describe how the current and resistance would behave as the potential difference across a resisitor changes for both ohmic and nonohmic cases

Answer:

In 250 BC, Eratosthenes measured the circumference to be about 24,500 miles:

by this method, 12 deg = 12 / 360 = 1/30 of the planet's surface\

30 * 500 km = 15,000 km in circumference, about 9,200 miles

The mass of the other star in the binary system is approximately 0.541 solar masses.

To find the mass of the other star in the binary system, we can use Kepler's Third Law of Planetary Motion, which can be applied to binary star systems. The law states that the square of the orbital period (\(T\)) is proportional to the cube of the semi-major axis (\(a\)) of the orbit. Mathematically, this can be expressed as\(\(T^2 = \frac{4\pi^2}{G(M_1 + M_2)}a^3\), where \(M_1\) and \(M_2\)\)  are the masses of the stars,\(\(G\)\) is the gravitational constant, and other variables have their usual meanings.

Given that one star has a mass of 1.00 solar masses, we can substitute the known values into the equation and solve for\(\(M_2\)\). Rearranging the equation, we have\(\(M_2 = \frac{4\pi^2}{G}(\frac{a^3}{T^2}) - M_1\)\).

Plugging in the values for\(\(a\) (1.34E+10 km) and \(T\) (400 years)\), and using the appropriate unit conversions, we can calculate the mass of the other star,\(\(M_2\\), to be approximately 0.541 solar masses.

Therefore, the mass of the other star in the binary system is approximately 0.541 solar masses.

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The minimum energy required to ionize a hydrogen atom in the n = 3 state is 12.09 eV.

What is hydrogen atom? A hydrogen atom is an atom of the chemical element hydrogen. It is made up of one proton and one electron, making it the simplest and most abundant element in the universe. In a hydrogen atom, the electron is bound to the proton by an electromagnetic force.The Rydberg formula is used to compute the energy required to ionize a hydrogen atom in the nth energy level. An ionization process occurs when an electron is removed from the outermost shell of an atom. The Rydberg formula is:

1/wavelength = R\((1/n1^2 - 1/n2^2)\) Where n1 and n2 are the initial and final energy levels, respectively.

R is the Rydberg constant, which is equal to

\(1.097 x 10^7 m^-1.\)

If we substitute the values given in the problem into the Rydberg formula, we can solve for the minimum energy required to ionize a hydrogen atom in the

n = 3 state:1/wavelength = R\((1/n1^2 - 1/n2^2)1/0 - 1/9 = 1.097 x 10^7 m^-1(1/9 - 1/4)\)Solving for the wavelength,

we get: wavelength = 972.5 nm The minimum energy required to ionize a hydrogen atom in the n = 3 state can now be determined by using the energy equation:

E = hc/wavelength E = \((6.626 x 10^-34 J s)(2.998 x 10^8 m/s)/(972.5 x 10^-9 m)E = 2.044 x 10^-18\)

JConverting Joules to electron volts (eV), we get:

\(E = 2.044 x 10^-18 J/(1.602 x 10^-19 J/eV)E = 12.09 eV\)

Therefore, the minimum energy required to ionize a hydrogen atom in the n = 3 state is 12.09 eV.

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Be sure to adhere to all safety precautions while performing the test, such as wearing protective gear, being cautious of the surroundings, and working in teams.

Ensure that you have obtained the necessary permits and permissions from relevant authorities to test the water at the reservoir. Be sure to adhere to all safety precautions while performing the test, such as wearing protective gear, being cautious of the surroundings, and working in teams.

Before leaving for the dam, planning is essential. It will allow you to test the water quality in a systematic and thorough way. The following are some important aspects that must be considered:

Test Equipment Selection - Choose suitable test equipment to test the water quality at various depths. This equipment must be calibrated, cleaned and checked for accuracy prior to testing. Also, take spare parts or additional equipment just in case something happens.Time and Duration - The testing time and duration should be planned to take into account all factors that may affect the test. This includes the weather, time of day, depth of the water and other environmental factors.

Depending on the location, the test should be done during the same period to ensure consistency.

Transportation - Make sure that the test equipment is safely transported and can be easily moved around the test site. Consider factors like access points, terrain, and the distance from the test site to the base. Make sure you take enough fuel for transportation.

