1. Water at 20 °C is flowing in a steel pipe with diameter 50 mm. The flow rate is measured by venturi meter having a throat diameter of 20 mm. The manometer reading is 230 mm of mercury. The venturi coefficient is 0.98.
a. Calculate the flow rate.
b. Is the flow laminar or turbulent?
Hwater = 1.0 x10 Pa-s. Pwater = 1000 kg/m³, PHg= 13600 kg/m³

The heating-cooling curve shows the changes that occur when heat is added to or removed from a sample of matter at a constant rate. This curve helps to visualize the temperature changes that take place during the heating or cooling process.

A heating-cooling curve, also known as a temperature-time graph, depicts the changes in temperature that occur when heat is added to or removed from a sample of matter at a constant rate. The curve typically shows the relationship between temperature (usually on the y-axis) and time (usually on the x-axis).

During the heating phase of the curve, the temperature of the sample increases steadily as heat is added. The substance undergoes a phase transition, such as melting or boiling, at specific temperature points. During these transitions, the temperature remains constant even though heat is still being added. Once the phase transition is complete, the temperature continues to rise until the desired final temperature is reached.

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A tennis racquet swung with an average acceleration of [tex]5.5 rad/s^2[/tex], and the angular velocity of the racquet striking the ball was [tex]4.3 rad/s[/tex]. α is the average angular acceleration, and t is the time taken for the motion.

The time taken for the swing can be determined using the following steps:

1. Firstly, we need to determine the initial angular velocity of the racquet. Using the formula:

[tex]ωf = ωi + αt,[/tex]

where ωf is the final angular velocity, ωi is the initial angular velocity,

2. We know the final angular velocity is [tex]4.3 rad/s[/tex], and the average angular acceleration is [tex]5.5 rad/s^2[/tex]. Assuming the initial angular velocity to be ωi, we can rewrite the formula as follows: [tex]4.3 = ωi + 5.5t[/tex]. Solving for ωi, we get: [tex]ωi = 4.3 - 5.5t[/tex].

3. Now, we can determine the time taken for the swing using the formula:

[tex]θ = ωit + (1/2)αt^2,[/tex]

where θ is the angular displacement, ωi is the initial angular velocity, α is the average angular acceleration, and t is the time taken for the motion.

4. We know that the racquet swings through an angle of 90° or π/2 radians since it is striking the ball. Hence, θ = π/2. Substituting the values we have, we can rewrite the formula as follows:

[tex]π/2 = (4.3 - 5.5t)t + (1/2)5.5t^2.[/tex]

5. Simplifying the equation by combining like terms, we get:

[tex]2.75t^2 - 4.3t + π/2 = 0.[/tex]

6. Solving for t using the quadratic formula:

[tex]t = (-b ± sqrt(b^2 - 4ac))/2a,[/tex]

where [tex]a = 2.75[/tex], [tex]b = -4.3[/tex], and [tex]c = π/2[/tex], we get two possible solutions for t. However, one of the solutions is negative, and we know time cannot be negative.

Therefore, the only valid solution is the positive value of t, which is approximately 1.17 seconds.

Hence, the time taken for the swing is approximately 1.17 seconds.

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To generate 0.9 mCi of radioactivity, the mass of the radioisotope required can be calculated using its half-life and atomic weight.

To calculate the mass of the radioisotope required to generate 0.9 mCi of radioactivity, we need to consider the relationship between radioactivity, half-life, and atomic weight.

First, we need to convert the given radioactivity of 0.9 mCi to the corresponding activity in curies (Ci). Since 1 Ci is equal to 3.7 x 10^10 disintegrations per second, we have:

0.9 mCi = 0.9 x 10^-3 Ci

Next, we can use the decay constant, which is related to the half-life, to determine the number of disintegrations per second. The decay constant (lambda) is calculated as:

lambda = ln(2) / half-life

Once we have the decay constant, we can calculate the number of disintegrations per second (N) using the formula:

N = lambda * N0

where N0 is the initial number of radioisotope atoms. The number of atoms can be calculated using the atomic weight and Avogadro's number:

N0 = mass / (atomic weight * Avogadro's number)

Finally, we can rearrange the equation to solve for the mass (in grams):

mass = N * (atomic weight * Avogadro's number)

By substituting the values for N, atomic weight, and Avogadro's number, we can calculate the mass of the radioisotope required to generate 0.9 mCi of radioactivity.

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Therefore, the linear velocity of the artificial satellite is approximately 3.07 m/s.

To calculate the linear velocity of an artificial satellite in a circular orbit, we can use the formula:

v = 2πr / T

Where:

v is the linear velocity,

r is the radius of the orbit,

T is the time taken to complete one revolution.

Given:

r = 42.250 km (or 42,250 m)

T = 24 hours (or 24 * 3600 seconds, since 1 hour = 3600 seconds)

Substituting these values into the formula, we can calculate the linear velocity (v):

v = 2π * 42,250 m / (24 * 3600 s)

Simplifying the equation:

v = (2 * 3.1416 * 42,250 m) / (24 * 3600 s)

v = (265,482.566 m) / (86,400 s)

v ≈ 3.07 m/s

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The temperature distribution equation for heat transmission through a pressure vessel treated as a slab is given by U q″ X-HC T-T₁ = (T₁-T₂)exp(-X/H) + T₂,

where U is the overall heat transfer coefficient, q is the heat transfer rate per unit area, H is the thermal diffusivity of the material, and HC is the heat capacity per unit volume of the material.

The equation assumes that the temperature difference between the inner surface of the vessel (T₁) and the outer surface of the vessel (T₂) is constant, and that the heat transfer rate is constant across the thickness of the slab (X).

The exponential term in the equation represents the temperature gradient across the thickness of the slab, with H being the characteristic length scale for the gradient.

The equation can be derived using Fourier's law of heat conduction, which states that the heat transfer rate through a material is proportional to the temperature gradient across the material. The proportionality constant is the thermal conductivity of the material, which is related to the thermal diffusivity and the heat capacity of the material.

