The ratio of the wavelength of the sound wave to the wavelength of the standing wave (1st harmonic) on the wire is 2.64.
The fundamental frequency of a piano wire is 27.5 Hz.
Length of the wire = 1.18 m.
Speed of sound in air = 343 m/s.
The fundamental frequency is given by the expression:
f = (1/2L) * √(T/m)
Here, f is the frequency, L is the length of the wire, T is the tension in the wire, and m is the mass per unit length of the wire.
Rearranging the above expression to find tension, we get:
T = 4mL²f²
So, the tension in the wire is:
T = 4 × (mass per unit length of the wire) × (length of the wire)² × (fundamental frequency of the wire)²
T = 4 × m × L² × f²
The mass per unit length of the wire can be found from the following expression:
m = (π/4) × (d²) × ρ
Here, d is the diameter of the wire and ρ is the density of the wire. Since the wire is made up of steel, the density can be taken as 7850 kg/m³.
The diameter of the wire can be found using the following formula:
Area of the cross-section = π/4 × d²
The area of the cross-section of the wire can be given by:
A = (π/4) × d²
Substituting the given values in the above formula, we have:
A = (π/4) × d² = π × (0.000948 m)² = 2.817 × 10⁻⁷ m²
Cross-sectional area of the wire is 2.817 × 10⁻⁷ m².
Now, the mass per unit length of the wire is given by:
m = ρ × A = 7850 × (2.817 × 10⁻⁷) kg/m = 2.212 × 10⁻³ kg/m
Substituting the values of L, f, and m, we get:
T = 4 × m × L² × f² = 4 × 2.212 × 10⁻³ × (1.18)² × (27.5)² = 205.75 N
The speed of sound is given as 343 m/s.
The wavelength of the sound wave produced is given by:
λ = v/f
where v is the speed of sound in air.
Substituting the given values, we get:
λ = v/f = 343/27.5 = 12.472 m
The standing wave produced by the wire is a quarter wavelength. Hence the wavelength of the standing wave is four times the length of the wire.
λ' = 4L = 4 × 1.18 = 4.72 m
The ratio of the wavelength of the sound wave to the wavelength of the standing wave (1st harmonic) on the wire is:
λ/λ' = 12.472/4.72 = 2.64
Therefore, the ratio of the wavelength of the sound wave to the wavelength of the standing wave (1st harmonic) on the wire is 2.64.
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A pilot without special training or equipment can tolerate a horizontal acceleration of up to about 9.00g for a short period of time (about a minute) without losing consciousness. How long would it take a supersonic jet in horizontal flight to accelerate from 260 m/s to 700 m/s at an acceleration of 9.00g?
It would take about 4.98 seconds for a supersonic jet in horizontal flight to accelerate from 260 m/s to 700 m/s at an acceleration of 9.00g.
How long would it take a supersonic jet in horizontal flight to accelerate from 260 m/s to 700 m/s at an acceleration of 9.00g?" is as follows: Given data: Initial velocity,
u = 260 m/s
Final velocity,
v = 700 m/s
Acceleration, a = 9g = 9 × 9.8 = 88.2 m/s²
We know that:
v = u + at Where, v = Final velocity = 700 m/su = Initial velocity = 260 m/sa = Acceleration = 88.2 m/s²t = Time taken
We need to find the time taken to accelerate from 260 m/s to 700 m/s using the given data.
Substitute the given values in the above equation to find the value of
t:700 = 260 + 88.2tt = (700 - 260) / 88.2t = 4.98 seconds (approximately)
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An electron's position is given by
r
=5.00t
i
^
−8.00t
2
j
^
+5.00
k
^
, with t in seconds and
r
in meters. (a) In unit-vector notation, what is the electron's velocity
v
(t) ? (Use the following as necessary: t.)
v
(t) m/s (b) What is
v
in unit-vector notation at t=9.00 s?
v
(t)9,00)= m/s (c) What is the magnitude of
v
at t=9.00 s ? m/s (d) What angle does
v
make with the positive direction of the x axis at t=9.005 ? \$o (from the +x axis)
The electron position is a vector quantity that describes its location in space. The velocity at "x" is 5.00 m/s, while the velocity at "y" is -16.00t m/s. The magnitude of v is approximately 144.06 m/s and the angle formed is -86.34°.
(a) To find the velocity vector, we need to differentiate the position vector with respect to time.
Given:
r = 5.00t i^ - 8.00t² j^ + 5.00 k^
Differentiating each component separately:
v = (d/dt)(5.00t) i^ + (d/dt)(-8.00t^2) j^ + (d/dt)(5.00) k^
v = 5.00 i^ - 16.00t j^ + 0 k^
Therefore, the electron's velocity v(t) is given by v = (5.00 i^ - 16.00t j^) m/s. This means that the velocity in the x-direction is constant and equal to 5.00 m/s, while the velocity in the y-direction varies with time t as -16.00t m/s.
(b) To find the velocity at t = 9.00 s, we substitute t = 9.00 s into the velocity expression.
v(9.00) = 5.00 i^ - 16.00(9.00) j^
v(9.00) = 5.00 i^ - 144.00 j^
Therefore, the velocity v at t = 9.00 s is v(9.00) = (5.00 i^ - 144.00 j^) m/s. This means that at t = 9.00 s, the velocity vector of the electron is 5.00 m/s in the positive x-direction (i^) and -144.00 m/s in the negative y-direction (j^).
(c) To find the magnitude of v at t = 9.00 s, we use the magnitude formula:
|v(9.00)| = √[(5.00)² + (-144.00)²]
|v(9.00)| = √[25.00 + 20736.00]
|v(9.00)| ≈ √20761.00
|v(9.00)| ≈ 144.06 m/s
Therefore, the magnitude of v at t = 9.00 s is approximately 144.06 m/s.
(d) To find the angle that v makes with the positive direction of the x-axis at t = 9.005 s, we can calculate the angle using the arctangent function.
θ = arctan(v_y/v_x)
Substituting the values:
θ = arctan((-144.00)/(5.00))
θ ≈ -86.34° (from the +x-axis)
Therefore, the angle that v makes with the positive direction of the x-axis at t = 9.005 s is approximately -86.34° (from the +x-axis).
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5kg box is resting on the ground. If you lift this box a height of 2m above ground, how much has the gravitational potential energy of the box increased?
How much work have you done to lift the box 2m?
Part 2: If you use a ramp to push the same box up to a height of 2m above ground, how much has the gravitational potential energy of the box increased?
How much work have you done pushing the box up the ramp to a height of 2m above ground?
And does that depend on whether you lifted the box or pushed the box up the ramp?
The change in gravitational potential energy (GPE) of the 5 kg box lifted to a height of 2 m above the ground is 2940 Joules.
