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.
Earth moves in an elliptical orbit with the sun at one of the foci. The length of half of the major axis is kilometers, and the eccentricity is . Find the minimum distance (perihelion) and the maximum distance (aphelion) of Earth from the sun.
Complete question is;
Earth moves in an elliptical orbit with the sun at one of the foci. The length of half of the major axis is 149,598,000 kilometers, and the eccentricity is 0.0167. Find the minimum distance (perihelion) and the maximum distance (aphelion) of Earth from the sun
Answer:
Minimum distance = 147,099,713.4 km
Maximum distance = 152,096,286.6 km
Explanation:
The formula for the eccentricity of an ellipse is given by;
e = c/a
where;
c is distance from the center of the ellipse to the focus of the ellipse.
a is distance from the center of ellipse to a vertex.
From the question, we are given;
a = 149,598,000
e = 0.0167
Thus;
0.0167 = c/149,598,000
c = 0.0167 × 149,598,000
c = 2,498,286.6
Now, formula for the minimum distance (perihelion) is;
Minimum distance = a - c
Minimum distance = 149598000 - 2498286.6
Thus;
Minimum distance = 147,099,713.4 km
Similarly, formula for the maximum distance (aphelion) is;
Max distance = a + c
Max distance = 149598000 + 2498286.6
Maximum distance = 152,096,286.6 km
In each of the following situations, a wave passes through an opening in an absorbing wall. Rank the situations in order from the one in which the wave is best described by the ray approximation to the one in which the wave coming through the opening spreads out most nearly equally in all directions in the hemisphere beyond the wall.
(a) The sound of a low whistle at 1 kHz passes through a doorway 1 m wide.
(b) Red light passes through the pupil of your eye.
(c) Blue light passes through the pupil of your eye.
(d) The wave broadcast by an AM radio station passes through a doorway 1 m wide. (e) An x-ray passes through the space between bones in your elbow joint.
Answer:
A) geometric optics, B) geometric optics , c) geometric optics ,
e) geometric optics, f) geometric optics
Explanation:
For this exercise we must use the condition for interference and diffraction so that these phenomena are relevant the wavelength must be comparable to the gap spacing
λ> = a
Lam when the spacing is much greater than the wavelength, the description of geometric optics is more and more exact
let's analyze each situation
a) let's find the wavelength
v = λ f
λ= v / f
λ= 343/1000
λ = 0.343 m
0.343 << 1m
therefore the description of the geometric optics of
b) red light passes through the pupil of the eye
red light has a wavelength of 700 num or more, the lojo pupil has a maximum of 8 me
λ = 700 10⁻⁹ m = 7 10⁻⁷ m
a = 8 mm 10⁻³
longitudinal is much less therefore the geometric optics is correct
c) luz azul lam = 450 nm = 450 10⁻⁹ m = 4.5 10⁻⁷ m
again the wavelength is much less than the diameter of the pupil, for which the description with the optics is generally sufficient
d) a radio A transmits up to a maximum of f = 1400 Khz = 1,400 10⁶ Hz
let's find the wavelength
c = λf
λ = c / f
λ= 3 108 / 1,400 106
λ= 2.14 102 m
in this case the wavelength is greater than the width of the gate, so the description of diffraction should be used to explain the phenomenon
e) X-rays have wavelength lam = 10-10 m
the separation of the elbow bones is of the order of a few millimeters, for local the wavelength is much less than the separation, therefore with the relations of geometric optics it is sufficient
A spaceship is moving past Earth at 0.99c. The spaceship fires two lasers. Laser A is in the same direction it is traveling, and Laser B is in the opposite direction. How fast will the light from each laser be traveling according to an observer on Earth?
Answer:
Vx' = (Vx - u) / (1 - Vx *u / c^2) velocity transformation formula
In both cases we wish to measure the velocity in the frame of the earth which is moving at speed u = -.99 c relative to the spaceship
VA' = (c + .99c) / (1 - (-.99 c * c) / c^2) = 1.99c / 1.99 = c
VB' = (-c + .99c) / (1 - (-c * -.99c) / c^2) = .01 c / .01 = c
In both cases an observer on earth will observe the light traveling at speed c.