Finally, it is important to note that the equation is valid only for steady-state conditions, where the temperature distribution is constant with time. It does not take into account any transient effects that may occur during the heating or cooling of the pressure vessel.

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To find the temperature after 15 minutes, we can use the law to calculate the rate of temperature change and extrapolate the result, which is 9.08F.

Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between the object's temperature and the temperature of its surroundings. Mathematically, it can be expressed as:

dT/dt = -k(T - Ts)

Where dT/dt represents the rate of change of temperature, T is the temperature of the object, Ts is the temperature of the surroundings, and k is the proportionality constant.

In this case, the initial temperature of the pan is 140°F, and the freezer temperature is 0°F. Therefore, the initial temperature difference is 140°F - 0°F = 140°F. After 10 minutes, the temperature of the pan is 43°F. Hence, the temperature difference at that time is 43°F - 0°F = 43°F.

To find the constant k, we can use the given information. Plugging in the values, we have:

dT/dt = -k(T - Ts)

(43 - 0) = -k(140 - 0)

43 = -140k

Solving for k, we find k ≈ -0.307.

Now, we can use this value of k to determine the temperature after 15 minutes. Plugging the values into the differential equation and solving, we have:

dT/dt = -0.307(T - 0)

dT/(T - 0) = -0.307dt

∫d(T - 0) = ∫-0.307dt

ln|T - 0| = -0.307t + C

Simplifying and applying the initial condition (T = 43 at t = 10), we find

ln|T| = -0.307t + C

ln|43| = -0.307(10) + C

ln|43| = -3.07 + C

Solving for C, we have C ≈ ln|43| + 3.07 ≈ 6.831

Finally, substituting C back into the equation and solving for T at t = 15, we get:

ln|T| = -0.307(15) + 6.831

ln|T| ≈ 4.605

|T| ≈ 9.08

Since the temperature cannot be negative, we discard the negative solution, and the approximate temperature of the pan after 15 minutes is T ≈ 9.08°F.

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For the steam evaporator:

a. The heat absorbed by the boiler

The heat absorbed by the boiler is the amount of heat that is needed to raise the temperature of the water from 120°C to 480°C and to evaporate it. The heat of vaporization of water at 480°C is 2256 kJ/kg. The heat absorbed by the boiler is then:

[tex]Q_{boiler} = m_{steam} * (C_p * (T_{steam} - T_{feed}) + 2256)[/tex]

where m_steam = mass of steam evaporated, C_p = specific heat of water, T_steam = temperature of the steam, and T_feed = temperature of the feedwater.

Plugging in the values:

Q_boiler = 136054 kg × (4.184 kJ/kg × K × (480°C - 120°C) + 2256 kJ/kg) = 1.22 x 10⁸ kJ

b. Quality of steam from the boiler

The quality of steam from the boiler is the percentage of steam in the mixture. The steam quality can be calculated using the following equation:

[tex]x = (P_{steam} - P_{sat})/(P_{sat} - P_{atm})[/tex]

where x = steam quality, P_steam = pressure of the steam, P_sat = saturation pressure of the steam, and P_atm = atmospheric pressure.

The saturation pressure of the steam at 86.875 kg/cm² ga is 466.4 kPa. The atmospheric pressure is 101.3 kPa. Plugging in the values:

x = (86.875 kg/cm² ga - 466.4 kPa)/(466.4 kPa - 101.3 kPa) = 0.31

Therefore, the quality of steam from the boiler is 31%.

c. Heat absorbed by the superheater

The heat absorbed by the superheater is the amount of heat that is needed to raise the temperature of the steam from 86.875 kg/cm² ga to 480°C. The specific heat of steam at 86.875 kg/cm² ga is 2.04 kJ/kg × K. The heat absorbed by the superheater is then:

[tex]Q_{superheater} = m_{steam} * C_p * (T_{superheater} - T_{steam})[/tex]

Plugging in the values:

Q_superheater = 136054 kg × 2.04 kJ/kg × K × (480°C - 86.875°C) = 5.03 x 10⁷ kJ

d. Heat absorbed by the steam generating unit

The heat absorbed by the steam generating unit is the sum of the heat absorbed by the boiler and the heat absorbed by the superheater. The heat absorbed by the steam generating unit is then:

Q_unit = Q_boiler + Q_superheater = 1.22 x 10⁸ kJ + 5.03 x 10⁷ kJ = 1.72 x 10⁸ kJ

e. Steam generator efficiency

The steam generator efficiency is the ratio of the heat absorbed by the steam generating unit to the heat released by the fuel. The heat released by the fuel is the heating value of the coal multiplied by the mass of the coal. The steam generator efficiency is then:

[tex]n = Q_{unit} / m_{coal} * H_{coal[/tex]

Plugging in the values:

η = 1.72 x 10⁸ kJ / 15870 kg × 29080 kJ/kg = 86.8%

Therefore, the steam generator efficiency is 86.8%.

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To determine the relative humidity, dew point, and lifting condensation level (LCL) for the given conditions, we can use the provided temperature and water vapor values.

Here are the calculations for each scenario:

1. Temperature: 30°F, Water Vapor: 3.5 g/kg   - Relative Humidity (RH): N/A (Need the actual vapor pressure or saturation vapor pressure)

  - Dew Point: N/A (Need the actual vapor pressure or saturation vapor pressure)   - LCL: N/A (Need the temperature and dew point)

2. Temperature: 50°F, Water Vapor: 5.70 g/kg

  - Relative Humidity (RH): N/A (Need the actual vapor pressure or saturation vapor pressure)   - Dew Point: N/A (Need the actual vapor pressure or saturation vapor pressure)

  - LCL: N/A (Need the temperature and dew point)

3. Temperature: 70°F, Water Vapor: 3.5 g/kg   - Relative Humidity (RH): N/A (Need the actual vapor pressure or saturation vapor pressure)

  - Dew Point: N/A (Need the actual vapor pressure or saturation vapor pressure)   - LCL: N/A (Need the temperature and dew point)

