To solve this problem we must find the values of the equivalent resistances in both section 1 and section 2. Later we will calculate the total current and the total voltage. With the established values we can find the values of the currents in the 3 Ohms resistance and the power there.
The equivalent resistance in section 1 would be
[tex]R_{eq1} = \frac{(3\Omega)(6\Omega)}{(3+6)\Omega}[/tex]
[tex]R_{eq1} = 2\Omega[/tex]
The equivalent resistance in section 2 would be
[tex]R_{eq2} = R_{eq1} +4\Omega[/tex]
[tex]R_{eq2} = 6\Omega[/tex]
Now the total current will be,
[tex]I_t = \frac{V_t}{R_{eq2}}[/tex]
[tex]I_t = \frac{12V}{6\Omega}[/tex]
[tex]I_t = 2.0A[/tex]
Finally the total Voltage will be,
[tex]V = IR_{eq1}[/tex]
[tex]V = (2.0A)(2.0\Omega)[/tex]
[tex]V = 4V[/tex]
Since the voltage across the 3 and 6 Ohms resistor is the same, because they are in parallel, the current in section 3 would be
[tex]I_{3.0\Omega} = \frac{V}{R}[/tex]
[tex]I_{3.0\Omega} = \frac{4.0V}{3.0\Omega}[/tex]
[tex]I_{3.0\Omega} = 1.3A[/tex]
Finally the power ratio is the product between the current and the voltage then,
[tex]P_{3.0\Omega} = I_{3.0\Omega} V[/tex]
[tex]P_{3.0\Omega} = (1.3A)(4.0V)[/tex]
[tex]P_{3.0\Omega} = 5.3W[/tex]
Therefore the correct answer is D.
Final answer:
To find the power dissipated in the 3.0-Ω resistor, calculate the total resistance and current, then apply Ohm's law to find the voltage across and current through the resistor. Finally, use the power formula P = V²/R, resulting in a power dissipation of (a) 12 W for the 3.0-Ω resistor.
Explanation:
The question deals with finding the power dissipated in the 3.0-Ω resistor. To answer this, we first need to find the total resistance of the circuit and the current through the circuit. We then apply this current to the parallel resistors to find the voltage across and the current through the 3.0-Ω resistor before calculating its power dissipation.
First, calculate the resistance of the resistors in parallel. Using the formula for resistors in parallel:
1/Rparallel = 1/R₁ + 1/R₂
1/Rparallel = 1/3.0 + 1/6.0
1/Rparallel = 1/3 + 1/6 = 2/6 + 1/6 = 3/6
1/Rparallel = 1/2
Rparallel = 2.0 Ω
Now, add the series resistor to find the total resistance:
Rtotal = Rparallel + Rseries
Rtotal = 2.0 + 4.0 = 6.0 Ω
Then, using Ohm's law, calculate the total current from the battery:
I = V/Rtotal
I = 12V/6.0Ω
I = 2.0 A
The current through the 3.0-Ω resistor in parallel is the same as the total current, so we use Ohm's law V = IR to find the voltage across the 3.0-Ω resistor:
V3.0-Ω = 2.0 A × 3.0 Ω = 6.0 V
Finally, calculate the power dissipated by the 3.0-Ω resistor:
P = V2/R
P = (6.0 V)2/3.0 Ω
P = 36 W/3.0 Ω
P = 12 W
Therefore, the power dissipated in the 3.0-Ω resistor is 12 W, which corresponds to choice (a).
A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1∘ from the vertical. The magnitude of the magnetic field B changes in time according to the equation B(t)=(3.75 T)+(2.75 Ts)t+(−7.05 Ts2)t2 If the radius of the wire loop is 0.210 m, find the magnitude of the induced emf in the loop when t=5.63 s.
Explanation:
Given that,
A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1∘ from the vertical, [tex]\theta=15.1^{\circ}[/tex]
The magnitude of the magnetic field B changes in time according to the equation :
[tex]B(t)=3.75+2.75 t-7.05 t^2[/tex]
Radius of the loop, r = 0.21 m
We need to find the magnitude of the induced emf in the loop when t=5.63 s. The induced emf is given by :
[tex]\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=\dfrac{-d(BA\cos \theta)}{dt}[/tex]
B is magnetic field
A is area of cross section
[tex]\epsilon=A\dfrac{-dB}{dt}\\\\\epsilon=\pi r^2\dfrac{-d(3.75+2.75 t-7.05 t^2)}{dt}\times \cos\theta\\\\\epsilon=\pi r^2\times(2.75-14.1t)\times \cos\theta[/tex]
At t = 5.63 seconds,
[tex]\epsilon=-\pi (0.21)^2\times(2.75-14.1(5.63))\times \cos(15.1)\\\\\epsilon=10.25V[/tex]
So, the magnitude of induced emf in the loop when t=5.63 s is 10.25 V.
