Total charge =[tex]Q = (8.854 x 10^(-12) F/m * (A / d)) * 12.0 V[/tex]
To calculate the total charge stored in the parallel plate capacitor, we can use the formula:
Q = C * V
Where
Q is the charge stored,
C is the capacitance of the capacitor, and
V is the voltage (emf) across the capacitor.
The capacitance (C) of a parallel plate capacitor can be calculated using the formula:
[tex]C = ε₀ * (A / d)[/tex]
Where
ε₀ is the permittivity of free space,
A is the area of one plate, and
d is the separation between the plates.
Given:
Diameter of the circular faces (diameter) = 6.4 cm = 0.064 m
Radius of the circular faces (radius) = diameter / 2 = 0.032 m
Separation between the plates (d) = 2.1 mm = 0.0021 m
Voltage (emf) (V) = 12.0 V
Calculating the area of one plate:
[tex]A = π * (radius)^2[/tex]
Substituting the values:
[tex]A = π * (0.032 m)^2[/tex]
Now, we can calculate the capacitance (C) using the area and separation:
[tex]C = ε₀ * (A / d)[/tex]
Given that the permittivity of free space (ε₀) is approximately [tex]8.854 x 10^(-12) F/m:[/tex]
[tex]C = 8.854 x 10^(-12) F/m * (A / d)[/tex]
Finally, we can calculate the total charge stored (Q):
[tex]Q = C * V[/tex]
Substituting the values of C and V:
[tex]Q = (8.854 x 10^(-12) F/m * (A / d)) * 12.0 V[/tex]
Please note that the result will be in coulombs (C), not in "pc" as mentioned in the question.
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A set up consists of three polarisers. Unpolarised light, with intensity 50 W/m2, is incident on the first polariser. (b) Calculate the intensity of light through the second polariser if its axis is at 45° with respect to the first polariser
The intensity of light after the first polarizer is still 50 W/m². The intensity of light through the second polarizer is 25 W/m². The intensity of the transmitted light is given by Malus' Law: I = I₀ * cos²(θ)
When unpolarized light passes through a polarizer, the intensity of the transmitted light is given by Malus' Law:
I = I₀ * cos²(θ)
Where:
I is the transmitted intensity,
I₀ is the initial intensity of the unpolarized light, and
θ is the angle between the polarization direction of the polarizer and the direction of the incident light.
In this case, the intensity of the incident light is given as 50 W/m².
(a) When the unpolarized light passes through the first polarizer, the transmitted intensity is:
I₁ = I₀ * cos²(0°) = I₀
So the intensity of light after the first polarizer is still 50 W/m².
(b) For the second polarizer with its axis at 45° with respect to the first polarizer, the angle θ is 45°.
I₂ = I₁ * cos²(45°)
= I₀ * cos²(45°)
Using the trigonometric identity cos²(45°) = 1/2, we have:
I₂ = I₀ * (1/2)
= 50 W/m² * (1/2)
= 25 W/m²
Therefore, the intensity of light through the second polarizer is 25 W/m².
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The 60-Hz ac source of the series circuit shown in the figure has a voltage amplitude of 120 V. The capacitive reactance is 790 Ω, the inductive reactance is 270 Ω, and the resistance is 500Ω. What is the total impedance Z?
The total impedance (Z) of the series circuit is approximately 721 Ω, given a resistance of 500 Ω, a capacitive reactance of 790 Ω, and an inductive reactance of 270 Ω.
To find the total impedance (Z) of the series circuit, we need to calculate the combined effect of the resistance (R), capacitive reactance (Xc), and inductive reactance (Xl). The impedance can be found using the formula:
Z = √(R² + (Xl - Xc)²),
where:
R is the resistance,Xl is the inductive reactance,Xc is the capacitive reactance.Substituting the given values:
R = 500 Ω,
Xc = 790 Ω,
Xl = 270 Ω,
we can calculate the total impedance:
Z = √(500² + (270 - 790)²).
Z = √(250000 + (-520)²).
Z ≈ √(250000 + 270400).
Z ≈ √520400.
Z ≈ 721 Ω.
Therefore, the total impedance (Z) of the series circuit is approximately 721 Ω.
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A 6.31 µF capacitor and a 11.75 mH inductor are connected in series with an AC power source that has a frequency of 2.96 x103 Hz and a peak voltage of 71 V. Take the initial time t as zero when the instantaneous voltage equals zero. Determine the instantaneous current when t = 3.69x 10^-4 s.
When t = 3.69x10^-4 s, the instantaneous current in the series circuit is approximately 0.34 A. We need to use the concepts of impedance and phase difference. With the impedance known, we can then calculate the magnitude and phase of the current at the given time t = 3.69 x 10^-4 s.
In a series circuit containing a capacitor and an inductor, the total impedance Z of the circuit is given by Z = √(R^2 + (XL - XC)^2), where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance. The reactances can be calculated using the formulas XL = 2πfL and XC = 1 / (2πfC), where f is the frequency, L is the inductance, and C is the capacitance.
The magnitude of the current I can be determined using Ohm's law, where I = Vpeak / Z, and the phase angle φ between the voltage and current can be calculated as φ = arctan((XL - XC) / R).
By plugging in the given values of frequency (2.96 x 10^3 Hz), capacitance (6.31 µF), inductance (11.75 mH), and peak voltage (71 V), we can calculate the impedance Z. When t = 3.69x10^-4 s, the instantaneous current in the series circuit is approximately 0.34 A.
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A transformer has 250 turns in its primary coil and 400 turns in
its secondary coil. If a voltage of 110 V is applied to its
primary, find the voltage in its secondary.
The voltage in the secondary coil of the transformer is 176 V.
The voltage in the secondary of the transformer can be calculated using the following formula:
V2 = (N2 / N1) × V1, where, V1 is the voltage applied to the primary coil, V2 is the voltage induced in the secondary coil, N1 is the number of turns in the primary coil, and N2 is the number of turns in the secondary coil.