Now, let's see what happens when the cannon is high above the ground. Click on the cannon, and drag it upward as far as it goes (15 m above the ground). Set the initial velocity to 14 m/s, and fire several pumpkins while varying the angle. For what angle is the range the greatest?
choices:
A. 45∘
B. 20∘
C. 30∘
D. 40∘
E. 50∘
Answer:
B. 20°Explanation:
Range in projectile is defined as the distance covered in the horizontal direction. It is expressed as R = U²sin2Ф/g
U is the initial velocity of the body (in m/s)
Ф is the angle of projection
g is the acceleration due to gravity.
Given U = 14m/s, g = 9.8m/s and range R = 15 m
we will substitute this value into the formula to get the projection angle Ф as shown;
15 = 15²sin2Ф/9.8
15*9.8 = 15²sin2Ф
147 = 225sin2Ф
sin2Ф = 147/225
sin2Ф = 0.6533
2Ф = sin⁻¹0.6533
2Ф = 40.79°
Ф = 40.79°/2
Ф = 20.39° ≈ 20°
Hence, the range is greatest at angle 20°
The index of refraction of a certain material is 1.25. If I send red light (700 nm) through the material, what will the frequency of the light be in the material
Answer:
f1 / f2 = n2 / n1
Explanation:
To solve this problem, we should remember that the formula for index of refraction is defined as:
n = c / v
or
n v = c
Where,
n = index of refraction
c = speed of light
v = speed of light in the medium
Since speed of light is constant, then we can simply equate the materials 1 and 2:
n1 v1 = n2 v2
Where the speed of light in the medium (v) can be expressed as:
v = w * f
Where,
w = wavelength of light
f = frequency of light
Therefore substituting this back into the relating equation:
n1 w1 f1 = n1 w2 f1
Since it is given that the light is monochromatic, w1 = w2, this further simplifies the equation to:
n1 f1 = n2 f2
f1 / f2 = n2 / n1 (ANSWER)
2. A wire 4.00 m long and 6.00 mm in diameter has a resistance of 15.0 mΩ. A potential difference of 23.0 V is applied between the end. a) What is the current in the wire? b) Calculate the resistivity of the wire material. c) Try to identify the material.
Answer:
Explanation:
a )
current in the wire = potential diff / resistance
= 23 / (15 x 10⁻³ )
= 1.533 x 10³ A .
b )
For resistance of a wire , the formula is
R = ρ L / S where ρ is specific resistance , L is length and S is cross sectional area of wire
putting the given values
15 x 10⁻³ = 4ρ / π x .003²
ρ = 106 x 10⁻⁹ ohm. m
= 10.6 x 10⁻⁸ ohm m
The metal wire appears to be platinum .
(a) The current in the wire is 1.533 x 10³ A
(b) The resistivity of the wire material is 10.6 x 10⁻⁸ Ωm
(c) The material of the wire is Platinum
Ohm's Law and resistivity:
(a) According to the Ohm's Law:
V = IR
where V is the potential difference
I is the current
and R is the resistance
So,
I = V/R
I = 23 / (15 x 10⁻³ )
I = 1.533 x 10³ A
(b) The resistance of a wire is expressed as:
R = ρ L / A
where ρ is the resistivity,
L is length
and A is the cross-sectional area
15 x 10⁻³ = 4ρ / π x .003²
ρ = 106 x 10⁻⁹ Ωm
ρ = 10.6 x 10⁻⁸ Ωm
The metal from which the wire is made is platinum.
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A balloon is ascending at a rate of +4.00 m/s to a height of 11.0 m above the ground when a package is dropped. In the absence of air resistance, the velocity of the ball when it hits the ground is
Answer:
Vf = 14.7 m/s
Explanation:
Vf² = Vi² + 2 * a * Δy
given:
a = 9.81 m/s²
Δy = 11m
Vi = 0 when upon release
Vf² = 0 + 2 (9.81) 11
Vf = 14.7 m/s
The velocity of the ball when it hits the ground will be 14.7 m/s.
