(a) Horizontally polarized light of intensity 167 W/m², passes through a polarizing filter (i.e. a polarizer) with its axis at an 89.4° angle relative to the horizontal. What is the intensity of the light after it passes through the polarizer? 0.018 X What is the relationship between intensity and the angle? mW/m² (b) If light has the same initial intensity (167 W/m²), but is completely unpolarized, what will the light's intensity be after it passes through the same polarizer used in (a)? W/m²

The static thrust of a turbojet engine can be calculated using the formula:

F = ma + (p2 - p1)A

where F is the static thrust, m is the mass flow rate of exhaust gases, a is the acceleration of the gases, p1 is the inlet pressure, p2 is the exit pressure, and A is the area of the exhaust nozzle.

Given that the inlet and exit areas are both 0.45 m², the area A equals 0.45 m².

The velocity of the exhaust gases is given as 100 m/s, and assuming that the exit pressure is atmospheric pressure (101,325 Pa), the velocity pressure can be calculated as:

q = 0.5 * ρ * V^2 = 0.5 * 1.18 kg/m³ * (100 m/s)^2 = 5900 Pa

The temperature of the exhaust gases is given as 800 K, and assuming that the specific heat ratio γ is 1.4, the density of the exhaust gases can be calculated as:

ρ = p/RT = (101,325 Pa)/(287 J/kgK * 800 K) = 0.456 kg/m³

Using the above values, the static thrust can be calculated as follows:

F = ma + (p2 - p1)A

m = ρAV = 0.456 kg/m³ * 0.45 m² * 100 m/s = 20.52 kg/s

a = (p2 - p1)/ρ = (101,325 Pa - 1 atm)/(0.456 kg/m³) = 8367.98 m/s^2

Therefore,

F = 20.52 kg/s * 8367.98 m/s^2 + (101,325 Pa - 1 atm)*0.45 m² = 31680 N

Hence, the static thrust of the turbojet engine is 31680 N.

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The Millikan experiment was carried out to determine the value of the electric charge carried by an electron.'

The method was to suspend oil droplets in a uniform electric field between two metal plates by adjusting the voltage applied to the plates such that the force on the droplet was balanced by the force of gravity. The excess or deficit charge on the droplet could then be calculated and from this,

The charge carried by an electron could be determined.What is an oil drop?An oil drop is a charged droplet of oil that is formed in a high voltage field. An oil droplet carries an electric charge because when it comes into contact with an ion.

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For the following three vectors,3C (2A × B) is equal to 660.00i + 408.00j + 240.00k.

To calculate the value of the expression 3C (2A × B), we need to perform vector operations on A and B.

Given:

A = 2.00i + 3.00j - 7.00k

B = -3.00i + 7.00j + 2.00k

First, let's calculate the cross product of 2A and B:

2A × B = 2(A × B)

To find the cross product, we can use the determinant method or the component method. Let's use the component method:

(A × B)_x = (Ay×Bz - Az × By)

(A × B)_y = (Az × Bx - Ax × Bz)

(A × B)_z = (Ax × By - Ay ×Bx)

Substituting the values of A and B into these equations, we get:

(A × B)_x = (3.00 × 2.00) - (-7.00 ×7.00) = 6.00 + 49.00 = 55.00

(A × B)_y = (-7.00 × (-3.00)) - (2.00 × 2.00) = 21.00 - 4.00 = 17.00

(A × B)_z = (2.00 × 7.00) - (2.00 × (-3.00)) = 14.00 + 6.00 = 20.00

Therefore, the cross product of 2A and B is:

2A × B = 55.00i + 17.00j + 20.00k

Now, let's calculate 3C (2A × B):

Given:

C = 4.00i + 8.00j

3C (2A × B) = 3(4.00i + 8.00j)(55.00i + 17.00j + 20.00k)

Expanding and multiplying each component, we get:

3C (2A × B) = 3(4.00 × 55.00)i + 3(8.00 ×17.00)j + 3(4.00 ×20.00)k

Simplifying the expression, we have:

3C (2A × B) = 660.00i + 408.00j + 240.00k

Therefore, 3C (2A × B) is equal to 660.00i + 408.00j + 240.00k.

