Answers
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The radiation pressure due to a bulb that emits 25 W of EM radiation at a distance of 6.5 cm from the center of the bulb and the light is completely absorbed is 1.5707x10⁻⁶ N/m².
What is the Radiation pressure?Radiation pressure was defined as the mechanical pressure exerted upon any surface due to the exchange of momentum between the object and the electromagnetic field.
Radiation pressure always includes the Momentum of light or electromagnetic radiation of any wavelength that can be absorbed, reflected, or otherwise emitted by matter on any scale.
E.g: Black-body radiation
Given the values are,
Wattage of bulb = W = 25 W
distance = d = 6.5 cm = 0.065 m
To know the Radiation Pressure,
It can be given by
P = I/c
Where, c = 299792458 m/s is the speed of light,
I is the intensity of radiation and given by
I = W/4πd²
Where W is the Wattage of bulb and d is the distance
I = 25/4π*0.065²
I = 470.872 w/m²
so, the radiation pressure becomes
P = 470.872/299792458
P = 1.5707x10⁻⁶ N/m²
Therefore, the radiation pressure due to a 25 W bulb at a distance of 6.5 cm is 1.5707x10⁻⁶ N/m²
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Related Questions
In your words, describe how momentum is related to energy.
Answers
Answer:
you need momentum in order to release energy. For example, if you need to push something heavy and you get a running head start, then it will be easier.
Explanation:
Si se deja caer una piedra desde un helicóptero en reposo, entonces al cabo de 20 s cual será la rapidez y la distancia recorrida por la piedra
Answers
Answer:
La piedra alcanza una rapidez de 196.14 metros por segundo y una distancia recorrida de 1961.4 metros en 20 segundos.
Explanation:
Si se excluye los efectos del arrastre por la viscosidad del aire, la piedra experimenta un movimiento de caída libre, es decir, que la piedra es acelerada por la gravedad terrestre. La distancia recorrida y la rapidez final de la piedra pueden obtenerse con la ayuda de las siguientes ecuaciones cinemáticas:
[tex]v = v_{o} + g\cdot t[/tex]
[tex]y - y_{o} = v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}[/tex]
Donde:
[tex]v[/tex], [tex]v_{o}[/tex] - Rapideces final e inicial de la piedra, medidas en metros por segundo.
[tex]t[/tex] - Tiempo, medido en segundos.
[tex]g[/tex] - Aceleración gravitacional, medida en metros por segundo al cuadrado.
[tex]y[/tex]. [tex]y_{o}[/tex] - Posiciones final e inicial de la piedra, medidos en metros.
Si [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{o} = 0\,m[/tex], entonces:
[tex]v = 0\,\frac{m}{s} +\left(-9.807\,\frac{m}{s^{2}} \right) \cdot (20\,s)[/tex]
[tex]v = -196.14\,\frac{m}{s}[/tex]
[tex]y-y_{o} = \left(0\,\frac{m}{s} \right)\cdot (20\,s) + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (20\,s)^{2}[/tex]
[tex]y-y_{o} = -1961.4\,m[/tex]
La piedra alcanza una rapidez de 196.14 metros por segundo y una distancia recorrida de 1961.4 metros en 20 segundos.
Suppose you are looking into the end of a long cylindrical tube in which there is a uniform magnetic field pointing away from you. If the magnitude of the field is decreasing with time the direction of the induced magnetic field is
Answers
Answer:
If the magnitude of the field is decreasing with time the direction of the induced magnetic field is CLOCKWISE
Explanation
This is because If the magnetic field decreases with time, the electric field will be produced in order to oppose the change in line with lenz law. Thus The right hand rule can be applied to find that the direction of electric field is in the clockwise direction.
In your own words, discuss how energy conservation applies to a pendulum. Where is the potential energy the most? Where is the potential energy the least? Where is kinetic energy the most? Where is kinetic energy the least?
Answers
Answer:
Explanation:
Energy conservation applies to the swinging of pendulum . When the bob is at one extreme , it is at some height from its lowest point . So it has some gravitational potential energy . At that time since it remains at rest its kinetic energy is zero or the least . As it goes down while swinging , its potential energy decreases and kinetic energy increases following conservation of mechanical energy . At the At the lowest point , its potential energy is least and kinetic energy is maximum .