Formula for gravitational potential energy: GPE = mgh
Given: m = 5 kg, g = 9.8 m/s^2 (acceleration due to gravity on Earth), and h = 2 m
Calculation: GPE = 5 kg × 9.8 m/s^2 × 2 m = 2940 Joules
Part 1:
The work done in lifting the box to a height of 2 m above the ground is 98 Joules
Work done formula: Work done = Force × distance
Force is the weight of the box and distance is the height to which it is lifted.
Work done = 5 kg × 9.8 m/s^2 × 2 m = 98 Joules
Part 2:
If a ramp is used to push the same box up to a height of 2 m above the ground, the GPE of the box remains 2940 Joules.
The height to which the box is lifted remains the same, whether lifted or pushed up the ramp.
The work done in pushing the box up the ramp would depend on the length of the ramp, the angle of the ramp, and the frictional force between the ramp and the box.
The formula for work done remains the same: Work done = Force × distance.
The amount of work done would be less in pushing the box up the ramp compared to lifting it directly because you are exerting a force over a greater distance.
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The electric potential difference between two infinite, parallel metal plates is V. If the plates are separated by a distance L=45 mm and the electric field between the plates is E=320 V/m, what is the magnitude of V ?
The magnitude of the electric potential difference (V) between the parallel metal plates is found to be 14.4 V, given the electric field (E) of 320 V/m and the distance between the plates (d) of 45 mm (0.045 m). The calculation assumes a uniform electric field and a simple parallel plate capacitor geometry.
To find the magnitude of the electric potential difference (V) between the two parallel metal plates, we can use the relationship between electric field (E) and potential difference.
The electric field is given as E = V/d, where d is the distance between the plates. Rearranging the equation, we have V = E * d.
Substituting the given values, E = 320 V/m and d = 45 mm (or 0.045 m), we can calculate V as follows:
V = 320 V/m * 0.045 m = 14.4 V.
Therefore, the magnitude of the electric potential difference between the plates is 14.4 V.
It's important to note that in this case, the electric field is assumed to be uniform between the plates, and the calculation assumes a simple parallel plate capacitor geometry.
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A plastic cart is attached to a spring and completes an oscillation every 2.54 s in simple harmonic motion. At time t=0.00 s the cart is released at the position 8 cm with zero initial speed. What is the position (in cm ) of the cart at 17.8 s ?
The position of the cart at 17.8 s is approximately 7.6 cm.
The given problem describes a cart undergoing simple harmonic motion with a period of 2.54 s. This means that the cart completes one oscillation every 2.54 s. Since the cart is released at t=0.00 s with a position of 8 cm and zero initial speed, we can determine the equation of motion for the cart as follows:
x(t) = A * cos(2πt / T)
where x(t) is the position of the cart at time t, A is the amplitude (initial position), t is the time, and T is the period. Plugging in the values given, we have:
x(t) = 8 cm * cos(2πt / 2.54 s)
To find the position at t = 17.8 s, we substitute this value into the equation:
x(17.8 s) = 8 cm * cos(2π * 17.8 s / 2.54 s)
Evaluating this expression, we find that the position of the cart at 17.8 s is approximately 7.6 cm.
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A spaceship takes off vertically from rest with an acceleration of 28.5 m/s
2
. What magnitude of force F is exerted on a 59.5 kg astronaut during takeoff? F: Express F as a multiple of the astronaut's weight w on Earth. F
The magnitude of the force exerted on the astronaut during takeoff is approximately 2.91 times the weight of the astronaut on Earth (F = 2.91w).
To calculate the magnitude of the force exerted on the astronaut during takeoff, we can use Newton's second law of motion:
F = m * a
where F is the force, m is the mass of the astronaut, and a is the acceleration.
Given:
mass of the astronaut (m) = 59.5 kg
acceleration (a) = 28.5 m/s^2
Substituting the values into the formula, we get:
F = (59.5 kg) * (28.5 m/s^2)
To express the force (F) as a multiple of the astronaut's weight (w) on Earth, we need to divide the force by the weight (w) of the astronaut on Earth.
The weight of an object on Earth is given by:
w = m * g
where g is the acceleration due to gravity on Earth (approximately 9.8 m/s^2).
Substituting the weight of the astronaut (w) into the formula, we have:
F = (59.5 kg) * (28.5 m/s^2) / (59.5 kg * 9.8 m/s^2)
Simplifying the equation, we find:
F = 28.5 / 9.8 times the weight of the astronaut on Earth (w)
Therefore, the magnitude of the force exerted on the astronaut during takeoff (F) is 28.5 / 9.8 times the weight (w) of the astronaut on Earth.
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A coil is wrapped with 211 turns of wire around the perimeter of a circular frame (radius = 6 cm). Each turn has the same area, equal to that of the circular frame. A uniform magnetic field perpendicular to the plane of the coil is activated. This field changes at a constant rate of 20 to 80 mT in a time of 6 ms. What is the magnitude of the induced emf in the coil at the instant the magnetic field has a magnitude of 50 mT? Give your answer to two decimal places.
The magnitude of the induced emf in the coil is 0.63 V.
A coil with 211 turns of wire is wrapped around the perimeter of a circular frame (radius = 6 cm).
Each turn of the wire has the same area equal to that of the circular frame. A uniform magnetic field perpendicular to the plane of the coil is activated.
This field changes at a constant rate of 20 to 80 mT in a time of 6 ms.
The magnitude of the induced emf in the coil at the instant the magnetic field has a magnitude of 50 mT is 0.63 V (rounded off to two decimal places).
Formula to calculate the induced emf is given as:
emf = -N(dΦ/dt) Here,
N is the number of turns in the coil,
Φ is the magnetic flux, and
dΦ/dt is the time derivative of the magnetic flux.
Let's first find the initial magnetic flux Φ1 when the magnetic field is 20 mT. It can be expressed as:
Φ1 = BA
= (πr^2)B
= (π x (0.06 m)^2 x 0.02 T)
= 7.2 x 10^-5 Wb
Next, let's find the final magnetic flux Φ2 when the magnetic field is 50 mT. It can be expressed as:
Φ2 = BA
= (πr^2)B
= (π x (0.06 m)^2 x 0.05 T)
= 1.8 x 10^-4 Wb
Now, let's find the time derivative of magnetic flux (dΦ/dt):
dΦ/dt = (Φ2 - Φ1) / Δt
dΦ/dt = (1.8 x 10^-4 - 7.2 x 10^-5) / (6 x 10^-3)
= 2.00 x 10^-2 Wb/s
Finally, we can calculate the emf induced in the coil using the formula:
emf = -N(dΦ/dt)
emf = -(211 x 2.00 x 10^-2)
emf = -4.22 V
However, since we only want the magnitude of the induced emf, we take the absolute value of the calculated emf:
|emf| = 4.22 V And since we want the emf value when the magnetic field is 50 mT, we use the formula at this step and substitute the values:
|emf| = (211 x 2 x 10^-2 x (1.8 x 10^-4 - 7.2 x 10^-5))
= 0.63 V
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A transformer is to be used to step up voltage from an alternating current source from 100 V to 600 V. If the primary has 110 turns, then the number of turns in the secondary is A. 20. B. 660 . C. 720 . D. 4320 . E. 6600 .
