Answer:
Explanation:
Given that the inputted angle of incidence is accepted
Angle of incidence θi = 76.5°
But angle of refraction is not acceptable
Angle of refraction θr = 76.5°
We are told that the value is too high
Then, θr < 76.5°
We want to calculate index of refraction n?
The acceptable value of refraction index of acrylic is 1.49
So the true value is 1.49
So, let calculate the measure value
Refractive index is given as
n = Sin(i) / Sin(r)
Then,
n = Sin(76.5) / Sin(76.5)
Then, n = 1
Now, percentage error of the refractive index,
Percentage error is
%error= |true value—measure value| / true value × 100
The true value is 1.49
The measure value is 1
Then,
%error = ( |1.49—1| / 1.49 ) × 100
% error = ( 0.49 / 1.49 ) × 100
%error = 0.3289 × 100
%error = 32.89%
Answer:
Explanation:
32.89%
A space transportation vehicle releases a 470-kg communications satellite while in a circular orbit 350 km above the surface of the Earth. A rocket engine on the satellite boosts it into an orbit 2350 km above the surface of the Earth. How much energy does the engine have to provide for this boost?
Answer:
E = 3.194 x 10⁹ J = 3.194 GJ
Explanation:
The formula for the absolute potential energy is:
U = - GMm/2r
where,
G = Gravitational Constant = 6.67 x 10⁺¹¹ N m²/kg²
M = mass of Earth = 5.972 x 10²⁴ kg
m = mass of satellite = 470 kg
r = distance between the center of Earth and satellite
Thus, the energy required from engine will be difference between the potential energies.
E = U₂ - U₁
E = - GMm/2r₂ - (- GMm/2r₁)
E = (GMm/2)(1/r₁ - 1/r₂)
where,
r₁ = Radius of Earth + 350 km = 6371 km + 350 km = 6721 km = 6.721 x 10⁶ m
r₂=Radius of Earth + 2350 km=6371 km + 2350 km= 8721 km = 8.721 x 10⁶ m
therefore,
E = [(6.67 x 10⁺¹¹ N m²/kg²)(5.972 x 10²⁴ kg)(470 kg)/2](1/6.721 x 10⁶ m - 1/8.721 x 10⁶ m)
E = 3.194 x 10⁹ J = 3.194 GJ
) So we are in a Universe with no center and no edge, but it is expanding... and it might be infinite. And it is all space and time and mass and energy, but was born out of nothingness. What do you think about this? Does it seem sensible and natural or do you find it odd and confusing? Can you think of any better explanation for what we see? I would like your answer to be at least five (5) sentences long.
Final answer:
The multiverse concept along with the universe's infinite nature and lack of a center challenges our understanding, yet represents current scientific explanations supported by observational evidence and theoretical models.
Explanation:
The concept of the multiverse, suggesting that our universe is just one among countless others, is a challenging yet intriguing idea. The nature of cosmic expansion, driven by mass and dark energy, adds complexity to our universe's fate, with models predicting expansion forever or eventual contraction. While it may seem sensible to accept that the Big Bang led to a universe with no center or edge, and infinite potential, it's natural to find this overwhelming due to its departure from human intuition. These concepts push the boundaries of our understanding and stimulate philosophical and metaphysical discussions, demonstrating the dynamic involvement of spacetime. As opinions on what constitutes the 'center' or 'boundary' of an infinite universe vary, the scientific community continues to explore why the universe is structured as it is, why certain constants have their values, and what 'accidents' may have been necessary for existence as we perceive it.
In an experiment of a simple pendulum, measurements show that the pendulum has length し 0.397 ± 0.006 m, mass M-0.3172 ± 0.0002 kg, and period Tem-1.274 ± 0.005 s. Take g = 9.8 m/s of error (use the error propagation method you learned in Lab 1). and the predicted value Ttheo-
i. Use the measured length L to predict the theoretical pendulum period Ttheo with a range
ii. Compute the percentage difference (as defined in Lab 1) between the measured value Texp
Answer:
1.)1.265+or minus 0.0006m
2).0.71%
Explanation:
See attached file
For the same setup as in the previous question, what is the rotational kinetic energy of the spinning cube after the string unwinds L=0.30m? Use Energy methods. ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- Be careful with this problem. Both the spinning cube picks up kinetic energy as it spins up and the hanging mass gains kinetic energy as the string unwinds. Rotational kinematics will be necessary e.g.
Answer:
Kinetic energy = 0.145 J
Explanation:
See the attached file for explanation.
Rotational kinetic energy of a spinning cube can be calculated using the formula KErot = 0.5*Iw². This scenario involves kinetic energy gained by both the cube and the unwinding string's hanging mass. The specific energy amount depends on certain parameters that are currently absent.
Explanation:The question involves the calculation of the rotational kinetic energy of a spinning cube. The rotational kinetic energy (KErot) of an object with moment of inertia (I) and angular velocity (w) is given by the formula KErot = 0.5*Iw².