In this way , there is conservation of mechanical energy .
You are using a hydrogen discharge tube and high quality red and blue light filters as the light source for a Michelson interferometer. The hydrogen discharge tube provides light of several different wavelengths (colors) in the visible range. The red light in the hydrogen spectrum has a wavelength of 656.3 nm and the blue light has a wavelength of 434.0 nm. When using the discharge tube and the red filter as the light source, you view a bright red spot in the viewing area of the interferometer. You now move the movable mirror away from the beam splitter and observe 158 bright spots. You replace the red filter with the blue filter and observe a bright blue spot in the interferometer. You now move the movable mirror towards the beam splitter and observe 114 bright spots. Determine the final displacement (include sign) of the moveable mirror. (Assume the positive direction is away from the beam splitter.)
Answers
Answer:
final displacement = +24484.5 nm
Explanation:
The path difference when 158 bright spots were observed with red light (λ1 = 656.3 nm) is given as;
Δr = 2d2 - 2d1 = 150λ1
So, 2d2 - 2d1 = 150λ1
Dividing both sides by 2 to get;
d2 - d1 = 75λ1 - - - - eq1
Where;
d1 = distance between the fixed mirror and the beam splitter
d2 = position of moveable mirror from splitter when 158 bright spots are observed
Now, the path difference between the two waves when 114 bright spots were observed is;
Δr = 2d'2 - 2d1 = 114λ1
2d'2 - 2d1 = 114λ1
Divide both sides by 2 to get;
d'2 - d1 = 57λ1
Where;
d'2 is the new position of the movable mirror from the splitter
Now, the displacement of the moveable mirror is (d2 - d'2). To get this, we will subtract eq2 from eq1.
(d2 - d1) - (d'2 - d1) = 75λ1 - 57λ2
d2 - d1 - d'2 + d1 = 75λ1 - 57λ2
d2 - d'2 = 75λ1 - 57λ2
We are given;
(λ1 = 656.3 nm) and λ2 = 434.0 nm.
Thus;
d2 - d'2 = 75(656.3) - 57(434)
d2 - d'2 = +24484.5 nm
You throw a stone vertically upward with a speed of 26.0 m/s. (a) How fast is it moving when it reaches a height of 15.0 m? (b) How much time is required to reach this height when it's falling down? a. 19.5 m/s , b. 4.51 s a. 17.9 m/s , b. 0.620 s a. 19.5 m/s , b. 0.800 s a. 17.9 m/s , b. 4.28 s a. 380 m/s , b. 8 s
Answers
Answer:
ok well
Explanation:
teghe
Answer:
v = 19.5 m/s
t = 4.51 s
Explanation:
a)
given:
height is 15m from the ground
initial velocity Vi = 26 m/s
acceleration a or g = 9.81 m/s²
formula: Vf² = Vi² + 2aΔy
26² = Vi² + 2 (9.81) 15
Vi = 19.5 m/s
now you can calculate the time by using the equations below:
Δy = 1/2 (Vi + Vf) t
Vf = Vi + a t
Δy = Vi t + 1/2 a t
time must be 4.51 s
An electric field can be created by a single charge or a distribution of charges. The electric field a distance from a point charge has magnitude E = k|q'|/r^2.
The electric field points away from positive charges and toward negative charges. A distribution of charges creates an electric field that can be found by taking the vector sum of the fields created by individual point harges. Note that if a charge is placed in an electric field created by q', q will not significantly affect the electric field if it is small compared to q'. Imagine an isolated positive point charge with a charge Q (many times larger than the charge on a single electron).
