Answer:
e. Both the acceleration and net force on the car point inward.
Explanation:
If no net force acts on the car, the car must drive in a straight line, at constant speed.
As the acceleration is defined as the rate of change of the velocity vector, this means that it can produce either a change in the magnitude of the velocity (the speed) or in the direction.
In order to the car can follow a circular trajectory, it must be subjected to an acceleration, that must go inward, trying to take the car towards the center of the circle.
The net force that causes this acceleration, aims inward, and is called the centripetal force.
It is not a different type of force, it can be a friction force, a tension force, a normal force, etc., as needed.
The net force on a car driving around a circular track at steady speed is zero, resulting in zero acceleration. Hence, the correct option is c.
The net force on the car is zero. When a car drives at a steady speed around a circular track, the centripetal force required to keep it moving in a circle balances out the outward inertia. Therefore, there is no net force on the car. This leads to zero acceleration (choice c).
A 10.0 cm long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wire and rail is 0.330 ohms . Pulling the wire at a steady speed of 4.00 m/s causes 4.40 W of power to be dissipated in the circuit.a) How big is the pulling force?b) What is the strength of the magnetic field?
Answer
given,
length of wire = 10 cm = 0.1 m
resistance of the wire = 0.330 ohms
speed of pulling = 4 m/s
Power = 4.40 W
a) Force of pull = ?
[tex]P = F_{pull} v[/tex]
[tex]F_{pull} =\dfrac{P}{v}[/tex]
[tex]F_{pull} =\dfrac{4.40}{4}[/tex]
[tex]F_{pull} =1.1\ N[/tex]
b) using formula
[tex]P = \dfrac{B^2l^2v^2}{R}[/tex]
where B is the magnetic field
v is the pulling velocity
R is the resistance of the wire
[tex]B =\sqrt{\dfrac{PR}{l^2v^2}}[/tex]
[tex]B =\sqrt{\dfrac{F_{pull} \times v R}{l^2v^2}}[/tex]
[tex]B =\sqrt{\dfrac{F_{pull} \times R}{l^2v}}[/tex]
[tex]B =\sqrt{\dfrac{1.1 \times 0.33}{0.1^2\times 4}}[/tex]
[tex]B =\sqrt{9.075}[/tex]
B = 3.01 T
The pulling force in this magnetic field is equal to 1.1 Newton.
Given the following data:
Length of wire = 10.0 cm.Total resistance = 0.330 Ohms.Speed = 4.00 m/s.Power = 4.40 Watts.How to calculate the pulling force.Mathematically, the pulling force in a magnetic field is given by this formula:
[tex]F = \frac{Power}{Speed} \\\\F=\frac{4.40}{4.00}[/tex]
F = 1.1 Newton.
To determine the strength of the magnetic field, we would apply this formula:
[tex]B=\sqrt{\frac{PR}{L^2V^2} }\\\\B=\sqrt{\frac{FVR}{L^2V^2} }\\\\B=\sqrt{\frac{FR}{L^2V} }[/tex]
Where:
B is the magnetic field.P is the power.V is the velocity.L is the length.R is the resistance.F is the pulling force.Substituting the given parameters into the formula, we have;
[tex]B=\sqrt{\frac{1.1 \times 0.330}{0.1^2 \times 4.00} } \\\\B=\sqrt{\frac{0.363}{0.01 \times 4.00} }\\\\B=\sqrt{9.075}[/tex]
B = 3.012 T.
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Suppose you are selling apple cider for two dollars a gallon when the temperature is 3.3 degree C. The coefficient of volume expansion of the cider is 280*10^-6(C degree)^-1. How much more money (in pennies) would you make per gallon be refilling the container on a day when the temperature is 32 degrees C? Ignore the expansion of the container. Round your answer to 0.1 penny.
Answer:
1.6 penny
Explanation:
[tex]V_0[/tex] = Original volume = 1 gal (Assumed)
[tex]\Delta T[/tex] = Change in temperature
[tex]\beta[/tex] = Coefficient of volume expansion = [tex]280\times 10^{-6}\ /^{\circ}[/tex]
Change in volume is given by
[tex]\Delta_V=\beta V_0\Delta T\\\Rightarrow \Delta_V=280\times 10^{-6}\times 1\times (32-3.3)\\\Rightarrow \Delta_V=0.008036[/tex]
New volume would be
[tex]1+0.008036=1.008036\ gal[/tex]
The amount of money earned extra would be
[tex]0.008036\times 2=0.016072\ \$[/tex]
1.6 penny more would be earned if the temperature is 32°C
Final answer:
By refilling a container of apple cider at 32 degrees C instead of 3.3 degrees C, you would make approximately 1.6 pennies more per gallon due to thermal expansion of the cider.
Explanation:
To calculate how much more money you would make per gallon by refilling the container of apple cider when the temperature is 32 degrees C, as opposed to 3.3 degrees C, you need to determine the change in volume due to thermal expansion.
The formula for volume expansion is ΔV = βV₀ΔT, where ΔV is the change in volume, β is the coefficient of volume expansion, V₀ is the initial volume, and ΔT is the change in temperature.
The initial temperature T1 is 3.3°C, and the final temperature T2 is 32°C, thus ΔT = T2 - T1 = 32°C - 3.3°C = 28.7°C. The coefficient of volume expansion of the cider, given as β, is 280 x 10^-6 (C°)^-1.
Assuming that the initial volume V₀ of the cider is 1 gallon, the change in volume ΔV would be:
ΔV = 280 x 10^-6 x 1 x 28.7 = 0.008036 gallons
To convert gallons to liters, we use the fact that 1 gallon is approximately 3.78541 liters. So, the increase in volume in liters would be:
ΔV (liters) = 0.008036 x 3.78541 = 0.0304 liters
Since there are approximately 3.78541 liters in a gallon, and knowing that the price for one gallon is two dollars, we can calculate the additional revenue (in pennies) as follows:
Extra revenue = ΔV (liters) / 3.78541 x 200 pennies = 0.0304 / 3.78541 x 200 ≈ 1.6 pennies
Therefore, you would make approximately 1.6 pennies more per gallon by refilling the container at 32°C compared to 3.3°C.
The diffusion constant for oxygen diffusing through tissue is 1.0 × 10-11 m2/s. In a certain sample oxygen flows through the tissue at 2.0 × 10-6 kg/s. If the thickness of the tissue is doubled, then what is the rate of oxygen flow through the tissue?
Answer:
m' = 1 x 10⁻⁶ kg/s
Explanation:
Given that
Diffussion constant = 1 x 10⁻¹¹
Mass flow rate ,m = 2 x 10⁻⁶ kg/s
The diffusion is inversely proportional to the thickness of the membrane and therefore when the thickness is doubled, the mass flow rate would become half.