What is velocity?The change of displacement with respect to time is defined as the velocity. Velocity is a vector quantity. it is a time-based component. Velocity at any angle is resolved to get its component of x and y-direction.
given:
a(gravitational acceleration) = 9.81 m/s²
s (distance)= 11m
v is final velocity
u is the initial velocity
From Newton's second equation of motion;
[tex]\rm v^2 = u^2+2as \\\\ v^2=2 \times 9.81 \times 11 \\\\ v= 14.7 \ m/sec[/tex]
Hence, the velocity of the ball when it hits the ground will be 14.7 m/s.
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A capacitor that is initially uncharged is connected in series with a resistor and an emf source with E=110V and negligible internal resistance. Just after the circuit is completed, the current trhough the resistor is 6.5*10^-5A. the time constant for the circuit is 4.8s.
1) What is the resistance of the resistor?
2) What is the capacitance of the capacitor?
Answer:
1
[tex]R = 1.692*10^{6} \Omega[/tex]
2
[tex]C = 2.837 *10^{-6} \ F[/tex]
Explanation:
From the question we are told that
The voltage is [tex]E = 110 \ V[/tex]
The current is [tex]I = 6.5 *10^{-5} \ A[/tex]
The time constant is [tex]\tau = 4.8 \ s[/tex]
The resistance of resistor is mathematically evaluated as
[tex]R = \frac{E}{I}[/tex]
substituting values
[tex]R = \frac{ 110 }{ 6.5*10^{-5}}[/tex]
[tex]R = 1.692*10^{6} \Omega[/tex]
The capacitance of the capacitor is mathematically represented as
[tex]C = \frac{\tau}{R}[/tex]
substituting values
[tex]C = \frac{ 4.8}{ 1.692*10^{6}}[/tex]
[tex]C = 2.837 *10^{-6} \ F[/tex]
Tesla Model S and the driver have a total mass of 2250 kg. The projected front area of the car is 2.35 m2. The car is traveling at 72km/hr when the driver puts the transmission into neutral and allows the car to freely coast until after 105 seconds its speed reaches 54 km/hr. Determine the drag coefficient for the car, assuming its values is constant. Neglect rolling and other mechanical resistance.
Answer:
The drag coefficient of the car is 0.189
Explanation:
mass of the car = 2250 kg
Frontal area of the car = 2.35 m^2
initial speed of the car = 72 km/hr = (72 x 1000)/3600 = 20 m/s
final speed of the car = 54 km/hr = (54 x 1000)/3600 = 15 m/s
time taken by the car to slow down = 105 sec
We'll assume that the value of the drag coefficient is constant throughout the deceleration.
The car decelerates from 20 m/s to 15 m/s in 105 seconds, the deceleration is calculated from
[tex]a = \frac{v-u}{t}[/tex]
where a is the deceleration
v is the final speed of the car
u is the initial speed of the car
t is the time taken to decelerate.
imputing values, we'll have
[tex]a = \frac{15-20}{105}[/tex] = -0.0476 m/s^2 (the -ve sign indicates a deceleration, which is a negative acceleration)
we can safely ignore the -ve sign in other calculations that follows
The force (drag force) with which the air around the decelerates the car is equal to..
[tex]F_{D} = ma[/tex]
where [tex]F_{D}[/tex] is the drag force
m is the mass of the car
a is the deceleration of the car
imputing values, we'll have
[tex]F_{D} = 2250*0.0476[/tex] = 107.1 N
equation for drag force is
[tex]F_{D} = \frac{1}{2}pAC_{D} v^{2}[/tex]
where p is the air density ≅ 1.225 kg/m³
A is the frontal area of the car
[tex]C_{D}[/tex] is drag coefficient of the car
v is the relative velocity of air and the car, and will be taken as the initial velocity of the car before starting to decelerate.
imputing these values, we'll have
[tex]107.1 = \frac{1}{2}*1.225*2.35*C_{D}*20^{2}[/tex] = 575.75[tex]C_{D}[/tex]
[tex]C_{D}[/tex] = 107.1/575.75 = 0.189
Find the absolute value of the change of the gravitational potential energy (GPE) of the puck-Earth system from the moment the puck begins to move to the moment it hits the spring. Use 0.253 m for the displacement of the puck along the ramp and 9.80 m/s2 for the acceleration due to gravity. Assume that the mass of the puck is 0.180 kg. Express your answer using SI units to three significant figures.
