In a circle with radius of 10 millimeters, find the area of a sector whose central angle is 102°. Use 3.14 for π a. 177.93 mm^2b. 88.97 mm^2 c. 314 mm^2 d. 355.87 mm^2

Answer:

The three elements we apply an ac voltage of 1 v of frequency of 1000 hz and given that r=100ohm l=8.0*10^-3 and c =1.0 *10^ -6f  the resonance frequency of the circuit is 1591 Hz.

Explanation:

The resonance frequency of an RLC circuit can be calculated using the formula:

f_res = 1 / (2 * pi * sqrt(L * C))

where f_res is the resonance frequency, L is the inductance, and C is the capacitance.

Plugging in the given values, we get:

f_res = 1 / (2 * pi * sqrt(8.0*10^-3 * 1.0*10^-6))

f_res = 1591 Hz (rounded to three significant figures)

Therefore, the resonance frequency of the circuit is 1591 Hz.

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In Part A, the electron's speed is given as 4.00 km/s and the force acting on it due to the electric field is 5.50×10−15N. To find the acceleration produced by this force,

we can use the equation F = ma, where F is the force, m is the mass of the electron, and a is the acceleration. As the mass of the electron is very small,

we can use the equation a = F/m. Therefore, the acceleration produced by this force in Part A is:

a = F/m = (5.50×10−15N) / (9.11×10−31kg) = 6.04×10^15 m/s^2

In Part B, the force acting on the electron is parallel to its velocity. This means that the force does not change the direction of the electron's motion, but only its speed.

As the electron is moving with a constant velocity, we can assume that its acceleration is zero. This means that the force acting on the electron must be balanced by another force,

such as a magnetic force, that prevents the electron from changing its direction of motion. Therefore, the acceleration produced by the force in Part B is zero.

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Given the production function q = 2kl and the input prices w = $4 and r = $4, we can use the following optimization problem to determine the optimal quantities of labor (l) and capital (k) that will be utilized to produce 40 units of output:

Maximize q = 2kl subject to the budget constraint wL + rK = C, where C is the cost of production.

Plugging in the given values, we have:

Maximize 2kl subject to 4L + 4K = C

We can rewrite the budget constraint as K + L = C/4, which tells us that the cost of production is equal to the total expenditure on labor and capital. We can then solve for K in terms of L: K = C/4 - L.

Substituting this into the production function, we get:

q = 2k(C/4 - L) = (C/2)k - kl

To maximize output, we need to take the partial derivatives of q with respect to both k and l and set them equal to zero:

∂q/∂k = C/2 - l = 0 --> l = C/2

∂q/∂l = C/2 - k = 0 --> k = C/2

Plugging these values back into the budget constraint K + L = C/4, we get:

C/2 + C/2 = C/4 --> C = 4

Therefore, the optimal quantities of labor and capital are:

l = C/2 = 2 units

k = C/2 = 2 units

So, to produce 40 units of output, we need 2 units of labor and 2units of c apital.

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To express the rate of climb of a mountain trail with no units, you can simply state it as a ratio or fraction: 1/8.33. This means that for every 8.33 units traveled horizontally, the trail ascends 1 unit vertically.

The rate of climb of 120 meters per kilometer can be expressed with no units as a ratio or fraction: 1/8.33. This ratio signifies that for every 8.33 units traveled horizontally (in any unit of distance), the trail ascends 1 unit vertically (in any unit of elevation). By removing the specific units (meters per kilometer), we create a dimensionless quantity that can be used universally. This allows for easier comparison and understanding of the rate of climb, regardless of the specific units used to measure distance and elevation.

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The magnitude of the magnetic field at the center of the cross section of the windings is 3.95 x 10^-3 T.

To solve this problem, we can use the equation B = (μ0 * n * I) / (2 * r), where B is the magnetic field, μ0 is the permeability of free space (4π x 10^-7 T m/A), n is the number of turns per unit length (in this case, it's just the total number of turns divided by the mean circumference of the ring), I is the current, and r is the mean radius of the ring.

First, we need to find the mean circumference and mean radius of the ring. The mean diameter is given as 14.5 cm, so the mean radius is 7.25 cm. The mean circumference is 2πr, which is approximately 45.5 cm.

Next, we can calculate n by dividing the total number of turns (615) by the mean circumference (45.5 cm) to get 13.5 turns/cm.