So new flow rate m'
[tex]m'=\dfrac{m}{2}[/tex]
[tex]m'=\dfrac{2\times 10^{-6}}{2}\ kg/s[/tex]
m' = 1 x 10⁻⁶ kg/s
A planet is in an elliptical orbit around a distant star. At periastron (the point of closest approach to the star), the planet is rp=4.50×108 km from the star and is moving with a speed of vp=18.5 km/s . When the planet is at apastron (the point of greatest distance from the star), it is ra=9.10×108 km from the star. How fast is the planet moving at apastron? va=? km/s
Answer:
9.15 km/s
Explanation:
rp = 4.5 x 10^8 km
vp = 18.5 km/s
ra = 9.10 x 10^8 km
va = ?
According to the conservation of angular momentum constant.
Let m be the mass of planet
m x rp x vp = m x ra x va
4.5 x 10^8 x 18.5 = 9.10 x 10^8 x va
va = 9.15 km/s
To calculate the speed of the planet at apastron, we can use Kepler's second law and the given values of rp, vp, and ra. Plugging in the values, we find that the planet is moving at a speed of 0.92 km/s at apastron.
Explanation:To calculate the speed of the planet at apastron, we can use Kepler's second law, which states that the area swept out by a planet in equal time intervals is constant. At periastron, the planet is moving fastest, so we can use the equation:
A1 = A2
where A1 is the area swept out at periastron and A2 is the area swept out at apastron.
Since the areas are equal, we can set up the following equation:
0.5 * rp * vp = 0.5 * ra * va
where rp is the distance at the periastron, vp is the velocity at the periastron, ra is the distance at the apastron, and VA is the velocity at the apastron. We can rearrange this equation to solve for va:
va = (rp * vp) / ra
Plugging in the given values, we get:
va = (4.50 x 10^8 km * 18.5 km/s) / (9.10 x 10^8 km)
va = 0.92 km/s
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Consider a particle in a square well. It has mass m and the length of the box is I.Consider the following situations (1) the particle of mass m in the ground state (2) the particle of mass m in the first excited state (3) a particle of mass m in the ground state (4) a particle of mass m in a box of length 2 (5) a particle of mass m in a box of length 21 in the first excited state Which is the proper order from lowest energy to highest energy? (Use«. > to reflect the relative values; for example (1) (3)く(4) 5). 2) Select the correct answer O 1)(2) (3)(4)(5)
Answer:
Explanation:
The energy of the particle in a square power well is given by the Schrödinger equation, with
E₁ = RA (h²/ 8m L²) n²
With h the Planck constant the mass of the particle, L the length of the box and n an integer starting 1
We find the energy for each situation presented
1) For the fundamental state n = 1
E₁ = RA (h/ 8m L²)
2) the first excited state corresponds to n = 2
E₂ = RA (h/ 8m L²) 2²
E₂ = E₁ 4
3) E₁ = RA (h/ 8m L²)
4) The length of the box is L = 2L
E’= RA (h’ / 8m (2L)²)
E’= RA (h’ / 8m L²) Ra ¼
E’= E₁ / 2
5) L = 2l first excited state
n = 2
E ’’ = RA (h/ 8m (2L)²2) 2²
E ’’ = RA (h ’/ 8m (L)²) 4/2
E ’’ = E₁ 2
All energies are in relation to the fundamental state (E1), milking from least to greatest
E1 / 2 <E1 = E1 <2E1 <4E1
4 <3 = 1 <5 <2
Final answer:
The correct order from lowest to highest energy for a particle in a square well, considering both quantum state and box length, is: (5) < (4) < (3) < (2) < (1). This order is based on the quantum energy level formula E_n = n^2 * pi^2 * h-bar^2 / (2 * m * L^2), where energy increases with the square of the principal quantum number and decreases with the square of the box length.
Explanation:
To determine the proper order from lowest energy to highest energy for the particle in a square well, we must consider the principles of quantum mechanics, particularly the energy levels of a particle in a one-dimensional box. The energy levels are quantized and are given by the expression En = n2π2ħ2/(2mL2), where ħ is the reduced Planck constant, m is the mass of the particle, L is the length of the box, and n is the principal quantum number (n = 1 for the ground state, n = 2 for the first excited state, and so on). Given this, here's the reasoning for the order:
A particle in the ground state (n = 1) in a box of length L has the lowest energy.A particle in the first excited state (n = 2) in a box of the same length L will have a higher energy due to the n2 dependency.A particle in the ground state in a box of length 2L will have a lower energy than the particle in the ground state in a box of length L, because the energy is inversely proportional to the square of the box length.A particle in the box of length 2L in the first excited state will still have lower energy than the first excited state in a box of length L, but higher than the ground state in the box of length 2L.Finally, a particle in the first excited state in a box of length 21L will have the lowest energy of all, due to the large increase in box length.Therefore, the order from lowest energy to highest energy is as follows:
(5) < (4) < (3) < (2) < (1)
A "biconvex" lens is one in which both surfaces of the lens bulge outwards. Suppose you had a biconvex lens with radii of curvature with magnitudes of |R1|=10cm and |R2|=15cm. The lens is made of glass with index of refraction nglass=1.5. We will employ the convention that R1 refers to the radius of curvature of the surface through which light will enter the lens, and R2 refers to the radius of curvature of the surface from which light will exit the lens.Part AIs this lens converging or diverging?Part BWhat is the focal length f of this lens in air (index of refraction for air is nair=1)?Express your answer in centimeters to two significant figures or as a fraction.
A biconvex lens with the given parameters is a converging lens. Using the Lens Maker's Equation with the radii of curvature and index of refraction for glass and air, the focal length of the lens is calculated to be approximately 12 cm.
Explanation:A biconvex lens, where both surfaces of the lens bulge outwards, will bend light rays such that they converge at a focal point. With the parameters given (|R1|=10cm, |R2|=15cm, and nglass=1.5), we can deduce that this lens is a converging lens.
Part A: Since a biconvex lens makes parallel rays of light converge at a point after passing through the lens, it is classified as a converging lens.
Part B: To calculate the focal length (f) of the lens, we use the Lens Maker's Equation:
First, we convert the radii of curvature to the appropriate signs as per the lensmaker's convention (positive for convex surfaces when the outside medium is air). R1 = +10cm and R2 = -15cm, since the light exits from the second surface.Next, we plug the values into the equation (1/f) = (nglass - nair) ((1/R1) - (1/R2)) to get the reciprocal of the focal length.Carrying out the calculation with the data given (nglass=1.5, nair=1, R1=+10cm, and R2=-15cm), we get:
(1/f) = (1.5 - 1) ((1/10cm) - (1/(-15cm)))
(1/f) = 0.5 * (0.1cm⁻¹ + 0.0667cm⁻¹)
(1/f) = 0.5 * 0.1667cm⁻¹
(1/f) = 0.08335cm⁻¹
Therefore, the focal length f is the reciprocal of 0.08335cm⁻¹ which is approximately:
f ≈ 12cm
A beam of x-rays with wavelength λ = 0.300 nm is directed toward a sample in which the x-rays scatter off of electrons that are effectively free. The wavelength of the outgoing electrons is measured as a function of scattering angle, where a scattering angle of 0 means the direction of the x-rays was unchanged when passing through the sample. When looking at all possible scattering angles, what are the longest and shortest wavelengths that the scattered x-rays can have?