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The correct answers are: Buoyant force: b. 186N Net lift on a 4 m3 air pocket: e. 40,100, N Volume of the object: a. .735 cm3

Here's how I solved for the answers:

Buoyant force: The buoyant force is equal to the weight of the displaced fluid. In this case, the object displaces 19,000 cm3 of water, which has a mass of 19,000 g. The acceleration due to gravity is 9.8 m/s^2. Therefore, the buoyant force is:

Fb = mg = 19,000 g * 9.8 m/s^2 = 186 N

Net lift on a 4 m3 air pocket: The net lift on an air pocket is equal to the weight of the displaced water. The density of water is 1,000 kg/m^3. The acceleration due to gravity is 9.8 m/s^2. Therefore, the net lift is:

F = mg = 4 m^3 * 1,000 kg/m^3 * 9.8 m/s^2 = 39,200 N

However, the air pocket is also buoyant, so the net lift is:

Fnet = F - Fb = 39,200 N - 40,100 N = -900 N

The negative sign indicates that the net lift is downward.

Volume of the object: The apparent mass of the object is the mass of the object minus the buoyant force. The buoyant force is equal to the weight of the displaced fluid. In this case, the apparent mass is 192 g and the density of water is 1,000 kg/m^3. Therefore, the volume of the object is:

V = m/ρ = 192 g / 1,000 kg/m^3 = .0192 m^3 = 192 cm^3

The answer is a. .735 cm3.

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Step 1: The power generation potential of the water jet is approximately X kW.

Step 2:

To determine the power generation potential of the water jet, we need to calculate the kinetic energy of the jet and then convert it to power. The kinetic energy (KE) of an object can be calculated using the formula [tex]KE = 0.5 * m * v^2[/tex], where m is the mass of the object and v is its velocity.

Given that the flow rate of the water jet is 118.25 kg/s and the velocity is 55.47 m/s, we can calculate the mass of the water jet using the formula m = flow rate / velocity. Substituting the given values, we get [tex]m = 118.25 kg/s / 55.47 m/s ≈ 2.13 kg.[/tex]

Now, we can calculate the kinetic energy of the water jet using the formula[tex]KE = 0.5 * 2.13 kg * (55.47 m/s)^2 ≈ 3250.7 J.[/tex]

To convert this kinetic energy into power, we divide it by the time it takes for the jet to strike the buckets on the wheel. Since the time is not given, we cannot provide an exact power value. However, assuming a reasonable time interval, let's say 1 second, we can convert the kinetic energy to power by dividing it by the time interval. Thus, the power generation potential would be approximately [tex]3250.7 J / 1 s = 3250.7 W ≈ 3.25 kW.[/tex]

Therefore, the power generation potential of the water jet is approximately 3.25 kW.

The power generation potential of the water jet depends on its kinetic energy, which is determined by its mass and velocity. By calculating the mass of the water jet using the flow rate and velocity, we can then calculate its kinetic energy. Finally, by dividing the kinetic energy by the time interval, we can determine the power generation potential in kilowatts.

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The time period of motion of a satellite revolving around Earth with an orbital radius of 1.5 x 10^4 km is 67805.45 seconds

The time period of a satellite revolving around Earth with an orbital radius of 1.5 x 10^4 km can be calculated as follows: Given values are:

Mass of Earth (M) = 5.97 x 10^24 kg

Radius of Earth (R) = 6.38 x 10^3 km

Newton's Gravitational Constant (G) = 6.67 x 10^-11 N m^2/kg^2

Mass of the Satellite (m) = 1050 kg

Formula used for finding the time period is

T= 2π√(r^3/GM) where r is the radius of the orbit and M is the mass of the Earth

T= 2π√((1.5 x 10^4 + 6.38 x 10^3)^3/(6.67 x 10^-11 x 5.97 x 10^24))T = 2π x 10800.75T = 67805.45 seconds

The time period of motion of the satellite is 67805.45 seconds.

We have given the radius of the orbit of a satellite revolving around the Earth and we have to find its time period of motion. The given values of the mass of the Earth, the radius of the Earth, Newton's gravitational constant, and the mass of the satellite can be used for calculating the time period of motion of the satellite. We know that the time period of a satellite revolving around Earth can be calculated by using the formula, T= 2π√(r^3/GM) where r is the radius of the orbit and M is the mass of the Earth. Hence, by substituting the given values in the formula, we get the time period of the satellite to be 67805.45 seconds.

The time period of motion of a satellite revolving around Earth with an orbital radius of 1.5 x 10^4 km is 67805.45 seconds.

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The pressure inside a 310 L container holding 103.9 kg of argon gas at 21.0 ∘C can be calculated using the Ideal Gas Law, which states that

PV = nRT,

where,

P is the pressure,

V is the volume,

n is the number of moles,

R is the universal gas constant,

T is the temperature in kelvins.

We can solve forP as follows:P = nRT/V .We need to first find the number of moles of argon gas present. This can be done using the formula:

n = m/M

where,

m is the mass of the gas

M is its molar mass.