Now we can plug in all the values into the equation and solve for B:

B = (4π x 10^-7 T m/A) * (13.5 turns/cm) * (0.640 A) / (2 * 0.0725 m)
B = 3.95 x 10^-3 T

Therefore, the magnitude of the magnetic field at the center of the cross section of the windings is 3.95 x 10^-3 T.

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(a) The work necessary to bring a 101 kg object to a height of 992 km above the surface of the Earth is 986 MJ. (b) The extra work needed to launch the object into circular orbit at a height of 992 km above the surface of the Earth is 458 MJ.

To bring an object to a height of 992 km above the surface of the Earth, we need to do work against the force of gravity. The work done is given by the formula;

W = mgh

where W is work done, m is mass of the object, g is acceleration due to gravity, and h is the height above the surface of the Earth.

Using the given values, we have;

m = 101 kg

g = 9.81 m/s²

h = 992 km = 992,000 m

W = (101 kg)(9.81 m/s²)(992,000 m) = 9.86 × 10¹¹ J

Converting J to MJ, we get;

W = 986 MJ

Therefore, the work necessary to bring a 101 kg object to a height of 992 km above the surface of the Earth is 986 MJ.

To launch the object into circular orbit at this height, we need to do additional work to overcome the gravitational potential energy and give it the necessary kinetic energy to maintain circular orbit. The extra work done is given by the formula;

W = (1/2)mv² - GMm/r

where W is work done, m is mass of the object, v is velocity of the object in circular orbit, G is gravitational constant, M is the mass of the Earth, and r is the distance between the object and the center of the Earth.

We can find the velocity of the object using the formula:

v = √(GM/r)

where √ is the square root symbol. Substituting the given values, we have;

v = √[(6.67 × 10⁻¹¹ N·m²/kg²)(5.97 × 10²⁴ kg)/(6,371 km + 992 km)] = 7,657 m/s

Substituting the values into the formula for work, we have;

W = (1/2)(101 kg)(7,657 m/s)² - (6.67 × 10⁻¹¹ N·m²/kg²)(5.97 × 10²⁴ kg)(101 kg)/(6,371 km + 992 km)

W = 4.58 × 10¹¹ J

Converting J to the required units, we get;

W = 458 MJ

Therefore, the extra work needed to launch the object into circular orbit at a height of 992 km above the surface of the Earth is 458 MJ.

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--The given question is incomplete, the complete question is

"(a) Calculate the work (in MJ) necessary to bring a 101 kg object to a height of 992 km above the surface of the Earth.__ MJ (b) Calculate the extra work (in MJ) needed to launch the object into circular orbit at this height of 992 km above the surface of the Earth .__MJ."--

The proton needs to be fired toward the lead target with an initial kinetic energy of 25.2 MeV.

To impact a lead of accelerators nucleus with 20 MeV of kinetic energy, a proton must be fired at the nucleus with a specific amount of initial kinetic energy. In this case, the required initial kinetic energy is 25.2 MeV.

To understand why this is the case, it's important to consider the nature of the nuclear reactions that occur when a proton impacts a nucleus. In order for the proton to penetrate the nucleus, it must have enough kinetic energy to overcome the electrostatic repulsion between the positively charged proton and the positively charged nucleus. This kinetic energy is determined by the velocity of the proton as it approaches the nucleus.

The specific amount of initial kinetic energy required to achieve the desired kinetic energy of the proton upon impact depends on a number of factors, including the mass of the target nucleus and the desired kinetic energy of the proton upon impact.

In this case, the 207 Pb nucleus is relatively heavy, which means that the proton must be fired with a higher initial kinetic energy in order to achieve the desired kinetic energy upon impact. The exact value of 25.2 MeV is calculated based on the mass of the lead nucleus and the desired kinetic energy of the proton upon impact.

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Spring force decreases by a factor of 3/2, and elastic potential energy decreases by a factor of 9/4.

The force exerted by a spring is given by Hooke's Law, F = -kx, where F is the force, x is the distance the spring is compressed or stretched, and k is the spring constant. If x is decreased by a third, then the force decreases proportionally by a factor of 3/2. So the spring force decreases by a factor of 3/2.

The elastic potential energy stored in a spring is given by the formula U = (1/2)kx^2. If x is decreased by a third, then the potential energy stored in the spring decreases by a factor of (1/2)k(1/3x)^2 = (1/18)kx^2. So the elastic potential energy decreases by a factor of 9/4.