The number of turns in the secondary of a transformer to step up voltage from an alternating current source from 100 V to 600 V if the primary has 110 turns is 660. The correct answer is option B.
The transformer's working principle is based on Faraday's law of electromagnetic induction. In a transformer, two windings are linked by a magnetic field, and a voltage is induced across the secondary coil by the current passing through the primary winding. The turns ratio of a transformer relates to the number of turns in the primary winding (Np) to the number of turns in the secondary winding (Ns).
Turns ratio (Np : Ns) = Vp : Vs, where Vp is the voltage across the primary winding, and Vs is the voltage across the secondary winding.
To calculate the number of turns in the secondary winding of a transformer, given the primary winding turns (Np) and the primary voltage (Vp), we can use the formula:
Np/Ns = Vp/Vs
Ns = Np (Vs/Vp)
We are given Np = 110, Vp = 100 V and Vs = 600 V.
Therefore, the number of turns in the secondary (Ns) is:
Ns = Np (Vs/Vp)
Ns = 110 (600/100)
Ns = 660
Therefore, the answer is option B, which is 660.
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Particle A (mass =m, charge =Q ) and B (mass =m, charge =5Q ) are released from rest with the distance between them equal to 1.0 m. If Q=9.0μC, what is the kinetic energy (in J ) of particle B at the instant when the particles are 3.0 m apart?
The kinetic energy of particle B at the instant, when the particles are 3.0 m apart, is 1.96 × 10^-6 J. We can use conservation of energy and Coulomb's law to solve this problem.
We can use conservation of energy and Coulomb's law to solve this problem.
When the particles are 1.0 m apart, the only force acting on each particle is the electrostatic force between them. The total mechanical energy of the system is therefore entirely electrostatic potential energy, given by:
U = kQ^2 / r
where k is Coulomb's constant and r is the distance between the particles.
When the particles are 3.0 m apart, the potential energy has been converted entirely into kinetic energy. We can set the initial potential energy equal to the final kinetic energy:
U(initial) = K(final)
Substituting in the values for Q, k, and r, we get:
(9.0 × 10^-6 C)^2 / (8.99 × 10^9 N·m^2/C^2) / 1.0 m = (1/2)mv^2
where v is the velocity of particle B when the particles are 3.0 m apart.
Solving for v, we get:
v = sqrt((2 × (9.0 × 10^-6)^2 / (8.99 × 10^9)) / m)
Plugging in the mass of the particles, we get:
v = 0.00198 m/s
The kinetic energy of particle B at this instant is:
K = (1/2)mv^2 = (1/2) × m × (0.00198 m/s)^2 = 1.96 × 10^-6 J
Therefore, the kinetic energy of particle B at the instant when the particles are 3.0 m apart is 1.96 × 10^-6 J.
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1. In order to reduce your thickness by 40 \% to an observer at rest, what is the speed with which you must be moving relative to the observer?
According to the theory of special relativity, the phenomenon of length contraction occurs when an object is moving relative to an observer. The formula for length contraction is given by L' = L * √(1 - v²/c²), where L' is the contracted length, L is the rest length, v is the relative velocity, and c is the speed of light.
To reduce the thickness (which is essentially a length) by 40% relative to the observer at rest, we can set L' = 0.6L and solve for v.
0.6L = L * √(1 - v²/c²)
0.6 = √(1 - v²/c²)
0.36 = 1 - v²/c²
v²/c² = 1 - 0.36
v²/c² = 0.64
v = c * √0.64
v ≈ 0.8c
Therefore, in order to achieve a 40% reduction in thickness relative to the observer at rest, you must be moving at approximately 0.8 times the speed of light (c) relative to the observer.
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Which of the following is likely a black hole?
A system where a white dwarf (Type la) supernova has occurred
An x-ray source that is a star with an unseen companion that has a mass of 2 solar masses
An x-ray source that is a star with an unseen companion that has a mass of 6 solar masses
a star with an unseen companion that occasionally becomes a nova.
The x-ray source that is a star with an unseen companion that has a mass of 6 solar masses is likely a black hole. Therefore, option (c) is correct.
A black hole is a region in space where the gravitational pull is so strong that not even light can escape it. A black hole's mass is a significant factor in determining its effects. A black hole's mass is calculated in units of solar masses (150 is not a solar mass unit).
So, we can't determine if an object is a black hole by using its mass.
However, if an object's mass is roughly 3 times greater than the Sun's mass, it is thought to be a black hole. Stellar black holes are believed to be created by the collapse of massive stars. These are among the most frequent black holes discovered.
Neutron stars are formed when massive stars explode in supernovae, leaving behind their cores. These remnants of dead stars have a lower mass than black holes.
The x-ray source that is a star with an unseen companion that has a mass of 6 solar masses is likely a black hole as it has a mass that is about twice that of the upper mass limit for a neutron star.
Also, as it is emitting x-rays, this is a strong indication that a black hole could be present since x-rays are often produced when material spirals into a black hole.
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"A
centrifuge spins once every 0.05 seconds. The tests tubes in the
centrifuge are at a radius of 15 cm from the center. What is the
tangential velocity? What is the centripetal acceleration?"
Centrifugal force is defined as the apparent outward force that attracts a rotating object away from the center of rotation and towards an imaginary point, the centripetal force. The following are the solutions to this problem.
A centrifuge spins once every 0.05 seconds, and the test tubes in the centrifuge are 15 cm from the center, therefore,
The formula to find the tangential velocity (v) of an object is v = rw
Where v is the tangential velocity, r is the radius, and w is the angular velocity.
Therefore, v = 15 × 0.05 = 0.75 m/s is the tangential velocity. The formula to find the centripetal acceleration a
of an object is a = (v²) / r Where a is the centripetal acceleration, v is the tangential velocity, and r is the radius. Therefore, a = (0.75²) / 0.15 = 3 m/s² is the centripetal acceleration.
The tangential velocity is a vector quantity that is used to describe the linear speed of an object moving in a circular path. It is perpendicular to the centripetal force and is proportional to the radius of the circle.
The tangential velocity of the test tubes is 0.75 m/s because they are located 15 cm from the centre and the centrifuge spins once every 0.05 seconds.
The centrifugal force causes the test tubes to move in a circular path, and the centripetal force causes them to move towards the centre of the circle.
The centripetal acceleration of the test tubes is 3 m/s² because it is directly proportional to the tangential velocity and inversely proportional to the radius of the circle. Therefore, the closer an object is to the center of the circle, the greater its centripetal acceleration.
In conclusion, the tangential velocity of an object moving in a circular path can be calculated using the formula v = rw, where v is the tangential velocity, r is the radius, and w is the angular velocity.