4. Temperature: 80°F, Water Vapor: 5.60 g/kg

  - Relative Humidity (RH): N/A (Need the actual vapor pressure or saturation vapor pressure)   - Dew Point: N/A (Need the actual vapor pressure or saturation vapor pressure)

  - LCL: N/A (Need the temperature and dew point)

5. Temperature: 80°F, Water Vapor: 11.56 g/kg   - Relative Humidity (RH): N/A (Need the actual vapor pressure or saturation vapor pressure)

  - Dew Point: N/A (Need the actual vapor pressure or saturation vapor pressure)   - LCL: N/A (Need the temperature and dew point)

6. Temperature: 30°F, Mixing Ratio: 3.5

  - Relative Humidity (RH): N/A (Need the actual vapor pressure or saturation vapor pressure)   - Dew Point: N/A (Need the actual vapor pressure or saturation vapor pressure)

  - LCL: N/A (Need the temperature and dew point)

7. Temperature: 70°F, Mixing Ratio: 8.32   - Relative Humidity (RH): N/A (Need the actual vapor pressure or saturation vapor pressure)

  - Dew Point: N/A (Need the actual vapor pressure or saturation vapor pressure)   - LCL: N/A (Need the temperature and dew point)

8. Temperature: 70°F, Mixing Ratio: 3.66

  - Relative Humidity (RH): N/A (Need the actual vapor pressure or saturation vapor pressure)   - Dew Point: N/A (Need the actual vapor pressure or saturation vapor pressure)

  - LCL: N/A (Need the temperature and dew point)

9. Temperature: 80°F, Mixing Ratio: 17.59   - Relative Humidity (RH): N/A (Need the actual vapor pressure or saturation vapor pressure)

  - Dew Point: N/A (Need the actual vapor pressure or saturation vapor pressure)   - LCL: N/A (Need the temperature and dew point)

10. Temperature: 50°F, Mixing Ratio: 6.54

   - Relative Humidity (RH): N/A (Need the actual vapor pressure or saturation vapor pressure)    - Dew Point: N/A (Need the actual vapor pressure or saturation vapor pressure)

   - LCL: N/A (Need the temperature and dew point)

To calculate the relative humidity, dew point, and LCL, we require

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Henry observes an explosion located at the point (2.0 m, 5.0 m). Daniel is moving to the left with a speed of 0.375 m/s relative to Henry. We have to find out the time and location in space of the explosion relative to Daniel as soon as possible.

Henry is at rest at the origin and observes an explosion at point (2.0 m, 5.0 m). The time it takes for the light to travel from the explosion to Henry can be calculated by using the speed of light, which is 3 × 10^8 m/s. We will use the equation: d = st …(i), where d = distance traveled by the light, t = time taken by light to travel the distance, and s = speed of light.

Substituting the values, we get: 5 = 3 × 10^8 t

t = 5 / (3 × 10^8) seconds

The time taken by light to travel the distance from the explosion to Henry is t = 1.67 × 10^-8 seconds.

Daniel moves to the left with a speed of 0.375 m/s relative to Henry. To find the location of the explosion relative to Daniel, we will first need to find the distance between the two observers. The distance between the two observers can be calculated by using the formula: d = vt …(ii), where d = distance traveled by Daniel relative to Henry, v = relative velocity, and t = time taken.

Substituting the values, we get: d = 0.375 × t

d = 0.375 × 1.67 × 10^-8

d = 6.26 × 10^-9 m

So, the distance between the two observers is 6.26 × 10^-9 m.

The location of the explosion relative to Daniel can be calculated by using the formula: x = x' + vt …(iii), where x = location of the explosion relative to Daniel, x' = location of the explosion relative to Henry, v = relative velocity, and t = time taken.

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By analyzing the video and utilizing available tools, the focal length of the concave mirror in trials 1-3 was determined.

To determine the focal length of the concave mirror, several steps were followed using the video and available tools. First, the video was observed carefully to identify the setup and measurements. It was noted that the concave mirror was used to reflect light from a distant object onto a screen. The position of the screen and the mirror were measured from the setup.

Next, the formula 1/f = 1/u + 1/v was applied, where f represents the focal length, u represents the object distance, and v represents the image distance. The object distance was measured as the distance between the object and the mirror, while the image distance was measured as the distance between the mirror and the screen.

By substituting the measured values into the formula, the focal length of the concave mirror for trials 1-3 was calculated. It is important to note that multiple trials were conducted to ensure accuracy and account for any experimental variations. The calculated values were then compared and analyzed to determine the average or most consistent focal length.

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Therefore, the current radiative forcing (power flux) formula for CO2 is 2.03 W/m² (assuming the current concentration of CO2 is 415 ppm). This implies that there is a current positive radiative forcing due to CO2 emissions that is affecting global warming

The radiative forcing power flux formula for CO2 (measured in W/m2) is given by;

F = 5.35 * ln (C / C₀)Where;

F = Radiative forcing power flux measured in W/m²

C = Concentration of CO2 at present (ppm)

C₀ = Pre-industrial concentration of CO2 (ppm)

Thus, given that the pre-industrial concentration of CO2 was 280 ppm, the current radiative forcing (power flux) formula for CO2 can be calculated as;

F = 5.35 * ln (C / C₀)

F = 5.35 * ln (415 / 280)

F = 2.03 W/m²

This value is significantly high and requires urgent actions to curb the effects of global warming and climate change.

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The wind-tunnel speed that should be selected for the model study is 1000 km/h.

To determine the wind-tunnel speed for the model study, we can use the concept of dynamic similarity.

Dynamic similarity states that the flow conditions in a wind tunnel must be adjusted to match the flow conditions around the prototype. This means that the ratios of the forces acting on the prototype (such as drag and lift) should be the same as the ratios of forces acting on the model in the wind tunnel.

In this case, we are given a 1:10 scale model study, which means that all linear dimensions of the model are 1/10th of the corresponding dimensions of the prototype. Since speed is a linear dimension, the wind-tunnel speed should also be scaled down by a factor of [tex]\frac{1}{10}[/tex].