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In the Charges and Fields PhET simulation (HTML 5 version), you can change the following aspects of the simulation: add positive or negative charges, adjust the strength of charges, measure electric field and potential and display field lines and equipotential lines.

1. Add positive or negative charges: You can place positive or negative point charges on the grid to create different electric fields.
2. Adjust the strength of charges: You can modify the strength of the point charges, influencing the electric field's intensity.
3. Measure electric field and potential: You can use the electric field and electric potential sensors to measure the field's strength and potential at various points in the simulation.
4. Display field lines and equipotential lines: You can toggle the display of electric field lines and equipotential lines to visualize the electric field and potential created by the charges.
Remember to experiment with different combinations of charges and their strengths to explore various electric field scenarios.

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If the Fed wants to lower the federal funds rate, it should buy government securities in the open market. This will increase the amount of money available in the banking system, leading to a decrease in the federal funds rate.

Selling government securities in the open market would have the opposite effect and raise the federal funds rate. Increasing the reserve ratio would require banks to hold more reserves and would also raise the federal funds rate. Increasing the discount rate would make borrowing from the Fed more expensive, which could indirectly increase the federal funds rate.

If the Fed wants to lower the federal funds rate, it should d. buy government securities in the open market.

By purchasing government securities, the Fed increases the supply of money in the economy. This results in a lower federal funds rate as banks have more funds available for lending, leading to increased demand for loans and lower borrowing costs.

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Answer:Main answer: The torque about the origin is (-131 + 263 + 26k) Nm.

Supporting explanation: The torque (τ) is defined as the cross product of the force (F) and the position vector (r) from the point of application to the axis of rotation. Therefore, we need to first find the position vector from the origin to the point of application of the force.

r = (-4.00i - 7.00j + 5.00k) m

Taking the cross product of r and F gives the torque:

τ = r × F

 = (-4.00i - 7.00j + 5.00k) × (2.00i - 3.00j + 4.00k) N

 = (8k - 15j)i + (16i + 20k)j + (-12i + 6j)k Nm

 = (-131 + 263 + 26k) Nm

Therefore, the torque about the origin is (-131 + 263 + 26k) Nm.

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According to the kinetic molecular theory of gases, the volume of the gas particles, which can be atoms or molecules, is considered to be negligible compared to the volume of the container that they occupy. The gas particles are assumed to be point masses.

This assumption is based on the fact that at normal temperatures and pressures, the space between gas particles is much larger than the size of the particles themselves. Therefore, the particles can be treated as point masses without significantly affecting the overall behavior of the gas.

The kinetic molecular theory of gases provides a useful framework for understanding the behavior of gases at the molecular level, and helps to explain many of the observed properties of gases, such as their pressure, volume, temperature, and the relationships between them, such as the ideal gas law.

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(a) 64.7° is the acceptance angle (in degrees) for the oil immersion objective

(b) 30° is the acceptance angle (in degrees) for the air immersion objective.

Part (a): The acceptance angle for the oil immersion objective can be calculated using the formula α1 = sin⁻¹(NA1/n), where NA1 is the numerical aperture of the objective and n is the refractive index of the medium between the specimen and the objective. Here, NA1 = 1.40 and n = 1.51 (refractive index of oil). Substituting these values, we get α1 = sin⁻¹(1.40/1.51) = 64.7°.
Part (b): The acceptance angle for the air immersion objective can be calculated using the formula α2 = sin⁻¹(NA2/n), where NA2 is the numerical aperture of the objective and n is the refractive index of the medium between the specimen and the objective. Here, NA2 = 0.5 and n = 1 (refractive index of air). Substituting these values, we get α2 = sin⁻¹(0.5/1) = 30°.
In summary, the acceptance angle for the oil immersion objective is 64.7°, while the acceptance angle for the air immersion objective is 30°. This difference in acceptance angle is due to the fact that oil has a higher refractive index than air, which allows for greater light refraction and therefore a larger acceptance angle.

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In which direction is the centripetal acceleration directed on a particle that is moving along a circular trajectory?

Centripetal acceleration is always directed towards the center of the circular path in which the particle is moving. This inward direction ensures that

the particle constantly changes its velocity as it moves along the circular trajectory, even if its speed remains constant.

The centripetal acceleration is responsible for maintaining the particle's circular motion by continuously altering its direction.