The EMF generated at time t = 5.63 is 10.18V.
Magnetic flux and EMF:Given a horizontal circular wire loop with a radius, r = 0.21m.
A time-dependent magnetic field B(t) = 3.75 + 2.75t -7.05t².
At an angle of θ = 15.1° to the area of the loop.
The magnetic flux passing through the loop is given by:
Ф = B(t)Acosθ
where A = πr² is the are of the loop.
Since the magnetic field is time-dependent, the magnetic flux through the loop changes with time, therefore an EMF is generated in the loop, given by:
[tex]E=-\frac{d\phi}{dt}\\\\E =-\frac{dB(t)}{dt}Acos\theta\\\\E=-\pi r^2cos\theta\frac{d}{dt}[ 3.75 + 2.75t -7.05t^2] \\\\E=\pi r^2cos\theta[14.10t-2.75]\\\\[/tex]
At time t = 5.63s
[tex]E=3.14\times(021)^2\times cos15.1\times[14.1\times5.63-2.75][/tex]
E = 10.18V
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Consider two straight wires lying on the x-axis, separated by a gap of 4 nm. The potential energy in the gap is about 3 eV higher than the energy of a ondu tion ele tron in either wire. What is the probability that a ondu tion ele tron in one wire arriving at the gap will pass through the gap into the other wire?
Answer: 1.3 ×10^-31
Explanation:
the required probability is P = e^(-2αL)
Firstly, evaluate (-2αL)
α= 1/hc √2mc^2 (U - E)
h= modified planck's constant
where,
(-2αL)
= -(2L)/(h/2π ) ×√2mc^2 (U - E)
= -(2L) / (hc^2/π )×√2mc^2 (U - E)
(hc^2/2pi) = 197*eV.nm (standard constant)
2*L = 8 nm
mc^2 = 0.511×10^6 eV
Where m = mass electron
C= speed of light
(-2αL) = [-8nm/(197 eV.nm)] × (1.022× 10^6 eV*×3 eV)^0.5
(-2αL) = -71.1
Probability = e^(-2αL) = e^-71.1 = 1.3 ×10^-31
Therefore, the probability that a ondu tion ele tron in one wire arriving at the gap will pass through the gap into the other wire 1.3 ×10^-31.
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
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
A loud factory machine produces sound having a displacement amplitude in air of 1.00 μmμm, but the frequency of this sound can be adjusted. In order to prevent ear damage to the workers, the maximum pressure amplitude of the sound waves is limited to 10.0 PaPa. Under the conditions of this factory, the bulk modulus of air is 1.30×105 PaPa . The speed of sound in air is 344 m/s
What is the highest-frequency sound to which this machine can be adjusted without exceeding the prescribed limit?
Answer:
f =3.4*10^3 hz
Since f is in the range of (20 Hz - 20,000 Hz] which is the range of audible frequencies, the frequency is audible.
Explanation:
The relation that describes the pressure amplitude for a sound wave is
P_MAX = B*k*A (1)
Where the bulk modulus of the air is B = 1.30 x 10^5 Pa and the displacement amplitude of the waves produced by the machine is 1.00 μmμm.
Using (1) we can calculate k then we can use k to determine the wavelength A of the wave, and remember that λ = 2π/k.
So, substitute into (1) with 10 Pa for P_max, (1.30 x 10^5 Pa) for B and
1 x 10^-6 m for A
10 Pa = (1.30 x 10^5 Pa) x k x (1 x 10^-6 m)
k = 62.5 m^-1
We can use the following relation to calculate the wavelength
λ = 2π/k.
λ = 0.100 m
Finally, the relation between the wavelength and the frequency of a sound
wave is given by the following equation
f = v/ λ
=344/0.100 m
f =3.4*10^3 hz
Since f is in the range of (20 Hz - 20,000 Hz] which is the range of audible frequencies, the frequency is audible.
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
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 .
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 )
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.
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.
A space cowboy wants to eject from his spacecraft 1.00×105 km after passing a space buoy, as seen by spectators at rest with respect to the buoy. To do this, the cowboy sets a timer on his craft that will start as he passes the buoy. He plans to cruise by the buoy at 0.217c.How much time should he allow between passing the buoy and ejecting?options:1.50 s1.70 s1.30 s1.60 s1.55 s
Answer:
option D.
Explanation:
Given,
distance, d = 1 x 10⁵ km
speed , v = 0.217 c
time of dilation , T₀= ?