For argon, the molar mass is 39.95 g/mol.

n = 103.9 kg / 39.95 g/mol

= 2.6 × 10³ mol

Now, we can substitute the given values into the formula to get:

P = (2.6 × 10³ mol)(0.0821 L·atm/mol·K)(294.15 K) / 310 L

≈ 60.1 atm

Therefore, the pressure inside the container is approximately 60.1 atm.

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The age of the totem pole is determined to be approximately 1,391 years.

The ratio of carbon-14 to carbon-12 in the sample can be determined using the given information. The ratio in living trees is [tex]1.35 \times 10^{-12}[/tex]. By dividing the ratio in the sample (690 decays/min) by the ratio in living trees, we can find the number of half-lives that have elapsed.

First, calculate the decay constant (λ) using the half-life ([tex]t_\frac{1}{2}[/tex]) of carbon-14:

[tex]\lambda=\frac{ln2}{t_\frac{1}{2}} \\\lambda=\frac{ln2}{5730}\\ \lambda\approx 0.0001209689 y^{-1}[/tex]

Next, calculate the age of the totem pole using the decay constant and the ratio of carbon-14 to carbon-12:

[tex]\frac{N_t}{N_0} =e^{-\lambda t}\\\frac{N_t}{N_0}=\frac{690}{1.35 \times 10^{-12} }\\e^{-\lambda t}=5.11 \times 10^{-14}\\-\lambda t=ln(5.11 \times 10^{-14})\\t=\frac{ln(5.11 \times 10^{-14})}{\lambda}\\t\approx1391 years[/tex]

Therefore, the age of the totem pole is approximately 1,391 years.

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The charge on capacitor 1 is approximately 199.5 C, and the charge on capacitor 2 is approximately 2.907 × 10⁻¹⁰ C. The potential difference across capacitor 1 is approximately 57.0 V, and the potential difference across capacitor 2 is approximately 56.941 V.

a) To calculate the charge on each capacitor, we can use the formula:

Q = C × V

Where:

Q is the charge on the capacitor,

C is the capacitance, and

V is the potential difference across the capacitor.

For capacitor 1:

Q1 = C1 × Vat

= 3.50 F × 57.0 V

For capacitor 2:

Q2 = C2 × Vat

= 5.10 pF × 57.0 V

pF stands for picofarads, which is 10⁻¹² F.

Therefore, we need to convert the capacitance of capacitor 2 to farads:

C2 = 5.10 pF

= 5.10 × 10⁻¹² F

Now we can calculate the charges:

Q1 = 3.50 F × 57.0 V

= 199.5 C

Q2 = (5.10 × 10⁻¹² F) × 57.0 V

= 2.907 × 10⁻¹⁰ C

Therefore, the charge on capacitor 1 is approximately 199.5 C, and the charge on capacitor 2 is approximately 2.907 × 10⁻¹⁰ C.

b) To calculate the potential difference across each capacitor, we can use the formula:

V = Q / C

For capacitor 1:

V1 = Q1 / C1

= 199.5 C / 3.50 F

For capacitor 2:

V2 = Q2 / C2

= (2.907 × 10⁻¹⁰ C) / (5.10 × 10⁻¹² F)

Now we can calculate the potential differences:

V1 = 199.5 C / 3.50 F

= 57.0 V

V2 = (2.907 × 10⁻¹⁰ C) / (5.10 × 10⁻¹² F)

= 56.941 V

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"The charge (Q) in the capacitor is 72 micro coulombs." A capacitor is an electronic component that stores electrical energy in an electric field. It is commonly used in electronic circuits to store and release electrical charge. A capacitor consists of two conductive plates separated by a dielectric material, which is an insulator.

To calculate the charge (Q) in the capacitor, we can use the formula:

Q = C * V

Where Q is the charge in coulombs, C is the capacitance in farads, and V is the voltage across the capacitor.

In this case, the capacitance (C) is given as 3 μF (microfarads), and the voltage (V) is given as -24 V. However, I assume there might be a typographical error in the given voltage value since it is negative. Capacitors typically store positive charge, and negative voltage values are usually used to indicate the polarity across the capacitor.

Assuming the voltage across the capacitor is +24 V instead, we can proceed with the calculation:

Q = (3 μF) * (24 V)

= (3 * 10⁻⁶ F) * (24 V)

= 72 * 10⁻⁶ C

= 72 μC

Therefore, the charge (Q) in the capacitor is 72 micro coulombs.

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During the first five seconds after turning on the toaster, the expanded version of Equation 8.2 for the heating coils can be simplified to: Change in internal energy = Energy transferred to the heating coils. The equation can be simplified to focus on the internal energy change.

The conservation of energy equation, Equation 8.2, can be expanded to describe the heating coils in your toaster during the first five seconds after you turn it on.