In this case, the cube picks up kinetic energy as it spins up, and the hanging mass also gains kinetic energy as the string unwinds. The total distance the string unwinds (L) also plays a role in the final rotational kinetic energy of the spinning cube.
However, the lack of specific parameters like the mass and angular velocity of the cube, as well as the mass of the hanging object in your question, makes it challenging to give a numerical answer.
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12028You are using a rope to lift a 14.5 kg crate of fruit. Initially you are lifting the crate at 0.500 m/s. You then increase the tension in the rope to 160 N and lift the crate an additional 1.25 m. During this d motion, how much work is done on the crate by the tension force
Ans: 237.5J
Explanation;
WC= T*d
T* dcostheta. But theta=0
= 190* 1.25
= 237.5J
See attached file for diagram
Final answer:
The work done on the crate by the tension force when lifting it an additional 1.25 m with a force of 160 N is 200 joules.
Explanation:
The question is asking about the work done on a 14.5 kg crate by tension force when lifting the crate an additional 1.25 m with a tension force of 160 N.
To calculate the work done by the tension force, we use the formula:
Work = Force x Distance x cosθ
Since the force and displacement are in the same direction (theta = 0 degrees), the cosine term is 1, and the equation simplifies to:
Work = 160 N x 1.25 m
The work done on the crate by the tension force is:
Work = 200 joules (J)
o study torque experimentally, you apply a force to a beam. One end of the beam is attached to a pivot that allows the beam to rotate freely. The pivot is taken as the origin or your coordinate system. You apply a force of F = Fx i + Fy j + Fz k at a point r = rx i + ry j + rz k on the beam. show answer No Attempt 33% Part (a) Enter a vector expression for the resulting torque, in terms of the unit vectors i, j, k and the components of F and r. τ = | γ θ i j k d Fx Fy Fz g m n rx ry rz ( ) 7 8 9 HOME ↑^ ^↓ 4 5 6 ← / * 1 2 3 → + - 0 . END √() BACKSPACE DEL CLEAR Grade Summary Deductions 0% Potential 100% Submissions Attempts remaining: 3 (4% per attempt) detailed view Hints: 4% deduction per hint. Hints remaining: 2 Feedback: 5% deduction per feedback. No Attempt No Attempt 33% Part (b) Calculate the magnitude of the torque, in newton meters, when the components of the position and force vectors have the values rx = 0.76 m, ry = 0.035 m, rz = 0.015 m, Fx = 3.6 N, Fy = -2.8 N, Fz = 4.4 N. No Attempt No Attempt 33% Part (c) If the moment of inertia of the beam with respect to the pivot is I = 442 kg˙m2, calculate the magnitude of the angular acceleration of the beam about the pivot, in radians per second squared. All content © 2020 Expert TA, LLC
Answer:
(a) Resulting torque = (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k
(b) Magnitude of resulting torque = = 3.99 Nm
(c) angular acceleration = = 0.009027 rad/s²
Explanation:
Given Data;
I = 442 kg˙m2
rx = 0.76 m,
ry = 0.035 m,
rz = 0.015 m,
Fx = 3.6 N,
Fy = -2.8 N,
Fz = 4.4 N
F = Fx i + Fy j + Fz ------------------------------1
r = rx i + ry j + rz k ------------------------------2
(a) Torgue is given by the formula;
T = r * F ------------------------------------3
Putting equation 1 and 2 into equation 3, we have;
Torque= r x F
= (rx i +ry j +rz k) x (Fx i + Fy j +Fz k )
= (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k
Therefore,
Resulting torque = (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k
b)
Putting given values into the above expression, we have
torque = (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k
=(0.035*4.4 - (0.015*-2.8))i +(0.015*3.6 - 0.76*4.4)j+(0.76* -2.8 - 0.035*3.6)k
= (0.154 +0.041) i + (0.054 - 3.344) j + (-2.128 -0.126) k
= (0.196) i - (3.29) j + (-2.254) k
Magnitude of resulting torque = √(0.196² + 3.29² +2.254²
=√15.943031
= 3.99 Nm
c) Angular acceleration is given by the formula;
angular acceleration = torque/moment of inertia
= 3.99/ 442
= 0.009027 rad/s²
Interactive Solution 28.5 illustrates one way to model this problem. A 7.11-kg object oscillates back and forth at the end of a spring whose spring constant is 61.6 N/m. An observer is traveling at a speed of 2.79 × 108 m/s relative to the fixed end of the spring. What does this observer measure for the period of oscillation?