Test Plan - Develop a test plan that outlines the testing procedure, test objectives, testing schedule, and personnel requirements. This plan will be your guide to ensure that the testing is done properly and within a specific time frame.

Record Keeping - During the test, record the data from all the measurements taken. Record-keeping is crucial for future analysis and comparisons. It is also important to have a detailed map of the site to track the exact location of each measurement taken.

Finally, ensure that you have obtained the necessary permits and permissions from relevant authorities to test the water at the reservoir. Be sure to adhere to all safety precautions while performing the test, such as wearing protective gear, being cautious of the surroundings, and working in teams.

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Answer:

40.8 m North West.

Explanation:

Please see attached photo for diagram.

In the attached photo, R represent the displacement of the dog.

The displacement (R) of the dog can obtained by using the pythagoras theory as illustrated below:

R² = 35² + 21²

R² = 1225 + 441

R² = 1666

Take the square root of both side.

R = √1666

R = 40.8 m North West.

Therefore, the displacement of the dog is 40.8 m North West.

Answer:

Explanation:

3.011 x 10^24 atoms

Answer:

jskskskssksksksskssksksksksks

The correct answer to the focal length of the corrective contact lens needed by a nearsighted person whose far point is 60 cm is 60 cm.

The focal length can be calculated by using the lens formula: 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance.

In this case, the far point is 60 cm, which means that the person can see objects clearly only if they are within 60 cm from the eye. Therefore, the image distance (v) is 60 cm. The object distance (u) is infinity, since the person is trying to see distant objects clearly.

Putting these values into the lens formula, we get:

1/f = 1/60 - 1/infinity

1/f = 1/60

f = 60 cm

Therefore, the focal length of the corrective contact lens needed by this person is 60 cm.

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The device which is used to measure potential difference between two points in a circuit is called a voltmeter.

What is a voltmeter?A voltmeter is a type of electrical instrument that is used to measure electrical potential difference between two points in a circuit. It is a type of electromechanical measuring instrument. It typically measures voltage in volts and it is represented by the symbol "V" in a circuit diagram. It is often used in parallel with the component that is being tested. The device which is used to measure potential difference between two points in a circuit is called a voltmeter.

Overall, a voltmeter is an essential tool for any electronics or electrical engineer. It enables engineers and technicians to measure and monitor voltage levels in circuits and devices, and it provides important information for designing and troubleshooting electrical systems.

In conclusion, the device that is used to measure potential difference between two points in a circuit is called a voltmeter. It is a type of electromechanical measuring instrument that measures voltage in volts and it is represented by the symbol "V" in a circuit diagram. Voltmeters are essential tools for measuring and monitoring voltage levels in circuits and devices, and they are used extensively in electronics and electrical engineering.

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A ball dropped from 3m rebounds to 2m. Velocity upon reaching the floor is 7.67 m/s, leaving the floor is -7.644 m/s. The average acceleration magnitude is \(1.3 m/s^2\), a downward direction.

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Round-nosed bullets with low velocities are specifically designed for improved accuracy and safety in target shooting and hunting scenarios. The round-nosed shape reduces air resistance, allowing for stable trajectory and accuracy at lower velocities. They are also safer for shooting in close quarters and reduce the risk of over-penetration.

Round-nosed bullets with low velocities are specifically designed for certain purposes in firearms. These bullets are commonly used in target shooting and hunting scenarios. The round-nosed shape of the bullet helps to reduce air resistance, allowing it to maintain a stable trajectory and accuracy at lower velocities. This makes them suitable for shooting at shorter distances or when precision is required.

Additionally, the low velocity of these bullets reduces the risk of over-penetration, making them safer for shooting in close quarters or in situations where there may be a risk of unintended collateral damage. The round-nosed design also helps to transfer energy more efficiently upon impact, which can be beneficial for hunting applications.

Overall, round-nosed bullets with low velocities offer improved accuracy and safety in specific shooting scenarios.

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(a)The pressure at B is 0.1248 atm.

(b)The temperature at C is 727.1 K.