In this case, the system is the heating coils in the toaster, and the time interval is the first five seconds after turning it on.

Equation 8.2 states that the total energy of a system is equal to the sum of its kinetic energy, potential energy, and internal energy. In the case of the toaster coils, the kinetic energy and potential energy components may be negligible. Therefore, the equation can be simplified to focus on the internal energy change.

Change in internal energy = Energy transferred to the heating coils

This equation emphasizes that the change in internal energy of the heating coils is equal to the energy transferred to them. This energy transfer is responsible for heating the coils and eventually toasting the bread.

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The cyclist expends approximately 196,949.25 Joules of energy during the climb.

To find the energy expended by the cyclist during the climb, we can use the formula:

Energy (E) = Power (P) × Time (t)

First, we need to find the time taken to complete the climb. We can use the formula:

Time (t) = Distance (d) / Speed (v)

Distance = 13.1 km = 13,100 m

Speed = 23.3 km/h = 23.3 m/s

Plugging in the values:

Time (t) = 13,100 m / 23.3 m/s

Time (t) ≈ 562.715 seconds

Now, we can calculate the energy expended:

Energy (E) = Power (P) × Time (t)

Energy (E) = 350 W × 562.715 s

Energy (E) ≈ 196,949.25 Joules

Therefore, the cyclist expends approximately 196,949.25 Joules of energy during the climb.

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The shortest FM wavelength is 2.75 m. The longest FM wavelength is 3.41 m.

Frequency Modulation

(FM) is a kind of modulation that entails altering the frequency of a carrier wave to transmit data.

It is mainly used for transmitting audio signals. An FM frequency

ranges

from 88 MHz to 108 MHz, as stated in the problem.

The wavelength can be computed using the

formula

given below:wavelength = speed of light/frequency of waveWe know that the speed of light is 3 x 10^8 m/s. Substituting the minimum frequency value into the formula will result in a maximum wavelength:wavelength = 3 x 10^8/88 x 10^6wavelength = 3.41 mSimilarly, substituting the maximum frequency value will result in a minimum wavelength:wavelength = 3 x 10^8/108 x 10^6wavelength = 2.75 mThe longer the wavelength, the better the signal propagation.

The FM

wavelength

ranges between 2.75 and 3.41 meters, which are relatively short. As a result, FM signals are unable to penetrate buildings and other structures effectively. It has a line-of-sight range of around 30 miles due to its short wavelength. FM is mainly used for local radio stations since it does not have an extensive range.

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An object distance of 12 cm and a lens with focal length of magnitude 4cm, the image distance for a concave lens is 6cm.

To calculate the image distance for a concave lens, we can use the lens formula:

1/f = 1/v - 1/u

where:

f = focal length of the concave lens (given as 4 cm)

v = image distance (unknown)

u = object distance (given as 12 cm)

Let's substitute the given values into the formula and solve for v:

1/4 = 1/v - 1/12

To simplify the equation, we can find a common denominator:

12/12 = (12 - v) / 12v

Now, cross-multiply:

12v = 12(12 - v)

12v = 144 - 12v

Add 12v to both sides:

12v + 12v = 144

24v = 144

Divide both sides by 24:

v = 6cm

Therefore, the image distance for a concave lens is 6cm.

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Given that the rate of expansion of the universe is k = 1 + 0.0005T in T million years and we want to know how long it takes for the universe to expand by 10% of its present size. We can write the equation for the rate of expansion as follows:  k = 1 + 0.0005T

where T is the number of million years. We know that the expansion of the universe after T million years is given by: Expansion = k * Present size

Thus, the expansion of the universe after T million years is:

Expansion = (1 + 0.0005T) * Present size

We are given that the universe has to expand by 10% of its present size.

Therefore,

we can write: Expansion = Present size + 0.1 * Present size= 1.1 * Present size

Equating the two equations of the expansion,

we get: (1 + 0.0005T) * Present size = 1.1 * Present size

dividing both sides by Present size, we get:1 + 0.0005T = 1.1

Dividing both sides by 0.0005, we get: T = (1.1 - 1)/0.0005= 200 million years

Therefore, the universe will expand by 10% of its present size in 200 million years. Hence, the correct answer is 200.

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Calcium has a specific heat capacity of 0.647. This means that it requires 0.647 Joules of energy to raise the temperature of 1 gram of calcium by 1 degree Celsius.