1. There is a single electron at a distance from the point charge. On which of the following quantities does the force on the electron depend?
a. the distance between the positive charge and the electron
b. the charge on the electron
c. the mass of the electron
d. the charge of the positive charge
e. the mass of the positive charge
f. the radius of the positive charge
g. the radius of the electron
2. For the same situation as in Part A, on which of the following quantities does the electric field at the electron's position depend?
a. the distance between the positive charge and the electron
b. the charge on the electron
c. the mass of the electron
d. the charge of the positive charge
e. the mass of the positive charge
f. the radius of the positive charge
g. the radius of the electron
Answers
Answer:
a) true.
b) True
c) False. In the equation above the mass does not appear
d) True
e) False. Mass does not appear in the equation
f) False. The load even when distributed in the space can be considered concentrated in the center
Explanation:
1. The electric force is given by the relation
F = k Q e / r2
where k is the Coulomb constant, Q the charge used, e the charge of the electron and r the distance between the two.
The strength depends on:
a) true.
b) True
c) False. In the equation above the mass does not appear
d) True
e) False. Mass does not appear in the equation
f) False. The load even when distributed in the space can be considered concentrated in the center
two.
a) True
b) Treu
c) Fail
f) false
For a single electron located at a distance from a positive charge, we have:
1. The force on the electron depends on the distance between it and the positive charge (option a) and the charge of both particles (option b and d).
2. The electric field at the electron's position depends on the distance between the positive charge and it (option a) and the charge of the positive particle (option d).
Part 1
The force on a single electron at a distance from the point charge is given by Coulomb's law:
[tex] F = \frac{Kq_{1}q_{2}}{r^{2}} [/tex] (1)
Where:
K: is the Coulomb's constant q₁: is the charge of the positive chargeq₂: is the charge of the electrond: is the distance between the positive charge and the electronAs we can see in equation (1), the force on the electron by the positive charge depends on both charges q₁ and q₂, and the distance, so the correct options are:
a. The distance between the positive charge and the electron
b. The charge on the electron
d. The charge of the positive charge
The other options (c, e, f, and g) are incorrect because the electric force does not depend on the particles' masses or their radii.
Part 2The electric field (E) at a distance "r" from a point charge is given by:
[tex] E = \frac{Kq_{1}}{r^{2}} [/tex] (2)
From equation (2), we can see that the electric field is directly proportional to the charge and inversely proportional to the distance of interest (r).
The electric field at the electron's position is given by the one produced by the positive charge, so the correct options are:
a. The distance between the positive charge and the electron
d. The charge of the positive charge
The other options (b, c, e, f, and g) are incorrect because the electric field is independent of the mass of the charges involved and their radii.
Therefore, the correct options for part 1 are a, b, and d and for part 2 are a and d.
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Kevin is a black high school senior. While walking home from a sporting event at school, he sees a police car and decides to take another street to avoid it. He worries that the police will stop and question him even though he has not done anything wrong. Which theory explains this thought process? Dramaturgy Social construction of reality Social exchange theory Ethnomethodology
Answers
Answer:
Ethnomethodology theory
Explanation:
Take note of the fact that we are told Kevin worries that the police will stop and question him even though he has not done anything wrong.
This statement shows us that Kevin already understood his society from past experiences, and thus he tries to avoid social interactions with particular member of his society (the police) who may be show discrimination towards him.
A generator rotates at 95 Hz in a magnetic field of 0.025 T. It has 550 turns and produces an rms voltage of 170 V and an rms current of 60.0 A.
Required:
a. What is the peak current produced?
b. What is the area of each turn of the coil?
Answers
Answer:
Peak current= 84.86 A
Area of each turn = 0.029 m^2
Explanation:
The peak value of current can be obtained from Irms= 0.707Io. Where Io is the peak current.
Hence;
Irms= 60.0A
Io= Irms/0.707
Io = 60.0/0.707
Io= 84.86 A
Vrms= 0.707Vo
Vo= Vrms/0.707= 170/0.707 = 240.45 V
From;
V0 = NABω
Where;
Vo= peak voltage
N= number of turns
B= magnetic field
A= area of each coil
ω= angular velocity
But ω= 2πf = 2×π×95= 596.9 rads-1
Substituting values;
A= Vo/NBω
A= 240.45/550×0.025×596.9
A= 0.029 m^2
Asteroid A has 3.5 times the mass and 2.0 times the velocity of Asteroid B. If
Asteroid B has a kinetic energy of 2,300,000 J then what is the kinetic energy of
Asteroid A?