Answer:
Explanation:
The problem relates to Compton Effect in which electrons are scattered due to external radiation . The electron is scattered out and photons relating to radiation also undergo scattering at angle θ .
The formula relating to Compton Effect is as follows
[tex]\lambda_f-\lambda_i=\frac{h}{m_0c} (1-cos\theta)[/tex]
Here [tex]\lambda_i[/tex] = 3 0 x 10⁻¹¹
For longest [tex]\lambda_f[/tex] θ =180°
[tex]\lambda_f[/tex] = [tex]\lambda_i + \frac{2\times h}{m_0c}[/tex]
= .3 x 10⁻⁹ + [tex]\frac{2\times6.6\times 11^{-34}}{9\times10^{-31}\times3\times10^8}[/tex]
= .348 nm
For shortest wavelength θ = 0
Putting this value in the given formula
[tex]\lambda_f=\lambda_i[/tex]
[tex]\lambda_f[/tex] = .3 nm
Model the concrete slab as being surrounded on both sides (contact area 24 m2) with a 2.1-m-thick layer of air in contact with a surface that is 5.0 ∘C cooler than the concrete. At sunset, what is the rate at which the concrete loses thermal energy by conduction through the air layer?
Final answer:
The rate at which the concrete loses thermal energy by conduction through the air layer can be calculated using Fourier's Law of Heat Conduction. The formula involves the thermal conductivity, area, temperature difference, and thickness of the air layer. However, without the thermal conductivity value for air, the calculation cannot be completed.
Explanation:
To calculate the rate at which the concrete slab loses thermal energy by conduction through the surrounding air layer at sunset, we can apply Fourier's Law of Heat Conduction. This law states that the heat transfer rate (Q) through a material is directly proportional to the temperature difference across the material (ΔT), the area through which heat is being transferred (A), and the thermal conductivity (k), and inversely proportional to the thickness of the material (L).
The formula to calculate the rate of heat loss is given by Q = k*A*(ΔT/L), where ΔT is the temperature difference between the two sides of the material, A is the contact area, k is the thermal conductivity of the material, and L is the thickness of the material.
Unfortunately, without the thermal conductivity value for air in the provided data, we cannot calculate the exact rate of heat loss for this specific scenario. Thermal conductivity is required to solve this problem, and it's typically obtained from tables in textbooks or scientific references.
A 2.95 g sample of an unknown chlorofluorocarbon (an organic compound containing chlorine and fluorine) is decomposed and produces 581 mL of chlorine gas at a pressure of 752 mmHg and a temperature of 298 K. What is the percent by mass of chlorine in the chlorofluorocarbon?
Answer:
mass percent of chlorine is 56.5%
Explanation:
Given data:
Volume V = 581 mL = 0.581 L
Pressure P = 752 mm of Hg = 752/760 = 0.9894 atm
Temperature T = 298 K
Molar gas constant R = 0.08206 atm.L/mol.K
from Ideal gas equation we have following relation
PV = nRT solving for n
[tex]n = \frac{PV}{RT}[/tex]
[tex]n = \frac{0.9894 \times 0.581}{0.08206 \times 298}[/tex]
n = 0.0235 mole
mas of [tex]Cl_2[/tex] = moles × molar mass of [tex]Cl_2[/tex]
[tex]= 0.0235 \times 70.906 = 1.66 g[/tex]
Mass percent of chlorine [tex]= \frac{mass\ of\ Cl}{mass\ of\ sample}[/tex]
[tex]=\frac{1.66}{2.95} = 0.565 = 56.5 %[/tex]
A 60-kg woman stands on the very end of a uniform board, which is supported one-quarter of the way from one end and is balanced. What is the mass of the board?
a. 60 kg
b. 30 kg
c. 20 kg
d. 15 kg
e. 120 kg
The correct option can be seen in Option A.
The diagrammatic expression of the question can be seen in the image attached below.
From the given question, we are being informed that the uniform board is balanced. As a result, the torque(i.e. a measurement about how significantly a force acts on a body for it to spin about an axis) acting on the right-hand side of the balance point should be equal to that of the left-hand side.
Mathematically;
[tex]\mathbf{\tau_{_{right}}= \tau_{_{left}}}[/tex]
Given that the mass of the woman = 60 kg
[tex]\mathbf{\tau =\dfrac{m\times g \times l}{\mu}}[/tex]
[tex]\mathbf{\tau_{left} =\dfrac{m\times g \times l}{\mu}}---(1)[/tex]
[tex]\mathbf{\tau_{_{right}} =\dfrac{60 \times g \times l}{\mu}}---(2)[/tex]
Equating both (1) and (2) together, we have:
[tex]\mathbf{\dfrac{m\times g \times l}{\mu} =\dfrac{60 \times g \times l}{\mu} }[/tex]
Dividing like terms on both side
mass (m) = 60 kg
As such, the correct option can be seen in Option A.
Thus, we can conclude that from the 60-kg woman who stands on the very end of a uniform board, the mass of the board on the other end is also 60 kg.
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Light emitting diode (LEDs) light bulbs have become required in recent years, but do they make financial sense? Suppose a typical 60-watt incandescent light bulb costs $.39 and lasts for 1,000 hours. A 15-watt LED, which provides the same light, costs $3.10 and lasts for 12,000 hours. A kilowatt hour of electricity costs $.115. A kilowatt-hour is 1,000 watts for 1 hour. However, electricity costs actually vary quite a bit depending on location and user type. An industrial user in West Virginia might pay $.04 per kilowatt-hour whereas a residential user in Hawaii might pay $.25. You require a return of 11 percent and use a light fixture 500 hours per year. What is the break-even cost per kilowatt-hour?
Answer:
The break even cost is $0.0063825
Explanation:
Break-even cost is the amount of money, or change in value, which equates to the amount at which an asset must be sold to equal the cost of acquiring it. For easier understanding it can be thought the amount of money for which a product or service must be sold to cover the costs of manufacturing or providing it.