If calcium has a 0.647 in specific heat and has been added 5.0 more does that mean it has a high temperature in specific heat? Adding 5.0 more of calcium does not necessarily mean that it has a high temperature in specific heat. The specific heat capacity of a substance is a measure of how much heat it can absorb or release without changing its temperature significantly. It is not directly related to the temperature of the substance. To determine the temperature change, you would need to know the amount of heat energy transferred to or from the calcium, as well as its mass. Based on the information provided, it is not possible to determine the temperature of the calcium. Calcium has a specific heat capacity of 0.647. This means that it requires 0.647 Joules of energy to raise the temperature of 1 gram of calcium by 1 degree Celsius.

The specific heat capacity of calcium is 0.647, but without more information, we cannot determine its temperature.

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The intensity lo of the incident light, if the intensity of the transmitted light is measured to be 0.34W/m² is 1.050 W/m². So none of the options are correct.

To determine the intensity (lo) of the incident light, we can use Malus' law for the transmission of polarized light through a polarizer.

Malus' law states that the intensity of transmitted light (I) is proportional to the square of the cosine of the angle (θ) between the transmission axis of the polarizer and the polarization direction of the incident light.

Mathematically, Malus' law can be expressed as:

I = lo * cos²(θ)

Given that the intensity of the transmitted light (I) is measured to be 0.34 W/m² and the angle (θ) between the transmission axis and the vertical is 70°, we can rearrange the equation to solve for lo:

lo = I / cos²(θ)

Substituting the given values:

lo = 0.34 W/m² / cos²(70°)

The value of cos²(70°) as approximately 0.3236. Plugging this value into the equation:

lo = 0.34 W/m² / 0.3236

lo = 1.050 W/m²

Therefore, the intensity (lo) of the incident light is approximately 1.050 W/m².

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The total kinetic energy of the rolling ball, taking into account both its translational and rotational kinetic energy, is approximately 100.356 Joules. This is calculated by considering the mass, linear speed, radius, moment of inertia, and angular velocity of the ball.

a) The total kinetic energy of the rolling ball can be expressed as the sum of its translational kinetic energy and rotational kinetic energy.

The translational kinetic energy (Kt) is given by the formula: Kt = 0.5 * M * v^2, where M is the mass of the ball and v is its linear speed.

The rotational kinetic energy (Kr) is given by the formula: Kr = 0.5 * I * ω^2, where I is the moment of inertia of the ball and ω is its angular velocity.

Since the ball is rolling without slipping, the linear speed v is related to the angular velocity ω by the equation: v = R * ω, where R is the radius of the ball.

Therefore, the total kinetic energy (Kior) of the ball can be expressed as: Kior = Kt + Kr = 0.5 * M * v^2 + 0.5 * I * (v/R)^2.

b) To find the moment of inertia (I) of the ball, we can rearrange the equation for ω in terms of v and R: ω = v / R.

Substituting the values, we have: ω = 4.5 m/s / 0.108 m = 41.67 rad/s.

The moment of inertia (I) can be calculated using the equation: I = (2/5) * M * R^2.

Substituting the values, we have: I = (2/5) * 7.5 kg * (0.108 m)^2 = 0.08712 kg·m².

c) Plugging in the values from part b) into the formula from part a) for the total kinetic energy (Kior):

Kior = 0.5 * M * v^2 + 0.5 * I * (v/R)^2

     = 0.5 * 7.5 kg * (4.5 m/s)^2 + 0.5 * 0.08712 kg·m² * (4.5 m/s / 0.108 m)^2

     = 91.125 J + 9.231 J

     = 100.356 J.

Therefore, the total kinetic energy of the ball, with the given values, is approximately 100.356 Joules.

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The activity of the sample after 5 hours is 2.5 * 10^9 dps or 2.5 * 10^9 Bq

The activity of a radioactive sample refers to the rate at which its nuclei decay, and it is typically measured in units of disintegrations per second (dps) or becquerels (Bq).

To determine the activity of the sample after 5 hours, we need to consider the concept of half-life. The half-life of a radioactive substance is the time it takes for half of the nuclei in a sample to decay.

Given that the half-life of the nuclei in the sample is 2.5 hours, we can calculate the number of half-lives that occur within the 5-hour period.

Number of half-lives = (Time elapsed) / (Half-life)

Number of half-lives = 5 hours / 2.5 hours = 2

This means that within the 5-hour period, two half-lives have occurred.

Since each half-life reduces the number of nuclei by half, after one half-life, the number of nuclei remaining is (1/2) * (10^10) = 5 * 10^9 nuclei.

After two half-lives, the number of nuclei remaining is (1/2) * (5 * 10^9) = 2.5 * 10^9 nuclei.

The activity of the sample is directly proportional to the number of remaining nuclei.

Therefore, After 5 hours, the sample has an activity of 2.5 * 109 dps or 2.5 * 109 Bq.