To further understand this concept, consider these steps:

1. As the particle moves along the circular path, it has both a linear velocity (tangential to the circle) and an angular velocity (change in angle per unit time).

2. The centripetal force, acting perpendicular to the linear velocity, is responsible for the change in direction of the particle as it moves.

3. The centripetal acceleration is the result of this centripetal force acting on the particle. It is given by the formula: a_c = (v^2) / r, where a_c is the centripetal acceleration,

v is the linear velocity, and r is the radius of the circular path.

4. Since the centripetal acceleration is always directed towards the center of the circle, it ensures that the particle remains in its circular trajectory.

In conclusion, the centripetal acceleration is directed towards the center of the circular path in which a particle moves.

This inward direction enables the particle to maintain its circular motion by continuously adjusting its velocity.

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Answer:

12×8=848

Explanation:

repell forces

The electric potential 15.0 cm from a 4.0 µC point charge is approximately 95930 V.

The electric potential (V) at a distance r from a point charge Q is given by:

V = kQ/r

where k is the Coulomb constant (k = 8.99 x 10^9 N·m^2/C^2).

In this case, we are given a point charge Q of 4.0 µC and a distance r of 15.0 cm (which is 0.15 m in SI units). Plugging these values into the equation, we get:

V = (8.99 x 10^9 N·m^2/C^2) x (4.0 x 10^-6 C) / (0.15 m)

Solving this expression, we get:

V ≈ 95930 V

Therefore, the electric potential 15.0 cm from a 4.0 µC point charge is approximately 95930 V.

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The center of Lake Superior is depressed by 5.2 meters due to the centrifugal force at a radius of 162 km and a latitude of 47°.

When a body rotates, objects on its surface are subject to centrifugal force which causes them to move away from the center.

In this case, Lake Superior is assumed to be at rest with respect to Earth and a circle of radius 162 km at a latitude of 47° is drawn around it.

Using the formula for centrifugal force, the depth that the center of the lake is depressed with respect to the shore is calculated to be 5.2 meters.

This means that the water at the center of Lake Superior is pushed outwards due to the centrifugal force, causing it to be shallower than the shore.

Understanding the effects of centrifugal force is important in many areas of science and engineering.

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The blackbody object that peaks at 1,000 nm (just outside the visible spectrum) would appear invisible to the human eye. The answer is b.

The visible spectrum for humans ranges from approximately 400 nm (violet) to 700 nm (red). A blackbody object's perceived color depends on its temperature and the wavelength at which it emits the most radiation. The peak wavelength of the radiation emitted by an object decreases as its temperature increases according to Wien's displacement law.

In this case, a blackbody object that peaks at 1,000 nm has a temperature of approximately 2,897 K. This is outside the range of temperatures that produce visible light.

Therefore, the object would not appear to have any color to the human eye. Instead, it would appear as a dark object, absorbing most of the visible light that strikes it. Hence, b is the right option.

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The change in energy (in joules) of the hot water due to the heat transfer when it is placed in contact with the cold water and allowed to reach equilibrium is -15,464 J.

The change in energy (in joules) of the hot water due to the heat transfer when it is placed in contact with the cold water and allowed to reach equilibrium can be calculated using the equation

Q = mcΔT

Where Q is the heat transferred, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature of the water.

For the hot water

m = 1.00 kg

c = 4,186 J/(kg·°C) (specific heat capacity of water)

ΔT = 41.5°C - Teq

Where Teq is the equilibrium temperature of the two bodies.

For the cold water

m = 1.00 kg

c = 4,186 J/(kg·°C) (specific heat capacity of water)

ΔT = Teq - 21°C

Because the heat transfer is from the hot water to the cold water, the magnitude of the heat transferred will be the same for both bodies. Therefore

mcΔT = mcΔT

(1.00 kg)(4,186 J/(kg·°C))(41.5°C - Teq) = (1.00 kg)(4,186 J/(kg·°C))(Teq - 21°C)

Simplifying this equation, we get

83.7 J/°C = Teq - 21°C + Teq - 41.5°C

Combining like terms, we get

2Teq - 62.5°C = 83.7 J/°C

Solving for Teq, we get

Teq = (83.7 J/°C + 62.5°C)/2

Teq = 73.1°C

Therefore, the change in energy (in joules) of the hot water due to the heat transfer when it is placed in contact with the cold water and allowed to reach equilibrium is

Qh = mcΔT = (1.00 kg)(4,186 J/(kg·°C))(41.5°C - 73.1°C) = -15,464 J

(Note that the negative sign indicates that the hot water loses energy, as expected.)