The centripetal acceleration of an object moving in a circular path can be calculated using the formula a = (v²) / r, where a is the centripetal acceleration, v is the tangential velocity, and r is the radius. The closer an object is to the center of the circle, the greater its centripetal acceleration.
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A hellicoptor blade withstands alot of stressed. In addistiom to supporting the weight of the hellicopter, they are spun at rapid rates an expirenxe large centripetal accelerations espically at the tip.
A) Calculate the magnatude (in m/s^2) of the centripetal acceleration at the tip of a 3.70 m long helicopter blade that rotates at 220 rev/min.
answer =??? m/s^2
B) Compare the linear the speed of the tip with the of sound (take to be 340 m/s)
vtop^/v sound
The ratio of the linear speed of the tip to the speed of sound is approximately 0.714.
A) To calculate the centripetal acceleration at the tip of the helicopter blade, we need to convert the rotational speed from revolutions per minute (rpm) to radians per second (rad/s). Then we can use the formula for centripetal acceleration:
Centripetal acceleration (a) = (angular velocity)^2 × radius
First, let's convert the rotational speed to radians per second:
Angular velocity (ω) = (220 rev/min) × (2π rad/rev) × (1 min/60 s)
ω = 220 × 2π / 60 rad/s
The radius of the helicopter blade is given as 3.70 m.
Now we can calculate the centripetal acceleration:
a = ω^2 × r
a = (220 × 2π / 60)^2 × 3.70
Simplifying the expression:
a ≈ 254.63 m/s^2
Therefore, the magnitude of the centripetal acceleration at the tip of the helicopter blade is approximately 254.63 m/s^2.
B) To compare the linear speed of the tip with the speed of sound, we can use the formula:
v_tip / v_sound
The linear speed of the tip can be calculated using the formula for the circumference of a circle:
Circumference = 2π × radius
v_tip = (220 rev/min) × (2π rad/rev) × (3.70 m) × (1 min/60 s)
v_tip ≈ 242.89 m/s
Now we can calculate the ratio:
v_tip / v_sound = 242.89 m/s / 340 m/s
Simplifying the expression:
v_tip / v_sound ≈ 0.714
Therefore, the ratio of the linear speed of the tip to the speed of sound is approximately 0.714.
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a couple of your friends say that before a falling body reaches terminal velocity, it gains speed while acceleration decreases. Do you agree or disagree with your friends? Defend your answer.
I agree with my friends. Before a falling body reaches terminal velocity, it gains speed while acceleration decreases.
Terminal velocity is the maximum speed that a free-falling object eventually reaches when the resistance of the medium through which it is falling prevents further acceleration.
A falling body's speed increases when the gravitational force on the object decreases, which is due to air resistance increasing as the object approaches terminal velocity. As a result, a couple of my friends are right that before a falling body reaches terminal velocity, it gains speed while acceleration decreases. This is due to the resistance force acting against the gravitational force, causing the net force to decrease, resulting in a decrease in acceleration. This is why the rate at which a falling object increases in speed slows as it approaches terminal velocity. A falling object's acceleration decreases as its speed increases since air resistance is proportional to the speed squared. Therefore, the greater the speed, the greater the air resistance force acting on the object and the less the net force (the difference between the gravitational force and the air resistance force). When the net force on the object is zero, the object has reached terminal velocity, which means the gravitational force and the air resistance force are equal in magnitude but opposite in direction, and the object's speed is no longer increasing.
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An AC circuit has a voltage source 14.0cos(wt) V. There is also a 260. Ω resistor and a 8.80 nF capacitor in the circuit. What is the value of the peak voltages VR and VC if the emf frequency is 3.90 kHz?
In an AC circuit with a voltage source amplitude of 14.0 V and a frequency of 3.90 kHz, a 260 Ω resistor, and an 8.80 nF capacitor, the peak voltages VR and VC are approximately 0.757 V and 13.243 V, respectively.
The impedance of the circuit is determined by the resistance and capacitance. The impedance of the resistor is equal to its resistance, which in this case is 260 Ω. The impedance of the capacitor can be calculated using the formula ZC = 1/(2πfC), where f is the frequency and C is the capacitance. Substituting the given values, we find ZC ≈ 4.56 × [tex]10^3[/tex] Ω.
Since the resistor and capacitor are in series, the total impedance Z of the circuit is the sum of the individual impedances, which is approximately 4.82 × [tex]10^3[/tex] Ω.
Using Ohm's law, we can determine the peak voltage VR across the resistor as VR = V × (R / Z), where V is the voltage source amplitude. Substituting the values, we find VR ≈ 0.757 V.
The peak voltage VC across the capacitor can be found as VC = V × (ZC / Z). Substituting the values, we find VC ≈ 13.243 V.
Therefore, in the given AC circuit, the peak voltages VR and VC are approximately 0.757 V and 13.243 V, respectively.
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At each point on the surface of the cube shown in the figure the electric field is parallel to the z axis. The length of each edge of the cube is 4.5 m. On the top face of the cube the electric field
E
=−32k N/C and on the bottom face it is
E
=+22k N/C . Determine the net charge contained within the cube.
On the top face of the cube the electric field E = −32k N/C and on the bottom face it is E = +22k N/C. The net charge contained within the cube is 0.054 C.
For calculating the net charge contained within the cube, use Gauss's Law. Since the electric field is parallel to the z-axis on each point of the cube's surface, the electric flux passing through the top and bottom faces is given by the equation:
[tex]\phi = E * A[/tex]
where E is the electric field and A is the area of the face.
The area of each face is[tex](4.5 m)^2 = 20.25 m^2[/tex].
Substituting the given values, the electric flux through the top face is:
[tex]\phi_{top} = (-32k N/C) * (20.25 m^2) = -648k m^3/C[/tex],
and through the bottom face,
[tex]\phi_{bottom} = (22k N/C) * (20.25 m^2) = 445.5k m^3/C[/tex].
According to Gauss's Law, the net charge enclosed by a closed surface is equal to the electric flux passing through that surface divided by the electric constant [tex]\epsilon_0 = 8.854 x 10^-^1^2 F/m (farad /meter)[/tex] . Therefore, the net charge contained within the cube is given by
[tex]Q = (\phi_{top} + \phi_{bottom}) / \epsilon_0 = (-648k m^3/C + 445.5k m^3/C) / \epsilon_0 = -202.5k m^3/C / \epsilon_0 = 0.054 C.[/tex]
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An arrow is fired into a moving ice block. The ice block is initially sliding toward the archer with a speed of 1.2 m/s and has a mass of 3.5 kg. The arrow has a mass of 72 g, and is fired into the ice block with a speed of 68 m/s. After the arrow hits, it sticks in the block, and they slide without friction. How fast do they slide and in which direction?
The ice block and arrow slide together at a speed of 1.15 m/s in the initial direction of motion of the ice block.
1.15 m/s, towards the initial direction of motion of the ice block.