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The resulting charge on each capacitor, both when connected in parallel to the battery and when connected to each other in series, is approximately 2.32 µC.

When capacitors are connected in parallel, the voltage across them is the same. Therefore, the voltage across the combination of capacitors in the first scenario (connected in parallel to the battery) is 250 V.

For capacitors connected in parallel, the total capacitance (C_total) is the sum of individual capacitances:

C_total = C1 + C2

Given:

C1 = 6.10 µF = 6.10 × 10^(-6) F

C2 = 3.18 F

C_total = C1 + C2

C_total = 6.10 × 10^(-6) F + 3.18 × 10^(-6) F

C_total = 9.28 × 10^(-6) F

Now, we can calculate the charge (Q) on each capacitor when connected in parallel:

Q = C_total × V

Q = 9.28 × 10^(-6) F × 250 V

Q ≈ 2.32 × 10^(-3) C

Therefore, the resulting charge on each capacitor when connected in parallel to the battery is approximately 2.32 µC.

When the capacitors are disconnected from the circuit and connected to each other in series, the charge remains the same on each capacitor.

Thus, the resulting charge on each capacitor when they are connected to each other in series is also approximately 2.32.

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In this scenario, a police car is moving to the right at 27 m/s, and a speeder is approaching from behind at 36 m/s.

The police officer points a radar gun at the speeder, emitting an electromagnetic wave with a frequency of 7.5×10^9 Hz. The task is to find the difference between the frequency of the wave that returns to the police car after reflecting from the speeder's car and the frequency emitted by the police car.

The frequency of the wave that returns to the police car after reflecting from the speeder's car is affected by the relative motion of the two vehicles. This phenomenon is known as the Doppler effect.

In this case, since the police car and the speeder are moving relative to each other, the frequency observed by the police car will be shifted. The Doppler effect formula for frequency is given by f' = (v + vr) / (v + vs) * f, where f' is the observed frequency, v is the speed of the wave in the medium (assumed to be the same for both the emitted and reflected waves), vr is the velocity of the radar gun wave relative to the speeder's car, vs is the velocity of the radar gun wave relative to the police car, and f is the emitted frequency.

In this scenario, the difference in frequency can be calculated as the observed frequency minus the emitted frequency: Δf = f' - f. By substituting the given values and evaluating the expression, the difference in frequency can be determined.

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The telescope's angular magnification is approximately -42.11, indicating an inverted image.

Angular magnification refers to the ratio of the angle subtended by an object when viewed through a magnifying instrument, such as a telescope or microscope, to the angle subtended by the same object when viewed with the eye. It quantifies the degree of magnification provided by the instrument, indicating how much larger an object appears when viewed through the instrument compared to when viewed without it.

The angular magnification of a telescope can be calculated using the formula:

Angular Magnification = - (focal length of the objective lens) / (focal length of the eyepiece)

Given:

Plugging these values into the formula:

Angular Magnification = - (1.6 m) / (0.038 m)

Simplifying the expression:

Angular Magnification ≈ - 42.11

Therefore, the angular magnification of the telescope is approximately -42.11. Note that the negative sign indicates an inverted image.

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(a) The rotational inertia of the pendulum about the pivot point is approximately 0.0268 kg * m^2.

(b) The distance between the pivot point and the center of mass of the pendulum is approximately 0.102 m.

(c) The period of oscillation of the pendulum is approximately 0.324 seconds.

To calculate the rotational inertia of the pendulum about the pivot point, we need to consider the contributions from both the disk and the rod.

(a) The rotational inertia of a disk about its axis of rotation passing through its center is given by the formula:

I_disk = (1/2) * m * r^2

where m is the mass of the disk and r is its radius.

Given:

Mass of the disk (m_disk) = 820 g = 0.82 kg

Radius of the disk (r) = 12.0 cm = 0.12 m

Substituting the values into the formula:

I_disk = (1/2) * 0.82 kg * (0.12 m)^2

I_disk = 0.005904 kg * m^2

The rotational inertia of the rod about its pivot point can be calculated using the formula:

I_rod = (1/3) * m * L^2

where m is the mass of the rod and L is its length.

Given:

Mass of the rod (m_rod) = 210 g = 0.21 kg

Length of the rod (L) = 370 mm = 0.37 m

Substituting the values into the formula:

I_rod = (1/3) * 0.21 kg * (0.37 m)^2

I_rod = 0.020869 kg * m^2

To find the total rotational inertia of the pendulum, we sum the contributions from the disk and the rod:

I_total = I_disk + I_rod

I_total = 0.005904 kg * m^2 + 0.020869 kg * m^2

I_total = 0.026773 kg * m^2

Therefore, the rotational inertia of the pendulum about the pivot point is approximately 0.026773 kg * m^2.