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Using AC current in an electromagnet affects the compass by causing it to oscillate or rapidly change direction.

This is because AC current alternates its direction of flow periodically. When the current flows through the electromagnet, it generates a magnetic field that changes direction along with the alternating current. As a result, the compass needle, which is sensitive to magnetic fields, will continuously change its direction in response to the fluctuating magnetic field created by the electromagnet.

In contrast to DC current, which produces a steady magnetic field, AC current creates a constantly changing magnetic field due to the alternating nature of the current. When an electromagnet is powered by AC current, its magnetic field will continuously change direction, causing the compass needle to rapidly change direction as well. This occurs because the compass needle aligns itself with the magnetic field generated by the electromagnet. The rapidly changing magnetic field can make it difficult to obtain a stable reading from the compass, as the needle will not settle in one direction.

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Therefore, the value of R that yields a damped frequency of 120 rad/s depends on the values of L and C in the circuit. We need more information about the specific values of these components in order to calculate R.

To find the value of R that yields a damped frequency of 120 rad/s, we need to use the formula for the damped frequency of a parallel RLC circuit:
d = 1/(LC - R2/4L2)
where d is the damped frequency, L is the inductance, C is the capacitance, and R is the resistance.
We can rearrange this formula to solve for R:
R = 2Lωd/√(1 - LCd2)
Substituting d = 120 rad/s and rounding to three significant figures, we get:
R = 2Lωd/√(1 - LCd2)
R = 2L(120 rad/s)/(1 - LC(120 rad/s)2)
R = 2L(120 rad/s)/(1 - (L/C)(14400))
R = 240L/√(1 - 14400L/C)
Therefore, the value of R that yields a damped frequency of 120 rad/s depends on the values of L and C in the circuit. We need more information about the specific values of these components in order to calculate R.

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If two concave lenses, each with f = -17 cm, are separated by 8.5 cm. An object is placed 35 cm in front of one of the lenses, then a) The final image distance is -23.2 cm. b) The magnification of the final image is 1.6.

a) We can use the thin lens equation to find the image distance and magnification for each lens separately, and then use the lensmaker's formula to combine the two lenses.

For each lens, the thin lens equation is:

1/f = 1/do + 1/di

where f is the focal length, do is the object distance, and di is the image distance.

Plugging in f = -17 cm and do = 35 cm, we get:

1/-17 cm = 1/35 cm + 1/di1

Solving for di1, we get:

di1 = -23.3 cm

The magnification for each lens is:

m1 = -di1/do = -(-23.3 cm)/35 cm = 0.67

Using the lensmaker's formula, we can find the combined focal length of the two lenses:

1/f = (n-1)(1/R1 - 1/R2 + (n-1)d/(nR1R2))

where n is the index of refraction, R1 and R2 are the radii of curvature of the two lens surfaces, and d is the thickness of the lens.

Since the two lenses are identical, we have R1 = R2 = -17 cm and d = 8.5 cm. Also, for simplicity, we can assume that the index of refraction is 1.

Plugging in these values, we get:

1/f = -2/R1 + d/R1²

Solving for f, we get:

f = -17.0 cm

So the combined focal length is still -17 cm.

We can now use the thin lens equation again, with f = -17 cm and di1 = -23.3 cm as the object distance for the second lens:

1/-17 cm = 1/-23.3 cm + 1/di2

Solving for di2, we get:

di2 = -13.8 cm

The magnification for the second lens is:

m2 = -di2/di1 = -(-13.8 cm)/(-23.3 cm) = 0.59

b) To find the total magnification, we multiply the individual magnifications:

m = m1 × m2 = 0.67 × 0.59 = 1.6

So the final image is upright and magnified, and its distance from the second lens is -13.8 cm, which means its distance from the first lens is:

di = di1 + d1 + di2 = -23.3 cm + 8.5 cm - 13.8 cm = -28.6 cm

Since the object is on the same side of the first lens as the final image, the image distance is negative, which means the image is virtual and on the same side of the lens as the object.