Answers
Answer:
K_A = 32.2 10⁶ J
Explanation:
In this exercise we must relate the quantities given to find the kinetic energy
Asteroid A data
m_A = 3.5 m_B
v_A = 2.0 v
they also give the value of the kinetic energy of asteroid A
K_B = 2.3 10⁶ J
the expression for scientific energy is
K = ½ m v²
let's replace
K_A = ½ m_a V_a2
K_A = ½ 3.5 m_B (2.0 v_B)^2
K_A = 3.5 2² (½ m_B v_B²)
K_A = 14 K_B
K_A = 32.2 10⁶ J
A small solid conductor with radius a is supported by insulating, nonmagnetic disks on the axis of a thin-walled tube with inner radius b. The inner andouter conductors carry equal currents i in oppositedirections.
Required:
a. Use Ampere's Law to find the magnetic field at any pointin the volume between the conductors.
b. Write the expression for the flux dΦB through anarrow strip of length l parallel to the axis , of width dr, at a distancer from the axis of the cableand lying in a plane containing the axis.
c. Integrate your expression from part B over the volumebetween the two conductors to find the total flux produced by acurrent i in the central conductor.
d. Use equation U=(1/2)LI2 to calculate the energy stored in the magnetic field for alength l of the cable.
Answers
Answer:
Pls see attached file
Explanation:
1. Water flows through a hole in the bottom of a large, open tank with a speed of 8 m/s. Determine the depth of water in the tank. Viscous effects are negligible.
Answers
Answer:
3.26m
Explanation:
See attached file
g A tube open at both ends, resonated at it's fundamental frequency, to a sound wave traveling at 330m/s. If the length of the tube is 4cm, find the frequency of the sound wave.
Answers
Answer:
frequency =4125Hz
Explanation:
L = 4cm = 0.04m
f =v/2L
f = 330/2 x 0.04
f = 4125Hz
At a certain instant the current flowing through a 5.0-H inductor is 3.0 A. If the energy in the inductor at this instant is increasing at a rate of 3.0 J/s, how fast is the current changing
Answers
Answer:
The current is changing at the rate of 0.20 A/s
Explanation:
Given;
inductance of the inductor, L = 5.0-H
current in the inductor, I = 3.0 A
Energy stored in the inductor at the given instant, E = 3.0 J/s
The energy stored in inductor is given as;
E = ¹/₂LI²
E = ¹/₂(5)(3)²
E = 22.5 J/s
This energy is increased by 3.0 J/s
E = 22.5 J/s + 3.0 J/s = 25.5 J/s
Determine the new current at this given energy;
25.5 = ¹/₂LI²
25.5 = ¹/₂(5)(I²)
25.5 = 2.5I²
I² = 25.5 / 2.5
I² = 10.2
I = √10.2
I = 3.194 A/s
The rate at which the current is changing is the difference between the final current and the initial current in the inductor.
= 3.194 A/s - 3.0 A/s
= 0.194 A/s
≅0.20 A/s
Therefore, the current is changing at the rate of 0.20 A/s.
The rate at which the current is changing is;
di/dt = 0.2 A/s
We are given;
Inductance; L = 5 H
Current; I = 3 A
Rate of Increase of energy; dE/dt = 3 J/s
Now, the formula for energy stored in inductor is given as;
E = ¹/₂LI²
Since we are looking for rate at which current is changing, then we differentiate both sides of the energy equation to get;
dE/dt = LI (di/dt)
Plugging in the relevant values gives;
3 = (5 × 3)(di/dt)
di/dt = 3/(5 × 3)
di/dt = 0.2 A/s
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A 7.0-kg shell at rest explodes into two fragments, one with a mass of 2.0 kg and the other with a mass of 5.0 kg. If the heavier fragment gains 100 J of kinetic energy from the explosion, how much kinetic energy does the lighter one gain?
Answers
Answer:
39.94m/s.Explanation:
Kinetic energy is expressed as KE = 1/2 mv² where;
m is the mass of the body
v is the velocity of the body.
For the heavier shell;
m = 5kg
KE gained = 100J
Substituting this values into the formula above to get the velocity v;
100 = 1/2 * 5 * v²
5v² = 200
v² = 200/5
v² = 40
v = √40
v = 6.32 m/s
Note that after the explosion, both body fragments will possess the same velocity.