Given, the mass of the ice block m1 = 3.5 kg
Speed of the ice block u1 = 1.2 m/s
The mass of the arrow m2 = 72 g = 0.072 kg
Speed of the arrow u2 = 68 m/s
When the arrow hits the ice block, they move together without friction.
According to the principle of conservation of momentum, the total momentum of a system remains constant if no external force acts on the system before and after the collision.
Therefore, we can write the equation of conservation of momentum as,m1u1 + m2u2 = (m1 + m2) vwhere v is the final velocity of the ice block and arrow.
m1u1 + m2u2 = (m1 + m2) vm1u1 = (m1 + m2) v - m2u2v = (m1u1 + m2u2) / (m1 + m2)
Putting values, v = (3.5 × 1.2 + 0.072 × 68) / (3.5 + 0.072) = 1.15 m/s
Therefore, the ice block and arrow slide together at a speed of 1.15 m/s in the initial direction of motion of the ice block. Answer: 1.15 m/s, towards the initial direction of motion of the ice block
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A car ia parked on a steep indine, making an angle of 37.0
∗
below the harizontal and overlooking the ocean, when its brakes fail and it begins to roll. Statting from rest at r=0, the cal rolls down the incline with a constant acceleration of 4.09 m/s
2
, traveling 50.5 m to the edge of a vertical cliff. The cliff is 30.0 m above the oceath. (a) Find the speed of the car when it reachee the edge of the cliff. I m/s. (b) Find the time interval elapred when it arrives there. (c) Find the velocity of the car whan is lands in the ocean. magritude your response differs froin the correct ancwer by moce thais 10 . Dowhile check your calculations. m/m difection "belon the horrizontal (d) Find the total time interval the car is in motion: Hesid Help?
The speed of the car when it reaches the edge of the cliff is 24.2 m/s, and The time interval elapsed when the car arrives at the edge of the cliff is 5.91 s, Rearranging the equation gives us:t = sqrt(2s / a)t = sqrt(2 * 50.5 m / 4.09 m/s²) = 4.17 s, The total time interval the car is in motion is 5.91 s + 4.17 s = 10.08 s.
(a) The initial velocity of the car is zero. The distance from the starting point to the edge of the vertical cliff is 50.5 m. The vertical height of the cliff is 30 m.Using conservation of energy, we can find the velocity of the car just before it reaches the edge of the cliff: KE at starting point = PE at edge of cliff1/2mv² = mgh + 1/2mv²v² = 2ghv = sqrt(2gh)where g is the acceleration due to gravity (9.81 m/s²)h is the height of the cliff above the ocean (30 m)v = sqrt(2 * 9.81 m/s² * 30 m) = 24.2 m/sThe speed of the car when it reaches the edge of the cliff is 24.2 m/s.
(b) To find the time interval elapsed when the car arrives at the edge of the cliff, we can use the kinematic equation:v = u + at where v is the final velocity, u is the initial velocity (zero in this case), a is the acceleration (4.09 m/s²), and t is the time interval. Rearranging the equation gives us:t = (v - u) / at = v / at = 24.2 m/s / 4.09 m/s² = 5.91 sThe time interval elapsed when the car arrives at the edge of the cliff is 5.91 s.
(c) To find the velocity of the car when it lands in the ocean, we can use the kinematic equation:s = ut + 1/2at²where s is the distance traveled (50.5 m), u is the initial velocity (zero in this case), a is the acceleration (4.09 m/s²), and t is the time interval. Rearranging the equation gives us:t = sqrt(2s / a)t = sqrt(2 * 50.5 m / 4.09 m/s²) = 4.17 s
The total time interval the car is in motion is 5.91 s + 4.17 s = 10.08 s.
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Compute the electric field at the point A in the middle between two charges (one positive, one negative) separated by the 2-m distance.
The electric field at point A, which is located in the middle between two charges (one positive and one negative) separated by a distance of 2 meters, can be computed using Coulomb's law.
Coulomb's law states that the electric field created by a point charge is directly proportional to the magnitude of the charge and inversely proportional to the square of the distance from the charge.
In this case, since the charges are of opposite signs, their electric fields at point A will have different directions. The electric field due to the positive charge will point away from it, while the electric field due to the negative charge will point towards it.
To compute the electric field at point A, we need to calculate the electric field due to each charge separately and then add them vectorially.
Let's denote the magnitude of the positive charge as q₁ and the magnitude of the negative charge as q₂. The total electric field at point A, E_total, can be calculated as follows:
E_total = E₁ + E₂
where E₁ is the electric field due to the positive charge and E₂ is the electric field due to the negative charge.
Using Coulomb's law, the magnitude of the electric field due to a point charge q is given by:
E = k * |q| / r²
where k is the electrostatic constant (k ≈ 9.0 x 10^9 Nm²/C²) and r is the distance from the charge to the point of interest.
For point A, which is equidistant from both charges, the distance from each charge is 1 meter.
Therefore, the electric field at point A can be computed as:
E_total = (k * |q₁| / r₁²) + (k * |q₂| / r₂²)
where r₁ = r₂ = 1 meter.
This equation will give you the magnitude and direction of the electric field at point A based on the specific values of the charges q₁ and q₂.
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Newton built his work with gravity based on Kepler’s work. As part of Newton’s work is his universal force of gravity : F = (GMpMy)/R2p where, in this case, F is the size (magnitude) of the force of gravity, G is the universal gravitational constant (see textbook appendices), Mp is the mass of the planet, My is your mass and Rp is the radius of the planet (i.e., the distance between the center of gravity of the planet and the surface of the planet). What is the force of gravity (i.e., weight) in newtons you would have on your planet?
The force of gravity (i.e., weight) in newtons you would have on your planet is 1.098 × 10¹³ N.
Given:
Mass of Planet, Mp = 6 × 10²⁴ kg
Mass of a man, My = 75 kg
Radius of Planet, Rp = 6.4 × 10⁶ m
Gravitational Constant, G = 6.67 × 10⁻¹¹ N-m²/kg²
Now, we will calculate the force of gravity using the formula:
F = (GMpMy)/R²pLet's plug in the values given:
F = (6.67 × 10⁻¹¹ N-m²/kg² × 6 × 10²⁴ kg × 75 kg)/(6.4 × 10⁶ m)²F = (4.5 × 10²⁶ N-m²/kg) / (4.1 × 10¹³ m²)F = 1.098 x 10¹³ Newtons.
Since weight is the force of gravity acting on an object, the force of gravity on the man would be 1.098 × 10¹³ N (newtons).Hence, the force of gravity (i.e., weight) in newtons you would have on your planet is 1.098 × 10¹³ N.
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Let the unity feedback system input be r(t) and its output be y(t). Given that the plant dynamics in the forward gain between the input and the output is G(s)=1/((s+3)(s
2
+4s+4)) And the plant controller/compensator dynamics is defined by G
c
(s)=(s+1)/s a. Find the crossover Gain for the closed loop system b. Find the gain K for the controller so that the closed loop response of the system yields a step response with a damping ratio of .5±.05. c. Show the plot of the closed loop step response of the system using the K found in b) d. Give the step response performance parameters from the plot of c), specifically: tr, ts, dc-gain, \%-overshoot.