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To find the particle's speed, we can use the de Broglie wavelength equation:

λ = h/p

where λ is the de Broglie wavelength, h is Planck's constant, and p is the momentum of the particle. We can rearrange this equation to solve for the momentum:

p = h/λ

Now we can use the momentum and the mass of the particle to find its speed:

v = p/m

where v is the speed and m is the mass.

Plugging in the given values, we get:

p = (6.626 x 10^-34 J s)/(7.25 x 10^-12 m) = 9.13 x 10^-23 kg m/s

v = (9.13 x 10^-23 kg m/s)/(6.68 x 10^-27 kg) = 1.37 x 10^4 m/s

Therefore, the particle's speed is 1.37 x 10^4 m/s.

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The speed of the wave on the string can be calculated by multiplying the frequency (60.00 Hz) with the wavelength (2.00 cm), which gives us a result of 120 cm/s.

To further explain, the speed of a wave is defined as the distance traveled by a wave per unit time. In this case, we have a frequency of 60.00 Hz, which means that the wave produces 60 cycles per second. The wavelength, on the other hand, is the distance between two consecutive points of the wave that are in phase with each other. So, with a wavelength of 2.00 cm, we know that the distance between two consecutive points that are in phase is 2.00 cm.

By multiplying these two values, we get the speed of the wave on the string, which is 120 cm/s. This means that the wave travels at a speed of 120 cm per second along the length of the string.

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Paper must be heated to a specific temperature (234°C) to begin reacting with oxygen because it needs enough energy to break down its complex structure and start the chemical reaction of combustion. Heating the paper over a flame provides the necessary energy to initiate this process.

Once the paper reaches its ignition temperature, the heat from the combustion reaction will continue to sustain the fire. Additionally, the heat causes the cellulose fibers in the paper to release volatile gases, which then ignite and contribute to the flame. Without sufficient heat, the paper would not reach its ignition temperature and would not begin to burn.

The paper must be heated to 234°C to start burning because that is its ignition temperature. At this temperature, the paper begins to react with oxygen, leading to combustion. Heating the paper to this point provides the necessary energy for the chemical reaction between the paper's molecules and the oxygen in the air. The flame acts as a heat source to raise the paper's temperature to its ignition point, allowing the burning process to commence.

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The approximate measurement for the angle of the 14th minimum in this diffraction pattern is 58.6 degrees.

We can use the single-slit diffraction formula to find the angle of the 14th minimum in this diffraction pattern. The formula is:

sin θ = mλ / b

where θ is the angle of the minimum, m is the order of the minimum (m = 1 for the first minimum, m = 2 for the second minimum, and so on), λ is the wavelength of the light, and b is the width of the slit.

Given:

m = 14 (order of the minimum)

λ = (unknown)

b = (unknown)

mλ for the 4th minimum = 5.9

We can find the wavelength of the light by using the known value of mλ for the fourth minimum:

sin θ4 = mλ / b

sin θ4 = (4λ) / b

λ = (b sin θ4) / 4

λ = (b sin (tan[tex]^(-1)[/tex](5.9 / 4))) / 4

λ = (b * 0.988) / 4

λ = 0.247b

Now we can use the value of λ to find the angle of the 14th minimum:

sin θ14 = mλ / b

sin θ14 = (14λ) / b

sin θ14 = 3.43λ / b

sin θ14 = 3.43(0.247b) / b

sin θ14 = 0.847

θ14 = sin[tex]^(-1)[/tex](0.847)

θ14 ≈ 58.6 degrees

Therefore, the angle of the 14th minimum in this diffraction pattern is approximately 58.6 degrees.

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The energy absorbed by 30.0 g of water in heating it from 0.0°C to 100.0°C is 12.7 kJ. Changing the rate at which heat is added from 50 J/s to 100 J/s does not affect this calculation since the energy required to raise the temperature of a substance is independent of the rate at which it is added.

In more detail, the energy absorbed in heating a substance is given by the equation Q = mCΔT, where Q is the energy absorbed, m is the mass of the substance, C is the specific heat capacity of the substance, and ΔT is the change in temperature. For water, the specific heat capacity is 4.18 J/g°C. Therefore, the energy absorbed in heating 30.0 g of water from 0.0°C to 100.0°C is:

Q = (30.0 g)(4.18 J/g°C)(100.0°C - 0.0°C) = 12,540 J = 12.7 kJ

Changing the rate at which heat is added, such as from 50 J/s to 100 J/s, does not affect the amount of energy required to raise the temperature of the water since the energy required is dependent only on the mass, specific heat capacity, and temperature change of the substance, and is independent of the rate at which it is added.