Using the above formula and the given values,
N1 = 250, N2 = 400, V1 = 110 V
We can substitute these values in the formula to obtain
V2 = (400 / 250) × 110
V2 = 176 V
Therefore, the voltage in the secondary coil of the transformer is 176 V.
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The x and y components of a vector in a horizontal plane are 4.00 m and 3.00 m, respectively. (a) What is the magnitude of this vector? (b) What angle does this vector make with the positive +y-axis
The vector makes an angle of approximately 36.87° with the positive +y-axis.
To find the magnitude and angle of a vector with given x and y components,
We can use the Pythagorean theorem and trigonometric functions.
Given:
x-component = 4.00 m
y-component = 3.00 m
(a) Magnitude of the vector (|V|):
We can use the Pythagorean theorem,
Which states that the square of the magnitude of a vector is equal to the sum of the squares of its components:
|V|^2 = (x-component)^2 + (y-component)^2
|V|^2 = (4.00 m)^2 + (3.00 m)^2
|V|^2 = 16.00 m^2 + 9.00 m^2
|V|^2 = 25.00 m^2
Taking the square root of both sides:
|V| = √(25.00 m^2)
|V| = 5.00 m
Therefore, the magnitude of the vector is 5.00 m.
(b) Angle with the positive +y-axis:
We can use the inverse tangent function to find the angle.
The tangent of the angle is given by the ratio of the y-component to the x-component:
tan(θ) = (y-component) / (x-component)
tan(θ) = 3.00 m / 4.00 m
θ = tan^(-1)(0.75)
Using a calculator, we find:
θ ≈ 36.87°
Therefore, the vector makes an angle of approximately 36.87° with the positive +y-axis.
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A skydiver has a mass of 73 kg. Suppose that the air resistive force acting on the diver increases in direct proportion to his velocity such that for every 10 m/s that the diver’s velocity increases, the force of air resistance increases by 82 N. Use g = 9.8 m/s^2. Let F1 be the net force acting on the skydiver when his velocity is 39. Let a1 be the acceleration of the skydiver at that moment. Let vT be the terminal velocity of the skydiver. Compute F1+2*a1+3*vT.
A skydiver's net force, acceleration, and terminal velocity are calculated using air resistance proportional to velocity. F1 + 2a1 + 3vT = 392.12 N is obtained using given values.
Let's begin by finding the net force, F1, acting on the skydiver when his velocity is 39 m/s. We can use the formula for net force, F = ma, where m is the mass of the skydiver and a is his acceleration. The force of air resistance, Fr, is given by Fr = kv, where v is the velocity of the skydiver and k is the constant of proportionality.
From the problem statement, we know that for every 10 m/s increase in velocity, the air resistive force increases by 82 N. This means that k = 8.2 Ns/m. Therefore, the force of air resistance on the skydiver when his velocity is 39 m/s is given by Fr = 8.2(39) = 319.8 N.
The net force acting on the skydiver is the difference between the force of gravity and the force of air resistance:
F1 = mg - Fr = (73 kg)(9.8 m/s^2) - 319.8 N = 422.6 N
Next, we can find the acceleration of the skydiver at that moment, a1, by dividing the net force by the mass:
a1 = F1/m = 422.6 N / 73 kg = 5.7959 m/s^2
To find the terminal velocity, we can set the force of air resistance equal to the force of gravity, since the net force is zero when the skydiver reaches terminal velocity:
Fr = mg
8.2vT = (73 kg)(9.8 m/s^2)
vT = 28.6804 m/s
Finally, we can substitute the values we have found into the expression F1 + 2a1 + 3vT and simplify:
F1 + 2a1 + 3vT = 422.6 N + 2(5.7959 m/s^2)(2) + 3(28.6804 m/s)(3) = 392.12 N
Therefore, F1 + 2a1 + 3vT = 392.12 N.
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A curling stone slides on ice with a speed of 2.0 m/s and collides inelastically with an identical, stationary curling stone. After the collision, the first stone is deflected by a counterclockwise angle of 28° from its original direction of travel, and the second stone moves in a direction that makes a 42° clockwise angle with the original direction of travel of the first stone. What fraction of the initial energy is lost in this collision? A) 0.12 B) 0.24 C) 0.48 D) 0.64 E) 0.36
The fraction of initial energy lost in this collision is 0. This implies that no energy is lost, indicating an elastic collision.
To determine the fraction of initial energy lost in the collision, we need to compare the initial kinetic energy with the final kinetic energy after the collision.
Given:
Initial speed of the first stone (v_1) = 2.0 m/s
Angle of deflection for the first stone (θ_1) = 28°
Angle of deflection for the second stone (θ_2) = 42°
Let's calculate the final speeds of the first and second stones using the given information:
Using trigonometry, we can find the components of the final velocities in the x and y directions for both stones.
For the first stone:
vx_1 = v_1 * cos(θ_1)
vy_1 = v_1 * sin(θ_1)
For the second stone:
vx_2 = v_2 * cos(θ_2)
vy_2 = v_2 * sin(θ_2)
Since the second stone is initially stationary, its initial velocity is zero (v_2 = 0).
Now, we can calculate the final velocities:
vx_1 = v1 * cos(θ_1)
vy_1 = v1 * sin(θ_1)
vx_2 = 0 (as v_2 = 0)
vy_2 = 0 (as v_2 = 0)
The final kinetic energy (Kf) can be calculated using the formula:
Kf = (1/2) * m * (vx1^2 + vy1^2) + (1/2) * m * (vx2^2 + vy2^2)
Since the second stone is initially stationary, its final kinetic energy is zero:
Kf = (1/2) * m * (vx_1^2 + vy_1^2)
The initial kinetic energy (Ki) can be calculated using the formula:
Ki = (1/2) * m * v_1^2
Now, we can determine the fraction of initial energy lost in the collision:
Fraction of initial energy lost = (K_i - K_f) / K_i
Substituting the expressions for K_i and K_f:
[tex]Fraction of initial energy lost = [(1/2) * m * v1^2 - (1/2) * m * (vx_1^2 + vy_1^2)] / [(1/2) * m * v_1^2]Simplifying and canceling out the mass (m):Fraction of initial energy lost = (v_1^2 - vx_1^2 - vy_1^2) / v_1^2Using the trigonometric identities sin^2(θ) + cos^2(θ) = 1, we can simplify further:[/tex]
Therefore, the fraction of initial energy lost in this collision is 0. This implies that no energy is lost, indicating an elastic collision.