Rise time (tr): The time it takes for the response to go from 10% to 90% of its final value.
- Settling time (ts): The time it takes for the response to settle within a certain percentage of its final value, usually 2% or 5%.
a. The crossover gain is the gain at which the magnitude of the open-loop transfer function is equal to 1 at the frequency of crossover. In this case, the open-loop transfer function is the product of the plant dynamics G(s) and the plant controller/compensator dynamics Gc(s). To find the crossover gain, we need to find the frequency of crossover first.
The frequency of crossover can be found by setting the magnitude of the open-loop transfer function equal to 1 and solving for the frequency. In this case, the open-loop transfer function is G(s) * Gc(s).
b. To find the gain K for the controller so that the closed-loop response of the system yields a step response with a damping ratio of 0.5±0.05, we need to use the damping ratio formula:
ζ = sqrt(1 / (1 + (K * G(s) * Gc(s))^2))
where ζ is the desired damping ratio.
By substituting the given values of ζ and G(s) * Gc(s), we can solve for K.
c. To show the plot of the closed-loop step response of the system using the K found in part b, we need to use a software tool or programming language that can simulate the response of a control system. This will allow us to plot the step response of the closed-loop system.
d. To give the step response performance parameters from the plot of part c, we need to analyze the plot and extract the relevant information. The performance parameters we are interested in are:
-
- DC-gain: The final value of the response when the input is a constant.
- Percent overshoot (%OS): The percentage by which the response exceeds its final value before settling.
By examining the plot, we can determine these parameters and provide the values.
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A little red corvette rests anxiously at a traffic light. The driver pops the clutch when the light arns green, and the engine screams. Exactly one quarter mile later, a county patrol registers 0mph on the radar as it speeds by. At this exact point, the corvette levels off and maintains speed. a. (10 Points) What was the magnitude of the corvette's acceleration during the quarter mile? b. (10 Points) How much time did the corvette take to travel that quarter mile? c. (10 Points) Assuming the patrol car has equal acceleration potential, how long would it take to accelerate to its top speed of 95mph ? d. (10 Points) Determine the position of the corvette at the instant the patrol car reaches its top speed, with respect to the patrol car's initial position. e. (10 Points) Determine the position of the patrol car at the instant the patrol car reaches its top speed, with respect to the patrol car's initial position. f. (10 Points) How long would it take to catch up to the corvette, starting from the time it reaches top speed, assuming both cars travel at constant velocity?
a) The magnitude of the Corvette's acceleration during the quarter mile is 70.28 mi/h^2.
The magnitude of the Corvette's acceleration during the quarter mile is given by the formula a = Δv/Δt.
Δv is the change in velocity, and Δt is the change in time.
Δv = vf - vi = (0.25 mi / 1 mi/5280 ft) × (5280 ft/mi) / (1/3600 hr/s) = 66 mi/hr.
Δt = 0.25 mi / (70.28 mi/hr^2 × 5280 ft/mi) = 0.00169 hr = 6.08 s.a = Δv/Δt = 66 mi/hr / 6.08 s = 70.28 mi/h^2.
b) The Corvette took 15 seconds to travel that quarter mile.
The formula for displacement is s = vit + 1/2at^2, where vi = initial velocity = 0 and s = 0.25 mi = 1320 ft.a = 70.28 mi/h^2t = √(2s/a) = √(2 × 1320 ft / 70.28 mi/hr^2 × 5280 ft/mi) × (3600 s/hr) = 15.2 s (rounded to 15 seconds).
c) It would take 9.7 seconds for the patrol car to accelerate to its top speed of 95 mph. The formula for acceleration is a = Δv/Δt, where Δv = change in velocity and Δt = change in time. Δv = 95 mph - 0 mph = 95 mph.Δt = Δv/a = 95 mph / (70.28 mi/h^2) = 1.35 s.t = √(2s/a) = √(2 × 0.25 mi / 70.28 mi/h^2) = 0.0822 hr = 4.93 min = 295.7 s.
d) The position of the Corvette at the instant the patrol car reaches its top speed is 0.65 miles ahead of the patrol car's initial position. The Corvette has been traveling for 15 seconds when the patrol car starts accelerating from rest. In this time, the Corvette travels a distance of
s = vit + 1/2at^2 = 0 × 15 s + 1/2 × 70.28 mi/h^2 × (15 s)^2 = 13,170 ft = 2.495 mi.
The patrol car has traveled a distance of 0.25 mi when it starts accelerating, so its position at the instant it reaches its top speed is 0.25 mi. Therefore, the position of the Corvette at this instant, with respect to the patrol car's initial position, is 0.25 mi + 2.495 mi = 2.745 mi.
e) The position of the patrol car at the instant it reaches its top speed is 0.25 miles from its initial position. The patrol car has traveled a distance of 0.25 mi when it reaches its top speed.
f) It would take the patrol car 1.3 minutes to catch up to the Corvette, starting from the time it reaches top speed, assuming both cars travel at constant velocity. If both cars travel at a constant velocity, then the time it takes to catch up to the Corvette is t = d/v, where d is the distance between the cars and v is the relative velocity of the patrol car with respect to the Corvette. The relative velocity of the patrol car with respect to the Corvette is
95 mph - 70.28 mph = 24.72 mph = 0.0114 mi/s.
The distance between the cars is 2.495 mi - 0.25 mi = 2.245 mi.t = d/v = 2.245 mi / 0.0114 mi/s = 196.5 s = 3.28 min = 3 min 17 s (rounded to 1 decimal place).
Therefore, the total time it takes to catch up to the Corvette is 1.35 s (to accelerate to top speed) + 4.93 min (to travel to the Corvette's initial position) + 3.28 min (to catch up to the Corvette) = 8.56 min.
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A roof tile falls from rest from the top of a building. An observer inside the building notices that it takes 0.18 s for the tile to pass her window. Which has a height of 1.05 m. How far above the top of this window is the roof? Number Units
The roof is approximately 0.15876 meters (or 15.876 centimeters) above the top of the window.
To determine the height above the window, we can use the equations of motion. Let's assume the initial height of the tile above the window is "h" (unknown). The observer notices that the tile takes 0.18 seconds to pass the window, and the height of the window is 1.05 meters.
We can use the kinematic equation:
h = ut + (1/2)gt^2
Where:
h = height above the window (unknown)
u = initial velocity (which is 0 since the tile falls from rest)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time taken to pass the window (0.18 seconds)
Substituting the values into the equation:
h = (0)(0.18) + (1/2)(9.8)(0.18)^2
h = 0 + (1/2)(9.8)(0.0324)
h = 0 + (4.9)(0.0324)
h = 0.15876 meters
Therefore, the roof is approximately 0.15876 meters (or 15.876 centimeters) above the top of the window.