Using the formula of time dilation
[tex]T=\frac{T_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/tex]
[tex]T_{0}=T \sqrt{1-\frac{v^{2}}{c^{2}}} [/tex]
[tex]=\left(\frac{d}{v}\right) \sqrt{1-\frac{v^{2}}{c^{2}}} [/tex]
[tex]=\left(\frac{1.00 \times 10^{8}}{0.217 \times 2.99 \times 10^{8}}\right) \sqrt{1-\frac{(0.217 c)^{2}}{c^{2}}} [/tex]
[tex]=1.54 \times 0.976 [/tex]
[tex]=1.50\ s[/tex]
Time he should be allowed between passing the buoy and ejecting is equal to 1.50 s.
The correct answer is option D.
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)
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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Which way do your feet need to push on the floor in order to
jump forward?
Answer: In order to walk on a floor (or any other surface), your foot must push backward on the floor (action force), so that the floor pushes you forward (reaction force).
Explanation:
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
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
Summarize ocean acidification in one sentence.
Answer:
The ocean absorbs a significant portion of carbon dioxide (CO2) emissions from human activities, equivalent to about one-third of the total emissions for the past 200 years from fossil fuel combustion, cement production and land-use change (Sabine et al., 2004). Uptake of CO2 by the ocean benefits society by moderating the rate of climate change but also causes unprecedented changes to ocean chemistry, decreasing the pH of the water and leading to a suite of chemical changes collectively known as ocean acidification. Like climate change, ocean acidification is a growing global problem that will intensify with continued CO2 emissions and has the potential to change marine ecosystems and affect benefits to society.
The average pH of ocean surface waters has decreased by about 0.1 unit—from about 8.2 to 8.1—since the beginning of the industrial revolution, with model projections showing an additional 0.2-0.3 drop by the end of the century, even under optimistic scenarios (Caldeira and Wickett, 2005).1 Perhaps more important is that the rate of this change exceeds any known change in ocean chemistry for at least 800,000 years (Ridgewell and Zeebe, 2005). The major changes in ocean chemistry caused by increasing atmospheric CO2 are well understood and can be precisely calculated, despite some uncertainty resulting from biological feedback processes. However, the direct biological effects of ocean acidification are less certain
image
1 “Acidification” does not mean that the ocean has a pH below neutrality. The average pH of the ocean is still basic (8.1), but because the pH is decreasing, it is described as undergoing acidification.
Page 2
Suggested Citation:"Summary." National Research Council. 2010. Ocean Acidification: A National Strategy to Meet the Challenges of a Changing Ocean. Washington, DC: The National Academies Press. doi: 10.17226/12904. ×
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and will vary among organisms, with some coping well and others not at all. The long-term consequences of ocean acidification for marine biota are unknown, but changes in many ecosystems and the services they provide to society appear likely based on current understanding (Raven et al., 2005).
In response to these concerns, Congress requested that the National Research Council conduct a study on ocean acidification in the Magnuson-Stevens Fishery Conservation and Management Reauthorization Act of 2006. The Committee on the Development of an Integrated Science Strategy for Ocean Acidification Monitoring, Research, and Impacts Assessment is charged with reviewing the current state of knowledge and identifying key gaps in information to help federal agencies develop a program to improve understanding and address the consequences of ocean acidification (see Box S.1 for full statement of task). Shortly after the study was underway, Congress passed another law—the Federal Ocean Acidification Research and Monitoring (FOARAM) Act of 2009—which calls for, among other things, the establishment of a federal ocean acidification program; this report is directed to the ongoing strategic planning process for such a program.
Although ocean acidification research is in its infancy, there is already growing evidence of changes in ocean chemistry and ensuing biological impacts. Time-series measurements and other field data have documented the decrease in ocean pH and other related changes in seawater chemistry (Dore et al., 2009). The absorption of anthropogenic CO2 by the oceans increases the concentration of hydrogen ions in seawater (quanti-
Explanation:
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.
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
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.
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.
In a physics lab, light with a wavelength of 540 nm travels in air from a laser to a photocell in a time of 16.5 ns. When a slab of glass with a thickness of 0.820 m is placed in the light beam, with the beam incident along the normal to the parallel faces of the slab, it takes the light a time of 21.5 ns to travel from the laser to the photocell.
What is the wavelength of the light in the glass? Use 3.00×10^8 m/s for the speed of light in a vacuum.
Answer:
189.47nm
Explanation:
We can solve this problem by taking into account the time that light takes in crossing the distance between the laser and the photocell, and the time in crossing the slab.