For the lighter shell;
mass = 2.0kg and v = 6.32m/s
KE of the lighter shell = 1/2 * 2 * 6.32²
KE of the lighter shell = 6.32²
KE of the lighter shell= 39.94m/s
Hence, the lighter one gains a kinetic energy of 39.94m/s.
The gain in the kinetic energy of the smaller fragment is 249.64 J.
The given parameters;
Mass of the shell, m = 7.0 kgMass of one fragment, m₁ = 2.0 kgMass of the second fragment, m₂ = 5.0 kgKinetic energy of heavier fragment, K.E₁ = 100 JThe velocity of the heavier fragment is calculated as follows;
[tex]K.E = \frac{1}{2} mv^2\\\\mv^2 = 2K.E\\\\v^2 = \frac{2K.E}{m} \\\\v= \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2 \times 100}{5} }\\\\v = 6.32 \ m/s[/tex]
Apply the principle of conservation of linear momentum to determine the velocity of the smaller fragment as;
[tex]m_1 u_1 + m_2 u_2 = v(m_1 + m_2)\\\\-6.32(5) \ + 2u_2 = 0(7)\\\\-31.6 + 2u_2 = 0\\\\2u_2 = 31.6\\\\u_2 = \frac{31.6}{2} \\\\u_2 = 15.8 \ m/s[/tex]
The gain in the kinetic energy of the smaller fragment is calculated as follows;
[tex]K.E_2 = \frac{1}{2} mu_2^2\\\\K.E_2 = \frac{1}{2} \times 2 \times (15.8)^2\\\\K.E_2 = 249.64 \ J[/tex]
Thus, the gain in the kinetic energy of the smaller fragment is 249.64 J.
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Two unknown resistors are connected together. When they are connected in series their equivalent resistance is 15 Ω. When they are connected in parallel, their equivalent resistance is 3.3 Ω. What are the resistances of these resistors?
Answers
Explanation:
Let x and y are two unknown resistors. When they are connected in series their equivalent resistance is 15 Ω. When they are connected in parallel, their equivalent resistance is 3.3 Ω.
For series combination,
[tex]x+y=15[/tex] ......(1)
For parallel combination,
[tex]\dfrac{1}{x}+\dfrac{1}{y}=3.3[/tex] ....(2)
We need to find the resistances of these resistors. Solving equation (1) and (2) we get :
x = 0.29 and y = 14.7
Hence, the resistances of these resistors are 0.29 ohms and 14.7 ohms.
A scientist is testing the seismometer in his lab and has created an apparatus that mimics the motion of the earthquake felt in part (a) by attaching the test mass to a spring. If the test mass weighs 13 N, what should be the spring constant of the spring the scientist use to simulate the relative motion of the test mass and the ground from part (a)?
Answers
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
[tex]a_{max} = 0.00246 \ m/s^2[/tex]
b
[tex]k =722.2 \ N/m[/tex]
Explanation:
From the question we are told that
The amplitude is [tex]A = 1.8 \ cm = 0.018 \ m[/tex]
The period is [tex]T = 17 \ s[/tex]
The test weight is [tex]W = 13 \ N[/tex]
Generally the radial acceleration is mathematically represented as
[tex]a = w^2 r[/tex]
at maximum angular acceleration
[tex]r = A[/tex]
So
[tex]a_{max} = w^2 A[/tex]
Now [tex]w[/tex] is the angular velocity which is mathematically represented as
[tex]w = \frac{2 * \pi }{T}[/tex]
Therefore
[tex]a_{max} = [\frac{2 * \pi}{T} ]^2 * A[/tex]
substituting values
[tex]a_{max} = [\frac{2 * 3.142}{17} ]^2 * 0.018[/tex]
[tex]a_{max} = 0.00246 \ m/s^2[/tex]
Generally this test weight is mathematically represented as
[tex]W = k * A[/tex]
Where k is the spring constant
Therefore
[tex]k = \frac{W}{A}[/tex]
substituting values
[tex]k = \frac{13}{0.018}[/tex]
[tex]k =722.2 \ N/m[/tex]
A positively charged particle has a velocity in the negative z direction at a certain point P. The magnetic force on the particle at this point is in the negative y direction. Which one of the following statements about the magnetic field at point P can be determined from this data?