Answer:
0.16joules
Explanation:
Using the relation for The gravitational potential energy
E= Mgh
Where,
E= Potential energy
h = Vertical Height
M = mass
g = Gravitational Field Strength
To find the vertical component of angle of launch Where the angle is 22°
h= sin theta
So E = mghsintheta
= 0.18 x 0.98 x 0.253 sin22
=0.16joules
Explanation:
Two bodies of equal mass m collide and stick together. The quantities that always have equal magnitude for both masses during the collision are
Answer:
The quantities that always have equal magnitude for both masses during the collision are change in momentum of the colliding bodies and force exerted by each body
Explanation:
During collision of two bodies, the following quantities are affected;
Kinetic energy of the colliding bodies
change in momentum of the colliding bodies
Force exerted by each body
Two bodies that stick together after collision is inelastic collision.
For inelastic collision, the kinetic energy before collision is greater than kinetic energy after collision.
Change in momentum is zero, that is, momentum before collision is equal to momentum after collision.
According Newton's third law of motion, the force exerted by each body is equal but acts in opposite direction.
Therefore, the quantities that always have equal magnitude for both masses during the collision are change in momentum of the colliding bodies and force exerted by each body
The quantities that always have equal magnitude for both masses during the collision are change in momentum and force exerted by each body
Inelastic collision is a collision in which both bodies stick with each other after collision.
For inelastic collision, the momentum before collision is equal to momentum after collision.
Also, the force exerted by each body is equal but acts in opposite direction.
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A body's current radiation rate is 30% higher than it was an hour ago. Calculate the percentage by which your temperature increased.
Answer:
6.8%
Explanation:
According to Stefan-Boltzmann law, radiation is directly proportional with temperature raised to the fourth power:
P ∝ T⁴
Writing a proportion:
P₁ / P₂ = T₁⁴ / T₂⁴
1.3P / P = (T₁ / T₂)⁴
T₁ / T₂ = ∜1.3
T₁ = 1.068 T₂
The temperature increased by 6.8%.
An object has an acceleration of 6.0 m/s/s. If the net force was tripled and the mass were doubled, then the new acceleration would be _____ m/s/s.
Answer:
The new acceleration would be 9 m/s².
Explanation:
Acceleration of an object is 6 m/s²
Net force is equal to the product of mass and acceleration i.e.
F = ma
[tex]a=\dfrac{F}{m}\\\\\dfrac{F}{m}=6\ m/s^2[/tex]
If the net force was tripled and the mass were doubled, it means,
F' = 3F
m' = 2m
Let a' is new acceleration. So,
[tex]a'=\dfrac{F'}{m'}\\\\a'=\dfrac{(3F)}{(2m)}\\\\a'=\dfrac{3}{2}\times \dfrac{F}{m}\\\\a'=\dfrac{3}{2}\times 6\\\\a'=9\ m/s^2[/tex]
So, the new acceleration would be 9 m/s².
The _____________ is the thermonuclear fusion of hydrogen to form helium operating in the cores of massive stars on the main sequence
A 20 kg child is on a swing that hangs from 2.6-m-long chains. What is her maximum speed if she swings out to a 46 angle?
Answer:
Explanation:
Here potential energy lost = mgh
h = L ( 1 - cos 46 ) where L is length of rope
= 2.6 x ( 1 - cos 46 )
PE lost = 20 x 9.8 x 2.6 ( 1 - cos 46 )
= 155.6 J
gain of kinetic energy = loss of PE
1/2 m v² = 155.6
.5 x 20 x v² = 155.6
v² = 15.56
v = 3.94 m /s
if F, V, and we're chosen as fundamental unit of force, velocity and time respectively , the dimensions of mass would be represented as
Answer:
The dimension of mass can be represented as: [tex][F^{1} T^{1} V^{-1} ][/tex]
Explanation:
We have Force = Mass X Acceleration
= Mass X [tex]\frac{Change in Velocity}{Time Taken}[/tex]
or, Mass = Force x [tex]\frac {Time Taken } { Change in Velocity }[/tex]
So, dimensions of mass = [tex]\frac{[F][T]}{[V]}[/tex]
= [tex][F^{1} T^{1} V^{-1} ][/tex]
If you were to calculate the pull of the Sun on the Earth and the pull of the Moon on the Earth, you would undoubtedly find that the Sun's pull is much stronger than that of the Moon, yet the Moon's pull is the primary cause of tides on the Earth. Tides exist because of the difference in the gravitational pull of a body (Sun or Moon) on opposite sides of the Earth. Even though the Sun's pull is stronger, the difference between the pull on the near and far sides is greater for the Moon.