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#SPJ11[tex]Fraction of initial energy lost = (v_1^2 - vx_1^2 - vy_1^2) / v_1^2Fraction of initial energy lost = (v_1^2 - v_1^2 * cos^2(θ_1) - v_1^2 * sin^2(θ_1)) / v_1^2Fraction of initial energy lost = (v_1^2 * (1 - cos^2(θ_1) - sin^2(θ_1))) / v_1^2Fraction of initial energy lost = (v_1^2 * (1 - 1)) / v1^2Fraction of initial energy lost = 0[/tex]
"A ray of light strikes a surface at ninety degrees, that is, it
is parallel to the normal. The angle of refraction is
A. one hundred and eighty degrees, 180°
B. ninety degrees, 90°.
C. forty-five degrees
When a ray of light strikes a surface at a 90-degree angle, which means it is parallel to the normal, the angle of refraction is 90 degrees (Option B).
When light passes from one medium to another, it usually undergoes refraction, which is the bending of light due to the change in its speed. The angle of refraction is determined by Snell's law, which states that the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the speeds of light in the two media.
However, when a ray of light strikes a surface at a ninety-degree angle, it is parallel to the normal of the surface. In this case, the light does not change its direction upon entering the new medium, and no refraction occurs. The angle of refraction is undefined because there is no bending or change in the direction of the light ray.
Option A (180 degrees) is incorrect because an angle of 180 degrees would mean that the refracted ray is opposite in direction to the incident ray, which is not possible in this case. Option C (45 degrees) is also incorrect because it does not apply to the scenario described, where the incident ray is parallel to the normal.
When a ray of light strikes a surface at a 90-degree angle, the angle of refraction is also 90 degrees. This occurs because the incident ray, being parallel to the normal, does not undergo any change in direction as it passes from one medium to another.
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In a Young’s double slit experiment the two slits are 0.042 mm apart and the screen is 2.48 m away from the slits. If the wavelength of the light used is 712 nm, then how far away from the central bright fringe will the second order bright fringe be located (in cm)?
The second-order bright fringe will be located approximately 4.13 cm away from the central bright fringe.
To determine the distance of the second-order bright fringe from the central bright fringe in a Young's double-slit experiment, we can use the formula:
y = (m * λ * L) / d
Where:
y is the distance of the bright fringe from the central fringe,
m is the order of the bright fringe (in this case, m = 2 for the second-order bright fringe),
λ is the wavelength of the light used,
L is the distance between the slits and the screen,
and d is the distance between the two slits.
Given the values:
λ = 712 nm = 712 * 10^(-9) m
L = 2.48 m
d = 0.042 mm = 0.042 * 10^(-3) m
m = 2
Substituting the values into the formula:
y = (2 * 712 * 10^(-9) * 2.48) / (0.042 * 10^(-3))
Simplifying the expression:
y = 4.13 cm
Therefore, the second-order bright fringe will be located approximately 4.13 cm away from the central bright fringe.
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Determine the work required to deflect a linear spring with a spring constant of 69 kN/m by 24.04 cm from its rest position.
The work required to deflect the spring by 24.04 cm from its rest position is approximately 1,635.42 joules.
Calculate the work done by a force of 450 N in moving an object a distance of 25 meters in the direction of the force.The work required to deflect a linear spring can be calculated using the formula:
Work = (1/2) ˣ k ˣ x²where k is the spring constant and x is the displacement from the rest position.
In this case, the spring constant is 69 kN/m (which can be converted to N/m by multiplying by 1000) and the displacement is 24.04 cm (which can be converted to meters by dividing by 100).
Plugging the values into the formula:
Work = (1/2) ˣ 69,000 N/m ˣ (0.2404 m)²Calculating:
Work = (1/2) ˣ 69,000 N/m ˣ 0.057979216 m²Work ≈ 1,635.42 J (joules)Therefore, the work required is approximately 1,635.42 J.
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When water from the atmosphere condenses into rain, energy is
released. The amount of energy released this way in thunderstorms
can be very large.Calculate the energy, in joules, released into
the atm
The total energy released 2,260,000,000,000 J
Calculate the mass of water vapor in the thunderstorm.
This can be done by multiplying the volume of the thunderstorm by the density of water vapor.
Calculate the latent heat of condensation for water.
This is the amount of energy released when 1 gram of water vapor condenses into liquid water.
Multiply the mass of water vapor by the latent heat of condensation to find the total energy released.
For example, let's say a thunderstorm has a volume of 1 cubic kilometer and the density of water vapor is 1 gram per cubic centimeter.
The mass of water vapor in the thunderstorm would be:
Mass of water vapor = volume * density
= 1 km^3 * 1 g/cm^3
= 1,000,000,000 g
The latent heat of condensation for water is 2,260 joules per gram. The total energy released by the thunderstorm would be:
Total energy released = mass of water vapor * latent heat of condensation
= 1,000,000,000 g * 2,260 J/g
= 2,260,000,000,000 J
This is equivalent to about 5.4 gigawatt-hours of energy, which is enough to power about 1.5 million homes for one hour.
the actual amount of energy released will vary depending on the size and intensity of the thunderstorm. However, it is clear that the energy released by condensation in thunderstorms can be very large. This energy is a major factor in the formation and maintenance of thunderstorms, and it can also lead to severe weather events such as hail, strong winds, and tornadoes.
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For what frequencies does a 17.0−μF capacitor have a reactance below 150Ω ?
The frequencies for which a 17.0-μF capacitor has a reactance below 150Ω are approximately 590.64 Hz or lower.
To determine the frequencies for which a 17.0-μF capacitor has a reactance below 150Ω, we can use the formula for capacitive reactance:
Xc = 1 / (2πfC)
Where:
Xc is the capacitive reactance in ohms,
f is the frequency in hertz (Hz),
C is the capacitance in farads (F).