a. Bx is positive
b. Bz is positive
c. By is negative
d. By is positive
e. Bx is negative
Answers
Answer:
a. Bx is positive
Explanation:
See attached file
A person can survive a feet-first impact at a speed of about 12 m/s (27 mi/h) on concrete, 15 m/s (34 mi/h) on soil, and 34 m/s (76 mi/h) on water. What is the reason for the different values for different surfaces
Answers
Answer:
Different surfaces have different impact force during collision which depends on the time it takes the person to come to rest after collision.
Explanation:
Given;
speed on concrete = 12 m/s (27 mi/h)
speed on soil = 15 m/s (34 mi/h)
speed on water = 34 m/s (76 mi/h)
The impact force on this person during collision is rate of change of momentum;
[tex]F = \frac{\delta P}{\delta t}[/tex]
During collision, the force exerted on this person depends on how long the collision lasts; that is, how long it takes for this person to come to rest after collision with each of the surfaces.
The longer the time of collision, the smaller the force exerted by each.
It takes shorter time for the person to come to rest on concrete surface than on soil surface, also it takes shorter time for the person to come to rest on soil surface than on water surface.
As a result of the reason above, the force exerted on the person during collision by the concrete surface is greater than that of soil surface which is greater than that of water surface.
An ideal air-filled parallel-plate capacitor has round plates and carries a fixed amount of equal butopposite charge on its plates. All the geometric parameters of the capacitor (plate diameter andplate separation) are now DOUBLED. If the original energy stored in the capacitor was U0, howmuch energy does it now store?
Answers
Answer:
U_f = (U_o)/2)
Explanation:
The capacitance of a given capacitor is given by the formula;
C = ε_o•A/d
While energy stored in plates capacitor is given as; U_o = Q²/2C
Now,we are told that that all the dimensions of the capacitor plate is doubled. Thus, we now have;
C' = ε_o•4A/2d
Hence, C' = 2C
so capacitance is now doubled
Thus, the final energy stored between the plates of capacitor is given as;
U_f = Q²/2C'
From earlier, we saw that C' = 2C.
Thus;
U_f = Q²/2(2C)
U_f = Q²/4C
Rearranging, we have;
U_f = (1/2)(Q²/2C)
From earlier, U_o = Q²/2C
Hence,
U_f = (1/2)(U_o)
Or
U_f = (U_o/2)
which of the following is a physical change?
A. a newspaper burns when placed in a fire.
B.an iron chair rusts when left outside
C.a sample of water boils and releases gas.
D.a plant changes carbon dioxide and water into sugar
Answers
For the cellar of a new house, a hole is dug in the ground, with vertical sides going down 2.10 m. A concrete foundation wall is built all the way across the 8.90 m width of the excavation. This foundation wall is 0.189 m away from the front of the cellar hole. During a rainstorm, drainage from the street fills up the space in front of the concrete wall, but not the cellar behind the wall. The water does not soak into the clay soil. Find the force that the water causes on the foundation wall. For comparison, the weight of the water is given by 2.10 m ✕ 8.90 m ✕ 0.189 m ✕ 1000 kg/m3 ✕ 9.80 m/s2 = 34.6 kN.