(c)The total work done by the gas in one cycle is -1979J

General calculation:

We can use the First Law of Thermodynamics to analyze the heat engine cycle:

ΔU = Q - W

where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system. For a complete cycle, ΔU = 0, so:

Q = W

We can also use the ideal gas law to relate the pressure, volume, and temperature of the gas:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the absolute temperature.

(a)How to find the pressure at B segment?

To find the pressure at B, we can use the fact that the segment AB is an isothermal expansion. This means that the temperature remains constant, so:

PV = nRT

PB = (nRT)/(2V) = (2.00 mol)(0.0821 L·atm/mol·K)(600 K)/(2V) = (0.0821 L·atm/mol)(600 K)/V

Since the pressure at A is 5.00 atm, we can use the fact that the temperature is constant to find the volume at A:

PV = nRT

VA = (nRT)/P = (2.00 mol)(0.0821 L·atm/mol·K)(600 K)/5.00 atm = 197.76 L

Since the volume at B is twice the volume at A, we have:

VB = 2VA = 395.52 L

Substituting into the expression for PB, we get:

PB = (0.0821 L·atm/mol)(600 K)/395.52 L = 0.1248 atm

Therefore, the pressure at B is 0.1248 atm.

To find the temperature at C, we can use the fact that the segment BC is an adiabatic expansion. This means that no heat is added or removed from the system, so:

\(PV^\gamma\)= constant

where γ is the ratio of specific heats (for a monoatomic gas, γ = 5/3). We can use the fact that the volume at C is equal to the volume at A to find the pressure at C:

\(PAV^\gamma = PCV^\gamma\)

PC =  \(PA(V/A)^\gamma\) = 5.00 atm\((1/2)^(^5^/^3^)\) = 1.556 atm

Since the segment BC is adiabatic, the temperature changes but no heat is added or removed from the system. Using the ideal gas law, we can relate the pressure, volume, and temperature:

PV = nRT

TC = (PCVC)/(nR) = (1.556 atm)(197.76 L)/(2.00 mol)(0.0821 L·atm/mol·K) = 727.1 K

Therefore, the temperature at C is 727.1 K.

The total work done by the gas in one cycle is the sum of the work done in each segment of the cycle:

W = WAB + WBC + WCD + WDA

For segment AB, the work done is:

WAB = -QAB = -∫PdV = -nRT∫(1/V)dV = -nRT ln(VB/VA) = -(2.00 mol)(0.0821 L·atm/mol·K)(600 K) ln(2) = -602 J

For segment BC, the work done is:

WBC = -QBC = -∫PdV = -nγRT∫(1/V)dV = -nγRT

We know that VB = 2VA and VC = 2VD, so we can express the ratio VB/VC in terms of VA/VD:

VB/VC = (2VA)/(2VD) = VA/VD

Substituting into the expression for WBC, we get:

WBC = -nγRT ln(VA/VD)

For segment CD, the work done is:

WCD = -QCD + PCDΔV = -nCpΔT + PCDΔV

where Cp is the specific heat at constant pressure, ΔT is the change in temperature, and ΔV is the change in volume. We know that the segment CD is isobaric, so ΔV = VB - VA = (2VA) - VA = VA. We can also use the ideal gas law to relate the pressure, volume, and temperature:

Substituting into the expression for WCD, we get:

WCD = -nCpΔT + (nRT/VD)VA = -nCp(TC - TD) + (nRT/VD)VA

For segment DA, the work done is:

WDA = -QDA + ΔU = -nCvΔT

where Cv is the specific heat at constant volume. We know that the segment DA is isovolumetric, so ΔV = 0. Using the First Law of Thermodynamics, we know that ΔU = 0 for a complete cycle, so:

QDA = -WDA = nCvΔT

Substituting into the expression for WDA, we get:

WDA = -nCvΔT

Adding up the work done in each segment, we get:

W = WAB + WBC + WCD + WDA

= -(2.00 mol)(0.0821 L·atm/mol·K)(600 K) ln(2)- (2.00 mol)(5/3)(0.0821 L·atm/mol·K)(727.1 K) ln(VA/VD)- (2.00 mol)(Cp)(TC - TD) + (2.00 mol)(0.0821 L·atm/mol·K)(600 K) ln(2)- (2.00 mol)(Cv)(TC - TA)

We know that Cp and Cv for a monoatomic gas are related by Cp = Cv + R, so we can express Cp in terms of Cv:

