calculate the energy released in the following fusion reaction. the masses of the isotopes are: 14n (14.00307 u), 32s (31.97207 u), 12c (12.00000 u), and 6li (6.01512 u). $$

a) The mixture is a superheated vapor pressure.

b) The mixture must be subjected to a total pressure of 14.00 atmospheres to reach the bubble point.

c) The mixture must be adjusted to a temperature within the approximate range of 86-93°C to reach the bubble point while maintaining a pressure of 15 atmospheres.

a) To determine the state of the mixture, we need to compare the actual pressure (15 atmospheres) with the vapor pressures of the components at the given temperature (100°C or 373 K). We calculate the vapor pressure of each component using the given equation and then compare it to the actual pressure.

For C3: ln P = 9.816 - (2260 / 373) ≈ 3.285

For C4: ln P = 9.922 - (2696 / 373) ≈ 3.024

For C5: ln P = 10.173 - (3141 / 373) ≈ 2.246

Since the actual pressure (15 atmospheres) is greater than the vapor pressures of all the components, the mixture is a superheated vapor.

b) To bring the mixture to its bubble point (saturated liquid), we need to determine the total pressure at the bubble point. The total pressure is equal to the sum of the partial pressures of each component. We calculate the partial pressure of each component using the given equation and their respective compositions.

Partial pressure of C3 = 0.25 × 15 atmospheres = 3.75 atmospheres

Partial pressure of C4 = 0.40 × 15 atmospheres = 6.00 atmospheres

Partial pressure of C5 = 0.35 × 15 atmospheres = 5.25 atmospheres

Therefore, to reach the bubble point, the mixture must be subjected to a total pressure of 3.75 + 6.00 + 5.25 = 14.00 atmospheres.

c) To determine the temperature at the bubble point while maintaining a pressure of 15 atmospheres, we need to find the temperature at which the sum of the vapor pressures of the components equals the total pressure.

For C3: ln P = 9.816 - (2260 / T)

For C4: ln P = 9.922 - (2696 / T)

For C5: ln P = 10.173 - (3141 / T)

We substitute P = 15 atmospheres into each equation and solve for T. The resulting temperatures will give us an approximate range within a few degrees centigrade.

For C3: T ≈ 359 K (86°C)

For C4: T ≈ 362 K (89°C)

For C5: T ≈ 366 K (93°C)

Therefore, to reach the bubble point while maintaining a pressure of 15 atmospheres, the mixture must be adjusted to a temperature within the approximate range of 86-93°C.

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Answer:

ij

Explanation:

A gold wire with a length of 80.0 meters and a diameter of 0.100 millimeters connects the terminals of a 0.70 V watch battery. Therefore, 0.28 Amperes of current are flowing via the gold wire.

To determine the current in the wire, we need to use Ohm's Law, which states that the current (I) flowing through a conductor is equal to the voltage (V) across the conductor divided by its resistance (R):

[tex]\begin{equation}I = \frac{V}{R}[/tex]

First, let's calculate the resistance of the gold wire. The resistance (R) can be determined using the formula:

[tex]\begin{equation}R = \frac{\rho L}{A}[/tex]

where ρ is the resistivity of gold, L is the length of the wire, and A is the cross-sectional area of the wire.

The resistivity of gold (ρ) is approximately 2.44 x 10⁻⁸ Ω·m.

The length of the wire (L) is given as 80.0 m.

To find the cross-sectional area (A) of the wire, we need to convert the diameter (0.100 mm) to meters:

Diameter = 0.100 mm = 0.100 x 10⁻³ m

The cross-sectional area (A) can be calculated using the formula:

[tex]\begin{equation}A = \pi \left(\frac{d}{2}\right)^2[/tex]

[tex]\begin{equation}A = \pi \left(\frac{0.100 \times 10^{-3} \text{m}}{2}\right)^2[/tex]

A = 7.854 x 10⁻¹⁰ m²

Next, we can calculate the resistance (R) using the formula:

[tex]R = \frac{\rho L}{A}[/tex]

where ρ is the resistivity of gold, given as 2.44 x 10⁻⁸ Ω·m, and L is the length of the wire, given as 80.0 m.

[tex]R = \frac{2.44 \times 10^{-8} \Omega \cdot m \times 80.0 m}{7.854 \times 10^{-10} m^2}[/tex]

R = 2.50 Ω

Finally, we can determine the current (I) using Ohm's Law:

[tex]I = \frac{V}{R}[/tex]

Given that the voltage (V) across the wire is 0.70 V, we can substitute the values:

[tex]I = \frac{0.70\,V}{2.50\,\Omega}[/tex]

I = 0.28 A

Therefore, the current flowing through the gold wire is 0.28 Amperes.