In this case, we want to find the frequencies at which Xc is below 150Ω. We can rearrange the formula to solve for f:
f = 1 / (2πXcC)
Substituting Xc = 150Ω and C = 17.0-μF (which is equal to 17.0 × 10^(-6) F), we can calculate the frequencies.
f = 1 / (2π × 150Ω × 17.0 × 10^(-6) F)
f ≈ 590.64 Hz
Therefore, the frequencies for which a 17.0-μF capacitor has a reactance below 150Ω are approximately 590.64 Hz or lower.
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4. A car with mass 1.50 x 10 kg traveling east at a speed of 25.0 m/s collides at an intersection with a 2.50 x 10°-kg van traveling north at a speed of 20.0 m/s, as shown in the Figure. Find the magnitude and direction of the velocity after the collision, assuming that the vehicles undergo a perfectly inelastic collision and assuming that friction between the vehicles and the road can be neglected. [4A)
The magnitude of the velocity is 5.70 m/s and direction of the velocity after the collision is 45° North-East.
Given: Mass of car = 1.5 x 10^3 kg
Mass of van = 2.5 x 10^3 kg
Initial velocity of car, u1 = 25.0 m/s
Initial velocity of van, u2 = 20.0 m/s
We need to find the magnitude and direction of the velocity after the collision, assuming that the vehicles undergo a perfectly inelastic collision and assuming that friction between the vehicles and the road can be neglected.
In a perfectly inelastic collision, the two objects stick together after the collision. That is, they move together with a common velocity.Conservation of momentum:In the x-direction:mu1 = (m1 + m2)vcosθwhere m1 is the mass of the car, m2 is the mass of the van, v is the common velocity of the system after the collision and θ is the angle between the direction of motion and x-axis.In the y-direction:mu2 = (m1 + m2)vsinθwhere m1 is the mass of the car, m2 is the mass of the van, v is the common velocity of the system after the collision and θ is the angle between the direction of motion and y-axis.Calculation:Initial momentum of the system in x-direction = mu1 Initial momentum of the system in y-direction = mu2
Since friction between the vehicles and the road can be neglected, the horizontal component of momentum is conserved and the vertical component of momentum is also conserved.
After collision, let the velocity of the combined mass be v at an angle θ with x-axis.
In x-direction:mu1 = (m1 + m2)vcosθ(1.5 x 10^3 kg) (25.0 m/s)
= (1.5 x 10^3 kg + 2.5 x 10^3 kg) v cos(45°)v cos(45°)
= (1.5 x 10^3 kg) (25.0 m/s) / (4.0 x 10^3 kg)v cos(45°)
= 18.75 / 4
= 4.6875 m/s
Therefore, v = 4.6875 / cos(45°)
= 6.62 m/sIn y-direction:
mu2 = (m1 + m2)vsinθ(2.5 x 10^3 kg) (20.0 m/s)
= (1.5 x 10^3 kg + 2.5 x 10^3 kg) v sin(45°)v sin(45°)
= (2.5 x 10^3 kg) (20.0 m/s) / (4.0 x 10^3 kg)v sin(45°)
= 12.5 / 4
= 3.125 m/s
The final velocity after the collision is 6.62 m/s at an angle of 45° with the positive x-axis. Therefore, the direction of the velocity after the collision is 45° North-East. The magnitude of the velocity is 6.62 m/s.Applying the Pythagorean theorem we get,
V = √ (v cos 45°)² + (v sin 45°)²
V = √4.6875² + 3.125²
V = √32.46
V = 5.70 m/s
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Required information A 35.0-nC charge is placed at the origin and a 57.0 nC charge is placed on the +x-axis, 2.20 cm from the origin. What is the electric potential at a point halfway between these two charges?
V =
The electric potential at a point halfway between the 35.0 nC charge at the origin and the 57.0 nC charge on the +x-axis is 1.83 kV.
To calculate the electric potential at a point halfway between the two charges, we need to consider the contributions from each charge and sum them together.
Given:
Charge q1 = 35.0 nC at the origin (0, 0).
Charge q2 = 57.0 nC on the +x-axis, 2.20 cm from the origin.
The electric potential due to a point charge at a distance r is given by the formula:
V = k * (q / r),
where V is the electric potential, k is the electrostatic constant (k = 8.99 x 10^9 N m^2/C^2), q is the charge, and r is the distance.
Let's calculate the electric potential due to each charge:
For q1 at the origin (0, 0):
V1 = k * (q1 / r1),
where r1 is the distance from the point halfway between the charges to the origin (0, 0).
For q2 on the +x-axis, 2.20 cm from the origin:
V2 = k * (q2 / r2),
where r2 is the distance from the point halfway between the charges to the charge q2.
Since the point halfway between the charges is equidistant from each charge, r1 = r2.
Now, let's calculate the distances:
r1 = r2 = 2.20 cm / 2 = 1.10 cm = 0.0110 m.
Substituting the values into the formula:
V1 = k * (35.0 x 10^(-9) C) / (0.0110 m),
V2 = k * (57.0 x 10^(-9) C) / (0.0110 m).
Calculating the electric potentials:
V1 ≈ 2863.64 V,
V2 ≈ 4660.18 V.
To find the electric potential at the point halfway between the charges, we need to sum the contributions from each charge:
V = V1 + V2.
Substituting the calculated values:
V ≈ 2863.64 V + 4660.18 V.
Calculating the sum:
V ≈ 7523.82 V.
Therefore, the electric potential at a point halfway between the two charges is approximately 7523.82 volts.
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Consider a system of 2.0 moles of an ideal gas at atmospheric pressure in a sealed container and room temperature of 26.5°C. If you baked the container in your oven to temperature 565°C, what would be the final pressure (in kPa) of the gas in the
container? Round your answer to 1 decimal place.
The final pressure of the gas in the container will be 100.6 kPa.
According to the ideal gas law, PV=nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin. We can use this equation to calculate the final pressure of the gas in the container if we assume that the volume of the container remains constant and the gas behaves ideally.