Wattage = W
Cost per kilo watt hour = C
Number of hours per year = H
Price per bulb/CFL = P
Discount rate = 11%
Life of bulb = 2 years
Price of bulb = $0.39
Wattage consumption of bulb per hours = 60
Life of CFL = 24 years
Price of CFL = $3.10
Wattage consumption of CFL per hour = 15
Calculate the Equated Annual Cost (EAC) of bulb
EAC = {- P - (W/1000 x H x C) x (PVIFA 11%, 2years)}/ (PVIFA 11%, 2years)
PVIFA 11%, 2years = Annuity PV Factor = [1 – {(1 + r)^(-n)}]/r, where r is the rate per period and n is the number per periods
PVIFA 11%, 2 years = [1 – {(1 + 0.11)^(-2)}]/0.11 = 1.712523 (for 2 years)
PVIFA 11%, 24 years = [1 – {(1 + 0.11)^(-24)}]/0.11 = 8.348136 (for 2 years)
Calculate the EAC of bulb
EAC = {- P - (W/1000 x H x C) x (PVIFA 11%, 2 years)}/ (PVIFA 11%, 2 years)
EAC = {- 0.39 - (60/1000 x H x C) x (1.712523)}/ (1.712523)
EAC = {-0.39 – (51.37570 x C)}/ 1.712523, consider this equation 1
Calculate the EAC of CFL
EAC = {- P - (W/1000 x H x C) x (PVIFA 11%, 24 years)}/ (PVIFA 11%, 24 years)
EAC = {- 3.10 - (15/1000 x 500 x C) x (8.348136)}/ (8.348136)
EAC = {-3.10 – (62.61102 x C)}/8.348137, consider this equation 2
Equate 1 and 2 to find the amount of C
{-0.39 – (51.37570 x C)}/ 1.712523 = {-3.10 – (62.61102 x C)}/8.348137
{-0.39 – (51.37570 x C) x 8.348137} = {-3.10 – (62.61102 x C) x 1.712523}
C = $0.0063825
Thus, the break- even cost per kilo – watt is $0.0063825
Why is fusion an appealing energy source?
Fusion products are generally not radioactive.
Extremely high temperatures are required.
The reaction can be confined by available structural materials.
Extremely high pressures are required.
To take place the process of nuclear fusion basically seeks to reach heavy nuclei through light nuclei. Reaching this process implies a release of energy that is what makes this process attractive because it is possible to obtain significant volumes of energy. The procedure to arrive at this process also implies a high cost concerning high temperatures and exorbitant pressures as it is necessary to be able to overcome the barrier of electrostatic repulsion.
This process does not generate any type of radioactive waste like other processes, therefore it is not as dangerous as nuclear fission. For this reason the correct answer is A. Fusion products are generally not radioactive.
During each cycle of operation a refrigerator absorbs 56 cal from the freezer compartment and expels 81 cal tothe room. If one cycle occurs every 10 s, how many minutes will it take to freeze 570 g of water, initially at 0°C?(Lv = 80 cal/g
Answer:
138.18 minutes
Explanation:
[tex]L_v[/tex] = Latent heat of water at 0°C = 80 cal/g
m = Mass of water = 570 g
Heat removed for freezing
[tex]Q=mL_v\\\Rightarrow Q=570\times 80\\\Rightarrow Q=45600\ cal[/tex]
Let N be the number of cycles and each cycle removes 56 cal from the freezer.
So,
[tex]55\times N=45600\\\Rightarrow N=\frac{45600}{55}[/tex]
Each cycle takes 10 seconds so the total time would be
[tex]\frac{45600}{55}\times \frac{10}{60}=138.18\ minutes[/tex]
The total time taken to freeze 138.18 minutes
A 0.20 kg particle moves along the x axis under the influence of a stationary object. The potential energy is given by U(x) = 8x2 + 2x4, where U is in joules and x is in meters. If the particle has a speed of 5.0 m/s when it is at x = 1.0 m, its speed when it is at the origin is:
Answer:11.18 m/s
Explanation:
Given
mass of particle m=0.2 kg
Potential Energy U(x) is given by
[tex]U(x)=8x^2+2x^4[/tex]
at x=1 m
[tex]U(1)=8+2=10 J[/tex]
kinetic energy at x=1 m
[tex]K.E.=\frac{1}{2}mv^2=\frac{1}{2}\times 0.2\times 5^2[/tex]
[tex]K.E.=2.5 J[/tex]
Total Energy =U+K.E.
[tex]Total=10+2.5=12.5 J[/tex]
at x=0, U(0)=0
as total Energy is conserved therefore K.E. at x=0 is equal to Total Energy
[tex]\frac{1}{2}\times 0.2\times v^2=12.5[/tex]
[tex]v^2=125[/tex]
[tex]v=\sqrt{125}[/tex]
[tex]v=11.18 m/s[/tex]
By conserving energy, we find that the speed of the particle at the origin is 25 m/s.
Explanation:The question is asking for the speed of the particle at the origin given its potential energy function and speed at x = 1.0 m. This can be solved using the principle of energy conservation which states that the total energy (kinetic energy + potential energy) of the particle is conserved unless acted upon by an outside force.
In this case, we first find the total energy of the particle at x = 1.0 m. The kinetic energy is (1/2)mv² = (1/2)*0.2 kg*(5 m/s)² = 2.5 J. The potential energy at x = 1.0 m, according to the given function, is U(1) = 8(1)² + 2(1)⁴ = 10 J. So, the total energy at x = 1.0 m is 2.5 J + 10 J = 12.5 J.
At the origin (x = 0), the potential energy is U(0) = 0. So, the kinetic energy at the origin is equal to the total energy, which is 12.5 J. From the kinetic energy, we can find the speed using the equation KE = (1/2)mv², which gives v = sqrt((2*KE)/m) = sqrt((2*12.5 J)/0.20 kg) = 25 m/s.
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A truck horn emits a sound with a frequency of 238 Hz. The truck is moving on a straight road with a constant speed. If a person standing on the side of the road hears the horn at a frequency of 220 Hz, then what is the speed of the truck? Use 340 m/s for the speed of the sound.
Answer:
[tex]v_s=27.8m/s[/tex]
Explanation:
If the person hearing the sound is at rest, then the equation for the frequency heard [tex]f[/tex] given the emitted frequency [tex]f_0[/tex], the speed of the truck [tex]v_s[/tex] and the speed of sound [tex]c[/tex] will be:
[tex]f=f_0\frac{c}{c+v_s}[/tex]
Where [tex]v_s[/tex] will be positive if the truck is moving away from the person, and negative otherwise. We then do:
[tex]\frac{f}{f_0}=\frac{c}{c+v_s}[/tex]
[tex]\frac{f_0}{f}=\frac{c+v_s}{c}=1+\frac{v_s}{c}[/tex]
[tex]v_s=c(\frac{f_0}{f}-1)=(340m/s)(\frac{238Hz}{220Hz}-1)=27.8m/s[/tex]
Steam is accelerated by a nozzle steadily from zero velocity to a velocity of 280 m/s at a rate of 2.5 kg/s. If the temperature and pressure of the steam at the nozzle exit are 400°C and 2 MPa, determine the exit area of the nozzle. Solve using appropriate software.