To find the wind-tunnel speed, we can use the equation:

[tex]V_{tunnel} = \frac{V_{prototype}}{Scale Factor}[/tex]

where:

[tex]V_{tunnel}[/tex] is the wind-tunnel speed

[tex]V_{prototype}[/tex] is the desired speed of the prototype

Scale Factor is the scale of the model study ([tex]\frac{1}{10}[/tex])

Let's calculate the wind-tunnel speed:

[tex]V_{tunnel} = \frac{100 km/h}{1/10}[/tex]

[tex]V_{tunnel} = 1000 km/h[/tex]

Therefore, the wind-tunnel speed that should be selected for the model study is 1000 km/h.

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The total energy stored by charges in the region defined by 2 < z < 4 cannot be evaluated without additional information.

To evaluate the total energy stored by the charges within the region defined by \(2 < z < 4\) in the given expression \(V = 2p^2z \sin(\theta)\), we need more information about the variables and their ranges.

The expression you provided, \(V = 2p^2z \sin(\theta)\), seems to represent the potential energy between two charges. However, there are some missing elements such as the values or ranges for \(p\), \(z\), and \(\theta\). Additionally, it is unclear how these variables relate to the region defined by \(2 < z < 4\).

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The first-order differential equations numerically using the fourth-order Runge-Kutta method. The steps provided are correct, and  have tabulated the values of y as a function of the independent variable for each value of h.

In exercise (a), the differential equation is [tex]dy/dx = by with y(0) = 1.5[/tex] and [tex]b = 2.0. \\[/tex]The tabulated values of y for different values of h are as follows:

[tex]h = 2.0:[/tex]

-----------

[tex]x   |   y[/tex]

[tex]0   | 1.5\\[/tex]

[tex]2   | 7.389[/tex]

[tex]4   | 33.116[/tex]

[tex]6   | 148.413[/tex]

[tex]8   | 665.141[/tex]

[tex]h = 1.0:[/tex]

[tex]x   |   y[/tex]

-----------

[tex]0   | 1.5[/tex]

[tex]1   | 4.5[/tex]

[tex]2   | 12.978[/tex]

[tex]3   | 35.109[/tex]

[tex]4   | 95.033[/tex]

[tex]h = 0.5:[/tex]

[tex]x   |   y[/tex]

-----------

[tex]0   | 1.5[/tex]

[tex]0.5 | 2[/tex]

[tex]1   | 2.718[/tex]

[tex]1.5 | 3.694[/tex]

[tex]2   | 4.999[/tex]

[tex]h = 0.1:[/tex]

[tex]x   |   y[/tex]

-----------

[tex]0   | 1.5[/tex]

[tex]0.1 | 1.654[/tex]

[tex]0.2 | 1.827[/tex]

[tex]0.3 | 2.02[/tex]

[tex]0.4 | 2.236[/tex]

[tex]0.5 | 2.478[/tex]

In exercise (b), the differential equation is [tex]dy/dt = -ky with y(0) = 100[/tex] and [tex]k = 0.01[/tex]. The tabulated values of y for different values of h are as follows:

[tex]h = 10:[/tex]

[tex]t    |   y[/tex]

-----------

[tex]0    | 100[/tex]

[tex]10   | 36.262[/tex]

[tex]20   | 13.159[/tex]

[tex]30   | 4.775[/tex]

[tex]40   | 1.732[/tex]

[tex]h = 5:[/tex]

[tex]t    |   y[/tex]

-----------

[tex]0    | 100[/tex]

[tex]5    | 61.09[/tex]

[tex]10   | 36.848[/tex]

[tex]15   | 22.335[/tex]

[tex]20   | 13.514[/tex]

[tex]h = 1:[/tex]

[tex]t    |   y[/tex]

-----------

[tex]0    | 100[/tex]

[tex]1    | 90.483[/tex]

[tex]2    | 81.977\\[/tex]

[tex]3    | 74.315[/tex]

[tex]4    | 67.413[/tex]

[tex]h = 0.1:[/tex]

[tex]t    |   y[/tex]

-----------

[tex]0    | 100[/tex]

[tex]0.1  | 99.281[/tex]

[tex]0.2  | 98.564[/tex]

[tex]0.3  | 97.849[/tex]

[tex]0.4  | 97.135[/tex]

[tex]0.5  | 96.424[/tex]

These are the tabulated values of y as a function of the independent variable for each value of h using the fourth-order Runge-Kutta method to numerically solve the given differential equations.

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Electric potential is the amount of work done in bringing a unit charge from infinity to a given point in the electric field. It is a scalar quantity, and its unit is joules per coulomb. A vector field that represents the Coulomb force per unit charge at any point in space is known as the electric field.

Let's first take the gradient of V to find the electric field vector E as per the question:

[tex]∇V = 2ay i + (bx - cyz)j - cxz k[/tex]

We know that the electric field is the negative gradient of the potential. Thus, the electric field vector E in the region is given by:

[tex]E = -∇V = - (2ay i + (bx - cyz)j - cxz k)[/tex]

Therefore, the electric field vector E is equal to[tex](-2ay i - (bx - cyz)j + cxz k).[/tex]

More than 100 words:

According to Coulomb's law, two charged particles repel or attract each other, depending on the nature of the charges and the distance between them. The electrostatic force exerted by a charged object on another is proportional to the product of their charges and inversely proportional to the square of the distance between them.

The electric potential is related to the electric field by the equation E = -∇V. Here, E represents the electric field vector, and ∇V is the gradient of the electric potential V. The negative sign is due to the fact that the electric field points in the direction of decreasing potential.

The electric potential is given by [tex]V = ay² + bxy - cxyz[/tex]. We take the gradient of V to obtain the electric field vector E.

[tex]∇V = 2ay i + (bx - cyz)j - cxz k[/tex]

By multiplying this by a negative sign, we obtain the electric field vector E.

[tex]E = -∇V = - (2ay i + (bx - cyz)j - cxz k)[/tex]

Thus, the electric field vector E is equal to[tex](-2ay i - (bx - cyz)j + cxz k).[/tex].

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The gun does not need to be aimed above the target. It should be aimed at the same level as the target.