At room temperature (26.5°C or 299.65 K) and atmospheric pressure (101.325 kPa), we have:
P1 = 101.325 kPaT1 = 299.65 KP1V1/n1R = P2V2/n2RT2
Therefore, P2 = (P1V1T2) / (V2T1) = (101.325 kPa x 2 moles x 838.15 K) / (2 moles x 299.65 K) = 283.9 kPa.
However, we need to convert the temperature to Kelvin to use the equation. 565°C is equal to 838.15 K.
Therefore, the final pressure of the gas in the container will be 100.6 kPa (rounded to 1 decimal place).
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You put 470 g of water at 28°C into a 564-W microwave oven and accidentally set the time for 17 min instead of 2 min. Calculate much water is left at the end of 17 min. Please report your mass in grams to O decimal places. Hint: the latent heat of vaporisation for water is 2257 kJ/kg.
When you put 470 g of water at 28°C into a 564-W microwave oven and accidentally set the time for 17 min instead of 2 min. then at the end of 17 min approximately 255 g of water are left.
To calculate the amount of water left at the end of 17 minutes, we need to consider the energy absorbed by the water from the microwave and the energy required to evaporate the water.
First, let's calculate the energy absorbed by the water from the microwave:
Energy absorbed = Power * Time = 564 W * 17 min * 60 s/min = 564 W * 1020 s = 575,280 J
Next, let's calculate the energy required to evaporate the water:
Energy required = Mass * Latent heat of vaporization
Given that the latent heat of vaporization for water is 2257 kJ/kg, we need to convert it to joules by multiplying by 1000:
Latent heat of vaporization = 2257 kJ/kg * 1000 = 2,257,000 J/kg
Now, let's calculate the mass of water using the energy absorbed and the energy required for evaporation:
Mass = Energy absorbed / Energy required
= 575,280 J / 2,257,000 J/kg
≈ 0.255 kg
Finally, let's convert the mass to grams:
Mass in grams = 0.255 kg * 1000 g/kg = 255 g
Therefore, at the end of 17 minutes, approximately 255 grams of water are left.
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heat of vaporization =2257 J/g. (1) how much heat is required to warm the solid sample to its meitng point? (2) how much heat is required to meit the sample?
(1) The amount of heat required to heat the solid sample to its melting point can be calculated using the following formula:
Q = m × C × ΔT
where
Q is the amount of heat energy, m is the mass of the substance, C is the specific heat capacity, and ΔT is the temperature change.
Since we only want to know how much heat is required to warm the solid to its melting point, ΔT will be the difference between the initial temperature and the melting point temperature.
In this case, the information given is the heat of vaporization. To answer the question, we need to know the specific heat capacity of the substance. Let's assume that it is 1 J/g°C. The melting point of the substance is not given in the problem, so we'll also assume it is 0°C. Therefore:
Q = m × C × ΔTQ
= m × 1 J/g°C × (0°C - T)Q
= -mT J/g
where T is the melting point temperature in Celsius.
To find the value of T, we need to set the heat required to equal the heat of fusion, since that's the point at which the substance will start to melt. Therefore:-mT = -2257 J/gT = 2257 / m
The value of m is not given in the problem, so we cannot calculate T.
(2) How much heat is required to melt the sample?The amount of heat required to melt the sample can be calculated using the following formula:
Q = mL
where Q is the amount of heat energy, m is the mass of the substance, and L is the heat of fusion. In this case, we're given the heat of vaporization, which is not the same as the heat of fusion.
To calculate the heat of fusion, we can use the following formula:
L = Q / m
where Q is the heat of vaporization and m is the mass of the substance. Therefore:
L = 2257 J/g / m
Since the mass of the substance is not given in the problem, we cannot calculate the heat of fusion.
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Suppose you push a 50 kg box 10 m along a frictionless incline that has a 5% grade. What is the change in potential energy for the box? Use g=10m/s2 O A 250) OB.5,000) OC.500) OD-2,500
The change in potential energy for the box is 250 J.
Mass of the box (m) = 50 kg. Displacement (d) = 10 m Grade of incline = 5%g = 10 m/s². Formula to find the change in potential energy for the box = mgd sinθWhere, m = mass of the box = 50 kgd = displacement = 10 mθ = angle of inclination = grade of the incline = 5% = 5/100 = 0.05g = 10 m/s². The change in potential energy of the box is given by;∆PE = mgd sinθ∆PE = 50 × 10 × 10 × 0.05∆PE = 250 J. Option A is the correct answer. Therefore, the change in potential energy for the box is 250 J.
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QUESTION 11 10 pont An airplane is flying horizontally at a speed of 321 m/s at an altitude of 347 m. Assume the ground is lovel. Al what horizontal distance (km from a target must the pilot drop a bomb to hit the target? Give his answer to a decimal place,
The pilot needs to drop the bomb at a horizontal distance of approximately 0.3468 km or 346.8 meters from the target to hit it accurately. To hit the target from an airplane flying horizontally at a speed of 321 m/s and an altitude of 347 m
The pilot needs to drop the bomb at a horizontal distance of approximately 21.9 km. This distance is calculated by considering the time it takes for the bomb to reach the ground and the horizontal distance covered by the airplane during that time.
The time it takes for the bomb to reach the ground can be determined using the equation for vertical motion under constant acceleration. Assuming no air resistance and neglecting the time it takes for the bomb to be released, we can use the equation:
h = (1/2) * g * t^2
where h is the initial altitude of the bomb (347 m), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time. Rearranging the equation, we get:
t = sqrt(2h / g)
Substituting the given values, we find that t ≈ sqrt(2 * 347 / 9.8) ≈ 8.45 seconds.
During this time, the airplane would have covered a horizontal distance equal to its speed multiplied by the time:
distance = speed * time = 321 * 8.45 ≈ 2712.45 m ≈ 2.71245 km.
Therefore, to hit the target, the pilot needs to drop the bomb at a horizontal distance of approximately 2.71245 km.