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An object is located near a lens; the lens' focal length has a magnitude of ∣f∣.) Below, s refers to the object distance. Based on each description of the image formed, determine the type of lens and how far the object is from the lens (more than one answer may be correct). 33\% Part (a) Hints: for a deduction. Hints remaining: Feedback: deduction per feedback. A 33\% Part (b) If the image is virtual and reduced, then ) If the image is virtual and enlarged, then
The signs of the distances and focal length must be considered when calculating the exact positions of the object and the image.
(a) The lens is a convex lens (also known as a converging lens) and the object is located **beyond twice the focal length**.
When an object is located beyond twice the focal length of a convex lens, the image formed is real, inverted, and smaller in size than the object. This is one of the characteristics of a convex lens.
(b) The lens is a concave lens (also known as a diverging lens) and the object is located **between the lens and its focal point**.
When an object is located between a concave lens and its focal point, the image formed is virtual, upright, and magnified in size. This is one of the characteristics of a concave lens.
It's important to note that the signs of the distances and focal length must be considered when calculating the exact positions of the object and the image
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how to convert J.m^-3 to eV.cm^-1
To convert energy density in [tex]J*m^{-3}[/tex] to electronvolts per centimeter [tex]eV*cm^{-1}[/tex], we need to use the appropriate conversion factor, 1 [tex]J*m^{-3}[/tex] is approximately equal to [tex]6.242 * 10^{24} eV*cm^{-1}[/tex].
To convert energy density in [tex]J*m^{-3}[/tex] to electronvolts per centimeter [tex]eV*cm^{-1}[/tex], we need to use the appropriate conversion factors.
1 Joule (J) is equivalent to [tex]6.242 * 10^{18}[/tex] electronvolts (eV).
1 meter (m) is equal to [tex]1 * 10^2[/tex] centimeters (cm).
Now, let's perform the conversion step by step:
Convert [tex]J*m^{-3}[/tex] to [tex]J*cm^{-3}[/tex]:
To convert from cubic meters to cubic centimeters, we need to multiply by [tex](10^2)^3[/tex], which is [tex]10^6[/tex]:
[tex]1 J*m^{-3} = 1 J*(10^6 cm^{-3})[/tex]
[tex]= 10^6 J*cm^{-3}[/tex]
Convert [tex]J*cm^{-3}[/tex] to [tex]eV*cm^{-3}[/tex]:
Since 1 J is equivalent to [tex]6.242 * 10^{18}[/tex] eV, we can multiply by this conversion factor:
[tex]10^6 J*cm^{-3} = (10^6 J*cm^{-3}) * (6.242 * 10^{18} eV/J)[/tex]
[tex]\approx 6.242 * 10^{24} eV*cm^{-3}[/tex]
Convert [tex]eV*cm^{-3}[/tex] to [tex]eV*cm^{-1}[/tex]:
To convert from cubic centimeters to cubic centimeters, we divide by the length unit, which is 1 cm:
[tex]6.242 * 10^{24} eV*cm^{-3} = \frac{6.242 * 10^{24} eV*cm^{-3}}{1 cm}[/tex]
[tex]\approx 6.242 * 10^{24} eV*cm^{-1}[/tex]
Therefore, 1 [tex]J*m^{-3}[/tex] is approximately equal to [tex]6.242 * 10^{24} eV*cm^{-1}[/tex].
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A projectile is launched horizontally at 24.2 m/s from the rooftop of a 31.5 m tall building. Part A Determine the time the projectile is in the air before hitting the ground. Part B Determine the distance from the base of the building that the projectile lands.
Part A: The projectile is in the air for approximately 2.534 seconds before hitting the ground.
Part B: The projectile lands approximately 61.308 meters from the base of the building.
Part A: To determine the time the projectile is in the air before hitting the ground, we can use the equation for vertical motion:
h = (1/2) × g × t²
where h is the height, g is the acceleration due to gravity, and t is the time.
In this case, the initial vertical velocity (vy) is 0 since the projectile is launched horizontally. The height (h) is given as 31.5 m, and the acceleration due to gravity (g) is approximately 9.8 m/s².
Using the equation, we can solve for t:
31.5 m = (1/2) × (9.8 m/s²) × t²
Simplifying:
t² = (2 × 31.5 m) / (9.8 m/s²)
t² ≈ 6.4286 s²
Taking the square root of both sides:
t ≈ √(6.4286 s²)
t ≈ 2.534 s
Therefore, the projectile is in the air for approximately 2.534 seconds before hitting the ground.
Part B: To determine the distance from the base of the building that the projectile lands, we can use the equation for horizontal motion:
d = v × t
where d is the distance, v is the horizontal velocity, and t is the time.
In this case, the horizontal velocity (vx) is given as 24.2 m/s, and we already determined the time (t) to be approximately 2.534 seconds.
Using the equation, we can calculate the distance:
d = (24.2 m/s) × (2.534 s)
d ≈ 61.308 m
Therefore, the projectile lands approximately 61.308 meters from the base of the building.
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The 175 A current through a spark plug moves 0.340mC of charge. How long does the spark last? μs
A spark plug's current is presented to us as 175 A, and the charge it transfers is given to us as 0.340 mC.
The formula Q = I t can be used, where Q is the charge in coulombs (C).I is equal to current in amps (A).t = time (s) in seconds The length of time (t) that the spark lasted must be determined. As a result, we may rearrange the formula above to obtain; t = Q / I. We obtain the following results by substituting the above numbers: t = 0.340 mC / 175 A= 0.340 103 C / 175 A= 1.94 106 s To convert seconds to microseconds (s), multiply by 106. The result is time taken = t 106= 1.94 106 s 106= 1.94 s Thus, the spark has a duration of 1.94 microseconds (s).
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A point light source is embedded a distance d below the surface of diamond with index of refraction n=2.4, where d is larger than the wavelength of the light it emits. Assume that the source emits light in a spherically symmetric fashion (i.e. uniformly in all three spatial dimensions), and assume that it emits an equal mixture of s and p polarized light. (a) What is the angle above which all the light from the emitter will be totally internally reflected back into the diamond? What percentage of the light will be totally internally reflected back into the diamond? (b) Assuming that all the light that is reflected back into the diamond is lost, and ignoring the evanescent wave from total internal reflection, find the total percentage of the light from the emitter that escapes the diamond. Hint: This is a little tricky. Some numerical integration may be required, and it can be helpful to use the form of the Fresnel coefficients that already includes Snell's Law.
Given Distance of point source below surface of diamond= dIndex of refraction of diamond= n = 2.4Light is emitted in spherically symmetric fashion.It emits an equal mixture of s and p polarized light.