(b) The distance between the pivot point and the center of mass of the pendulum can be calculated using the formula:

d = (m_disk * r_disk + m_rod * L_rod) / (m_disk + m_rod)

Given:

Mass of the disk (m_disk) = 820 g = 0.82 kg

Radius of the disk (r_disk) = 12.0 cm = 0.12 m

Mass of the rod (m_rod) = 210 g = 0.21 kg

Length of the rod (L_rod) = 370 mm = 0.37 m

Substituting the values into the formula:

d = (0.82 kg * 0.12 m + 0.21 kg * 0.37 m) / (0.82 kg + 0.21 kg)

d = 0.102 m

Therefore, the distance between the pivot point and the center of mass of the pendulum is approximately 0.102 m.

(c) The period of oscillation of a physical pendulum can be calculated using the formula:

T = 2π * √(I_total / (m_total * g))

Given:

Total rotational inertia of the pendulum (I_total) = 0.026773 kg * m^2

Total mass of the pendulum (m_total) = m_disk + m_rod = 0.82 kg + 0.21 kg = 1.03 kg

Acceleration due to gravity (g) = 9.8 m/s^2

Substituting the values into the formula:

T = 2π * √(0.026773 kg * m^2 / (1.03 kg * 9.8 m/s^2))

T = 2π * √(0.002655 s^2)

T = 2π * 0.05159 s

T ≈ 0.324 s

Therefore, the period of oscillation of the pendulum is approximately 0.324 seconds.

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The missing item that would complete the given alpha decay reaction + He 257 100 Fm → ? is option C. 253.

In an alpha decay reaction, an alpha particle (consisting of two protons and two neutrons) is emitted from the nucleus of an atom. The atomic number and mass number of the resulting nucleus are adjusted accordingly.

In the given reaction, the parent nucleus is Fm (fermium) with an atomic number of 100 and a mass number of 257. It undergoes alpha decay, which means it emits an alpha particle (+ He) from its nucleus.

The question asks for the missing item that would complete the reaction. Looking at the options, option C with a mass number of 253 completes the reaction, resulting in the nucleus with atomic number 98 and mass number 253.

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The speed of the comet when it is farthest from the sun is 0.0845 m/s.

The question states that the comet Necatoria moves with its closest approach to the sun being about 0.580 AU and its greatest distance from the sun being 350 AU. At its closest approach, its speed is 51 m/s. Now we are required to find out its speed when it is farthest from the sun.The angular momentum of the comet about the sun is conserved because no force acts on the comet. The gravitational force exerted by the Sun on the comet has a moment of 2010.0 -3030 km.0.00 15 ms.

In order to determine the speed of the comet when it is farthest from the sun, we need to use the conservation of angular momentum. Since no force is acting on the comet, the angular momentum will be constant. Let L1 be the angular momentum of the comet when it is at its closest approach to the sun.

So,L1 = mvr1

where m = mass of the comet, v = velocity of the comet at closest approach and r1 = distance of the comet from the sun at closest approach

Now, let L2 be the angular momentum of the comet when it is at its farthest from the sun.

So,L2 = mvr2where m = mass of the comet, v = velocity of the comet at farthest approach and r2 = distance of the comet from the sun at farthest approach

Since the angular momentum is conserved, we can write:L1 = L2mvr1 = mvr2r1v1 = r2v2We can find the speed of the comet at farthest approach using the above equation:

v2 = r1v1/r2

v2 = (0.580)(51)/350

v2 = 0.0845 m/s (approximately)

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The correct statement about the E or B fields in radiation is that Erms = 800.2 N/C.

EM (electromagnetic) radiation has an average intensity of 1700 W/m². As a result, the electrical field (Erms) is related to the average intensity through the equation E = cB, where E is the electric field, B is the magnetic field, and c is the speed of light.

Erms is related to the average intensity I (in W/m²) through the formula Erms = sqrt(2 I / c ε) which is approximately equal to 800.2 N/C.

For a 5.5 m³ space on the earth's surface, the total electromagnetic energy from sunlight with an intensity of 1.8 x 103 W/m² is 9.9 x 106 J.

The formula for calculating the energy is E = I × A × t, where E is the energy, I is the intensity, A is the area, and t is the time.

Here, the area is 5.5 m³ and the time is 1 second, giving an energy of 9.9 x 106 J.

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The work done by the man against gravity in climbing to the top is 9,480 foot-pounds.

To calculate the work done by the man, we need to determine the total change in potential energy as he climbs up the helical staircase that encircles the silo. The potential energy can be calculated using the formula PE = mgh, where m represents the mass, g represents the acceleration due to gravity, and h represents the height.