Explanation:
Given that,
Mass of the object, m = 7.11 kg
Spring constant of the spring, k = 61.6 N/m
Speed of the observer, [tex]v=2.79\times 10^8\ m/s[/tex]
We need to find the time period of oscillation observed by the observed. The time period of oscillation is given by :
[tex]t_o=2\pi \sqrt{\dfrac{m}{k}} \\\\t_o=2\pi \sqrt{\dfrac{7.11}{61.6}} \\\\t_o=2.13\ s[/tex]
Time period of oscillation measured by the observer is :
[tex]t=\dfrac{t_o}{\sqrt{1-\dfrac{v^2}{c^2}} }\\\\t=\dfrac{2.13}{\sqrt{1-\dfrac{(2.79\times 10^8)^2}{(3\times 10^8)^2}} }\\\\t=5.79\ s[/tex]
So, the time period of oscillation measured by the observer is 5.79 seconds.
Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degrees, respectively, to that of the first. If unpolarized light is incident on the stack, the light has intensity 60.0 w/cm
2
after it passes through the stack.
If the incident intensity is kept constant:
1) What is the intensity of the light after it has passed through the stack if the second polarizer is removed?
2) What is the intensity of the light after it has passed through the stack if the third polarizer is removed?
Answer:
A) I_f3 = 27.58 W/cm²
B) I_f2 = 102.26 W/cm²
Explanation:
We are given;
-The angle of the second polarizing to the first; θ_2 = 21°
-The angle of the third polarizing to the first; θ_1 = 61°
- The unpolarized light after it pass through the polarizing stack; I_u = I_3 = 60 W/cm²
A) Let the initial intensity of the beam of light before polarization be I_p
Thus, when the unpolarized light passes through the first polarizing filter, the intensity of light that emerges would be given as;
I_1 = (I_p)/2
According to Malus’s law,
I = I_max(cos²Φ)
Thus, we can say that;
the intensity of light that would emerge from the second polarizing filter would be given as;
I_2 = I_1(cos²Φ1) = ((I_p)/2)(cos²Φ1)
Similarly, the intensity of light that will emerge from the third filter would be given as;
I_3 = I_2(cos²Φ1) = ((I_p)/2)(cos²Φ1)(cos²(Φ2 - Φ1)
Thus, making I_p the subject of the formula, we have ;
I_p = (2I_3)/[(cos²Φ1)(cos²(Φ2 - Φ1)]
Plugging in the relevant values, we have;
I_p = (2*60)/[(cos²21)(cos²(61 - 21)]
I_p = 234.65 W/cm²
Now, when the second polarizer is removed, the third polarizer becomes the second and final polarizer so the intensity of light emerging from the stack would be given as;
I_f3 = (I_p/2)(cos²Φ2)
I_f3 = (234.65/2)(cos²61)
I_f3 = 27.58 W/cm²
B) Similarly, when the third polarizer is removed, the second polarizer becomes the final polarizer and the intensity of light emerging from the stack would be given as;
I_f2 = (I_p/2)(cos²Φ1)
I_f2 = (234.65/2)(cos²21)
I_f2 = 102.26 W/cm²
In the sport of curling, large smooth stones are slid across an ice court to land on a target. Sometimes the stones need to move a bit farther across the ice and other times players want the stones to stop a bit sooner. Suggest a way to increase the kinetic friction between the stone and the ice so that the stone stops more quickly. Next, suggest a way to decrease the kinetic friction between the stone and the ice so that the stone slides farther along the ice before coming to a stop
Answer:
To increase kinetic friction, the amount of fine water droplets sprayed before the game is limited.
To reduce kinetic friction. increase the amount of fine water droplets during pregame preparation and sweeping in front of the curling stones.
Explanation:
In curling sports, since the ice sheets are flat, the friction on the stone would be too high and the large smooth stone would not travel half as far. Thus controlling the amount of fine water droplets sprayed before the game is limited pregame is necessary to increase friction.
On the other hand, reducing ice kinetic friction involves two ways. The first way is adding bumps to the ice which is known as pebbling. Fine water droplets are sprayed onto the flat ice surface. These droplets freeze into small "pebbles", which the curling stones "ride" on as they slide down the ice. This increases contact pressure which lowers the friction of the stone with the ice. As a result, the stones travel farther, and curl less.
The second way to reduce the kinetic friction is sweeping in front of the large smooth stone. The sweeping action quickly heats and melts the pebbles on the ice leaving a film of water. This film reduces the friction between the stone and ice.
Final answer:
To increase kinetic friction in curling, roughen the ice or stone's bottom; to decrease it, smooth the ice or use lubricants, often achieved by sweeping.
Explanation:
To increase the kinetic friction between the curling stone and the ice, thereby causing the stone to stop more quickly, you could roughen the surface of the ice or the bottom of the stone. This creates more irregularities which bump against each other, leading to increased vibrations and energy conversion to heat and sound, thus slowing down the stone. Additionally, players could sweep less or not at all in front of the stone so that the ice surface remains rougher and produces more resistance.