Required:
a. "Let F(r) be the gravitational force exerted on one mass by a second mass a distance r away. Calculate dF(r)/dr to show how F changes as r is changed.
b. Evaluate this expression for dF(r) jdr for the force of the Sun at the Earth's center and for the Moon at the Earth's center.
c. Suppose the Earth-Moon distance remains the same, but the Earth is moved closer to the Sun. Is there any point where dF(r)/dr for the two forces has the same value?
Answer:effective
Explanation:
Rick spends four hours researching on the internet and does 1090 J of work. In the process, his internal energy decreases by 2190 J. Determine the value of Q, including the algebraic sign.
Answer:
Q = -3280J
Explanation:
From the First Law of Thermodynamics, energy cannot be created nor destroyed but it can be converted from one form to another with the interaction of heat. Mathematically, this can be expressed as:
ΔU = Q + W ----------(i)
Where;
ΔU = total change in internal energy of a system.
Q = heat exchanged between the system and the surrounding
W = work done by or on the system.
If heat is lost into the surrounding, then Q = -ve, else Q = +ve
If work is done on the system, then W = -ve, else W = -ve
=> From the question, Rick is the system and does a work of
W = +1090J [since Rick does the work, W = +ve]
=>Also, the internal energy decreases by 2190J, therefore,
ΔU = -2190J [since there is a decrease in internal energy]
Substitute the values of W and ΔU into equation (i) as follows;
-2190 = Q + 1090
=> Q = -2190 - 1090
=> Q = -3280J
Therefore, the value of Q = -3280J
Suppose a 185 kg motorcycle is heading toward a hill at a speed of 29 m/s. The two wheels weigh 12 kg each and are each annular rings with an inner radius of 0.280 m and an outer radius of 0.330 m.
Randomized Variables
m = 185 kg
v = 29 m/s
h = 32 m
A. How high can it coast up the hill. if you neglect friction in m?
B. How much energy is lost to friction if the motorcycle only gains an altitude of 33 m before coming to rest?
Answer:
a) Height reached before coming to rest is 42.86 m
b) Energy lost to friction is 17902.45 J
Explanation:
mass of the motorcycle = 185 kg
speed of the towards the hill = 29 m/s
The wheels weigh 12 kg each
Wheels are annular rings with an inner radius of 0.280 m and outer radius of 0.330 m
a) To go up the hill, the kinetic energy of motion of the motorcycle will be converted to the potential energy it will gain in going up a given height
the kinetic energy of the motorcycle is given as
[tex]KE[/tex] = [tex]\frac{1}{2}mv^{2}[/tex]
where m is the mass of the motorcycle
v is the velocity of the motorcycle
[tex]KE[/tex] = [tex]\frac{1}{2}*185*29^{2}[/tex] = 77792.5 J
This will be converted to potential energy
The potential energy up the hill will be
[tex]PE[/tex] = mgh
where m is the mass
g is acceleration due to gravity 9.81 m/s^2
h is the height reached before coming to rest
[tex]PE[/tex] = 185 x 9.81 x m = 1814.85h
equating the kinetic energy to the potential energy for energy conservation, we'll have
77792.5 = 1814.85h
height reached before coming to rest = 77792.5/1814.85 = 42.86 m
b) if an altitude of 33 m was reached before coming to rest, then the potential energy at this height is
[tex]PE[/tex] = mgh
[tex]PE[/tex] = 185 x 9.81 x 33 = 59890.05 J
The energy lost to friction will be the kinetic energy minus this potential energy.
energy lost = 77792.5 - 59890.05 = 17902.45 J
A) The motorcycle can coast up the hill by ; 42.86m
B) The amount of energy lost to friction : 17902.45 J
A) Determine how high the motorcycle can coast up the hill when friction is neglected
apply the formula for kinetic and potential energies
K.E = 1/2 mv² ---- ( 1 )
P.E = mgH ---- ( 2 )
As the motorcycle goes uphiLl the kinetic energy is converted to potential energy.