Final answer:
To determine the exit area of the nozzle, use the principle of conservation of mass and the equation for mass flow rate. Calculate the density using the Ideal Gas Law and substitute it into the equation for area.
Explanation:
To determine the exit area of the nozzle, we can use the principle of conservation of mass and the equation for mass flow rate:
Mass flow rate = density x velocity x area
Given that the mass flow rate is 2.5 kg/s and the velocity is 280 m/s, we can rearrange the equation to solve for the area:
Area = mass flow rate / (density x velocity)
However, we need to find the density of the steam at the nozzle exit. To do this, we can use the Ideal Gas Law:
Pressure x Volume = n x R x Temperature
Where pressure = 2 MPa, volume can be assumed to be the volume of the nozzle exit, R is the gas constant, and temperature is 400°C converted to Kelvin.
Once we have the density, we can substitute it into the equation for the area to find the exit area of the nozzle.
The exit area of the nozzle is approximately [tex]\( 0.00140 \text{ m}^2 \) or \( 1.40 \text{ mm}^2 \)[/tex].
The continuity equation for a steady-state flow is given by:
[tex]\[ \dot{m} = \rho \cdot A \cdot v \][/tex]
To find the density [tex]\( \rho \)[/tex], we need to use the ideal gas law, which is a good approximation for steam under these conditions:
[tex]\[ P = \rho \cdot R \cdot T \][/tex]
where:
- P is the absolute pressure at the nozzle exit (2 MPa or 2000 kPa),
- R is the specific gas constant for steam (0.4615 kJ/kg·K),
- T is the absolute temperature at the nozzle exit (400°C + 273.15 = 673.15 K).
Rearranging the ideal gas law to solve for [tex]\( \rho \)[/tex]:
[tex]\[ \rho = \frac{P}{R \cdot T} \][/tex]
Now, we can substitute the density [tex]\( \rho \)[/tex] back into the continuity equation to solve for the exit area A:
[tex]\[ A = \frac{\dot{m}}{\rho \cdot v} \][/tex]
Substituting the values we have:
[tex]\[ \rho = \frac{2000 \text{ kPa}}{0.4615 \text{ kJ/kg·K} \cdot 673.15 \text{ K}} \] \[ \rho = \frac{2000}{310.56} \text{ kg/m}^3 \] \[ \rho \approx 6.44 \text{ kg/m}^3 \][/tex]
Now, we can find the exit area A:
[tex]\[ A = \frac{2.5 \text{ kg/s}}{6.44 \text{ kg/m}^3 \cdot 280 \text{ m/s}} \] \[ A = \frac{2.5}{1787.2} \text{ m}^2 \] \[ A \approx 0.00140 \text{ m}^2 \][/tex]
(7%) Problem 5: A thermos contains m1 = 0.73 kg of tea at T1 = 31° C. Ice (m2 = 0.095 kg, T2 = 0° C) is added to it. The heat capacity of both water and tea is c = 4186 J/(kg⋅K), and the latent heat of fusion for water is Lf = 33.5 × 104 J/kg. dho32@student.mtsac.edu
The final temperature of the mixture is approximately 29.91°C.
The heat exchange between the tea and the ice. The temperature of the final mixture will be somewhere between 0°C and 31°C, and we need to determine that final temperature. The heat transfer can be calculated using the principle of conservation of energy:
Qin = Qout
The heat gained by the ice as it melts is given by:
Qice = m2 ⋅ Lf
The heat gained by the tea as it cools down is given by:
Qtea = m1 ⋅ c ⋅ (T1 - Tfinal)
The negative sign is used because the tea is losing heat. Setting these equal to each other and solving for Tfinal, we get:
m1 ⋅ c ⋅ (T1 - Tfinal) = m2 ⋅ Lf
Now, let's plug in the given values:
0.73 kg ⋅ 4186 J/(kg ⋅ K) ⋅ (31°C - Tfinal) = 0.095 kg ⋅ 33.5 × 10^4 J/kg
Now, solve for Tfinal:
0.73 ⋅ 4186 ⋅ (31 - Tfinal) = 0.095 ⋅ (33.5 × 10^4)
3050.78 ⋅ (31 - Tfinal) = 3182.5
94602.78 - 3050.78 ⋅ Tfinal = 3182.5
-3050.78 ⋅ Tfinal = -91420.28
Tfinal = 91420.28/3050.78
Tfinal ≈ 29.91°C
So, the final temperature of the mixture is approximately 29.91°C.
A parallel plate capacitor is connected to a battery that maintains a constant potential difference between the plates. If the plates are pulled away from each other, increasing their separation, what happens to the amount of charge on the plates?
a. The amount of the charge decreases, because the capacitance increases.
b. Nothing happens; the amount of charge stays the same.
c. The amount of the charge increases, because the capacitance increases.
d. The amount of the charge increases, because the capacitance decreases.
e. The amount of the charge decreases, because the capacitance decreases.
When the separation between the plates of a parallel plate capacitor is increased, the amount of charge on the plates decreases due to the decrease in capacitance (option e), with the voltage remaining constant.
When parallel plate capacitor plates are pulled away from each other while connected to a battery maintaining a constant potential difference, the capacitance decreases. This is because the capacitance is inversely proportional to the distance between the plates. As the capacitance decreases, the charge on the plates also decreases since the voltage (V) remains constant, and the relation between charge (Q), capacitance (C), and voltage (V) is given by Q = CV. Therefore, the amount of charge on the plates decreases because the capacitance decreases (option e).
Which one of the following phrases best describes the electric potential of a charged particle?
A) the total force exerted on or by the charged particle
B) the force per unit charge
C) the potential energy of the particle relative to infinity
D) the potential energy per unit charge
E) the potential energy per unit force on the particle
Answer:
D.The potential energy per unit charge
Explanation:
Electric potential of a charged particle:
It is scalar quantity because it has magnitude but it does not have direction.
It is the amount of work done required to move a unit positive charge from reference point to specific point in the electric field without producing any acceleration.
Mathematical representation:
[tex]V=\frac{W}{Q_0}[/tex]
Where W= Work done
[tex]Q_0[/tex]= Unit positive charge
Other formula to calculate electric field:
[tex]V=\frac{KQ}{r}[/tex]
Where K=[tex]\frac{1}{4\pi \epsilon_0}[/tex]
It can be defined as potential energy per unit charge.
Hence, option D is true.
A 45.0-kg girl is standing on a 166-kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless surface. The girl begins to walk along the plank at a constant velocity of 1.48 m/s to the right relative to the plank. (Let the direction the girl is moving in be positive. Indicate the direction with the sign of your answer.)