To decrease the kinetic friction and allow the stone to slide farther across the ice before coming to a stop, you could smooth the surface of the ice or apply a substance to the bottom of the stone to make it more slippery. Sweeping the ice in front of the stone is a common technique used to achieve this; it melts and smooths the ice surface, reducing the irregularities that cause friction and allowing the stone to glide more easily.
Consider the inelastic collision. Two lumps of matter are moving directly toward each other. Each lump has a mass of 1.500 kg 1.500 kg and is moving at a speed of 0.930 c 0.930c . The two lumps collide and stick together. Answer the questions, keeping in mind that relativistic effects cannot be neglected in this case. What is the final speed v f vf of the combined lump, expressed as a fraction of c ?
Final answer:
In an inelastic collision, two lumps of matter collide and stick together. The final speed of the combined lump is 0, meaning it comes to a complete stop after the collision.
Explanation:
In an inelastic collision, two lumps of matter collide and stick together. The final speed of the combined lump can be determined by applying the principles of conservation of momentum and the relativistic addition of velocities.
Let's calculate the final speed of the combined lump using the given information:
Mass of each lump (m) = 1.500 kg
Speed of each lump (v) = 0.930c
Using the relativistic addition of velocities formula:
vf = (v1 + v2) / (1 + (v1 * v2 / c2))
Plugging in the given values:
vf = (0.930c + (-0.930c)) / (1 + (0.930c * (-0.930c) / c2))
vf = 0
Therefore, the final speed (vf) of the combined lump is 0, which means it comes to a complete stop after the inelastic collision.
This part of the problem applies to any road surface, so the value of h is not known. Suppose you are driving at 70 miles per hour but, because of approaching darkness, you wish to slow to a speed that will cut your emergency stopping distance in half. What should your new speed be?
49.5 miles per hour will be the new speed.
This problem can be resolved using one-dimensional kinematics: v² = v₀² - 2 a d
Let's use this equation to find the beginning speed of the vehicle. In the first example, the distance is half (d´ = ½ d), the vehicle's acceleration is the same, and we find the fine velocity zero 0 = v₀² - 2 a d a = v₀² / 2d.
v₀´² = 2 a d = 2 (v₀² / 2d) 0 = v₀´² - 2 a d´ d´ v₀´ = v₀ √ 1/2
Let's compute: v₀ = 49.5 miles per hour; v₀´ = 70 ra 0.5.
Consequently, 49.5 miles per hour will be the new speed.
Below you can see the energy levels of the Helium atom. The right axis is a quantum number related to angular momentum (do not worry too much about that). An electron is in the state 2s and after a little while it decays back to the ground state. What is the energy of the photon emitted?
Answer:
ΔE = 20 eV
Explanation:
In a Helium atom we have two electrons in the s layer, so they can accommodate one with the spin up and the other with the spin down, give us a total spin of zero (S = 0) this state is singlet, in general this very stable states,
When you transition to the 1s state to complete the two electrons allowed by layers
ΔE = -5 - (-25) = 20 eV
this is the energy of the transition,
It should be mentioned that there can also be transitions with the two spins of the same orientation, but in this case the energy is a little different due to the electron-electron repulsion, this state is called ortho helium S = 1
1.) A metal sphere is held in a fluid flowing with a temperature of 30°C and velocity 2.5 m/s. The sphere has a radius of 10 mm and a constant surface temperature of 60°C. Find the drag force on the sphere and the rate of heat transfer from the sphere for: a.)Water b.)Air at 1 bar pressure Discuss what factors influence the answers.
Answer:
Explanation:
The pictures attached herewith shows the explanation and i hope it all helps. Thank you
A proton is accelerated from rest through a potential difference of 2.5 kV and then moves perpendicularly through a uniform 0.60-T magnetic field. What is the radius of the resulting path
Answer:
1.2cm
Explanation:
V=(2ev/m)^1/2
=(2*1.6*10^19 x2500/ 1.67*10^27)^1/2
=6.2x10^5m/s
Radius of resulting path= MV/qB
= 1.67*10^-27x6.92*10^6/1.6*10^-16 x0.6
=0.012m
=1.2cm
Answer:
0.012m
Explanation:
To find the radius of the path you can use the formula for the radius od the trajectory of a charge that moves in a constant magnetic field:
[tex]r=\frac{mv}{qB}[/tex]
m: mass of the proton 9.1*10^{-31}kg
q: charge of the proton 1.6*10^{-19}C
B: magnitude of the magnetic field 0.60T
v: velocity of the proton
In order to use the formula you need to calculate the velocity of the proton. This can be made by using the potential difference and charge, that equals the kinetic energy of the proton:
[tex]qV=E_k=\frac{1}{2}mv^2\\\\v=\sqrt{\frac{2qV}{m}}=\sqrt{\frac{2(1.6*10^{-19}C)(2.5*10^{3}V)}{1.67*10^{-27}kg}}=6.92*10^{5}\frac{m}{s}[/tex]
Then, by replacing in the formula for the radius you obtain:
[tex]r=\frac{(1.67*10^{-27}kg)(6.92*10^5\frac{m}{s})}{(1.6*10^{-19}C)(0.60T)}=0.012m[/tex]
hence, the radius is 0.012m
Two massive, positively charged particles are initially held a fixed distance apart. When they are moved farther apart, the magnitude of their mutual gravitational force changes by a factor of n. Which of the following indicates the factor by which the magnitude of their mutual electrostatic force changes?
a. 1/n2
b. 1/n
c. n
d. n2
The electrostatic force between two positively charged particles changes by a factor of 1/n² when they are moved further apart, based on Coulomb's Law.