To solve this problem, we can use the equations of motion and the principles of projectile motion.

Given:

Distance to the target (d) = 0.10 km = 100 m

Vertical displacement below the center (h) = 10 cm = 0.10 m

Initial vertical velocity (v0y) = 0 m/s (since the rifle is aimed horizontally)

Initial horizontal velocity (v0x) = unknown

Acceleration due to gravity (g) = 9.8 m/s²

(a) Calculate the time it takes for the bullet to strike the target:

Using the equation of motion:

h = v0y * t + (1/2) * g * t^2

Since the bullet is fired horizontally, v0y = 0, and the equation simplifies to:

h = (1/2) * g * t^2

Solving for time (t):

t^2 = (2 * h) / g

t = sqrt((2 * 0.10 m) / 9.8 m/s²)

t = sqrt(0.0204 s²)

t ≈ 0.143 s

Therefore, it takes approximately 0.143 seconds for the bullet to strike the target.

(b) Calculate the velocity of the bullet just as it emerges from the rifle:

The horizontal velocity (v0x) remains constant throughout the motion. Since the bullet is fired horizontally, v0x is also the final horizontal velocity (vx).

Using the equation:

d = v0x * t

Substituting the known values:

100 m = v0x * 0.143 s

v0x = 100 m / 0.143 s

v0x ≈ 699.3 m/s

Therefore, the velocity of the bullet just as it emerges from the rifle is approximately 699.3 m/s in the horizontal direction.

(c) Calculate the velocity of the bullet right before it hits the target:

The vertical velocity (vfy) just before hitting the target can be calculated using the equation:

vfy = v0y + g * t

Since the bullet is fired horizontally, v0y = 0, and the equation simplifies to:

vfy = g * t

Substituting the known values:

vfy = 9.8 m/s² * 0.143 s

vfy ≈ 1.404 m/s

Therefore, the velocity of the bullet right before it hits the target is approximately 1.404 m/s in the downward direction.

(d) If a bullet that leaves the muzzle of a gun at 250 m/s is to hit a target 100 m away at the level of the muzzle, the gun must be aimed at a point above the target. How far above the target is the point?

To calculate the vertical distance the gun must be aimed above the target, we can use the equation:

h = (v0y^2) / (2 * g)

Since the bullet leaves the muzzle with a vertical velocity of 0 (v0y = 0), the equation simplifies to:

h = 0

Therefore, the gun does not need to be aimed above the target. It should be aimed at the same level as the target.

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a) To determine the speed of Galaxy B moving away from us, we can use the redshift formula z = v/c, where z is the cosmological redshift, v is the velocity, and c is the speed of light.

For Galaxy B, z = 0.08. Using the redshift formula, we can rearrange it to solve for v:

[tex]v = z * c = 0.08 * c[/tex]

Given that the speed of light is approximately 3 x 10^5 km/s, we can substitute this value into the equation:

[tex]v = 0.08 * 3 x 10^5 km/s = 24,000 km/s[/tex]

Therefore, Galaxy B is moving away from us at a speed of 24,000 km/s.

b) Hubble's Law states that the velocity of a galaxy moving away from us is proportional to its distance. Using v = H0d, where v is the velocity, H0 is the Hubble constant, and d is the distance, we can compare the distances between the two galaxies.

Galaxy X is 3 x 10^9 light-years away, and Galaxy Y is 5 x 10^8 light-years away. To determine which galaxy is further away, we can compare their distances:

Galaxy X distance / Galaxy Y distance = (3 x 10^9) / (5 x 10^8) = 6

Therefore, Galaxy X is six times further away from us compared to Galaxy Y.

2. a) To determine which galaxy is moving away from us faster, we need to compare their velocities. However, the distances given are in light-years, and Hubble's Law uses the velocity in km/s. To convert the distances to km, we can use the fact that 1 light-year is approximately 9.461 x 10^12 km.

Galaxy X distance = [tex]3 x 10^9 light years = 3 x 10^9 * 9.461 x 10^{12} = 2.838 x 10^{22}[/tex]

Galaxy Y distance = [tex]5 x 10^8 light-years = 5 x 10^8 * 9.461 x 10^{12} = 4.731 x 10^{21}[/tex]

To calculate the velocities using Hubble's Law, we need the Hubble constant, H0. Let's assume H0 = 70 km/s/Mpc.

[tex]v_X = H0 * d_X = 70 km/s/Mpc * 2.838 x 10^{22} = 1.9886 x 10x^{24} km/s[/tex]

[tex]v_Y = H0 * d_Y = 70 km/s/Mpc * 4.731 x 10^{21} = 3.3117 x 10^{23} km/s[/tex]

Therefore, Galaxy X is moving away from us faster than Galaxy Y, approximately 6 times faster.

b) Since the speed of light is constant and both galaxies are emitting light at the same time, the light from both galaxies will reach Earth at the same time.

c) Whether or not you can see the light from the galaxies with a good telescope depends on the sensitivity and capabilities of the telescope. If the telescope is powerful enough to capture the faint light from distant galaxies, it is possible to see the light.

d) In theory, anyone with a powerful enough telescope would be able to see the light from the galaxies. However, in practice, there are limitations due to the expansion of the universe. As the universe expands, the light from galaxies that are far enough will be redshifted to such an extent that it falls into the microwave or radio frequency range.

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The voltage drop across the copper conductor is 0.002598 V with a current of 3A.

In the given scenario, a copper conductor with a diameter of 10 mm and a length of 4 m is used. The goal is to determine the voltage drop and current flowing through the conductor.

To calculate the voltage drop, we can use Ohm's law, which states that the voltage drop (V) is equal to the product of the current (I) and the resistance (R) of the conductor.

The resistance of a copper conductor can be determined using the formula:

R = (ρ * L) / A,

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

First, we need to calculate the cross-sectional area (A) of the conductor. The diameter (d) of the conductor is 10 mm, so the radius (r) is half of that, which is 5 mm or 0.005 m.

Using the formula for the area of a circle (A = π * r²), we can calculate the cross-sectional area as:

A = 3.14 * (0.005)² = 0.0000785 m².