By using the values of c and 16.5ns we can calculate the value of d
[tex]d=(3*10^{8}m/s)(16.5*10^{-9}s)=4.95m[/tex]
We have to compute the time that light takes in crossing d-0.820m:
[tex]4.95m-0.820m=4.13m\\\\t=\frac{4.13m}{3*10^8m/s}=6.22*10^{-8}s=13.7ns[/tex]
Now, we can calculate the speed of the light in the slab by using the time difference between 21.5 ns and 13.7ns:
[tex]\Delta t=21.5ns-13.7ns=7.8ns\\\\v_l=\frac{0.82m}{7.8ns}=1.05*10^8m/s[/tex]
Then, the index of refraction will be:
[tex]n=\frac{c}{v_l}=2.85[/tex]
Finally, we have that:
[tex]\frac{\lambda_2}{\lambda_1}=\frac{n_1}{n_2}\\\\\lambda_2=\lambda_1 \frac{n_1}{n_2}=(540nm)\frac{1.00}{2.85}=189.47nm[/tex]
hope this helps!!
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.
instrument used to measure the length of a football field
Answer:
If you need to measure much longer lengths - for example the length of a football pitch - then you could use a trundle wheel. You use it by pushing the wheel along the ground. It clicks every time it measures one metre.
You could employ a trundle wheel if you need to estimate considerably greater distances, like the size of a football field. By moving the wheels along the surface, you can use it. It makes a click sound each time it measures a meter.
What is football?Football, commonly known as association soccer, is a sport in which two groups of 11 players attempt to advance the ball into the goal of the other team by using any component of their bodies other than their hands and arms.
Only the goalie is allowed to handle the ball, and only within the towns near the goal that is designated as the penalty area. The team with the most goals scored wins.
According to the number of players and spectators, football is the most watched sport in the world.
The sport may be played practically everywhere, from official football fields (pitches) to gyms, streets, school playground, parks, or beaches, thanks to its basic rules and necessary equipment. The Federation Internationale Football Association is the governing organization of football.
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"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.
Problem: 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 wi. 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)MR2.
Answer:
The final angular velocity is [tex]w_f = \frac{MR^2}{MR^2+ mr^2} w[/tex]
Explanation:
From the question we are told that
The mass of the first disk is m
The radius of the first disk is r
The mass of second disk is M
The radius of second disk is R
The speed of rotation is w
The moment of inertia of second disk is [tex]I = \frac{1}{2} MR^2[/tex]
Since the first disk is at rest initially
The initial angular momentum would be due to the second disk and this is mathematically represented as
[tex]L_i = Iw = \frac{1}{2} MR^2 w[/tex]
Now when the first disk is then dropped the angular momentum of the whole system now becomes
[tex]L_f = (I_1 + I_2 ) w_f= ( \frac{1}{2} MR^2 + \frac{1}{2} m^2 r^2) w_f[/tex]
This above is because the formula for moment of inertia is the same for every disk
According to the law conservation of angular momentum
[tex]L_f = L_i[/tex]
[tex]( \frac{1}{2} MR^2 + \frac{1}{2} m^2 r^2) w_f = \frac{1}{2} MR^2 w[/tex]
=> [tex]w_f = \frac{\frac{1}{2} MR^2 w }{( \frac{1}{2} MR^2 + \frac{1}{2} m^2 r^2)}[/tex]
[tex]w_f = \frac{MR^2}{MR^2+ mr^2} w[/tex]
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
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.
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.
A ray of light is incident on a flat surface of a block of polystyrene, with an index of refraction of 1.49, that is submerged in water. The ray is split at the surface, where the angle of refraction of the transmitted ray is 19.1°. What is the angle of reflection (in degrees) of the reflected ray?
Answer:
21.5°
Explanation:
Given,
Refractive index of water, n₁ = 1.33
Refractive index of polystyrene, n₂ = 1.49
Angle of reflection = ?
Angle of refraction = 19.1°
Using Snell's law
n₁ sin θ₁ = n₂ sin θ₂
1.33 x sin θ₁ = 1.49 x sin 19.1°
sin θ₁ = 0.366
θ₁ = 21.5°
According to law of reflection angle of incidence is equal to angle of reflection.
Angle of reflection = 21.5°
Final answer:
The angle of reflection of the reflected ray in polystyrene that is submerged in water will be equal to the angle of refraction, so it is 19.1°.
Explanation:
The question is about the reflection and refraction of light as it strikes a flat surface of a polystyrene block that is in water. According to the Law of Reflection, the angle of reflection is equal to the angle of incidence.
Since we know the angle of refraction is 19.1°, and by Snell's Law (n₁ × sin(i) = n₂ × sin(r)), we can infer that the angle of incidence is also 19.1° because the boundary is between two different mediums (water and polystyrene), resulting in the light being split into reflected and refracted rays.
Therefore, the angle of reflection of the reflected ray is also 19.1°, identical to the angle of refraction given in the question, because the angle of incidence equals the angle of reflection.