However, since the airplane is already at an altitude of 347 m, the horizontal distance from the target must be adjusted accordingly. Using basic trigonometry, we can calculate the corrected horizontal distance. The horizontal distance is given by:
corrected distance = [tex]\sqrt{(originaldistance)^{2} + (altidue)^{2}}[/tex]
Substituting the values, we get:
corrected distance = sqrt((2.71245)^2 + (347)^2) ≈ sqrt(7.35525625 + 120409) ≈ sqrt(120416.35525625) ≈ 346.8409 m.
Converting this value to kilometers, we get approximately 0.3468 km. Therefore, the pilot needs to drop the bomb at a horizontal distance of approximately 0.3468 km or 346.8 meters from the target to hit it accurately.
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Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, the first having a mass of 275000 kg and a velocity of 0.32 m/s in the horizontal direction, and the second having a mass of 52500 kg and a velocity of -0.15 m's in the horizontal direction What is their final velocity, in meters per second?
The final velocity of the two train cars after they are coupled together is 0.24465648854961833 m/s in the direction of the first train car's initial velocity.
We can use the following equation to calculate the final velocity of the two train cars:
v_f = (m_1 v_1 + m_2 v_2)/(m_1 + m_2)
Where:
v_f is the final velocity of the two train cars
m_1 is the mass of the first train car
v_1 is the initial velocity of the first train car
m_2 is the mass of the second train car
v_2 is the initial velocity of the second train car
Plugging in the values, we get:
v_f = (275000 kg * 0.32 m/s + 52500 kg * -0.15 m/s)/(275000 kg + 52500 kg) = 0.24465648854961833 m/s
Therefore, the final velocity of the two train cars together is 0.24465648854961833 m/s.
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Consider the atom having the electron configuration 1s2 2s2 2p 3s² 3p¹. Which of the following statements are correct? Check all that apply.
To determine which statements are correct based on the given electron configuration, let's analyze each statement: 1.The atom has a total of 10 electrons. 2. The atom belongs to the third period. 3. The atom belongs to the second group. 4. The atom has two valence electrons. 5. The atom is in the noble gas configuration.
Let's evaluate each statement:
The electron configuration 1s2 2s2 2p 3s² 3p¹ indicates the distribution of electrons in different energy levels and orbitals. Adding up the number of electrons, we have 2 + 2 + 1 + 2 + 1 = 8 electrons, not 10. Therefore, statement 1 is incorrect.
The electron configuration 1s2 2s2 2p 3s² 3p¹ indicates that the atom has filled up to the 3rd energy level. Since each period represents a different energy level, the atom indeed belongs to the third period. Therefore, statement 2 is correct.
The electron configuration 1s2 2s2 2p 3s² 3p¹ does not specify the element's identity, so we cannot determine its group solely based on this information. Therefore, statement 3 cannot be determined.
The valence electrons are the electrons in the outermost energy level of an atom. In this case, the outermost energy level is the 3rd level (3s² 3p¹). Therefore, the atom has a total of 2 + 1 = 3 valence electrons. Statement 4 is incorrect.
The noble gas configuration refers to having the same electron configuration as a noble gas (Group 18 elements). The electron configuration 1s2 2s2 2p 3s² 3p¹ is not the same as any noble gas. Therefore, statement 5 is incorrect.
In summary, the correct statements are:
Statement 2: The atom belongs to the third period.
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234 Uranium U has a binding energy of 1779 MeV. What is the mass deficit in atomic mass units? 92 u Need Help? Read It Master It
The mass deficit of Uranium-234 with a binding energy of 1779 MeV is equivalent to approximately 0.0054 atomic mass units.
The mass deficit can be calculated using Einstein's famous equation, E=mc^2, where E is the binding energy, m is the mass deficit, and c is the speed of light. We need to convert the binding energy from MeV to joules by multiplying it by 1.602 × 10^-13, which is the conversion factor between MeV and joules. So, the binding energy in joules is 1779 MeV * 1.602 × 10^-13 J/MeV = 2.845 × 10^-10 J.
Next, we divide the binding energy by the square of the speed of light (c^2) to find the mass deficit:
m = E / c^2 = 2.845 × 10^-10 J / (3 × 10^8 m/s)^2
Calculating this expression gives us the mass deficit in kilograms. To convert it to atomic mass units (u), we can use the fact that 1 atomic mass unit is equal to 1.66 × 10^-27 kg. So, the mass deficit in kilograms divided by this conversion factor will give us the mass deficit in atomic mass units:
m (u) = m (kg) / (1.66 × 10^-27 kg/u)
Performing the calculations, we find that the mass deficit is approximately 0.0054 atomic mass units for Uranium-234 with a binding energy of 1779 MeV.
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The sonar unit on a boat is designed to measure the depth of fresh water ( = 1.00 x 103 kg/m3, Bad = 2.20 x 109 Pa). When the boat moves into salt water ( = 1025 kg/m3, Bad = 2.37 x 109 Pa), the sonar unit is no longer calibrated properly. In salt water, the sonar unit indicates the water depth to be 7.96 m. What is the actual depth (in m) of the water?
The actual depth of the water in saltwater is 240.3 m.
The sonar unit on a boat is designed to measure the depth of fresh water, but when the boat moves into salt water the sonar unit is no longer calibrated properly.
Given, Depth indicated by sonar in saltwater=7.96 m
Density of freshwater =1.00 x 10³ kg/m³
Density of saltwater =1025 kg/m³
Pressure of freshwater=2.20 x 10⁹ Pa
Pressure of saltwater=2.37 x 10⁹ Pa.
To find out the actual depth of water in m we need to use the relationship between pressure and depth which is given as follows : ρgh = P
where ρ is the density of the fluid
g is the acceleration due to gravity
h is the depth of the fluid
P is the pressure of the fluid in N/m²
For freshwater, ρ = 1.00 x 10³ kg/m³ and P = 2.20 x 10⁹ Pa and
For saltwater, ρ = 1025 kg/m³ and P = 2.37 x 10⁹ Pa.