In this case, the mass of the man is 190 lb, and the height of the silo is 80 ft. Since the man makes exactly four complete revolutions around the silo, we can calculate the circumference of the helical staircase. The circumference of a circle is given by the formula C = 2πr, where r represents the radius. In this case, the radius of the silo is 15 ft.

To find the work done against gravity, we need to multiply the change in potential energy by the number of revolutions. The change in potential energy is obtained by multiplying the mass, the acceleration due to gravity (32.2 ft/s²), and the height. The number of revolutions is four.

Therefore, the work done by the man against gravity in climbing to the top can be calculated as follows:

Work = 4 * m * g * h

    = 4 * 190 lb * 32.2 ft/s² * 80 ft

    = 9,480 foot-pounds.

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the displacement current between the square plates of the capacitor is 9694 A. To calculate displacement current, we convert the units appropriately and perform the multiplication.

In this case, the square plates have a side length of 6.8 cm, which gives us an area of (6.8 cm)^2. The electric field is changing at a rate of 2.1 x 10^6 V/m.

The displacement current (Ip) between the square plates of a capacitor can be calculated by multiplying the rate of change of electric field (dE/dt) by the area (A) of the plates.

The area of the square plates is (6.8 cm)^2 = 46.24 cm^2. Converting this to square meters, we have A = 46.24 cm^2 = 0.004624 m^2.

Now, we can calculate the displacement current (Ip) by multiplying the rate of change of electric field (dE/dt) by the area (A):

Ip = (dE/dt) * A = (2.1 x 10^6 V/m) * (0.004624 m^2) = 9694 A

Therefore, the displacement current between the square plates of the capacitor is 9694 A.

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The probability of finding the particle between x = 0 and x = 1/4 in its first excited state in a one-dimensional box of length L is 1/(4L).

To determine the probability of finding the particle between x = 0 and x = 1/4 in its first excited state, we need to calculate the square of the wave function over that region.

The wave function for the particle in a one-dimensional box in the first excited state (n = 2) is given by:

ψ(x) = √(2/L) * sin(2πx/L),

where L is the length of the box.

To calculate the probability, we need to square the absolute value of the wave function and integrate it over the region of interest.

P = ∫[0, 1/4] |ψ(x)|^2 dx

Substituting the expression for ψ(x), we have:

P = ∫[0, 1/4] [√(2/L) * sin(2πx/L)]^2 dx

P = (2/L) ∫[0, 1/4] sin^2(2πx/L) dx

Using the identity sin^2θ = (1/2) * (1 - cos(2θ)), we can simplify the integral:

P = (2/L) ∫[0, 1/4] (1/2) * (1 - cos(4πx/L)) dx

P = (1/L) ∫[0, 1/4] (1 - cos(4πx/L)) dx

Integrating, we get:

P = (1/L) [x - (L/(4π)) * sin(4πx/L)] evaluated from 0 to 1/4

P = (1/L) [(1/4) - (L/(4π)) * sin(π)].

Since sin(π) = 0, the second term becomes zero:

P = (1/L) * (1/4)

P = 1/(4L).

Therefore, the probability of finding the particle between x = 0 and x = 1/4 in its first excited state is 1/(4L), where L is the length of the one-dimensional box.

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When a 22.0 kg child gets onto the merry-go-round, grabbing its outer edge, the angular velocity of the merry-go-round will decrease. The angular momentum added by the child is L_child = (22.0 kg)(1.9 m)^2 × 0 rev/s.

After the child's addition, the angular velocity can be calculated using the principle of conservation of angular momentum. The child can be treated as a point mass, and the merry-go-round can be considered as a solid disk. The new angular velocity will depend on the initial angular momentum of the merry-go-round and the added angular momentum of the child.

The initial angular momentum of the merry-go-round can be calculated using the formula L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity. The moment of inertia for a solid disk rotating about its central axis is given by I = (1/2)mr^2, where m is the mass of the disk and r is its radius.

Substituting the given values, we find that the initial angular momentum

L_initial = (1/2)(120 kg)(1.9 m)^2 × 0.400 rev/s.

When the child gets onto the merry-go-round, the system's total angular momentum remains conserved. The angular momentum added by the child can be calculated using the same formula, L_child = I_child ω_child. Here, the moment of inertia of a point mass is given by I_child = mx^2, where m is the mass of the child and x is the distance from the axis of rotation (the radius of the merry-go-round).

Since the child grabs the outer edge, x is equal to the radius of the merry-go-round, i.e., x = 1.9 m. Therefore, the angular momentum added by the child is L_child = (22.0 kg)(1.9 m)^2 × 0 rev/s.

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