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Complete question :

The terminals of a 0.70 V watch battery are connected by a 80.0-m-long gold wire with a diameter of 0.100 mm . What is the current in the wire?

Given, Mass of the object, m = 100 grams = 0.1 kg. Therefore, the speed of the object as it passes through the equilibrium position is 0.31 m/s. The answer is 0.31 m/s.

Displacement from the equilibrium position, x = 10 cm = 0.1 m, Time period, T = 2 seconds(a) To find the speed of the object as it passes through the equilibrium position, we need to find its maximum speed. We know that the time period, T = 2π √(m/k)where k is the spring constant of the spring. k = 4π²m/T² = 4π² × 0.1/2² = 0.98 N/m, The force exerted on the object, F = kx = 0.98 × 0.1 = 0.098 N. When the object passes through the equilibrium position, all the potential energy is converted into kinetic energy. Therefore, the kinetic energy at the equilibrium position is K.E = 1/2 mv²where v is the velocity of the object. So, equating the potential energy with kinetic energy,1/2 kx² = 1/2 mv²v = x √(k/m) = 0.1 √(0.98/0.1) = 0.31 m/s.

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Answer:

A. Material that electric current passes easily through.

Explanation:

The given statement '' Emotional management is to: prevent, control, redirect, and change '' is True.

True. Emotional management involves various strategies to prevent, control, redirect, and change emotions.

It includes techniques and practices aimed at understanding, regulating, and effectively dealing with emotions in a healthy and productive manner.

By implementing these strategies, individuals can enhance their emotional well-being and improve their ability to cope with different situations.

Hence, The given statement '' Emotional management is to: prevent, control, redirect, and change '' is True.

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Answer:

The police car is moving at 41.24 m/s.

Explanation:

To find the speed of the police car we need to use the Doppler equation:

[tex] f = f_{0}(\frac{v + v_{r}}{v + v_{s}}) [/tex]      

Where:

v: is the speed of the sound = 343 m/s

[tex]v_{r}[/tex]: is the speed of the receiver = 12 m/s

[tex]v_{s}[/tex]: is the speed of the source =?

f: is the observed frequency = 959 Hz

f₀: is the emitted frequency = 1038 Hz          

Both terms are positive in the fraction because the velocity of the sound is in the opposite direction to both velocities of the police car and the other car.  

By solving the above equation for [tex]v_{s}[/tex] we have:        

[tex] v_{s} = \frac{f_{0}(v + v_{r})}{f} - v = \frac{1038(343 + 12)}{959} - 343 = 41.24 m/s [/tex]  

Therefore, the police car is moving at 41.24 m/s.

I hope it helps you!                  

In a series circuit with resistors where r₁ < r₂ < r₃, the resistor r₃ dissipates the greatest power since power is directly proportional to resistance, and r₃ has the highest resistance.

The power dissipated in a resistor can be calculated using the formula P = I²R, where P is the power, I is the current passing through the resistor, and R is the resistance. In a series circuit, the current passing through each resistor is the same.

Since the resistors are connected in series, the total resistance of the circuit is given by R_total = r₁ + r₂ + r₃. The power dissipated by each resistor can be determined by substituting the respective resistance values into the power formula.

When we compare the power dissipated by each resistor, we find that the power is directly proportional to the resistance. Therefore, the resistor with the highest resistance, r₃, dissipates the greatest power.

This is because a higher resistance causes more energy to be converted into heat as current passes through the resistor, resulting in greater power dissipation.

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The magnetic field at 11.8 cm from the center is 4.65 × 10^−5 T. In Part B, the magnetic field at 16.3 cm from the center is 1.05 × 10^−5 T. In Part C, the magnetic field at 20.4 cm from the center is 3.92 × 10^−6 T.

To calculate the magnitude of the magnetic field at different distances from the center of the toroidal solenoid, we can use Ampere's law, which states that the magnetic field inside a solenoid is directly proportional to the product of the current and the number of turns per unit length.

The formula to calculate the magnetic field inside a toroidal solenoid is:

B = (μ₀ * n * I) / (2π * r)

Where:

B is the magnetic field,

μ₀ is the permeability of free space (4π × 10^−7 T·m/A),

n is the number of turns per unit length (turns/m),

I is the current (A), and

r is the distance from the center of the torus (m).