Cp = Cv + R = (3/2)R + R = (5/2)R

Substituting and simplifying, we get:

W = (2.00 mol)(0.0821 L·atm/mol·K)(600 K) ln(2)- (2.00 mol)(5/3)(0.0821 L·atm/mol·K)(727.1 K) ln(VA/VD)- (2.00 mol)(5/2)(0.0821 L·atm/mol·K)(727.1 K)+ (2.00 mol)(5/2)(0.0821 L·atm/mol·K)(600 K)

W = -966.2 J - 4957 J - 7476 J + 5154 J

    = -1979 J

Therefore, the total work done by the gas in one cycle is -1979 J

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Strong x-ray beams can be produced when cathode rays strike a metal anode.

This process is known as the production of bremsstrahlung radiation. When high-speed electrons, also called cathode rays, are accelerated and then collide with a metal target, they are abruptly decelerated, and the kinetic energy lost is converted into X-ray photons.

The resulting X-ray beam produced can be strong and intense, and its properties depend on the energy of the incident electrons and the material of the target.

Gamma rays moving through a magnetic field, alpha rays passing through a thin metal foil, or beta rays being absorbed by bones do not directly produce strong X-ray beams.

In summary, the correct answer is A) cathode rays striking a metal anode can produce strong X-ray beams.

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Answer:

shirt is little attached to the body, it can come off and fly away

Explanation:

In electrostatics, charges of different signs attract and charges of the same sign repel.

In this case, when a negative charge is placed on it, both the inventor and the shirt are charged, therefore there is a repulsive force, also there is an attraction between the positive charge of the roof attracts the negative charge, such as the shirt. of weak the two forces not greater than the resistance of the walk.

As the shirt is little attached to the body, it can come off and fly away

Therefore, the electric field magnitude at the point \(\(r = 21.16\)\) mm is\(\(3.82 \times 10^7\) N/C.\)

1. Electric Field due to the Infinite Cylindrical Conductor:

  The electric field inside and outside a uniformly charged cylindrical conductor is given by:  

\(\[ E_1 = \frac{{\lambda_1}}{{2\pi\epsilon_0}} \left(\frac{1}{{r_a}} - \frac{1}{{r_b}}\right) \]\)

  Where:

  - \(\( E_1 \)\) is the electric field due to the cylindrical conductor.

  -\(\( \lambda_1 \)\) is the linear charge density of the conductor.

  - \(\( r_a \)\) is the inner radius of the conductor.

  - \(\( r_b \)\) is the outer radius of the conductor.

  - \(\( \epsilon_0 \)\) is the permittivity of free space \((\( 8.854 \times 10^{-12} \, \text{C}^2/\text{N}\cdot \text{m}^2 \)).\)

  Substituting the given values:

\(\[ E_1 = \frac{{102 \times 10^{-9}}}{{2\pi \times 8.854 \times 10^{-12}}} \left(\frac{1}{{48.5 \times 10^{-3}}} - \frac{1}{{76.3 \times 10^{-3}}}\right) \]\)

2. Electric Field due to the Infinite Line Charge:

  The electric field due to an infinite line charge is given by:

\(\[ E_2 = \frac{{\lambda_2}}{{2\pi\epsilon_0 r}} \]\)

  Where:

  - \(\( E_2 \)\) is the electric field due to the line charge.

  - \(\( \lambda_2 \)\)is the linear charge density of the line charge.

  - r is the distance from the line charge to the point of interest.

  Substituting the given values:

 \(\[ E_2 = \frac{{-10.4 \times 10^{-9}}}{{2\pi \times 8.854 \times 10^{-12} \times 21.16 \times 10^{-3}}} \]\)

3. Total Electric Field at the Point:

  Since the electric fields due to the cylindrical conductor and line charge are in opposite directions, we subtract their magnitudes to find the net electric field at the point of interest:  

\(\[ E_{\text{total}} = \sqrt{{E_1}^2 + {E_2}^2} \]\)

Sure! Let's calculate the electric field magnitude at the point \(r = 21.16\) mm using the given values.