Answers
Answer:
The force on the foundation wall is [tex]F_f = 191394 \ N[/tex]
Explanation:
From the question we are told that
The depth of the hole's vertical side is [tex]d = 2.10 \ m[/tex]
The width of the hole is [tex]b = 8.90 \ m[/tex]
The distance of the concrete wall from the front of the cellar is [tex]c = 0.189 \ m[/tex]
Generally the area which the water from the drainage covers is mathematically represented as
[tex]A = d * b[/tex]
substituting values
[tex]A = 2.10 * 8.90[/tex]
[tex]A = 18.69 \ m^2[/tex]
Now the gauge pressure exerted on the foundation wall is mathematically evaluated as
[tex]P_g = \rho * d_{avg} * g[/tex]
Here is the average height foundation wall where the pressure of the water is felt and it is evaluated as
[tex]d_{avg} = \frac{h_1 + h_2 }{2}[/tex]
where [tex]h_1[/tex] at the height at bottom of the hole which is equal to [tex]h_1 = 0[/tex]
and [tex]h_2[/tex] is the height at the top of the hole [tex]h_2 = d = 2.10[/tex]
[tex]d_{avg} = \frac{0 + 2.10 }{2}[/tex]
[tex]d_{avg} = 1.05[/tex]
Where [tex]\rho[/tex] is the density of water with constant value [tex]\rho = 1000 \ kg/m^3[/tex]
substituting values
[tex]P_g = 1000 * 1.05 * 9.8[/tex]
[tex]P_g = 10290 \ Pa[/tex]
Then the force exerted by the water on the foundation wall mathematically represented as
[tex]F_f = P_g * A[/tex]
substituting values
[tex]F_f = 10290 * 18.69[/tex]
[tex]F_f = 191394 \ N[/tex]
The intensity of sunlight at the Earth's distance from the Sun is 1370 W/m2. (a) Assume the Earth absorbs all the sunlight incident upon it. Find the total force the Sun exerts on the Earth due to radiation pressure. N (b) Explain how this force compares with the Sun's gravitational attraction.
Answers
Answer:
F= 3.56e22N
Explanation:
Using the force of radiation acting on the earth which is
force = radiation pressure x area = (intensity/c)xpi R^2
force = 1370W/m^2 x pi x( 6.37x10^6m)^2/3x10^8m/s
force = 5.82x10^8 N
But the sun's gravitational attraction means the magnitude of the solar gravitational force on earth: If that's the case, the answer is approx 10^22 N:
F=GMm/r^2
G=6.67x10^(-11)=6.67e-11
M=mass sun = 2x10^30kg=2e30
m=mass earth = 6x10^24kg
r=earth sun distance = 1.5x10^11m
F=(6.6e-11)(2e30)(6e24)/(1.5e11)^2 =
F= 3.56e22N
Write a numerical expression for the emissive intensity (in W/m^2.sr) coming out of a tiny hole in an enclosure of surface temperature 1000K and emissivity 0.6:
Answers
Answer:
6.0 × [tex]10^{11}[/tex] W/[tex]m^{2}[/tex]
Explanation:
From Wien's displacement formula;
Q = e A[tex]T^{4}[/tex]
Where: Q is the quantity of heat transferred, e is the emissivity of the surface, A is the area, and T is the temperature.
The emissive intensity = [tex]\frac{Q}{A}[/tex] = e[tex]T^{4}[/tex]
Given from the question that: e = 0.6 and T = 1000K, thus;
emissive intensity = 0.6 × [tex](1000)^{4}[/tex]
= 0.6 × 1.0 × [tex]10^{12}[/tex]
= 6.0 × [tex]10^{11}[/tex] [tex]\frac{W}{m^{2} }[/tex]
Therefore, the emissive intensity coming out of the surface is 6.0 × [tex]10^{11}[/tex] W/[tex]m^{2}[/tex].
An automotive air conditioner produces a 1-kW cooling effect while consuming 0.75 kW of power. What is the rate at which heat is rejected from this air conditioner
Answers
Answer:
The rejected by the air conditioning system is 1.75 kilowatts.
Explanation:
A air conditioning system is a refrigeration cycle, whose receives heat from cold reservoir with the help of power input before releasing it to hot reservoir. The First Law of Thermodynamics describes the model:
[tex]\dot Q_{L} + \dot W - \dot Q_{H} = 0[/tex]
Where:
[tex]\dot Q_{L}[/tex] - Heat rate from cold reservoir, measured in kilowatts.
[tex]\dot Q_{H}[/tex] - Heat rate liberated to the hot reservoir, measured in kilowatts.
[tex]\dot W[/tex] - Power input, measured in kilowatts.
The heat rejected is now cleared:
[tex]\dot Q_{H} = \dot Q_{L} + \dot W[/tex]
If [tex]\dot Q_{L} = 1\,kW[/tex] and [tex]\dot W = 0.75\,kW[/tex], then:
[tex]\dot Q_{H} = 1\,kW + 0.75\,kW[/tex]
[tex]\dot Q_{H} = 1.75\,kW[/tex]
The rejected by the air conditioning system is 1.75 kilowatts.