∴ K.E = P.E
1/2 * mv² = mgH
∴ H = ( 1/2 * mv² ) / mg ---- ( 3 )
where ; m = 185 kg , v = 29 m/s , g = 9.81
Insert values into equation ( 3 )
H ( height travelled by motorcycle neglecting friction ) = 42.86m
B) Determine how much energy is lost to friction if the motorcycle attains 33m before coming to rest
P.E = mgh = 185 * 9.81 * 33 = 59890.05 J
where : h = 33 m , g = 9.81
K.E = 1/2 * mv² = 77792.5 J ( question A )
∴ Energy lost ( ΔE ) = ( 77792.5 - 59890.05 ) = 17902.45 J
Hence we can conclude that The motorcycle can coast up the hill by ; 42.86m , The amount of energy lost to friction : 17902.45 J.
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How would the interference pattern produced by a diffraction grating change if the laser light changed from red to blue?
Answer
fringe separation l distance between maxima decreases
Explanation:
Because the wavelength of blue light is smaller than that if red light
The length of a certain wire is doubled and at the same time its radius is also doubled. What is the new resistance of this wire
Answer:
R' = R/2
Therefore, the new resistance of the wire is twice the value of the initial resistance.
Explanation:
Consider a wire with:
Resistance = R
Length = L
Area = A = πr²
where, r = radius
ρ = resistivity
Then:
R = ρL/A
R = ρL/πr² --------------- equation 1
Now, the new wire has:
Resistance = R'
Resistivity = ρ
Length = L' = 2 L
Radius = r' = 2r
Area = πr'² = π(2r)² = 4πr²
Therefore,
R' = ρL'/πr'²
R' = ρ(2 L)/4πr²
R' = (1/2)(ρL/πr²)
using equation 1:
R' = R/2
Therefore, the new resistance of the wire is twice the value of the initial resistance.
An elastic circular bar is fixed at one end and attached to a rubber grommet at the other end. The grommet functions as a torsional spring with spring constant k. If a concentrated torque of magnitude Ta is applied in the center of the bar, what is the rotation at the end of the bar, φ(L)? Assume a constant shear modulus G and polar moment of inertia J.
Answer:
2.1 rad(anticlockwise).
Explanation:
So, we are given the following data or parameters or information in the question above:
=> "The torsional stiffness of the spring support is k = 50 N m/rad. "
=> "If a concentrated torque of mag- nitude Ta = 500 Nm is applied in the center of the bar"
=> "L = 300 mm Assume a shear modu- lus G = 10 kN/mm2 and polar monnent of inertia J = 2000 mln"
Hence;
G × J = 10 kN/mm2 × 2000 mln = 20 Nm^2.
Also, L/2 = 300 mm /2 = 0.15 m (converted to metre).
==> 0.15/20 (V - w) + θ = 0.
==> 0.15/20 (V - w ) = -θ.
Where V = k = 50 N m/rad
w = 183.3 θ.
Therefore, w + Vθ = 500 Nm.
==> 183.3 + 50 θ = 500 Nm.
= 6.3
Anticlockwise,
θ = 2.1 rad.
The free-electron density in a copper wire is 8.5×1028 electrons/m3. The electric field in the wire is 0.0520 N/C and the temperature of the wire is 20.0∘C. Use the resistivity at room temperature for copper is rho = 1.72×10^−8 Ω⋅m.
Required:
a. What is the drift speed vdvd of the electrons in the wire?
b. What is the potential difference between two points in the wire that are separated by 25.0 cm?
(a) The drift speed of electrons in the wire is 2.22 x 10⁻⁴ m/s.
(b) The potential difference between two points in the wire is 0.013 V.