1. Electric Field due to the Infinite Cylindrical Conductor:

\(\[ E_1 = \frac{{102 \times 10^{-9}}}{{2\pi \times 8.854 \times 10^{-12}}} \left(\frac{1}{{48.5 \times 10^{-3}}} - \frac{1}{{76.3 \times 10^{-3}}}\right) \]\\ Simplifying the expression: \[ E_1 = 2.19 \times 10^6 \, \text{N/C} \]\)

2. Electric Field due to the Infinite Line Charge:

 \(\[ E_2 = \frac{{-10.4 \times 10^{-9}}}{{2\pi \times 8.854 \times 10^{-12} \times 21.16 \times 10^{-3}}} \]\\ Simplifying the expression: \[ E_2 = -3.81 \times 10^7 \, \text{N/C} \]\)

3. Total Electric Field at the Point:

 \(\[ E_{\text{total}} = \sqrt{{E_1}^2 + {E_2}^2} \]\\ Substituting the calculated values: \[ E_{\text{total}} = \sqrt{{(2.19 \times 10^6)^2 + (-3.81 \times 10^7)^2}} \] Simplifying the expression: \[ E_{\text{total}} = 3.82 \times 10^7 \, \text{N/C} \]\)

Therefore, the electric field magnitude at the point \(\(r = 21.16\)\)mm is\(\(3.82 \times 10^7\) N/C.\)

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The frequencies of the normal modes for small oscillations of the wheel with the mass m located at its center are ω₁ = √(k/m) and ω₂ = √(k/m + (kr²)/(2*m²)).

To find the frequencies of the normal modes for small oscillations of the wheel with the mass m located at its center, we can use the method of Lagrangian mechanics.

Let θ be the angle through which the wheel has rotated and x be the displacement of the mass m from its equilibrium position. Then, the Lagrangian of the system can be written as:

L = T - V

where T is the kinetic energy of the system and V is the potential energy of the system.

The kinetic energy of the system is given by:

T = 0.5m(dx/dt)² + 0.5I(d²θ/dt²)²

where I is the moment of inertia of the wheel about its center, which is given by I = 0.5mr².

The potential energy of the system is given by:

V = 0.5kx²

where k is the spring constant.

Using Lagrange's equations, we can find the equations of motion for the system:

d/dt(∂L/∂(dθ/dt)) - ∂L/∂θ = 0

d/dt(∂L/∂(dx/dt)) - ∂L/∂x = 0

Substituting the expressions for T and V into the above equations and simplifying, we get:

mr(d²θ/dt²) + kx = 0

m(d²x/dt²) + mr(d²θ/dt²) = 0

These equations can be combined and written in matrix form as:

(d²/dt²)[x;θ] + (k/m)[1,-r;1,0]*[x;θ] = 0

This is a system of coupled differential equations, which can be solved using the method of normal modes. We assume a solution of the form:

[x;θ] = [A;B]*exp(iωt)

where A and B are constants and ω is the frequency of the normal mode.

Substituting the above solution into the matrix equation and solving for ω, we get:

det[(d²/dt²)I + (k/m)[1,-r;1,0]] = 0

where I is the identity matrix.

Expanding the determinant and simplifying, we get:

(d²/dt² + k/m)[(d²/dt² + k/(2m))² + (kr²)/(4*m²)] = 0

The two roots of the above equation correspond to the two normal modes of the system. The first root is:

ω₁ = √(k/m)

which corresponds to a simple harmonic motion of the mass m along the axis of the wheel.

The second root is:

ω₂ = √(k/m + (kr²)/(2*m²))

which corresponds to a combination of the simple harmonic motion of the mass m and the rotational motion of the wheel about its center.

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The proton and electron, being opposite charges, would accelerate toward each other. Thus, 3rd option is correct.

According to Coulomb's law, the force between two charged particles is given by:

F = (k * |q₁ * q₂|) / r²

where F is the force between the particles, k is the electrostatic constant, q₁ and q₂ are the magnitudes of the charges of the particles (in this case, the charge of the proton and the charge of the electron), and r is the distance between the particles.

In the case of a proton and an electron, the proton has a positive charge (+e) and the electron has a negative charge (-e), where e is the elementary charge. Since the charges are opposite in sign, the product q₁ * q₂ is negative.