Explanation:The subject matter of this question involves gravitational force and electrostatic force, which makes it a physics question. When two positively charged particles are moved further apart, the gravitational force between them changes by a factor of n. The electrostatic force between the two objects, however, obeys Coulomb's Law, which states that the force between two charges is inversely proportional to the square of the distance between them. Therefore, when the particles are moved further apart, the magnitude of their electrostatic force changes by a factor of 1/n². So, the correct answer to this question would be Option A: 1/n².
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The magnitude of the mutual electrostatic force between two charged particles that are moved farther apart changes by a factor of 1/n²
When two positively charged particles are moved farther apart, we know that the force between them, according to Coulomb's law, is inversely proportional to the square of the distance between them.
Therefore, the magnitude of their mutual electrostatic force will change by a factor of 1/n² when the distance between them is changed.
This is analogous to the gravitational force between two masses, which follows the same inverse-square law as electrostatic force does.
A 0.12 kg paper airplane flies in a straight line at a speed of 1.3 m/s. How much kinetic energy does the airplane have?
Answer:
0.1014 J
Explanation:
Considering the air resistance negligible kinetic energy is given by
K.E = (1/2) m v^2
K.E = 0.5 x 0.12 x (1.3)^2
K.E = 0.06 x 1.69
K.E = 0.1014 J
Final answer:
The 0.12 kg paper airplane flying at 1.3 m/s has a kinetic energy of 0.1014 joules, calculated using the formula KE = 0.5 × m × v^2.
Explanation:
The kinetic energy of a paper airplane can be calculated using the kinetic energy formula KE = 0.5 × m × v^2, where KE is kinetic energy, m is the mass of the object, and v is its velocity. For a 0.12 kg paper airplane flying at a speed of 1.3 m/s, the kinetic energy can be calculated as follows:
KE = 0.5 × 0.12 kg × (1.3 m/s)^2 = 0.1014 J
Therefore, the paper airplane has 0.1014 joules of kinetic energy.
"The International Space Station (ISS) orbits at a distance of 350 km above the surface of the Earth. (a) Determine the gravitational field strength that astronauts onboard the ISS would experience. (b) Calculate the orbital velocity of the ISS. (c) Determine the period of the ISS orbit around Earth"
Answer:
(a) g = 8.82158145[tex]m/s^2[/tex].
(b) 7699.990192m/s.
(c)5484.3301s = 1.5234 hours.(extremely fast).
Explanation:
(a) Strength of gravitational field 'g' by definition is
[tex]g = \frac{M_{(earth)} }{r^2} G[/tex] , here G is Gravitational Constant, and r is distance from center of earth, all the values will remain same except r which will be radius of earth + altitude at which ISS is in orbit.
r = 6721,000 meters, putting this value in above equation gives g = 8.82158145[tex]m/s^2[/tex].
(b) We have to essentially calculate centripetal acceleration that equals new 'g'.
[tex]a_{centripetal}=\frac{V^2}{r} =g[/tex] here g is known, r is known and v is unknown.
plugging in r and g in above and solving for unknown gives V = 7699.990192m/s.
(c) S = vT, here T is time period or time required to complete one full revolution.
S = earth's circumfrence , V is calculated in (B) T is unknown.
solving for unknown gives T = 5484.3301s = 1.5234hours.
A cleaver physics professor wants to create a situation where a block starts from rest at the top of a 31.0° inclined plane and encounters a spring at the bottom of the incline. The spring has a constant 3.4 kN/m and the block's mass is 33.0 kg. How far does the block travel before hitting the spring, if the spring was compressed 37 cm in it's initial collision?
Answer:
Explanation:
Let the length of inclined plane be L .
work done by gravity on the block
= force x length of path
= mg sinθ x L , m is mass of the block , θ is inclination of path
This in converted into potential energy of compressed spring
1/2 k x² = mgL sin31 , k is force constant . x is compression
.5 x 3400 x .37² = 33 x9.8 x sin31 L
L = 1.4
Length of incline = 1.4 m .
106 grams of liquid water are in a cylinder with a piston maintaining 1 atm (101325 Pa) of pressure. It is exactly at the boiling point of water, 373.15 K. We then add heat to boil the water, converting it all to vapor. The molecular weight of water is 18 g/mol and the latent heat of vaporization is 2260 J/g.