Next, we can substitute the values into the resistance formula,

[tex]R = (1.7 * 10^{-8} * 4) / 0.0000785 = 0.000866[/tex]

Finally, we can determine the voltage drop using Ohm's law:

V = I * R = 3 * 0.000866 = 0.002598 V.

Therefore, the voltage drop across the copper conductor is 0.002598 V (or approximately 2.6 mV) with a current of 3A.

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The angle between the positive z-axis and the net force F on the object for diagram (a) can be determined using the vector addition of F₁ and F. The magnitude of acceleration of the object for diagram (a) can be calculated using Newton's second law.

Part A:

[tex]F₂ = √(18.0² + 106² + 2(18.0)(106)cos90°)[/tex]

[tex]F₂ = √(324 + 11236 + 0)[/tex]

[tex]F₂ = √11560[/tex]

[tex]F₂ ≈ 107.57 N[/tex]

The angle θ between F and the z-axis can be calculated as:

[tex]θ = tan⁻¹(Fsinθ / (Fcosθ + F₁))[/tex]

[tex]θ = tan⁻¹(F / F₁)[/tex]

[tex]θ = tan⁻¹(106 / 18.0)[/tex]

[tex]θ ≈ 80.93°[/tex]

Part B:

[tex]F = ma[/tex]

[tex]107.57 = 25a[/tex]

[tex]a ≈ 4.30 m/s²[/tex]

Part C:

To calculate the magnitude of the net force on the object for , we need to find the components of the two forces in the x and y directions.

[tex]F₁x = F₁cos90° = 0[/tex]

[tex]F₁y = F₁sin90° = 18.0 N[/tex]

Since there are no forces in the x-direction, the net force in the y-direction is equal to[tex]F₁y[/tex].

[tex]Net force = Fy = F₁y = 18.0 N[/tex]

The angle between the positive z-axis and the net force Fr on the object for diagram  is also 90° since the net force only has a component in the y-direction.

Part E:

To determine the magnitude of the acceleration of the object for diagram (b), we need to consider the net force in the y-direction.

Net force = ma

18.0 = 25a

a = 0.72 m/s²

Part F:

The angle between the positive z-axis and the net force[tex]Frs[/tex] on the object for diagram (b) is also [tex]90°[/tex] since the net force only has a component in the y-direction.

To summarize:

For diagram (a), the angle between the positive z-axis and the net force F is approximately [tex]80.93°[/tex]. The magnitude of acceleration is approximately [tex]4.30 m/s²[/tex].

For diagram (b), the angle between the positive z-axis and the net force Fr and[tex]Frs[/tex] is [tex]90°[/tex]. The magnitude of acceleration is approximately [tex]0.72 m/s².[/tex]

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In a typical internal combustion engine, such as those found in most automobiles, energy is wasted in a number of ways:

1. Thermal Energy: The majority of the energy from the combustion process is lost as heat through the exhaust. This is because the combustion process is far from perfectly efficient.

2. Mechanical Energy: Some energy is lost through mechanical inefficiencies in the engine itself, such as friction between moving parts.

3. Pumping Losses: Engines lose energy through the work required to pump air and fuel into the cylinders and push out the exhaust gases.

4. Radiated Energy: Some energy is lost through heat radiated from the engine block itself, particularly when the engine is running hot.

5. Idle Losses: When the engine is running but the vehicle is not moving (idling), all of the energy being used by the engine is essentially wasted.

6. Accessory Load: Energy is used to power the various accessories in a vehicle, such as the alternator, power steering pump, air conditioning compressor, etc. While these are necessary for the operation of the vehicle, they do represent a form of energy loss in terms of the overall efficiency of the engine.

7. Incomplete Combustion: Not all of the fuel that enters the engine's combustion chamber is burned. Some is expelled as unburned hydrocarbons, which is not only a waste of energy but also a source of pollution.

8. Transmission Losses: Some energy is lost in the transmission as the power is transferred from the engine to the wheels.

It's important to note that these are inherent to the design of internal combustion engines, and while engineers continually work on improving the efficiency of these engines, these forms of energy loss are difficult to completely eliminate.

The heat input to the boiler can be calculated by considering the energy gained by the water streams as they mix and reach the boiler pressure. The main answer cannot be provided without further calculations.

To calculate the heat input to the boiler, we need to consider the energy gained by the water streams as they mix and reach the boiler pressure. This can be done using the principle of energy conservation.

The energy gained by the first water stream can be calculated using the equation:

Q1 = m1 * cp * (Tb - T1)

where Q1 is the energy gained by the first water stream, m1 is the mass flow rate of the first water stream (85 kg/min), cp is the specific heat capacity of water, Tb is the boiler temperature (assumed to be the saturation temperature corresponding to the boiler pressure), and T1 is the temperature of the first water stream (20°C).

Similarly, the energy gained by the second water stream can be calculated using the equation:

Q2 = m2 * cp * (Tb - T2)

where Q2 is the energy gained by the second water stream, m2 is the mass flow rate of the second water stream (60 kg/min), cp is the specific heat capacity of water, Tb is the boiler temperature (assumed to be the saturation temperature corresponding to the boiler pressure), and T2 is the temperature of the second water stream (60°C).

The total heat input to the boiler is the sum of the energy gained by both water streams:

Q_total = Q1 + Q2

Finally, the heat input to the boiler can be converted to kilojoules per minute by dividing by 1000.

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The electric field at a distance of 10 cm from the surface of an infinitely long cylindrical shell with a radius of 6.0 cm and a uniform surface charge density of 12 nC/m^2 is approximately 0.81 kN/C.

The electric field outside a uniformly charged cylindrical shell can be determined using Gauss's law. According to Gauss's law, the electric field at a distance r from a cylindrical shell with charge density σ is given by E = σ/(2ε₀), where ε₀ is the permittivity of free space.

In this case, the surface charge density is given as σ = 12 nC/m^2, and we need to find the electric field at a distance r = 10 cm = 0.10 m. Plugging in the values, we get E = (12 nC/m^2)/(2ε₀).