So, ρgh = P
⇒h = P/(ρg)
For freshwater, h = 2.20 x 10⁹/(1.00 x 10³ x 9.8) = 224.5 m
For saltwater , h = 2.37 x 10⁹/(1025 x 9.8) = 240.3 m
So, the actual depth is 240.3 m.
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An aluminum cylinder 30 cm deep has an internal capacity of 2.00L at 10 °C. It is completely filled with turpentine and then warmed to 80 °C. (a) If the liquid is then cooled back to 10 °C how far below the surface of the rim of the cylinder will the liquid be.( the coefficient of volume expansion for turpentine is 9.0 x 10 –4 °C-1. )
The distance below the surface of the rim of the cylinder will be approximately 30 cm, to two decimal places.
The volume of the aluminum cylinder = 2 L
Let the volume of turpentine = V1 at 10°C
Let the new volume of turpentine = V2 at 80°C
Coefficient of volume expansion of turpentine = β = 9.0 × 10⁻⁴/°C.
Volume expansion of turpentine from 10°C to 80°C = ΔV = V2 - V1 = V1βΔT
Let the distance below the surface of the rim of the cylinder be 'h'.
Therefore, the volume of the turpentine at 80°C is given by; V2 = V1 + ΔV + πr²h...(1)
From the problem, we have the Diameter of the cylinder = 2r = 4 cm.
So, radius, r = 2 cm. Depth, d = 30 cm
So, the height of the turpentine in the cylinder = 30 - h cm
At 10°C, V1 = 2L
From the above formulas, we have: V2 = 2 + (2 × 9.0 × 10⁻⁴ × 70 × 2) = 2.126 L
Now, substituting this value of V2 in Eq. (1) above, we have;2.126 = 2 + π × 2² × h + 2 × 9.0 × 10⁻⁴ × 70 × 2π × 2² × h = 0.126 / (4 × 3.14) - 2 × 9.0 × 10⁻⁴ × 70 h
Therefore, h = 29.98 cm
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You whirl a stone on a string in a horizontal circle of radius 1.25 m located 1.80 m above level ground. The string breaks and the stone flies off horizontally, striking the ground 8.00 m away. If the stone’s mass was 0.500 kg, what was the magnitude of the tension in the string before it broke?
The radius of the circle is given by r = 1.25 m. The height of the stone from the ground is 1.80 m. The horizontal distance the stone moves is 8.00 m. The mass of the stone is 0.500 kg.
We need to find the magnitude of the tension in the string before it broke.
Step 1: Finding the velocity of the stone when it broke away.
The velocity of the stone is given by the equation:v² = u² + 2as where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance covered by the stone.
Let u = 0, a = g, and s = 1.80 m, the equation becomes:
v² = 0 + 2g × 1.80 = 3.6gv = √(3.6g) m/s where g is the acceleration due to gravity.
Step 2: Finding the time the stone takes to travel 8.00 m.
The time the stone takes to travel 8.00 m is given by the equation:t = s/v = 8.00/√(3.6g) s.
Step 3: Find the magnitude of the tension in the string.
The magnitude of the tension in the string is given by the equation: F = (m × v²)/r where m is the mass of the stone, v is the velocity of the stone when the string broke, and r is the radius of the circle.
F = (0.500 × 3.6g)/1.25 = (1.8g)/1.25 = 1.44g = 1.44 × 9.81 = 14.1 N.
Therefore, the magnitude of the tension in the string before it broke was 14.1 N.
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A loop with radius r = 20cm is initially oriented perpendicular
to 1.2T magnetic field. If the loop is rotated 90o in 0.2s. Find
the induced voltage ε in the loop.
The induced voltage ε in the loop is equal to the rate of change of magnetic flux: ε = -dΦ/dt = -0.24π T/s
The induced voltage ε in the loop can be determined using Faraday's law of electromagnetic induction, which states that the induced voltage is equal to the rate of change of magnetic flux through the loop.
The magnetic flux Φ through the loop is given by the formula:
Φ = B * A * cosθ
Where B is the magnetic field strength, A is the area of the loop, and θ is the angle between the magnetic field and the normal to the loop.
In this case, the magnetic field B is 1.2T, the radius of the loop r is 20cm (0.2m), and the angle θ changes from 90 degrees to 0 degrees.
The area A of the loop is π *[tex]r^2[/tex] = π * (0.2[tex]m)^2[/tex] = 0.04π [tex]m^2[/tex].
The rate of change of magnetic flux is given by:
dΦ/dt = (Φf - Φi) / Δt
Where Φf is the final magnetic flux and Φi is the initial magnetic flux, and Δt is the time taken for the change.
Since the loop is initially perpendicular to the magnetic field, the initial magnetic flux is zero, and the final magnetic flux is:
Φf = B * A * cosθf = 1.2T * 0.04π [tex]m^2[/tex] * cos(0 degrees) = 1.2T * 0.04π [tex]m^2[/tex]
The time taken for the change is Δt = 0.2s.
Plugging these values into the formula, we get:
dΦ/dt = (1.2T * 0.04π [tex]m^2[/tex] - 0) / 0.2s
Simplifying, we find:
dΦ/dt = 0.24π T/s
The negative sign indicates that the induced voltage creates a current in the opposite direction to oppose the change in magnetic flux.
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Question 2 A simple pendulum is made from a ping-pong ball with a mass of 10 grams, attached to a 60 cm length of thread with a negligible mass. The force of air resistance on the ball is F = rx, in which r = 0.016 kg s-¹. (a) Show that the pendulum is underdamped. Find the angular frequency w and the period T of oscillation and compare to the natural (undamped) wo and To- (b) How long does it take for the amplitude of the pendulum's swing to decrease by a factor 1000? By what factor does the mechanical energy decreases in this time? (c) If a pendulum made with the same ping-pong ball were to critically damped by air resistance, what would its length have to be?