Therefore, the force between the proton and the electron is attractive, causing them to accelerate toward each other. This acceleration will continue until they collide or until external factors come into play (such as the presence of other particles or forces) that may alter their motion.

It's important to note that in a typical atomic or molecular system, electrons are usually bound to nuclei due to the attractive electrostatic forces between them. However, if an electron and a proton are initially separated and have no other influences, they will accelerate toward each other due to their opposite charges.

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Answer: FRANCE

Explanation:

Answer:

France

Explanation:

UK has 15 nuclear power plants and 2 under construction whereas France has 56 in operation and 1 under construction  

Conversion of electromagnetic (EM) energy from the sun into other forms of energy occurs for the following options:

a. Biofuels

c. Solar thermal power

e. Wind power (EM to air movements)

A- Biofuels: During photosynthesis, plants capture electromagnetic energy from the sun and convert it into chemical energy, stored in the bonds of organic molecules, such as glucose. This process allows for the conversion of EM energy to chemical energy in the form of biofuels.

c. Solar thermal power: Solar thermal power plants use mirrors or lenses to concentrate sunlight, which is then converted into heat energy. This thermal energy can be used to generate steam, which drives a turbine and produces mechanical energy.

e. Wind power: Wind turbines harness the kinetic energy of moving air, which is ultimately driven by the sun's uneven heating of the Earth's surface. The sun's energy heats the atmosphere, creating temperature and pressure gradients that result in wind currents.

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The molar mass of a gas at 78 c and 560 torr if 206 ng occupies 0.206 ul is 41.64 g/mol

The molar mass of a gas can be calculated using the ideal gas law equation and the given values of temperature, pressure, mass, and volume.

To calculate the molar mass of a gas, we can use the formula:

Molar mass = (mass of gas) / (number of moles of gas)

First, we need to determine the number of moles of gas. We can use the ideal gas law equation:

\(PV = nRT\)

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

We are given the temperature as 78 °C, which needs to be converted to Kelvin by adding 273.15:

T = 78 + 273.15 = 351.15 K

The pressure is given as 560 torr, and the volume is given as 0.206 µl.

Next, we can calculate the number of moles using the ideal gas law equation:

\(n = (PV) / (RT)\)

(0.5105 atm)(2.06×10⁻⁷ L) = n(0.0821 L-atm/mol-K)(318 K)

n = 4×10⁻⁹ mol

Molar mass = Mass/n = 2.06×10⁻⁷ g/4×10⁻⁹ mol

Molar mass = 41.64 g/mol

Now that we have the number of moles, we can calculate the molar mass by dividing the mass of the gas (given as 206 ng) by the number of moles.

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Answer: A. Objects 1 and 2 will not move, and objects 3 and 4 will accelerate upward.

Explanation:

Because objects 1 and 2 require more force in order to move, they will stay motionless.

Correct me if I am incorrect.

Answer:

The ball exerts a force of 400 N on the bat.

Explanation:

Given that,

A baseball player hits a ball with 400 N of force.

We need to find the force the ball exert on the bat.

We know that,

According to Newton's third law, when object 1 exerts a force on an object 2, then object 2 will exert a force on object 1 but in opposite direction.

So, the ball exerts a force of 400 N on the bat.

Given

Rate of metabolic rate,

Mass of water,

The initial temperature,

Explanation

The heat transferred in 43 minutes is given by

Conclusion

The temperature increase to 28.099 degC

Answer:

I would say there is friction against the floor, air resistance, and gravity.

Explanation:

The contact force acting on the toy is the frictional force, and the non-contact force acting on the toy is the gravitational force.

According to the question, a toy train is being pulled by a string along the floor.

The gravitational force acting on the toy train is the weight of the toy train, which is a non-contact force given by:

F  = mg

where m is the mass of the toy, and g is the acceleration due to gravity.

Now, the floor will exert a normal force against the toy, equal to the weight of the toy. Let the normal force be N.

Since the toy is in contact with the floor, there is a frictional force acting against the motion of the toy trains, which is a contact force and is given by:

f  = μN = μmg

where μ is the coefficient of friction between the toy and the floor.

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In terms of the thermal inversion phenomena, the temperature of air is decreasing within a limited layer in the lower atmosphere, and pollution is trapped near the ground.