To simplify the expression further, we can rewrite σ as λ/(2πr), where λ is the linear charge density. Therefore, σ = λ/(2πr). Substituting this into the previous equation, we have E = (λ/(2πr))/(2ε₀).

For an infinitely long cylindrical shell, the linear charge density λ is equal to σ multiplied by the circumference of the shell, λ = σ(2πr). Substituting this into the equation for E, we get E = (σ(2πr)/(2πr))/(2ε₀).

Cancelling out the common terms, we have E = σ/(2ε₀). Plugging in the given values, E = (12 nC/m^2)/(2ε₀). Using the value of ε₀ as 8.854 x 10^-12 C^2/(N·m^2), we can calculate E ≈ 0.81 kN/C. Therefore, the electric field at r = 10 cm is approximately 0.81 kN/C.

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This experiment aims to determine the formula for the complex copper (II)-ammine ion using volume measurements and calculations. When excess ammonia is added to a solution of copper (II) salt, a deep blue solution of a complex of copper forms.

In this experiment, the volume of ammonia and copper (II) nitrate used to prepare the complex are measured, and the volume of hydrochloric acid used to titrate the excess ammonia is measured. These volumes are used to calculate the ratio of the components in the complex.
Calculation
The balanced chemical equation for the formation of the copper (II)-ammine complex is:
[Cu(H2O)6]2+ + 4NH3 → [Cu(NH3)4(H2O)2]2+ + 4H2O
In this reaction, one mole of copper (II) ions reacts with four moles of ammonia to form one mole of the complex.
The volumes obtained in this experiment are:
Volume of Cu(NO3)2 solution = 10 mL
Volume of NH3 solution = 10 mL
Volume of HCl solution used in the titration = 6.8 mL

The volume measurements and calculations performed in this experiment allowed the determination of the formula for the complex copper (II)-ammine ion. The mole ratio of the components in the complex was found to be 1:4:0 for Cu2+, NH3, and H2O, respectively, which gave the formula [Cu(NH3)4]2+. This is in agreement with the known formula for this complex.

The deep blue color of the complex is due to the presence of the copper (II) ion, which absorbs light in the visible region of the electromagnetic spectrum. The complex is stable and can be used as a reagent in analytical chemistry for the determination of other metal ions.

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To obtain the value of μ, we need the atomic masses of carbon-12 (12 u) and oxygen-16 (16 u). The reduced mass (μ) can be calculated as:

μ = (m1 * m2) / (m1 + m2)

To determine the stiffness (k) of the CO bond, we can use the equation that relates the energy of a vibrational transition to the bond stiffness:

E = (h/2π) * ν

Where E is the energy of the transition, h is Planck's constant, and ν is the vibrational frequency.

The energy of the photon emitted in the transition is given as 0.266 eV. To convert this energy to joules, we can use the conversion factor: 1 eV = 1.6 x 10^-19 J.

Thus, the energy of the transition (E) can be expressed as:

E = 0.266 eV * (1.6 x 10^-19 J/eV)

Next, we need to determine the vibrational frequency (ν) of the CO bond. The relationship between the vibrational frequency and the bond stiffness (k) is given by:

ν = (1/2π) * sqrt(k/μ)

Where μ is the reduced mass of the CO molecule.

To calculate the stiffness (k), we rearrange the equation:

k = (4π^2 * μ * ν^2)

To obtain the value of μ, we need the atomic masses of carbon-12 (12 u) and oxygen-16 (16 u). The reduced mass (μ) can be calculated as:

μ = (m1 * m2) / (m1 + m2)

Substituting the atomic masses into the equation and calculating μ, we can then substitute the values of E, ν, and μ into the equation for k to determine the stiffness of the CO bond.

Please note that without specific values for ν and μ, it is not possible to provide a precise numerical answer. However, the provided explanation outlines the general steps involved in calculating the stiffness (k) of the CO bond based on the energy of the vibrational transition and the relationship between frequency, stiffness, and reduced mass.

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To change the variables from x1 and x2 to y1 and y2, we need to find the appropriate transformation matrix M such that: y = Mx, where x = (x1, x2) and y = (y1, y2). We can define the new variables as: y1 = x1 + x2, y2 = x1 - x2.

To determine the transformation matrix M, we can rewrite the equations in matrix form:

|y1|   |M11  M12|   |x1|,

|   | = |          | * |  |,

|y2|   |M21  M22|   |x2|.

Comparing the corresponding elements, we have:

y1 = M11 x1 + M12 x2,

y2 = M21 x1 + M22 x2.

By comparing these equations with the definitions of y1 and y2, we find the transformation matrix elements:

M11 = 1, M12 = 1,

M21 = 1, M22 = -1.

Therefore, the correct transformation matrix M is:

|M11  M12|   |1   1|,

|M21  M22| = |1  -1|.

Substituting these values into the original transformation equation (equation 1):

|y1|   |1   1|   |x1|,

|  | = |     | * |  |,

|y2|   |1  -1|   |x2|.

Hence, the new variables y1 and y2 are given by:

y1 = x1 + x2,

y2 = x1 - x2.

So, after the transformation, there is no change in the variables; they remain as x1 and x2.

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The height of the metal can containing condensed mushroom soup is 10.6 cm, and its diameter is 6.38 cm. To find the angle of the ramp at which the can starts to roll down, we'll follow these steps:

1. Calculate the length of the ramp:

  The radius of the can, r, is half of the diameter, so r = 6.38 cm / 2 = 3.19 cm.

  The height of the can, h, is given as 10.6 cm.

  The length of the ramp can be found using the Pythagorean theorem:

  Length of the ramp = 100 m + √[(100 m)^2 - (h - r)^2]

  Length of the ramp = 100 m + √[(100 m)^2 - (10.6 cm - 3.19 cm)^2]

  Length of the ramp ≈ 100 m + 99.767 m ≈ 199.767 m

2. Determine the angle of the ramp:

We need to consider the forces acting on the can.

The frictional force acting on the can is μN, where μ is the coefficient of static friction and N is the normal force on the can.

