/s[/tex]
Considering the motion of the 60kg mass as shown on the first and second uploaded image
Let the z represent the reading on the scale which is equivalent to the normal force acting on the mass.
Now at the topmost the reading of the scale would be
[tex]mg - z_{top} =\frac{mv^2}{r} =m\omega^2r[/tex]
Where mg is the gravitational force acting on the mass and [tex]\frac{mv^2}{r}[/tex] is the centripetal force keeping the mass from spiraling out of the circle
Now making z the subject of the formula
[tex]z_{top} = mg - \frac{mv^2}{r}[/tex]
[tex]= m(g- \frac{v^2}{r})[/tex]
[tex]= 60(9.8 - \frac{1.9^2}{9.7} )[/tex]
[tex]= 565.6\mu N[/tex]
Now at the bottom the scale would be
[tex]z_{bottom}-mg = \frac{mv^2}{r} = m \omega^2r[/tex]
This is because in order for the net force to be in the positive y-axis (i.e for the mass to keep moving in the Ferris wheel) the Normal force must be greater than the gravitational force.
Making z the subject
[tex]z_{bottom}= mg +m\omega^2r[/tex]
[tex]z =m(g+\omega^2r)[/tex]
[tex]z= 60(9.8 + (0.196)^2 *(9.7))[/tex]
[tex]= 610.356\ N[/tex]
A chlorine and bromine atom are adsorbed on a small patch of surface (see sketch at right). This patch is known to contain possible adsorption sites. The and bromine atoms have enough energy to move from site to site, so they could be on any two of them. Suppose the atom desorbs from the surface and drifts away. Calculate the change in entropy. Round your answer to significant digits, and be sure it has the correct unit symbol.
Answer:
ΔS = - 3.74 × 10⁻²³ J/K = - 3.7 × 10⁻²³ to 2 s.f
Explanation:
The change in entropy for a system with changing allowable Microsystems is given as
ΔS = K In (W/W₀)
K = Boltzmann's constant = 1.381 × 10⁻²³ J/K
W₀ = initial number of microstates
To calculate this, how many ways can 2 atoms occupy 16 available microstates with order important?
That is, ¹⁶P₂ = 16!/(16 - 2)! = 16 × 15 = 240 microstates.
W = the number of microstates for Chlorine atom when Bromine atom desorps = 16 microstates.
ΔS = K In (W/W₀)
ΔS = (1.381 × 10⁻²³) In (16/240)
ΔS = (1.381 × 10⁻²³) × 2.7081
ΔS = - 3.74 × 10⁻²³ J/K
Entropy change is positive when an atom desorbs from a surface and drifts away, allowing for more freedom of movement and increasing the system's entropy.
Entropy change is a measure of disorder in a system. When an atom desorbs from a surface and drifts away, the change in entropy is positive as it increases the freedom of movement of the atom. This occurs because there are more possible locations for the atom to occupy, leading to an increase in entropy.