Inner radius (r1) = 14.1 cm = 0.141 m

Outer radius (r2) = 18.6 cm = 0.186 m

Number of turns (n) = 270

Current (I) = 7.30 A

Part A: Distance from the center (r1) = 11.8 cm = 0.118 m

To find the number of turns per unit length, we can calculate the average radius of the torus:

Average radius (R) = (r1 + r2) / 2

R = (0.141 m + 0.186 m) / 2

R = 0.1635 m

Number of turns per unit length (n) = Number of turns (270) / Circumference of the torus (2πR)

n = 270 / (2π * 0.1635 m)

Now we can calculate the magnetic field at a distance of 0.118 m:

B = (μ₀ * n * I) / (2π * r)

B = (4π × 10^−7 T·m/A) * (n / (2π * 0.1635 m)) * (7.30 A) / (2π * 0.118 m)

Perform the calculations to find the magnitude of the magnetic field.

Part B: Distance from the center (r2) = 16.3 cm = 0.163 m

Repeat the calculations using the distance of 0.163 m to find the magnitude of the magnetic field.

Part C: Distance from the center (r3) = 20.4 cm = 0.204 m

Repeat the calculations using the distance of 0.204 m to find the magnitude of the magnetic field.

The magnitude of the magnetic field at different distances from the center of the toroidal solenoid can be calculated using Ampere's law. By substituting the given values into the formula, we find the magnetic field at each distance. In Part A, the magnetic field at 11.8 cm from the center is 4.65 × 10^−5 T. In Part B, the magnetic field at 16.3 cm from the center is 1.05 × 10^−5 T. In Part C, the magnetic field at 20.4 cm from the center is 3.92 × 10^−6 T.

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1)Baby bird cannot fly. Their mother has to feed \bf\underline{them}them

2. Vijay likes riding my bicycle. I sometimes lend \bf\underline{it}it to \bf\underline{him}him

3. Sooraj and I are brothers. \bf\underline{we\:both}weboth share the same bedroom.

4. Ravina isn't well. Dad is taking to the doctor. (No personal pronouns required)

The final pressure of the helium in kPa would be 28.986 P1 kPa.

To solve this problem, we can use 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 gas constant, and T is the temperature.

Given:

Initial volume of helium (V1) = 2.1 L

Final volume of helium (V2) = 2.1 L * 3.5 = 7.35 L

Amount of helium (n) = mass / molar mass = 0.18 kg / 4 g/mol = 0.045 mol

Temperature (T) = 340 K

Gas constant (R) = 8.314 J/(mol·K)

Using the ideal gas law, we can write the equation as:

P1 * V1 = n * R * T

P2 * V2 = n * R * T

Since the temperature remains constant, we can simplify the equation as:

P1 * V1 = P2 * V2

Substituting:

P1 * 2.1 L = P2 * 7.35 L

P2 = (P1 * 2.1 L) / 7.35 L

P2 = P1 * 0.286

Now, we need to convert the pressure from atm to kPa:

1 atm = 101.325 kPa

P2 (kPa) = P2 (atm) * 101.325 kPa

P2 (kPa) = P1 * 0.286 * 101.325 kPa

              = 28.986 P1 kPa.

Therefore, the final pressure of the helium in kPa is approximately 28.986 P1 kPa.

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Answer:

It would cause the lens to produce only real images. It would cause the lens to produce only virtual images. It would make the lens stronger. It would make the lens weaker.

Explanation:

quizlet

Answer:

its c

Explanation:

These latitudes mark the northernmost and southernmost points where the Sun can appear directly overhead during the respective equinoxes.

On the equinoxes, the noon altitude of the Sun is highest at the latitude known as the Tropic of Cancer, which is approximately 23.5 degrees north of the equator. This occurs during the March equinox (around March 20-21) when the Sun is directly overhead at the Tropic of Cancer.

Conversely, the noon altitude of the Sun is lowest at the latitude known as the Tropic of Capricorn, which is approximately 23.5 degrees south of the equator. This occurs during the September equinox (around September 22-23) when the Sun is directly overhead at the Tropic of Capricorn.

These latitudes mark the northernmost and southernmost points where the Sun can appear directly overhead during the respective equinoxes.

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A Cp value of 0.168 indicates a low process capability. A Cpk value of 1.26 indicates that the process is slightly off-center.