The missing part of the question is;
1) How much heat is required to boil the water?
2) Assume that the liquid water takes up approximately zero volume, and the water vapor takes up some final volume Vf. You may also assume that the vapor is an ideal gas. How much work did the vapor do pushing on the piston?
3) How much did the water internal energy change?
Answer:
A) Q = 239.55 KJ
B) W = 18.238 KJ
C) ΔU = 221.31 J
Explanation:
We are giving;
Mass; m = 106 g
Latent heat of vaporization; L = 2260 J/g.
Molecular weight of water; M = 18 g/mol
Pressure; P = 101325 Pa
Temperature; T = 373.15 K
A) Formula for amount of heat required is;
Q = mL
Q = 106 x 2260
Q = 239560J = 239.55 KJ
B) number of moles; n = m/M
n = 106/18
n = 5.889 moles
Now, we know from ideal gas equation that;
PV = nRT
Thus making V the subject, we have; V = nRT/P
However, here we are told V is V_f. Thus, V_f = nRT/P
R is gas constant = 8.314 J/mol·K
Plugging in the relevant values ;
V_f = (5.889 x 8.314 x 373.15)/101325
V_f = 0.18 m³
Now, at constant pressure, work done is;
W = P(ΔV)
W = P(V_f - V_i)
W = 101325(0.18 - 0)
W = 101325 x 0.18
W = 18238.5J = 18.238 KJ
C) Change in water internal energy is gotten from;
ΔU = Q - W
Thus, ΔU = 239.55 KJ - 18.238 KJ
ΔU = 221.31 J
Particles (mass of each = 0.40 kg) are placed at the 60-cm and 100-cm marks of a meter stick of negligible mass. This rigid body is free to rotate about a frictionless pivot at the 0-cm end. The body is released from rest in the horizontal position. What is the magnitude of the initial linear acceleration of the end of the body opposite the pivot?
Answer:
12 m/s ∧2
Explanation:
The picture attached explains it all and i hope it helps. Thank you
To find the initial linear acceleration, calculate the initial torque due to gravity, the moment of inertia, and then apply Newton’s second law for rotation to obtain the initial angular acceleration. The initial linear acceleration can then be found by multiplying the initial angular acceleration by the length of the meter stick.
Explanation:The question is asking for the initial linear acceleration of the end of the rigid body (meter stick) at the moment it is released.
The initial torque τ is equal to the gravitational forces acting on each particle times their respective distances from the pivot, summed up.
τ_initial = m*g*d1 + m*g*d2 = 0.4 kg * 9.8 m/s^2 * 0.6 m + 0.4 kg * 9.8 m/s^2 * 1.0 m
The moment of inertia I for the two-particle system can be calculated with the formula I = ∑mr^2 for each particle:
I = m*d1^2 + m*d2^2 = 0.4 kg * (0.6 m)^2 + 0.4 kg * (1.0 m)^2
According to Newton’s second law for rotation, the initial angular acceleration α is equal to the initial torque divided by the moment of inertia:
α = τ_initial / I
The initial linear acceleration a of the end of the body at the point opposite the pivot is equal to the product of the initial angular acceleration and the total length of the meter stick (1.0 m):
a = α * 1.0 m
Learn more about Rigid Body Dynamics here:https://brainly.com/question/12903545
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The maximum current output of a 60 Ω circuit is 11 A. What is the rms voltage of the circuit?
Answer:
660V
Explanation:
V=IR
V=?,I=11A,R=60w
V=60×11
=660V
A small mirror is attached to a vertical wall, and it hangs a distance of 1.70 m above the floor. The mirror is facing due east, and a ray of sunlight strikes the mirror early in the morning and then again later in the morning. The incident and reflected rays both lie in a plane that is perpendicular to both the wall and the floor. Early in the morning, the ray is observed to strike the floor at a distance of 3.75 m from the base of the wall. Later on in the morning, the ray strikes the floor at a distance of 1.18 m from the wall. The earth rotates at 15.0 degrees per hour. How much time (in hours) has elapsed between the two observations
Answer:
Around 2 hours
Explanation:
The answer is given in the pictures below. All the page numbers are circled on the top left corner of each page.
Final answer:
The question deals with calculating time based on the change in position of a reflected sunlight ray due to Earth's rotation, with the Earth rotating at a rate of 15 degrees per hour.
Explanation:
The question concerns the reflection of light from a mirror and the calculation of time based on the Earth's rotation rate. The sunlight's path changes throughout the morning due to the Earth's rotation, which moves at a consistent rate of 15 degrees per hour. Given the different positions where the ray of sunlight strikes the floor at different times, we can calculate the angle change and then the time elapsed.
To find the time elapsed, we first determine the angular change between the two observations. The change in position on the floor from 3.75 m to 1.18 m corresponds to a difference in angle when considering the mirror as the vertex of a right triangle with the wall. Next, we apply the Earth's rotational rate to find the corresponding time.