1. What is her velocity relative to the surface of the ice?
2. What is the velocity of the plank relative to the surface of ice?
Answer:
-0.31563 m/s
1.16437 m/s
Explanation:
[tex]m_1[/tex] = Mass of girl = 45 kg
[tex]m_2[/tex] = Mass of plank = 166 kg
[tex]v_1[/tex] = Velocity of girl relative to plank = 1.48 m/s
[tex]v_2[/tex] = Velocity of the plank relative to ice surface
In this system the linear momentum is conserved
[tex](m_1+m_2)v_2+m_1v_1=0\\\Rightarrow v_2=-\frac{m_1v_1}{m_1+m_2}\\\Rightarrow v_2=-\frac{45\times 1.48}{45+166}\\\Rightarrow v_2=-0.31563\ m/s[/tex]
Velocity of the plank relative to ice surface is -0.31563 m/s
Velocity of the girl relative to the ice surface is
[tex]v_1+v_2=1.48-0.31563=1.16437\ m/s[/tex]
Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface. Model a spherically symmetric planet, with the same radius as the earth, as having a density that decreases linearly with distance from the center. Let the density be 1.30×104 kg/m3 at the center and 2100 kg/m3 at the surface. Part A What is the acceleration due to gravity at the surface of this planet?
The acceleration due to gravity at the surface of a planet depends on its mass and radius, and assumes a uniform density. Since your model has a density that decreases linearly from the center to the surface, the exact value for gravity would require integration over the volume of the planet to account for mass distribution. This arrangement involves advanced calculus.
Explanation:The acceleration due to gravity at the surface of any planet, including Earth, is determined by a constant (G), the mass of the planet (M), and the radius of the planet (R). The formula is g = GM/R². However, this calculation assumes a uniform density throughout the planet, which is often not the case. In reality, like in your model where the density decreases linearly from the center to the surface, finding the precise acceleration due to gravity at the surface becomes more complicated and involves integration over the entire volume of the planet to account for how the mass is distributed.
Given that you provided the densities at the center and surface of the modeled planet, and these densities decrease linearly, one can utilize the formula for the linear density ρ(r) = ρ_center - r(ρ_center - ρ_surface)/R, where R is the radius of the planet, r is the distance from the center, and ρ_center and ρ_surface are the density at the center and surface, respectively. Then, integrate over the volume of the planet to find the total mass.
Once you have the mass, you can use the formula g = GM/R² again to find the acceleration due to gravity at the surface. However, this calculation goes beyond a basic understanding of gravity and requires knowledge of calculus. Without specific numbers for the mass and the integration result, I cannot provide the exact value for surface gravity in this case.
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The acceleration due to gravity at a planet's surface depends on the planet's radius, mass and the linear decrease of density from center to surface. The formula of this acceleration is G×M/r², considering that M is the planet's mass obtained by the product of volume and average density. However, as the density changes linearly, the force of gravity also decreases linearly from the center to the surface.
Explanation:To calculate the acceleration due to gravity at the surface of the planet, we have to consider the planet's radius, mass and density. Given the density at the center and surface, we can calculate the average density which is the total mass of the planet divided by the total volume. In this spherically symmetric planet model, we can use the formula for the volume of a sphere, which is 4/3πr³, with r being the Earth's radius. We consider that mass (M) equals density (ρ) times volume (V), and the force of gravity (F) is G×(M1×M2)/r², where G is the gravitational constant. In this case, M1 is the mass of the planet and M2 is the mass of the object where we want to know the acceleration, and r is the distance between the centers of the two masses, or in this case the radius of the planet. As force is also mass times acceleration, we can replace F in the formula with M2 times a (acceleration), and find that acceleration is G×M1/r². However, as the density changes linearly from the center to the surface, the force of gravity will also decrease linearly, affecting the acceleration.
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With the switch open, roughly what must be the resistance of the resistor on the right for the current out of the battery to be the same as when the switch is closed (and the resistances of the two resistors are 20 Ω and 10 Ω)?
A) 7ΩB) 15ΩC) 30ΩD) 5Ω
Answer:
The resistance must be 6.67[tex]\Omega[/tex]
Solution:
Resistance, [tex]R_{1} = 20\Omega[/tex]
Resistance, [tex]R_{2} = 10\Omega[/tex]
For the current to be the same when the switch is open or closed, the resistances must be connected in parallel as current is distributed in parallel with the same voltage across the circuit:
Thus in parallel:
[tex]\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}[/tex]
[tex]\frac{1}{R_{eq}} = \frac{1}{20} + \frac{1}{10}[/tex]
[tex]R_{eq} = 6.67\ \Omega[/tex]
Case 1: A 0.780-kg silver pellet with a temperature of 85 oC is added to 0.150 kg of water in a copper cup of unknown mass. The initial temperature of the water and the copper cup is 14 oC. The equilibrium temperature of the system (silver water copper cup) is measured to be 26.0 °C. Assume no heat is exchanged with the surroundings. The specific heats of silver, water and copper are: 234 J/(kg oC), 4186 J/(kg oC) and 387 J/(kg oC) , respectively. (a) Which substance releases heat
Answer: The silver pellet will release heat
Explanation:
Based on the case scenario, the silver pellet has a higher temperature that the system of water and copper cup and is thereby added to the system. Because of the higher kinetic energy of the molecules of silver in the silver pellet, some of energy will be released to the water and copper cup system because the system will aim to achieve thermal equilibrium.
A liquid of density 1290 kg/m 3 1290 kg/m3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the flow speed is 9.83 m/s 9.83 m/s and the pipe diameter d 1 d1 is 12.1 cm 12.1 cm . At Location 2, the pipe diameter d 2 d2 is 17.7 cm 17.7 cm . At Location 1, the pipe is 8.35 m higher than it is at location 2. Ignoring viscosity, calculate the difference between fluid pressure at location 2 and the fluid pressure at location 1.