Thermal inversion refers to a weather condition where the temperature of air usually decreases with altitude but shows a reversal, causing a layer of warm air to be trapped above a layer of cooler air near the ground.

This inversion layer acts as a lid, preventing the vertical mixing of air and trapping pollutants close to the surface. The cooler air beneath the inversion layer cannot rise and disperse the pollutants effectively, leading to poor air quality and potential health hazards.

This phenomenon commonly occurs during stable atmospheric conditions, such as clear nights with light winds or in urban areas surrounded by mountains, where cold air drains into valleys and forms a temperature inversion.

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The net electric force on particle q2 is 27.81 N.

The net force on the particle q2 is calculated by applying Coulomb's law of electrostatic force.

The force on particle q1 due to particle q2 is calculated as;

F ( 12 ) = kq₁q₂/r²

where;

F ( 12 ) = ( 9 x 10⁹ x 18 x 10⁻⁶ x 11.2 x 10⁻⁶ ) / ( 0.28² )

F ( 12 ) = 23.14 N

The force on particle q3 due to particle q2 is calculated as;

F (23) = kq₃q₂/r²

F (23) = (9 x 10⁹ x 5.67 x 10⁻⁶ x 11.2 x 10⁻⁶)/(0.35²)

F (23) = 4.67 N

The net electric force on particle q2 is calculated as follows;

F (net) = 23.14 N + 4.67 N

F (net) = 27.81 N

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Therefore, the object distance that creates a magnification of 5.05 in front of the concave mirror with a focal length of 89.5 cm is approximately 22.1 cm.

To determine the object distance that creates a magnification of 5.05 in front of a concave mirror with a focal length of 89.5 cm, we can use the mirror equation:

1/f = 1/di + 1/do

Where:

f = focal length of the mirror (given as -89.5 cm for a concave mirror)

di = image distance

do = object distance

Given:

f = -89.5 cm

magnification (m) = 5.05

Since the image is upright and magnified, the magnification (m) can be calculated using the following formula:

m = -di/do

Substituting the given magnification into the formula, we get:

5.05 = -di/do

Rearranging the equation, we can solve for di:

di = -5.05 * do

Now, substitute the values into the mirror equation:

1/-89.5 cm = 1/di + 1/do

Substitute di = -5.05 * do:

1/-89.5 cm = 1/(-5.05 * do) + 1/do

Simplify the equation:

-1/89.5 cm = (1 - 5.05) / do

-1/89.5 cm = -4.05 / do

Cross-multiplying:

-89.5 cm = -4.05 * do

Dividing both sides by -4.05:

do = -89.5 cm / -4.05

do ≈ 22.1 cm

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Explanation:

a) i) Calculation of the friction force:

The friction force can be determined using the equation:

friction force = coefficient of friction * normal force

The normal force is equal to the weight of the object, which can be calculated as:

normal force = mass * gravitational acceleration

where the gravitational acceleration is approximately 9.8 m/s².

normal force = 75 kg * 9.8 m/s² = 735 N

friction force = 0.4 * 735 N = 294 N

ii) Calculation of the acceleration of the body:

Now, we can calculate the acceleration using Newton's second law:

net force = mass * acceleration

Since the applied force and the friction force act in opposite directions, the net force can be calculated as:

net force = applied force - friction force

net force = 300 N - 294 N = 6 N

mass = 75 kg

6 N = 75 kg * acceleration

acceleration = 6 N / 75 kg = 0.08 m/s²

b) Explanation:

In part (a), we calculated the friction force to be 294 N and the acceleration of the body to be 0.08 m/s². The positive acceleration indicates that the body is moving in the direction of the applied force.

The friction force opposes the motion of the body and acts in the opposite direction to the applied force. In this case, the applied force of 300 N is greater than the friction force of 294 N. As a result, the net force acting on the body is 6 N in the direction of the applied force.

The small net force of 6 N, compared to the body's mass of 75 kg, results in a relatively low acceleration of 0.08 m/s². This indicates that the body will accelerate slowly in the direction of the applied force due to the presence of friction.

Overall, the friction force and the resulting acceleration of the body are determined by the coefficient of friction (μ) and the mass of the object. In this case, the body experiences a relatively high friction force, leading to a small acceleration.