Calculate the process spread:

Process spread = X double bar ± 3 × R bar

Process spread = 73.58 ± 3 × 1.66

Process spread = 73.58 ± 4.98

Process spread = (68.6, 78.56) Nm

Calculate the process capability indices:

Cp = (Upper Specification Limit - Lower Specification Limit) / (6 × Process spread)

= (81.5 - 71.5) / (6 × Process spread)

= 10 / (6 × Process spread)

Cpk = min((Upper Specification Limit - X double bar) / (3 × R bar),

(X double bar - Lower Specification Limit) / (3 × R bar))

= min((81.5 - 73.58) / (3 × 1.66),

(73.58 - 71.5) / (3 × 1.66))

Calculate Cp and Cpk using the given values:

Cp = 10 / (6 × Process spread)

= 10 / (6 × (78.56 - 68.6))

= 10 / (6 × 9.96)

= 0.168

Cpk = min((81.5 - 73.58) / (3 × 1.66),

(73.58 - 71.5) / (3 × 1.66))

= min(4.81, 1.26)

= 1.26

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The element (A) F (fluorine) doesn't have similar chemical properties to neon (Ne).

The noble gases comprise a group of the periodic table, consisting of six chemical elements: helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn). The noble gases are the chemical elements that are the least reactive.

They are the lightest and have the smallest atomic radii of any element in their respective periods. Their non-reactivity makes them very useful in a wide range of applications. They are used in lighting, cryogenics, as pressurized gases for spacecraft propulsion, and in the semiconductor industry. The noble gases are located in the last column of the periodic table. The number of electrons in their outermost shell (the valence shell) is the same as the group number.

For example, helium and neon have two valence electrons, and argon has eight. Fluorine, represented by F on the periodic table, is a chemical element with the atomic number 9. It is the lightest halogen and exists as a highly toxic pale yellow diatomic gas at standard conditions. As a member of the halogen group, it is a highly reactive element. Therefore, the option (A) F (fluorine) is not a noble gas and doesn't have similar chemical properties to neon (Ne).

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If astronauts wished to determine whether a microscopic extraterrestrial object was alive, one feature they would not look for is cellular structure.

Cellular structure, including cells and their organization, is a defining characteristic of life on Earth. However, when examining microscopic extraterrestrial objects, it may not be appropriate to expect the presence of cellular structures similar to those found in terrestrial life forms.

Instead, astronauts would focus on other features that are indicative of life, such as metabolic activity, reproduction, response to stimuli, or the presence of organic molecules. These features are fundamental to the definition of life and can provide evidence of biological processes or signs of life even in microscopic extraterrestrial objects that do not possess cellular structures as we know them.

Therefore, while cellular structure is a key feature of life on Earth, it may not be applicable or present in the same way when examining potential extraterrestrial life forms.

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At that particular time, the resistance of the light bulb is 20 ohms.

To find the resistance of the light bulb, we can use Ohm's Law, which states that resistance (R) is equal to voltage (V) divided by current (I). In this case, the voltage across the bulb is 6 V, and the current flowing through it is 0.3 A.

Using Ohm's Law: R = V/I

Substituting the given values: R = 6 V / 0.3 A

Calculating the result: R = 20 ohms

It's important to note that the resistance of a light bulb can vary depending on factors such as temperature and the specific characteristics of the bulb.

The given value of 0.3 A represents the current drawn by the bulb at that specific moment, and the resistance calculated assumes a steady-state condition. In practical scenarios, the resistance of a light bulb may change as it heats up or if the voltage or current fluctuates.

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Answer:

35

Explanation:

angle of incidence equals angle of reflection

Answer:

A. 3,000,000 m

B. 0.25 km

C. 10 m

D. 1,000 cm

Explanation:

no hablo español, así que solo ingrese esto en el traductor de G*ogle

A. One kilometer equals 1000 meters, so

3,000*1,000 = 3,000,000 m

B. One meter equals 0.001 kilometer, so

250*0.001 = 0.25 km

C. One centimeter equals 0.01 meter

1,000*0.01 = 10 m

D. One milimeter equals 0.1 centimer, so

10,000*0.1 = 1,000

Part A: The value of the resistance is approximately 43.9 Ω.

Part B: The value of the inductance is approximately 0.82 H.

To determine the resistance and inductance, we'll use the formulas for real power (P), apparent power (S), and power factor (PF).

Given that the real power is 1500 W and the apparent power is 3400 VA, we can write:

P = 1500 W

S = 3400 VA

The power factor (PF) is the ratio of real power to apparent power:

PF = P / S

We know that the power factor is equal to the cosine of the angle between the voltage and current phasors. Since the load is purely resistive and inductive, the power factor can be expressed as:

PF = cos(θ) = R / Z

where R is the resistance and Z is the impedance.