Current and Resistivity Concepts1.Electric charge is conserved. As a consequence, when current arrives at a junction of wires, the charges can take either of two paths out of the junction and the numerical sum of the currents in the two paths equals the current that entered the junction. Thus, current is which of the following?a)a scalarb)a vector c)neither a vector nor a scalar2.A cylindrical wire has a radius r and length l. If both r and l are doubled, the resistance of the wire does what?a)decreasesb)increases c)remains the same
Answer:
1. A). Scalar
2.C) remains de same
Explanation:
A plane monochromatic radio wave (? = 0.3 m) travels in vacuum along the positive x-axis, with a time-averaged intensity I = 45.0 W/m2. Suppose at time t = 0, the electric field at the origin is measured to be directed along the positive y-axis with a magnitude equal to its maximum value. What is Bz, the magnetic field at the origin, at time t = 1.5 ns? Bz = I got .04800 but that answer didnt work.
Answer:
The magnetic field [tex]B_Z[/tex] [tex]= - 6.14*10^{-7} T[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 0.3m[/tex]
The intensity is [tex]I = 45.0W/m^2[/tex]
The time is [tex]t = 1.5ns = 1.5 *10^{-9}s[/tex]
Generally radiation intensity is mathematically represented as
[tex]I = \frac{1}{2} c \epsilon_o E_o^2[/tex]
Where c is the speed of light with a constant value of [tex]3.0 *10^8 m/s[/tex]
[tex]E_i[/tex] is the electric field
[tex]\epsilon_o[/tex] is the permittivity of free space with a constant value of [tex]8.85*10^{-12} C^2 /N \cdot m^2[/tex]
Making [tex]E_o[/tex] the subject of the formula we have
[tex]E_i = \sqrt{\frac{2I}{c \epsilon_0} }[/tex]
Substituting values
[tex]E_i = \sqrt{\frac{2* 45 }{(3*10^8 * (8.85*10^{-12}) )} }[/tex]
[tex]= 184.12 \ V/m[/tex]
Generally electric and magnetic field are related by the mathematical equation as follows
[tex]\frac{E_i}{B_i} = c[/tex]
Where [tex]B_O[/tex] is the magnetic field
making [tex]B_O[/tex] the subject
[tex]B_i = \frac{E_i}{c}[/tex]
Substituting values
[tex]B_i = \frac{184.12}{3*10^8}[/tex]
[tex]= 6.14 *10^{-7}T[/tex]
Next is to obtain the wave number
Generally the wave number is mathematically represented as
[tex]n = \frac{2 \pi }{\lambda }[/tex]
Substituting values
[tex]n = \frac{2 \pi}{0.3}[/tex]
[tex]= 20.93 \ rad/m[/tex]
Next is to obtain the frequency
Generally the frequency f is mathematically represented as
[tex]f = \frac{c}{\lambda}[/tex]
Substituting values
[tex]f = \frac{3 *10^8}{0.3}[/tex]
[tex]= 1*10^{9} s^{-1}[/tex]
Next is to obtain the angular velocity
Generally the angular velocity [tex]w[/tex] is mathematically represented as
[tex]w = 2 \pi f[/tex]
[tex]w = 2 \pi (1* 10^9)[/tex]
[tex]= 2 \pi * 10^9 rad/s[/tex]
Generally the sinusoidal electromagnetic waves for the magnetic field B moving in the positive z direction is expressed as
[tex]B_z = B_i cos (nx -wt)[/tex]
Since the magnetic field is induced at the origin then the equation above is reduced to
[tex]B_z = B_i cos (n(0) -wt) = B_i cos ( -wt)[/tex]
x =0 because it is the origin we are considering
Substituting values
[tex]B_z = (6.14*10^{-7}) cos (- (2 \pi * 10^{9})(1.5 *10^{-9}))[/tex]
[tex]= - 6.14*10^{-7} T[/tex]
During the lab, you will build the following circuit. Draw arrows beside the resistors R1 , R2 and R3 to denote the direction of (conventional) current through the resistors. [3] A B R1 E R2 C R3 D Vs In each case, explain why the arrows point in the way that you have drawn. _______________________________________________________________________________ ____________________________________________________________________________[2]
Question: The question is incomplete. Diagram of the circuit was not added to your question. Find attached of the circuit diagram and the answer.
Answer:
For R1: Current moves from A to B
For R2: Current moves from B to E
For R3: Current moves from C to D
Explanation:
See the attached file for the explanation
An unusual lightning strike has a vertical portion with a current of –400 A downwards. The Earth’s magnetic field at that location is parallel to the ground and has a magnitude of 30 μT. In units of N (Newtons), the force exerted by the Earth’s magnetic field on the 25 m-long current is Group of answer choices 300 A, East. 0.30 A, West. 0 0.012 A, East. 0.012 A, West.