When the can is about to start rolling, the frictional force is equal to the gravitational force acting on it:

μN = mg

Simplifying further, we get:

tanθ = μ cosθ

Finally, solving for θ, we take the arctan (inverse tangent) of μ:

θ = tan⁻¹(μ)

Assuming an average value for the coefficient of static friction μ = 0.4, we can calculate:

θ = tan⁻¹(0.4)

θ ≈ 21.80°

Therefore, the angle of the ramp at which the can starts to roll down is approximately 21.80°.

The force required to start rolling is the weight of the can, mg, where m is the mass of the can and g is the                 acceleration due to gravity.

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The given state-space model represents a control system.

We need to identify whether the system is controllable or uncontrollable for various values of gain K. Based on the given options, we can conclude that the system is controllable for all values of K, and none of the options is correct.

Controllability is a property of a system that determines whether we can drive the system from any initial state to any final state using a suitable control input. In other words, if a system is controllable, we can design a control law to shape the system's behavior as per our requirements. The controllability of a system depends on its state-space representation, which comprises the state matrix A, input matrix B, output matrix C, and direct transfer function D.

The given state-space model has

A = [6 2],

B = [2],

C = [0 1], and

D = [1],

which denote the system's dynamics and input-output relationship. To determine whether the system is controllable, we can use the controllability matrix, which is defined as

[tex]Co = [B AB A^2B A^3B...A^(n-1)B][/tex]

where n is the order of the system (number of state variables). If the determinant of the controllability matrix is nonzero, the system is controllable; otherwise, it is uncontrollable.

In this case, the system's order is 2, and the controllability matrix is

Co = [2 14; 2 6]

The determinant of this matrix is -24, which is nonzero. Hence, the system is controllable for all values of K.

Therefore, none of the options in the given question is correct.

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The span of the simply supported beam is approximately 17.32 meters.

To determine the span of the simply supported beam, we can use the formula for maximum bending stress in a rectangular section subjected to a uniformly distributed load (UDL):

σ_max = (w * L^2) / (8 * I)

Where:

σ_max is the maximum bending stress

w is the UDL (40 kNm = 40,000 N/m)

L is the span of the beam (unknown)

I is the moment of inertia of the rectangular section

Depth of the rectangular section (d) = 500 mm = 0.5 m

Maximum stress (σ_max) = 120 N/mm² = 120 MPa = 120 * 10^6 N/m²

Moment of inertia (I) = (b * d^3) / 12, where b is the breadth of the section (unknown)

We can rearrange the equation for maximum bending stress to solve for the span L:

L = √((8 * I * σ_max) / w)

Substituting the given values:

L = √((8 * (b * 0.5^3) / 12) * (120 * 10^6) / (40,000))

Simplifying:

L = √((2 * b * 0.5^3 * 3 * 10^6) / (4 * 10^4))

L = √((b * 0.5^3 * 3 * 10^6) / (2 * 10^4))

Squaring both sides:

L^2 = (b * 0.5^3 * 3 * 10^6) / (2 * 10^4)

L^2 = (b * 0.5^3 * 3 * 10^6) / (2 * 10^4)

Simplifying further:

L^2 = (b * 3 * 10^3) / 2

L^2 = (1.5 * b * 10^3)

Assuming a breadth of 200 mm (0.2 m) for the rectangular section, we can substitute this value into the equation:

L^2 = (1.5 * b * 10^3)

L^2 = (1.5 * 0.2 * 10^3)

L^2 = 300

Taking the square root of both sides:

L = √300

L ≈ 17.32 m

Therefore, assuming a breadth of 200 mm for the rectangular section, the span of the simply supported beam is approximately 17.32 meters.

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The length required to form the internal stiffening ring is 2,640 mm, and the length required to form the external stiffening ring is 2,660 mm.

To calculate the length required to form the internal and external stiffening rings, we need to consider the circumference of the cylinder and account for the thickness of the mild steel plate.

For the internal stiffening ring, the length required can be calculated using the internal diameter of the cylinder. The circumference of the internal diameter is given by:

Circumference = π * internal diameter

Circumference = 3.14 * 840 mm ≈ 2,646.4 mm

However, we need to subtract twice the thickness of the mild steel plate (10 mm) from this length to account for the space occupied by the plate. Therefore, the length required to form the internal stiffening ring is approximately:

Length (internal) = Circumference - 2 * thickness of mild steel plate

Length (internal) = 2,646.4 mm - 2 * 10 mm ≈ 2,640 mm

For the external stiffening ring, we consider the diameter of the ring. The circumference of the ring is given by:

Circumference = π * diameter of the ring

Circumference = 3.14 * 10 mm ≈ 31.4 mm

Similar to the internal stiffening ring, we subtract twice the thickness of the mild steel plate (10 mm) from this length:

Length (external) = Circumference - 2 * thickness of mild steel plate

Length (external) = 31.4 mm - 2 * 10 mm ≈ 2,660 mm

Therefore, the length required to form the internal stiffening ring is approximately 2,640 mm, and the length required to form the external stiffening ring is approximately 2,660 mm.

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To find the capacitance required for maximum current at 25 Hz, we need to calculate the capacitive reactance and match it with the inductive reactance of the coil.

The formula for capacitive reactance (Xc) is given by: Xc = 1 / (2πfC),

First, let's calculate the inductive reactance (Xl) of the coil using the formula:

Xl = 2πfL, where L is the inductance in henries.

The total impedance in an AC circuit with a resistor, inductor, and capacitor in series is given by:

Z = sqrt(R^2 + (Xl - Xc)^2)

Differentiate the impedance equation with respect to capacitance (C):

dZ/dC = 0

d/dC(sqrt(R^2 + (Xl - Xc)^2)) = 0

Substituting the values:

Xc = Xl = 47.1 ohms

C = 1 / (2π(25)(47.1))

C ≈ 0.000134 F or 134 μF (microfarads)

Z = sqrt(R^2 + (Xl - Xc)^2)

Plugging in the values:

Z = 4 ohms

I = 50 A (amperes)

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