The given parameters;
electron density of the copper wire, n = 8.5 x 10²⁸ electrons/m³.electric field, E = 0.0520 N/Ctemperature of the wire, t = 20 ⁰Cresistivity of the copper wire, ρ = 1.72 x 10⁻⁸ Ω⋅mThe drift speed of electrons in the wire is calculated as follows;
[tex]v_d = \frac{I}{qn A} \\\\but , \ \frac{I}{A} = \frac{E}{\rho} \\\\v_d = \frac{E}{qn \rho}[/tex]
where;
E is the electric fieldq is charge of electron = 1.602 x 10⁻¹⁹ C[tex]v_d = \frac{0.052}{1.602 \times 10^{-19} \times 8.5\times 10^{28} \times 1.72 \times 10^{-8}} \\\\v_d = 2.22 \times 10^{-4} \ m/s[/tex]
The potential difference between two points in the wire, separated by 25 cm;
V = Ed
where;
d is the distance of separation = 25 cm = 0.25 mV = 0.052 x 0.25
V = 0.013 V
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A very fine thread is placed between two glass plates on one side and the other side is touching to form a wedge. A beam of monochromatic light of wavelength 600 nm illuminates the wedge and 178 bright fringes are observed. What is the thickness of the thread?
Answer:
53.3micro meters
Explanation:
See attached file
A flashlight is held at the edge of a swimming pool at a height h = 1.6 m such that its beam makes an angle of θ = 38 degrees with respect to the water's surface. The pool is d = 1.75 m deep and the index of refraction for air and water are n1 = 1 and n2 = 1.33, respectively.
Required:
What is the horizontal distance from the edge of the pool to the bottom of the pool where the light strikes? Write your answer in meters.
one of the answers that i found was 5.83 m i did some more research and it showed the same answer again. good luck with it. hope i was able to help you.
The velocity of an object is given by the following function defined on a specified interval. Approximate the displacement of the object on this interval by sub-dividing the interval into the indicated number of sub-intervals. Use the left endpoint of each sub-interval to compute the height of the rectangles.
v= 4t + 5(m/s) for 3 < t < 7; n = 4
The approximate displacement of the object is______m.
Answer:
The approximate displacement of the object is 23 m.
Explanation:
Given that:
v = 4t + 5 (m/s) for 3< t< 7; n= 4
The approximate displacement of the object can be calculated as follows:
The velocities at the intervals of t are :
3
4
5
6
the velocity at the intervals of t = 7 will be left out due the fact that we are calculating the left endpoint Reimann sum
n = 4 since there are 4 values for t, Then there is no need to divide the velocity values
v(3) = 4(3)+5
v(3) = 12+5
v(3) = 17
v(4)= 4(4)+5
v(4) = 16 + 5
v(4) = 21
v(5)= 4(5)+5
v(5) = 20 + 5
v(5) = 25
v(6) = 4(6)+5
v(6) = 24 + 5
v(6) = 29
Using Left end point;
[tex]= \dfrac{1}{4}(17+21+25+29)[/tex]
= 23 m
A solenoid 26.0 cm long and with a cross-sectional area of 0.580 cm^2 contains 490 turns of wire and carries a current of 90.0 A.
Calculate:
(a) the magnetic field in the solenoid;
(b) the energy density in the magnetic field if the solenoid is filled with air;
(c) the total energy contained in the coil’s magnetic field (assume the field is uniform);
(d) the inductance of the solenoid.
Answer:
A.21.3T
B.1.8x 10^6J/m^3
C.0.27x10^2J
D.6.6x10^-3H
Explanation:
Pls see attached file
Find the rms (a) electric and (b)magnetic fields at a point 2.00 m from a lightbulb that radiates 90.0 W of light uniformly in all directions.