A simple pendulum is made from a ping-pong ball with a mass of 10 grams, attached to a 60 cm length of thread with a negligible mass. The force of air resistance on the ball is F = rx, in which r = 0.016 kg s-¹.(a) The motion of a simple pendulum is given by the equation T = 2π\sqrt(l/g) where T is the period, l is the length of the pendulum and g is the acceleration due to gravity which is taken as 9.81 m s-². The undamped angular frequency w, is given by w = √(g/l). As the pendulum is underdamped, we can use the formula w' = w * √(1 - b²/4m²), where m is the mass, b is the damping coefficient, and w' is the damped angular frequency.
Therefore, m = 0.01 kg (mass of the ball), b = r (damping coefficient) and l = 60 cm = 0.6 m (length of the thread). Undamped angular frequency, w = √(g/l) = √(9.81/0.6) = 3.188 rad s-¹Damped angular frequency, w' = w * √(1 - b²/4m²) = 3.188 * √(1 - (0.016/4*0.01²)) = 3.131 rad s-¹Time period, T = 2π/w = 2π/3.131 = 2.003 s(b) The amplitude of the oscillation decreases by a factor of 1000, that is 1000 times the initial amplitude, so the amplitude ratio A/A₀ = 1/1000, where A₀ is the initial amplitude. Using the formula A = A₀e^-bt/2m,
where A is the amplitude after time t, we can solve for t.A/A₀ = e^-bt/2m1/1000 = e^-bt/2m-ln(1/1000) = -bt/2m= ln1000t = 2m/b * ln1000t = 2 * 0.01/0.016 * 6.9078t = 8.545 s
The mechanical energy E of the pendulum is given by E = ½mω²A². At any time t, the mechanical energy E is given by E = ½mω²A₀²e^-bt/m. Therefore, the factor by which the mechanical energy decreases isE/E₀ = (1/2)ω²e^-bt/m / (1/2)ω² = e^-bt/m = e^-0.016/0.01 * 8.545 = 0.300 or 30%(c) A critically damped system will have a damping coefficient b = 2m√(k/m) = 2m w = 2m√(g/l).Therefore, b = 2m√(g/l) = 2 * 0.01 * √(9.81/0.6) = 0.776 kg s-¹.The length of the pendulum for critical damping is given by l = g/b²m = 9.81/(0.776)² * 0.6 = 12.05 cm = 0.1205 m.
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The driver of a car wishes to pass a truck that is traveling at a constant speed of (about ). Initially, the car is also traveling at and its front bumper is 24. 0 m behind the truck’s rear bumper. The car accelerates at a constant then pulls back into the truck’s lane when the rear of the car is 26. 0 m ahead of the front of the truck. The car is 4. 5 m long and the truck is 21. 0 m
The car takes a certain amount of time to pass the truck and travels a certain distance during the maneuver.
In the given scenario, the car starts 24.0 m behind the truck and accelerates at a constant rate. The car then moves ahead of the truck until its rear is 26.0 m ahead of the truck's front. The lengths of the car and the truck are also provided. To determine the time it takes for the car to pass the truck, we can use the relative positions and velocities of the car and the truck. By calculating the time it takes for the car's rear to reach a position 26.0 m ahead of the truck's front, we can find the duration of the maneuver. Additionally, by subtracting the initial and final positions, taking into account the lengths of the car and the truck, we can determine the distance traveled by the car during the passing maneuver.
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What is the angular velocity of a tricycle wheel relative to the angular velocity of a bicycle wheel (what is w tricycle/w bicycle) if both wheels are traveling with the same translational speed? The bicycle has a wheel radius that is 3.00 times that of the tricycle wheel. Would it be safe to make a child tricycle/adult bicycle tandem?
The angular velocity of the tricycle wheel is three times that of the bicycle wheel (ω_tricycle / ω_bicycle = 3) and it would not be safe to make a child tricycle/adult bicycle tandem.
To determine the angular velocity ratio between the tricycle wheel and the bicycle wheel, we can use the relationship between linear speed, angular velocity, and the radius of a rotating object.
The linear speed of both wheels is the same since they are traveling at the same translational speed.
Let's denote the linear speed as v.
For the bicycle wheel, let's denote its radius as r_bicycle.
For the tricycle wheel, let's denote its radius as r_tricycle.
The relationship between linear speed and angular velocity is given by:
v = ω * r,
where v is the linear speed, ω (omega) is the angular velocity, and r is the radius of the rotating object.
For the bicycle wheel, we have:
v_bicycle = ω_bicycle * r_bicycle.
For the tricycle wheel, we have:
v_tricycle = ω_tricycle * r_tricycle.
Since both wheels have the same linear speed, we can set the two equations equal to each other:
v_bicycle = v_tricycle.
ω_bicycle * r_bicycle = ω_tricycle * r_tricycle.
We can rewrite this equation in terms of the angular velocity ratio:
ω_tricycle / ω_bicycle = r_bicycle / r_tricycle.
Given that the radius of the bicycle wheel is 3.00 times that of the tricycle wheel (r_bicycle = 3 * r_tricycle), we can substitute this into the equation:
ω_tricycle / ω_bicycle = (3 * r_tricycle) / r_tricycle.
ω_tricycle / ω_bicycle = 3.
Therefore, the angular velocity of the tricycle wheel is three times that of the bicycle wheel (ω_tricycle / ω_bicycle = 3).
Based on this, it would not be safe to make a child tricycle/adult bicycle tandem because the tricycle wheel would rotate at a much higher angular velocity than the bicycle wheel, potentially causing stability issues and safety concerns.
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A screen is placed 5 m from a single slit of width 0.0021 m, which is illuminated with light of wavelength 7.1.107 m. Consider that the angle is small. ] Which formula can be used to calculate the location of a minima on the viewing screen?
The formula that can be used to calculate the location of minima on the viewing screen for the single slit diffraction is;
x = mλL/d
Where,
x is the location of the minima on the viewing screen
λ is the wavelength of the incident light
m is an integer representing the order of the minima
L is the distance from the slit to the viewing screen
d is the width of the slit.
The formula is applicable when the angle is small since the angle of the diffraction pattern depends on the wavelength of light and the width of the slit. When the angle is small, the small angle approximation can be made, making sinθ ≈ tanθ ≈ θ, where θ is the angle of diffraction.
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