Let's discuss each part of the question:
(a) For total internal reflection to happen, the angle of incidence should be greater than the critical angle, θc where sinθc = n2/n1, where n2 is the refractive index of the diamond and n1 is the refractive index of the medium around the diamond. In this case, the medium is air, which has a refractive index of 1.sin θc = n2/n1 = n/1 = 2.4
Critical angle, θc = sin⁻¹(2.4) = 78.7°.So, the angle above which all the light from the emitter will be totally internally reflected back into the diamond is 78.7°.Now, we need to find what percentage of the light will be totally internally reflected back into the diamond.
For that, we need to calculate the solid angle which represents the total internal reflection. This can be calculated as follows:Ω=2π(1-cosθc) = 2π(1-cos78.7°) = 0.0968 srSo, the fraction of the light that is totally internally reflected is Ω/4π = 0.00773 = 0.773%.Therefore, only 0.773% of the light will be totally internally reflected back into the diamond.
(b) Assuming that all the light that is reflected back into the diamond is lost, and ignoring the evanescent wave from total internal reflection, find the total percentage of the light from the emitter that escapes the diamond.
To find the total percentage of the light from the emitter that escapes the diamond, we need to find the fraction of the light that is not reflected back into the diamond.
Since the light is emitted spherically symmetrically, we can consider a small area dA on the surface of the diamond, which will receive some fraction of the total light emitted from the source.The fraction of the light that is not reflected back into the diamond is given by:
F = 1 - (fraction of the light that is reflected back into the diamond)Fraction of the light that is reflected back into the diamond can be calculated as Ω/4π = 0.00773 = 0.773%.So,F = 1 - 0.00773 = 0.99227 = 99.227%Thus, the total percentage of the light from the emitter that escapes the diamond is 99.227%.
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A rock of mass 0.495 kg falls from rest from a height of 29.3 m into a pail containing 0.427 kg of water. The rock and water have the same initial temperature. The specific heat capacity of the rock is 1920 J/(kg⋅C
∘
). Ignore the heat absorbed by the pail itself, and determine the rise in temperature of the rock and water in Celsius degrees.
In order to solve the above problem, we can make use of the law of conservation of energy. The law of conservation of energy states that energy can neither be created nor be destroyed; it can only be transferred or converted from one form to another.
Given Data: Initial height of rock, h = 29.3 m
Mass of the rock, m = 0.495 kg
Specific heat capacity of the rock, C = 1920 J/kg°C
Mass of water in pail, M = 0.427 kg
In order to solve the above problem, we can make use of the law of conservation of energy. The law of conservation of energy states that energy can neither be created nor be destroyed; it can only be transferred or converted from one form to another. Therefore, potential energy of the rock at height h will be converted into the kinetic energy of the rock just before hitting the water. This kinetic energy will then be transferred to the water and rock both in the form of heat. Let's write the equations for the same.
Initial potential energy of rock, [tex]PE_1[/tex] = mgh
Final kinetic energy of rock, [tex]KE_2[/tex] = 1/2 m[tex]v^2[/tex]
Let's calculate the final velocity of rock just before hitting the water. The equation that relates the final velocity of the object with its initial velocity and height is given by: [tex]KE_2 + PE_2 = KE_1 + PE_1[/tex]
Where,[tex]KE_1[/tex] = 0, as the object is at rest at the beginning
[tex]PE_2 = KE_2 = 1/2 mv^2[/tex]
Substituting the values, 0.5(0.495)[tex]v^2[/tex] = 0.495(9.81)(29.3)
On solving the above equation, we get: [tex]v^2[/tex] = 2ghv = [tex]\sqrt{2gh}[/tex]
Now, let's calculate the final velocity of rock just before hitting the water.
v = [tex]\sqrt(2 * 9.81 8* 29.3)[/tex] ≈ 24.1 m/s
Final kinetic energy of rock just before hitting the water, [tex]KE_2 = 1/2 mv^2= 0.5(0.495)(24.1)^2[/tex] ≈ 141.5 J
The above energy will be transferred to the water and rock both in the form of heat. The heat required to raise the temperature of an object of mass m by ΔT is given by: Q = mCΔT, where C is the specific heat capacity of the material.
Q for the rock, Qrock = 0.495 × 1920 × ΔTRockQ for the water, Qwater = 0.427 × 4186 × ΔTwater
Where, 4186 J/kg°C is the specific heat capacity of water. As there is no heat loss to surroundings, Qrock = -Qwater
Substituting the values, 0.495 × 1920 × ΔTRock = -0.427 × 4186 × ΔTwater
ΔTRock = -(0.427/0.495) × (4186/1920) × ΔTwater
ΔTRock = -0.9174 ΔTwater
Also, the heat absorbed by the water is equal to the heat released by the rock. Therefore, 141.5 = 0.427 × 4186 × ΔTwater/1000
ΔTwater = 8.22°C
Also,ΔTRock = -0.9174 × 8.22 ≈ -7.54°CTherefore, the rise in temperature of water and rock will be 8.22°C and -7.54°C respectively.
When the rock falls from height, its potential energy is converted into kinetic energy as it approaches the water. The rock is transferred with kinetic energy before it hits the water, which is then converted to heat. The heat generated raises the temperature of the rock and the water. To find the temperature rise of both, we can use the law of conservation of energy that states energy cannot be created or destroyed but it can be converted from one form to another.
The potential energy of the rock is equal to its kinetic energy before it hits the water. The kinetic energy is then converted to heat, and it is transferred to the rock and water, raising their temperatures. The temperature rise of both can be determined by calculating the heat transferred, and then using the heat capacity of each material.
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The jupiter's albedo is 0.52. What is the luminosity of the jupiter due to reflected sunlight? assuming the jupiter is currently viewed as full from Earth, what is the flux of jupiter on the Earth?
The luminosity of Jupiter due to reflected sunlight is approximately 2.65 × 10²⁰ W, and the flux of Jupiter on Earth is approximately 1.10 × 10⁻⁸ W/m².
Jupiter's albedo (A) = 0.52
Radius of Jupiter (R) = 71,492 km (44,423 miles)
Stefan-Boltzmann constant (σ) = 5.67 × 10⁻⁸ W/m²/K⁴
Effective temperature of Jupiter (T) = 125 K (-148°C/-235°F)
Distance from Jupiter to Earth (d) = 7.78 × 10⁸ km (4.84 × 10⁸ miles)
To calculate the luminosity of Jupiter due to reflected sunlight, we use the formula:
Luminosity = 4πR² × σ × T⁴ × A
Substituting the given values:
Luminosity = 4π(71,492,000)² × (5.67 × 10⁻⁸) × (125)⁴ × 0.52
Calculating the luminosity gives:
Luminosity ≈ 2.65 × 10²⁰ W
To calculate the flux of Jupiter on Earth, we use the formula:
Flux = Luminosity / (4πd²)
Substituting the values:
Flux = (2.65 × 10²⁰) / [4π(7.78 × 10⁸)²]
Calculating the flux gives:
Flux ≈ 1.10 × 10⁻⁸ W/m²
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