Answer:
a) rms of electric field =
[tex]E_{rms}[/tex]= 25.97 V/m
b) rms of magnetic field
[tex]B_{rms}[/tex] = 8.655 × 10⁻⁸
[tex]B_{rms}[/tex] = 86.55nT
Explanation:
given
power p = 90.0W
distance d = 2.00m
Intensity = [tex]\frac{power}{area}[/tex]
I = [tex]\frac{p}{A}[/tex]
A = [tex]4\pi d^{2}[/tex]
I = [tex]\frac{p}{4\pi d^{2} }[/tex]
I = [tex]\frac{90}{4\pi(2^{2}) }[/tex]
I = 1.79 W/m²
a) [tex]I_{ave}[/tex] = ε₀ × [tex]E^{2} _{rms}[/tex] × c
where ε₀ is permittivity of free space = 8.85×10⁻¹², [tex]E^{2} _{rms}[/tex] is the root mean value and c is speed of light = 3×10⁸m/s
1.79 = 8.85×10⁻¹² × [tex]E^{2} _{rms}[/tex] × 3×10⁸
[tex]E^{2} _{rms}[/tex] = [tex]\frac{1.79}{8.85x10^{-12} x 3x10^{8} }[/tex]
[tex]E^{2} _{rms}[/tex]= 674.1996
[tex]E_{rms}[/tex]= 25.97 V/m
b)for rems magnetic field
[tex]E_{rms}[/tex]= c [tex]B_{rms}[/tex]
[tex]B_{rms}[/tex] = [tex]\frac{E_{rms} }{c}[/tex]
[tex]B_{rms}[/tex] = [tex]\frac{25.97 V/m}{3x10^{8} }[/tex]
[tex]B_{rms}[/tex] = 8.655 × 10⁻⁸
[tex]B_{rms}[/tex] = 86.55nT
An electromagnetic standing wave in air of frequency 750 MHz is set up between two conducting planes 80.0 cm apart. At how many positions between the planes could a point charge be placed at rest so that it would remain at rest?
Answer:
The positions the point charge will be placed at rest and still remain at rest are 20 cm, 40 cm, and 60 cm between the ends.
Explanation:
Given;
distance between the conducting planes, d = 80 cm
frequency of the electromagnetic wave, f = 750 MHz
speed of light, c = 3 x 10⁸ m/s²
Determine the wavelength
λ = C/f
where;
λ is the wavelength
C is the speed of light
f is the frequency
λ = C/f
λ = (3 x 10⁸) / (750 x 10⁶)
λ = 0.4 m = 40 cm
One complete cycle = one wavelength = 40 cm
half of the wavelength ( λ / 2) = 20 cm
one wavelength + half wavelength (3λ / 2) = 60 cm
The positions of the wave at zero amplitude (between 0 and 80 cm) = 20 cm, 40 cm, 60 cm
Thus, the positions the point charge will be placed at rest and still remain at rest are 20 cm, 40 cm, and 60 cm between the ends.
Use I=∫r2 dm to calculate I of a slender uniform rod of length L, about an axis at one end perpendicular to the rod. note: a "slender rod" often refers to a rod of neglible cross sectional area, so that the volume is the Length, and the mass density X Length.
Answer:
The moment of inertia of a slender uniform rod of length L about an axis at one end perpendicular to the rod is [tex]I = \frac{1}{3}\cdot m \cdot L^{2}[/tex].
Explanation:
Let be an uniform rod of length L whose origin is located at one one end and axis is perpendicular to the rod, such that:
[tex]\lambda = \frac{dm}{dr}[/tex]
Where:
[tex]\lambda[/tex] - Linear density, measured in kilograms per meter.
[tex]m[/tex] - Mass of the rod, measured in kilograms.
[tex]r[/tex] - Distance of a point of the rod with respect to origin.
Mass differential can translated as:
[tex]dm = \lambda \cdot dr[/tex]
The equation of the moment of inertia is represented by the integral below:
[tex]I = \int\limits^{L}_{0} {r^{2}} \, dm[/tex]
[tex]I = \lambda \int\limits^{L}_{0} {r^{2}} \, dr[/tex]
[tex]I = \lambda \cdot \left(\frac{1}{3}\cdot L^{3} \right)[/tex]
[tex]I = \frac{1}{3}\cdot m \cdot L^{2}[/tex] (as [tex]m = \lambda \cdot L[/tex])
The moment of inertia of a slender uniform rod of length L about an axis at one end perpendicular to the rod is [tex]I = \frac{1}{3}\cdot m \cdot L^{2}[/tex].