Given that,
Current, I = 400 A (downwards)
The Earth’s magnetic field at that location is parallel to the ground and has a magnitude of 30 μT.
We need to find the force exerted by the Earth’s magnetic field on the 25 m-long current. We know that the magnetic force is given by :
[tex]F=iLB\sin\theta[/tex]
Here, [tex]\theta=90^{\circ}[/tex]
[tex]F=400\times 25\times 30\times 10^{-6}\\\\F=0.3\ N[/tex]
The force is acting in a plane perpendicular to the current and the magnetic field i.e out of the plane.
Compressed gases aren't ideal. Let's consider a gas that's non-ideal only because the volume available to each of the N molecules is reduced because each other molecule occupies volume v. Instead of pV=NkT, we get: p(V-Nb)=NkT. Let b=1.2 × 10-28 m3. Let's look at 3moles of this gas at T=300K starting in 0.001 m3 volume. 1) What's the initial value of the pressure?
Answer:
P = 7482600 Pa = 7.482 MPa
Explanation:
We have the equation:
P(V - Nb) = NKT
here,
P = Initial Pressure = ?
V = Initial Volume = 0.001 m³
N = No. of moles = 3
b = constant = 1.2 x 10⁻²⁸ m³
T = Temperature = 300 k
k = Gas Constant = 8.314 J/mol . k
Therefore,
P[0.001 m³ - (3)(1.2 x 10⁻²⁸ m³)] = (3 mol)(8.314 J/mol. k)(300 k)
P = (7482.6 J)/(0.001 m³)
P = 7482600 Pa = 7.482 MPa
A 6.85-m radius air balloon loaded with passengers and ballast is floating at a fixed altitude. Determine how much weight (ballast) must be dropped overboard to make the balloon rise 114 m in 17.0 s. Assume a constant value of 1.2 kg/m3 for the density of air. Ballast is weight of negligible volume that can be dropped overboard to make the balloon rise.
Answer: 120 kg
Explanation:
Given
Radius of balloon, r = 6.85 m
Distance moved by the balloon, d = 114 m
Time spent in moving, t = 17 s
Density of air, ρ = 1.2 kg/m³
Volume of the balloon = 4/3πr³
Volume = 4/3 * 3.142 * 6.85³
Volume = 4/3 * 3.142 * 321.42
Volume = 4/3 * 1009.90
Volume = 1346.20 m³
Density = mass / volume ->
Mass = Density * volume
Mass = 1.2 * 1346.2
Mass = 1615.44 kg
Velocity = distance / time
Velocity = 114 / 17
Velocity = 6.71 m/s
If it starts from rest, 0 m/s, then the final velocity is 13.4 m/s
acceleration = velocity / time
acceleration = 13.4 / 17 m/s²
The mass dropped from the balloon decreases Mb and increases buoyancy
F = ma
mg = (Mb - m) * a
9.8 * m = (1615.44 - m) * 13.4/17
9.8m * 17/13.4 = 1615.44 - m
12.43m = 1615.44 - m
12.43m + m = 1615.44
13.43m = 1615.44
m = 1615.44 / 13.43
m = 120.29 kg
Young's experiment is performed with light of wavelength 502 nmnm from excited helium atoms. Fringes are measured carefully on a screen 1.40 mm away from the double slit, and the center of the 20th fringe (not counting the central bright fringe) is found to be 10.4 mmmm from the center of the central bright fringe. Part A What is the separation of the two slits? dd = nothing mmmm SubmitRequest Answer Provide Feedback
Answer:
Explanation:
wavelength of light λ = 502 x 10⁻⁹ m /s
screen distance D = 1.4 m
Slit separation d = ?
position of n the separation is given by the formula
x = n Dλ / d , n is order of fringe , x = distance of n th fringe
10.4 x 10⁻³ = 20 x 1.4 x 502 x 10⁻⁹ / d
d = 20 x 1.4 x 502 x 10⁻⁹ / 10.4 x 10⁻³
= 1351.54 x 10⁻⁶
= 1.35 x 10⁻³ m
1.35 mm.
A disk of mass m and radius r is initially held at rest just above a larger disk of mass M and radius R that is rotating at angular speed ωi. What is the final angular speed of the disks after the top one is dropped onto the bottom one and they stop slipping on each other? Note: the moment of inertia of a disk of mass M and radius R about an axis through its center and perpendicular to the disk is I = (1/2)MR .
Answer:
Explanation:
Moment of inertia of larger disk I₁ = 1/2 MR²
Moment of inertia of smaller disk I₂ = 1/2 m r ²
Initial angular velocity
We shall apply law of conservation of angular momentum .
initial total momentum = final angular momentum
I₁ X ωi = ( I₁ + I₂ )ωf
1/2 MR² x ωi = 1/2 ( m r² + MR² ) ωf
ωf = ωi / ( 1 + m r²/MR² )