Answer:
[tex]\Delta P=1060184.8946\ Pa[/tex]
[tex]P_1=124651.2383\ Pa[/tex]
Explanation:
Given:
density of liquid, [tex]\rho=1290\ kg.m^{-3}[/tex]speed of flow at location 1, [tex]v_1=9.83\ m.s^{-1}[/tex]diameter of pipe at location 1, [tex]d_1=0.121\ m[/tex]diameter of pipe at location 2, [tex]d_2=0.177\ m[/tex]height of pipe at location 1, [tex]z_1=8.35\ m[/tex]We know the Bernoulli's equation of in-compressible flow:
[tex]\frac{P}{\rho.g} +\frac{v^2}{2g} + z=constant[/tex] ........................(1)
Cross sectional area of pipe at location 2:
[tex]A_2=\pi \frac{d_2^2}{4}[/tex]
[tex]A_2=\pi\times \frac{0.177^2}{4}[/tex]
[tex]A_2=0.0246\ m^2[/tex]
Cross sectional area of pipe at location 1:
[tex]A_1=\pi \frac{d_1^2}{4}[/tex]
[tex]A_1=\pi\times \frac{0.121^2}{4}[/tex]
[tex]A_1=0.0115\ m^2[/tex]
Using continuity equation:
[tex]A_1.v_1=A_2.v_2[/tex]
[tex]0.0115\times 9.83=0.0246\times v_2[/tex]
[tex]v_2=4.5953\ m.s^{-1}[/tex]
Now apply continuity eq. on both the locations:
[tex]\frac{P_1}{\rho.g} +\frac{v_1^2}{2g} + z_1= \frac{P_2}{\rho.g} +\frac{v_2^2}{2g} + z_2[/tex]
[tex](P_2-P_1) = \rho.g [\frac{v_1^2}{2g} + z_1-\frac{v_2^2}{2g} ][/tex]
[tex]\Delta P=1290\times 9.8 [\frac{9.83^2}{19.6} + 8.35-\frac{4.5953^2}{19.6} ][/tex]
[tex]\Delta P=154266.016\ Pa[/tex]...................................Ans (a)
Now the mass flow rate at location 1:
[tex]\dot{m_1}=\rho\times \dot{V}[/tex]
[tex]\dot{m_1}=1290\times (0.0115\times 9.83)[/tex]
[tex]\dot{m_1}=145.828\ kg.s^{-1}[/tex]
Now pressure at location 1:
[tex]P_1=\frac{\dot{m_1}\times v_1}{A_1}[/tex]
[tex]P_1=\frac{145.828\times 9.83}{0.0115}[/tex]
[tex]P_1=124651.2383\ Pa[/tex] ...................................Ans (b)
The difference between fluid pressure at location 2 and fluid pressure at location 1 is mathematically given as
dP = 114 kPa
What is the difference between fluid pressure at location 2 and fluid pressure at location 1.?Question Parameter(s):
Generally, the Bernoulli's equation is mathematically given as
P + ρ*g*y + v² =pipe constant
Where
A1*v1 = A2*v2
π*(0.105/2)²*9.91 = π*(0.167/2)²*v2
v2 = 3.9 m/s
Therefore
P1 + ρ*g*y1 + v1² = P2 + ρ*g*y2 + v2²
dP = 1290*9.8*9.01 + 9.91² - 3.9²
dP = 114 kPa
In conclusion, difference between fluid pressure is
dP = 114 kPa
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A horizontal rod 0.300 m long carries a current through a uniform horizontal magnetic field of magnitude 6.40×10−2 T that points perpendicular to the rod. Part A If the magnetic force on this rod is measured to be 0.140 N , what is the current flowing through the rod?
Answer:
Current, 7.29 A
Explanation:
It is given that,
Length of the horizontal rod, L = 0.3 m
Magnetic field through a horizontal rod, [tex]B=6.4\times 10^{-2}\ T[/tex]
The magnetic force acting on the rod, F = 0.14 N
Let the current flowing through the rod is given by I. The magnetic force acting on an object in the uniform magnetic field is given by :
[tex]F=ILB\ sin\theta[/tex]
Here, [tex]\theta=90^{\circ}[/tex]
[tex]F=ILB[/tex]
[tex]I=\dfrac{F}{LB}[/tex]
[tex]I=\dfrac{0.14\ N}{0.3\ m\times 6.4\times 10^{-2}\ T}[/tex]
I = 7.29 A
So, the current flowing through the rod 7.29 A.
Final answer:
The current flowing in the rod is approximately 7.292 A when it is subject to a 0.140 N magnetic force within a 6.40 x [tex]10^-^2[/tex] T magnetic field. The calculation uses the formula F = I * L * B, and since the rod is perpendicular to the magnetic field, the angle θ is 90°, which simplifies the calculation.
Explanation:
The question asks for the current flowing through a rod that is experiencing a magnetic force due to an external magnetic field.
The force on a current-carrying conductor in a magnetic field is given by the equation F = I * L * B * sin(θ), where F is the force in newtons, I is the current in amperes, L is the length of the conductor in meters, B is the magnetic field in teslas, and θ is the angle between the direction of the current and the magnetic field. In this case, the rod is perpendicular to the magnetic field, so the angle θ is 90°, making sin(θ) equal to 1.
To find the current I, we rearrange the formula to be I = F / (L * B). Substituting the given values:
Force F = 0.140 N
Length L = 0.300 m
Magnetic field B = 6.40×[tex]10^-^2[/tex] T
The current I can thus be calculated as I = 0.140 N / (0.300 m * 6.40×[tex]10^-^2[/tex] T).
Performing the calculation, I equals approximately 7.292 A.
How much heat must be removed from 456 g of water at 25.0°C to change it into ice at - 10.0°C ? The specific heat of ice is 2090 J/(kg.K) and the latent heat of fusion of water is 33.5 x 10^4 J/kg .
Answer:209.98 kJ
Explanation:
mass of water [tex]m=456 gm[/tex]
Initial Temperature of Water [tex]T_i=25^{\circ}C[/tex]
Final Temperature of water [tex]T_f=-10^{\circ}C[/tex]
specific heat of ice [tex]c=2090 J/kg-K[/tex]
Latent heat [tex]L=33.5\times 10^4 J/kg[/tex]
specific heat of water [tex]c_{water}=4.184 KJ/kg-K[/tex]
Heat require to convert water at [tex]T=25^{\circ}C[/tex] to [tex]T=0^{\circ}C[/tex]
[tex]Q_1=0.456\times 4.184\times (25-0)=47.69 kJ[/tex]
Heat require to convert water at [tex]T=0^{\circ}[/tex] to ice at [tex]T=0^{\circ}[/tex]
[tex]Q_2=m\times L=0.456\times 33.5\times 10^4=152.76 kJ[/tex]
heat require to convert ice at [tex]T=0^{\circ} C\ to\ T=-10^{\circ} C[/tex]
[tex]Q_3=0.456\times 2090\times (0-(-10))=9.53 kJ[/tex]
Total heat [tex]Q=Q_1+Q_2+Q_3[/tex]
[tex]Q=47.69+152.76+9.53=209.98 kJ[/tex]
The hydraulic oil in a car lift has a density of 8.81 x 102 kg/m3. The weight of the input piston is negligible. The radii of the input piston and output plunger are 5.07 x 10-3 m and 0.150 m, respectively. What input force F is needed to support the 27800-N combined weight of a car and the output plunger, when (a) the bottom surfaces of the piston and plunger are at the same level, and (b) the bottom surface of the output plunger is 1.20 m above that of the input plunger?
Answer:
a. [tex]F_2=31.76N[/tex]
b. [tex]F_2=185.86N[/tex]
Explanation:
Given:
[tex]F_1=27800N[/tex]
[tex]r_1=5.07x10^{-3}m[/tex]
[tex]r_2=0.150 m[/tex]
[tex]p=8.81x10^2 kg/m^3[/tex]
Using the equation to find the force so replacing
a.