You are in the frozen food section of the grocery store and you notice that your hand gets cold when you place it on the glass windows of the display cases. Your friend says this is because coolness is transferred from the display case to your hand. What do you think?
Answers
Answer:
I think my friend got it all wrong, as coolness can not be transferred but heat was actually transferred between my hand and the glass windows
Explanation:
In thermodynamics, coolness can not be transferred, only heat can be transferred
Here is how the mechanism of why i felt cold works, my body gave out heat, hence there was heat transfer from a region of high to a low heat region, equilibrium was reached and I started feeling the coolness in my hands.
Rays that pass through a lens very close to the principle axis are more sharply focused than those that are very far from the axis. This spherical aberration helps us understand why:_______
Answers
Answer: it is easier to read in bright light than dim light.
Explanation:
The ray of light is the direction that is used by light in travelling through a medium. Rays that pass through a lens very close to the principle axis are more sharply focused than those that are very far from the axis.
Because of the fact that the rays are close to the principle axis, the spherical aberration helps us to understand the reason why it is easier for people to read in bright light than readin iin dim light.
A 0.2 kg rubber ball is dropped from the window of a building. It strikes the sidewalk below at 30 m/s and rebounds at 20 m/s. The magnitude of the change in momentum of the ball as a result of the collision with the sidewalk is _______.
Answers
Answer:
10 kgm/s
Explanation:
Change in momentum: This can be defined as the product of mass and change in velocity. The S.I unit of change in momentum is Kgm/s.
From the question,
ΔM = m(v-u)...................... Equation 1
Where ΔM = change in momentum, u = initial velocity, v = final velocity.
Note: Let upward direction be negative, and downward direction be positive.
Given: m = 0.2 kg, v = -20 m/s, u = 30 m/s
Substitute into equation 1
ΔM = 0.2(-20-30)
ΔM = 0.2(-50)
ΔM = -10 kgm/s.
The negative sign shows that the change in momentum is Upward
The magnitude of the change in momentum of the ball as a result of the collision with the sidewalk is -10 kg-m/s.
Given data:
The mass of rubber ball is, m = 0.2 kg.
The initial speed of ball is, u = 30 m/s.
The final rebounding speed of ball is, v = - 20 m/s ( Negative sign shows that during the rebounding, the ball changes its direction)
The momentum of any object is defined as the product of mass and change in velocity. The S.I unit of momentum is Kg-m/s. And the expression for the change in momentum is given as,
[tex]p= m ( v-u)[/tex]
Solving as,
[tex]p= 0.2 \times ( -20-30)\\\\p=-10 \;\rm kg.m/s[/tex]
Thus, we can conclude that the magnitude of the change in momentum of the ball as a result of the collision with the sidewalk is -10 kg-m/s.
Learn more about the change in momentum here:
https://brainly.com/question/904448
Recent technological developments like high-resolution satellite imagery and diagnostic positron emission tomography (PET scans) have refined and extended the camera’s capacity to provide information. Which passage assertion does this information support most strongly?
Answers
Answer:
D) Photography can be used to both control and benefit society.
Explanation:
High-resolution satellite imagery and diagnostic positron emission tomography (PET scans) have been used to both control and benefits the society in the sense that it has helped to take records of information of crime, traffic offenders such drunk drivers and over speeding drivers, e.t.c. it helps control by given their information and automatically penalizing them or ensuring the agency penalized them and also benefit the society by preventing people from committing crime thereby, protecting them from offenders.
Consider the following three objects, each of the same mass and radius:
(1) a solid sphere
(2) a solid disk
(3) a hoop
All three are released from rest at the top of an inclined plane. The three objects proceed down the incline undergoing rolling motion without slipping. Use work-kinetic energy theorem to determine which object will reach the bottom of the incline first.
a) 1, 2, 3
b) 2, 3, 1
c) 3, 1, 2
d) 3, 2, 1
e) All three reach the bottom at the same time.
Answers
Answer:
Explanation:a 1