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If r is a primitive root modulo m, then its inverse r(bar) is also a primitive root modulo m.

Let's assume that r is a primitive root modulo m. This means that the set of residues generated by r modulo m is a complete residue system, i.e., it covers all the numbers from 1 to [tex]m^{-1[/tex].

Now, let's consider the inverse of r, denoted as r(bar). By definition, r(bar) is the number such that:

r × r(bar) ≡ 1 (mod m).

To show that r(bar) is also a primitive root modulo m, we need to prove that the set of residues generated by r(bar) modulo m is also a complete residue system.

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Two charges of -25 pc and 36 pc are located inside a sphere of a radius of r = 0.25 m. The total electric flux through the surface of the sphere is 1.24 N[tex]m^{2}[/tex]/C.

We can use Gauss's law to calculate the electric flux through the surface of the sphere due to the enclosed charges

ϕ = qenc / ε0

Where ϕ is the electric flux, qenc is the total charge enclosed by the surface, and ε0 is the electric constant.

To calculate qenc, we need to first find the net charge inside the sphere

qnet = q1 + q2

qnet = -25 pc + 36 pc

qnet = 11 pc

Where q1 and q2 are the charges of -25 pc and 36 pc, respectively.

Now we can calculate the electric flux through the surface of the sphere:

ϕ = qenc / ε0

ϕ = qnet / ε0

ϕ = (11 pc) / ε0

Using the value of the electric constant, ε0 = 8.85 × [tex]10^{-12} C^{2} / Nm^{2}[/tex], we can calculate the electric flux

ϕ = (11 pc) / ε0

ϕ = (11 × [tex]10^{-12}[/tex] C) / (8.85 × [tex]10^{-12} C^{2} / Nm^{2}[/tex])

ϕ = 1.24 N[tex]m^{2}[/tex]/C

Therefore, the total electric flux through the surface of the sphere is 1.24 N[tex]m^{2}[/tex]/C.

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The total electric flux through the surface of the sphere is 9.80 × 10^9 pc.The total electric flux through the surface of the sphere can be calculated using Gauss's Law, which states that the total electric flux through a closed surface is proportional to the total charge enclosed by that surface. In this case, we have two charges of -25 pc and 36 pc located inside the sphere.

To calculate the total charge enclosed by the surface of the sphere, we need to find the net charge inside the sphere. The net charge is the algebraic sum of the two charges, which is 11 pc.

Now, using Gauss's Law, the total electric flux through the surface of the sphere can be calculated as follows:

Flux = Q/ε₀
Where Q is the total charge enclosed by the surface of the sphere and ε₀ is the permittivity of free space.

Substituting the values, we get:

Flux = (11 pc) / (4πε₀r²)
where r is the radius of the sphere, which is 0.25 m.

Simplifying the equation, we get:

Flux = (11 pc) / (4π × 8.85 × 10^-12 × 0.25²)
Flux = 9.80 × 10^9 pc

Therefore, the total electric flux through the surface of the sphere is 9.80 × 10^9 pc.

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The new temperature of the ideal gas after removing 3.6 J of thermal energy is approximately 12.1°C.

To calculate the new temperature, we'll use the formula for the change in internal energy of an ideal gas, which is ΔU = (3/2)nRΔT, where ΔU is the change in internal energy, n is the number of moles, R is the ideal gas constant, and ΔT is the change in temperature.

First, we need to determine the number of moles (n) from the given number of atoms (2.2 × 10²² atoms). Since 1 mole contains Avogadro's number (6.022 × 10²³) of atoms, we can find n by dividing the number of atoms by Avogadro's number:

n = (2.2 × 10²² atoms) / (6.022 × 10²³ atoms/mol) ≈ 0.0365 moles

Next, we need to find the change in internal energy (ΔU), which is -3.6 J since thermal energy is being removed from the gas.

Now, we can rearrange the formula ΔU = (3/2)nRΔT to solve for the change in temperature (ΔT):

ΔT = ΔU / [(3/2)nR] = -3.6 J / [(3/2)(0.0365 moles)(8.314 J/mol K)] ≈ -7.9°C

Since the initial temperature was 20°C, the new temperature is:

New Temperature = Initial Temperature + ΔT = 20°C -7.9°C ≈ 12.1°C.

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