[tex]F_1*A_2=F_2*A_1[/tex]
[tex]A=\pi*r^2[/tex]
[tex]F_2=F_1*\frac{A_2}{A_1}=27800*\frac{\pi*(5.07x10^{-3}m)^2}{\pi*(0.150m)^2}[/tex]
[tex]F_2=31.76N[/tex]
b.
[tex]F_2=F_1+F_p[/tex]
[tex]F_2=27800*\frac{\pi*(5.07x10^{-3}m)^2}{\pi*(0.150m)^2}+(8.81x10^2kg/m^3*9.8m/s^2*1.20m*\pi*(5.07x10^{-3})m^2)[/tex]
[tex]F_2=185.86N[/tex]
Part APart complete If the CD rotates clockwise at 500 rpm (revolutions per minute) while the last song is playing, and then spins down to zero angular speed in 2.60 s with constant angular acceleration, what is α, the magnitude of the angular acceleration of the CD, as it spins to a stop?
Answer:
20.13841 rad/s²
Explanation:
[tex]\omega_i[/tex] = Initial angular velocity = [tex]500\times \frac{2\pi}{60}\ rad/s[/tex]
[tex]\omega_f[/tex] = Final angular velocity = 0
t = Time taken = 2.6 s
[tex]\alpha[/tex] = Angular acceleration
Equation of rotational motion
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-500\times \frac{2\pi}{60}}{2.6}\\\Rightarrow \alpha=-20.13841\ rad/s^2[/tex]
The magnitude of the angular acceleration of the CD, as it spins to a stop is 20.13841 rad/s²
A parallel beam of light in air makes an angle of 43.5 ∘ with the surface of a glass plate having a refractive index of 1.68. You may want to review (Pages 1080 - 1086) . For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Reflection and refraction.
a. What is the angle between the reflected part of the beam and the surface of the glass? θθ = nothing ∘
b. What is the angle between the refracted beam and the surface of the glass? θθ = nothing ∘
a. The angle between the reflected part of the beam and the surface of the glass is [tex]\(43.5^\circ\).[/tex]
b. The angle between the refracted beam and the surface of the glass is [tex]\(64.5^\circ\).[/tex]
To solve this problem, we need to apply the laws of reflection and refraction. Let's address each part separately.
Part (a): Angle Between the Reflected Beam and the Surface of the Glass
The law of reflection states that the angle of incidence is equal to the angle of reflection. The angle of incidence is given as 43.5° with respect to the surface of the glass. However, angles in optics are typically measured with respect to the normal (a line perpendicular to the surface).
So, the angle of incidence with respect to the normal (which we'll call [tex]\(\theta_i\)[/tex] ) is:
[tex]\[ \theta_i = 90^\circ - 43.5^\circ = 46.5^\circ \][/tex]
Since the angle of incidence equals the angle of reflection:
[tex]\[ \theta_r = \theta_i = 46.5^\circ \][/tex]
Therefore, the angle between the reflected part of the beam and the surface of the glass is:
[tex]\[ 90^\circ - \theta_r = 90^\circ - 46.5^\circ = 43.5^\circ \][/tex]
So, the angle between the reflected beam and the surface of the glass is:
[tex]\[ 43.5^\circ \][/tex]
Part (b): Angle Between the Refracted Beam and the Surface of the Glass
For the refracted beam, we need to apply Snell's Law, which is:
[tex]\[ n_1 \sin(\theta_i) = n_2 \sin(\theta_t) \][/tex]
Where:
- [tex]\( n_1 \)[/tex] is the refractive index of the first medium (air), [tex]\( n_1 = 1.00 \)[/tex],
- [tex]\( \theta_i \)[/tex] is the angle of incidence with respect to the normal, [tex]\( \theta_i = 46.5^\circ \),[/tex]
- [tex]\( n_2 \)[/tex] is the refractive index of the second medium (glass), [tex]\( n_2 = 1.68 \)[/tex],
- [tex]\( \theta_t \)[/tex] is the angle of refraction with respect to the normal.
Using Snell's Law, we can solve for [tex]\(\theta_t\):[/tex]
[tex]\[ 1.00 \sin(46.5^\circ) = 1.68 \sin(\theta_t) \][/tex]
[tex]\[ \sin(\theta_t) = \frac{\sin(46.5^\circ)}{1.68} \][/tex]
Calculating [tex]\(\sin(46.5^\circ)\):[/tex]
[tex]\[ \sin(46.5^\circ) \approx 0.723 \][/tex]
So,
[tex]\[ \sin(\theta_t) = \frac{0.723}{1.68} \approx 0.430 \][/tex]
Now we find [tex]\(\theta_t\):[/tex]
[tex]\[ \theta_t = \sin^{-1}(0.430) \approx 25.5^\circ \][/tex]
The angle between the refracted beam and the surface of the glass is:
[tex]\[ 90^\circ - \theta_t = 90^\circ - 25.5^\circ = 64.5^\circ \][/tex]
So, the angle between the refracted beam and the surface of the glass is:
[tex]\[ 64.5^\circ \][/tex]
The volume of water in the Pacific Ocean is about 7.00 × 108 km3. The density of seawater is about 1030 kg/m3. For the sake of the calculations, treat the Pacific Ocean as a point like object (obviously a very rough approximation). 1) Determine the gravitational potential energy of the Moon–Pacific Ocean system when the Pacific is facing away from the Moon. (Express your answer to three significant figures.) Answer in Joules 2) Repeat the calculation when Earth has rotated so that the Pacific Ocean faces toward the Moon. (Express your answer to three significant figures.) Answer in Joules
The concepts used to solve this exercise are given through the calculation of distances (from the Moon to the earth and vice versa) as well as the gravitational potential energy.
By definition the gravitational potential energy is given by,
[tex]PE=\frac{GMm}{r}[/tex]
Where,
m = Mass of Moon
G = Gravitational Universal Constant
M = Mass of Ocean
r = Radius
First we calculate the mass through the ratio given by density.
[tex]m = \rho V[/tex]
[tex]m = (1030Kg/m^3)(7*10^8m^3)[/tex]
[tex]m = 7.210*10^{11}Kg[/tex]
PART A) Gravitational potential energy of the Moon–Pacific Ocean system when the Pacific is facing away from the Moon
Now we define the radius at the most distant point
[tex]r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m[/tex]
Then the potential energy at this point would be,
[tex]PE_1 = \frac{GMm}{r_1}[/tex]
[tex]PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}[/tex]
[tex]PE_1 = 9.05*10^{15}J[/tex]
PART B) when Earth has rotated so that the Pacific Ocean faces toward the Moon.
At the nearest point we perform the same as the previous process, we calculate the radius
[tex]r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m[/tex]
The we calculate the Potential gravitational energy,
[tex]PE_2 = \frac{GMm}{r_2}[/tex]
[tex]PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}[/tex]
[tex]PE_2 = 9.361*10^{15}J[/tex]