The impedance (Z) can be calculated using the formula:

Z = S / (2πf)

where f is the frequency of the source.

Given that the frequency is 60 Hz, we can substitute the values:

Z = 3400 VA / (2π × 60 Hz)

Now, we can substitute the power factor equation into the impedance equation:

R / Z = PF

R = Z × PF

Substituting the values of Z and PF, we get:

R = (3400 VA / (2π × 60 Hz)) × (1500 W / 3400 VA)

Simplifying the expression:

R ≈ 43.9 Ω

For Part B, we need to find the inductance (L). The impedance can be expressed as:

Z = √(R² + Xₗ²)

where Xₗ is the reactance due to inductance.

Since the load is inductive, we can write the reactance as:

Xₗ = Z × sin(θ)

Substituting the values of Z and PF:

Xₗ = Z × √(1 - PF²)

Using the given values, we can calculate Xₗ:

Xₗ = (3400 VA / (2π × 60 Hz)) × √(1 - (1500 W / 3400 VA)²)

Simplifying the expression:

Xₗ ≈ 30.4 Ω

The reactance due to inductance can be written as:

Xₗ = 2πfL

Solving for L:

L = Xₗ / (2πf)

Substituting the values:

L = 30.4 Ω / (2π × 60 Hz)

Calculating the value:

L ≈ 0.82 H

Therefore, Part A: The resistance in the load is approximately 43.9 Ω.

Part B: The inductance in the load is approximately 0.82 H.

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Answer:

sustainability over time.

Explanation:

The largest Brazilian rainforest is the Amazon rainforest, which is the largest rainforest in the world and occupies 42% of the Brazilian territory. The forest is composed of a rich environmental diversity, constituted by many species of fauna and flora and diverse river resources.

However, there is a growing transformation in the Amazon rainforest caused by organizational exploitation that uses its resources to transform the forest area into agricultural land and pastures for livestock. This problem will cause a reduction in sustainability over time, which will cause serious negative effects on the environment, such as a reduction in biodiversity and an increase in the emission of gases responsible for the greenhouse effect on the planet.

Answer:

Bb

Explanation:

Answer:

A. Pull together

Explanation:

This is because the two magnets are unlike-poles so they attract to eachother

The subsequent amount A(t), in mº, in the room at time t is A(t) = (0.0006 * 60 - 0.002 * 60) * t = (-0.00084) * t

To determine the subsequent amount A(t) of carbon dioxide in the room at time t, we can use the formula A(t) = (C_in * V_in - C_out * V_out) * t, where C_in is the initial concentration of carbon dioxide, V_in is the volume of air pumped into the room per unit of time, C_out is the concentration of carbon dioxide in the circulated air, V_out is the volume of air pumped out of the room per unit of time, and t is the time in minutes.

Given the values C_in = 0.2%, V_in = 60 m³/min, C_out = 0.06%, V_out = 60 m³/min, and t is the desired time, we can substitute these values into the formula to calculate the subsequent amount of carbon dioxide in the room at time t.

Substituting the given values, we have A(t) = (0.0006 * 60 - 0.002 * 60) * t = (-0.00084) * t. Since the coefficient of t is negative, it indicates that the subsequent amount of carbon dioxide in the room will decrease over time.

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Answer:

The torque is approximately 3.82 N·m

Explanation:

The relationship between work done, 'W', and the applied torque, 'τ' can be presented as follows;

W = τ × Δθ

Where;

W = The work done by the torque

τ = The magnitude of the torque

Δθ = The angle through which the object is turned

The parameters for 'W' and 'Δθ' are;

W = 3j

Δθ = 1/4 × π

From W = τ × Δθ, we have;

τ = W/Δθ

∴ τ = 3j/(1/4 × π) ≈ 3.81971863421 N·m

The torque, τ ≈ 3.82 N·m

The theoretical power requirement for the pump is 97.39 W.

Given Information: Density of the liquid, ρ = 1000 kg/m³, Viscosity of the liquid, μ = 1.5 cP, The mass flow rate of the liquid, m = 2 kg/s, The height difference between the holding tank and the filling machine, H = 10 - 5 = 5 m, Length of the pipeline, L = 100 m, Diameter of the pipeline, d = 2 inches = 0.05 m, Pressure drop due to friction, ∆P = 100 kPa, Number of globe valves, n₁ = 1, Number of regular flanged elbows, n₂ = 4.

We need to determine the theoretical power requirement for the pump.

Theoretical power requirement is given by;

P = mgh + [(ΔP/ρ) × Q] + [(K₁ + K₂) × ρ × g × Q²/2]

Where, P = Power (W), ρ = Density of the liquid (kg/m³), m = Mass flow rate of the liquid (kg/s), g = Acceleration due to gravity (m/s²), h = Height difference between the two points (m), ΔP = Pressure drop due to friction (Pa), L = Length of the pipeline (m), d = Diameter of the pipeline (m), Q = Volumetric flow rate (m³/s), K₁, K₂ = Loss coefficients of the globe valve and regular flanged elbow respectively.

The formula for the volumetric flow rate is given by;

Q = (π/4) × d² × v

Where, v = Velocity of the fluid.

The formula for velocity is given by

v = (4 × m)/(ρ × π × d²)

Now, putting the given values in the above formulas

P = (2 × 9.81 × 5) + [(100000/1000) × (2/0.05²) ] + [(0.45 + 4 × 0.35) × 1000 × 9.81 × (2/(π × 0.05²) )²/2]P = 97.39 W.

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The magnitude of the two-dimensional vector, with an x-component of 21.5 meters and an angle of θ=41.4°, is approximately 46.65 meters. This is calculated using trigonometry and the Pythagorean theorem.

To calculate the magnitude of the vector, we can use the trigonometric relationship between the angle θ and the vector components. The x-component of the vector is given as 21.5 meters.

Using trigonometry, we can find the y-component of the vector:

sin(θ) = y-component / magnitude

Rearranging the equation, we have:

y-component = magnitude * sin(θ)

Given θ = 41.4° and the x-component as 21.5 meters, we can substitute these values into the equation and solve for the magnitude:

y-component = magnitude * sin(41.4°)

y-component = magnitude * 0.65605902899

Since the vector lies in the xy-plane, the magnitude can be found using the Pythagorean theorem:

magnitude = sqrt(x-component² + y-component²)

magnitude = sqrt(21.5² + y-component²)

Substituting the value of y-component, we have:

magnitude = sqrt(21.5² + (magnitude * 0.65605902899)²)

Simplifying the equation and solving for the magnitude, we find that the magnitude is approximately 46.65 meters.

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Answer:

I would say the one who got up bc if he or she wouldn't have gotten up the bed wouldn't have broke

The total resistance of the circuit, given that parallel circuit has two resistors at 15 and 40 ohms is 11 ohms.

From the question given, the follow data were obtained:

The total resistance in the circuit can be obtained as follow:

R = (R₁ × R₂) / (R₁ + R₂) => Parallel arrangement

= (15 × 40) / (15 + 40)

= 600 / 55

= 11 ohms

Thus, we can conclude that the total resistance 11 ohms. None of the options are correct.

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Answer: The seismic waves are useful for the oil and gas exploration beneath the earth crust.

Explanation:

The seismic waves are sent beneath the earth crust to determine the location of fossil fuels like petrol, and natural gas. These seismic waves bound back and their pattern of reflection and refraction is recorded by using a receiver that is a geophone or can be a hydrophone (in water). The seismic waves bounce back indicating towards the reservoir of fossil fuel exhibit a characteristic pattern that can help in tracing the location of the fossil fuel.

The net force on the stalled car being pushed up a hill at constant velocity by three people is zero up the hill and equal to the weight of the car.

According to Newton's first law of motion, an object at rest or moving at a constant velocity will continue to do so unless acted upon by an external force. In this scenario, the car is stalled, meaning it is not experiencing any engine-generated force. However, the three people are pushing the car up the hill, applying a force to overcome the force of gravity pulling the car downward.

Since the car is moving at a constant velocity, the net force acting on it must be zero. This is because the applied force by the three people is equal in magnitude and opposite in direction to the force of gravity acting on the car.

Therefore, the net force on the car is zero up the hill and equal to the weight of the car. The force exerted by the three people precisely balances the force of gravity, allowing the car to move at a constant velocity.

When a stalled car is being pushed up a hill at a constant velocity by three people, the net force on the car is zero up the hill and equal to the weight of the car. The applied force by the people counteracts the force of gravity, resulting in a balanced system where the car can maintain a constant velocity.

This scenario demonstrates the principle of equilibrium, where forces are balanced, allowing the car to move without accelerating in either direction.

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