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ra1l [238]
4 years ago
14

The value 100 MW is equivalent to (a) 100×10^6 w (b) 100 x 10^-6 w (c) 100 x 10^-3 w (d) 100 x 10^3 w

Engineering
1 answer:
insens350 [35]4 years ago
5 0

Answer:

(a) 100×10⁶ W

Explanation:

Power is defined as the rate of doing work. Thus,

Power=\frac{Work\ done}{Time\ taken}

<u>The S.I. unit of Power is watt which is J/s.</u>

Mega is a unit prefix which denotes a factor of 1 million (10⁶). It is denoted by 'M'.

Thus,

<u>1 MW=10⁶ W</u>

The Conversion of 100 MW into W is shown below:

<u>100 MW = 100×10⁶ W</u>

The option that correct resembles is option (a)

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Answer:

Explanation:

The python code to generate this is quite simple to run.

i hope you understand everything written here, you can as well try out other problems to understand better.

First to begin, we import the package;

Code:

import pandas as pd

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plt.legend(loc="upper right")

plt.xticks(rotation=20)

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This should generate the data(plot) as seen in the uploaded screenshot.

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3 years ago
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Imagine you are making a pizza, you start of with the dough for the crust in a large ball. You then begin to roll out the pizza
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3 years ago
2. Ackermann's Function is a recursive mathematical algorithm that can be used to test how well a system optimizes its performan
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The python program is an implementation of the Ackermann function that a system optimizes its performance of recursion.

As per the question,

Here is an implementation of the Ackermann function in Python:

<em />

<em>def ackermann(m, n):</em>

<em>    if m == 0:</em>

<em>        return n + 1</em>

<em>    elif n == 0:</em>

<em>        return ackermann(m - 1, 1)</em>

<em>    else:</em>

<em>        return ackermann(m - 1, ackermann(m, n - 1))</em>

<em />

<em># get input values for m and n from the user</em>

<em>m = int(input("Enter an integer value for m: "))</em>

<em>n = int(input("Enter an integer value for n: "))</em>

<em />

<em># calculate and print the result of the Ackermann function</em>

<em>result = ackermann(m, n)</em>

<em>print("Ackermann ({},{}) = {}".format(m, n, result))</em>

This implementation follows the logic described in the prompt, using a recursive function to calculate the result of the Ackermann function for the given values of m and n.

To learn more about the Python Program click here:

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5 0
1 year ago
Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you
viktelen [127]

Answer:

\theta_1=15^o\\\theta_2=75^o

Explanation:

<u>Projectile Motion</u>

In projectile motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration (assuming no friction), and the acceleration in the vertical direction is always the acceleration of gravity. The basic formulas are shown below:

V_x=V_{ox}=V_ocos\theta

Where \theta is the angle of launch respect to the positive horizontal direction and Vo is the initial speed.

V_y=V_{oy}-gt=V_osin\theta-gt

The  horizontal and vertical distances are, respectively:

x=V_{o}cos\theta t

\displaystyle y=y_o+V_{o}sin\theta t-\frac{gt^2}{2}

The total flight time can be found as that when y = 0, i.e. when the object comes back to ground (or launch) level. From the above equation we find

\displaystyle t_f=\frac{2V_osin\theta}{g}

Using this time in the horizontal distance, we find the Range or maximum horizontal distance:

\displaystyle R=\frac{V_o^2sin2\theta}{g}

Let's solve for \theta

\displaystyle sin2\theta=\frac{R.g}{V_o^2}

This is the general expression to determine the angles at which the projectile can be launched to hit the target. Recall the angle can have to values for fixed positive values of its sine:

\displaystyle \theta_1=\frac{asin\left(\frac{R.g}{V_o^2}\right)}{2}

\displaystyle \theta_2=\frac{180^o-asin\left(\frac{R.g}{V_o^2}\right)}{2}

Or equivalently:

\theta_2=90^o-\theta_1

Given Vo=37 m/s and R=70 m

\displaystyle \theta_1=\frac{asin\left(\frac{70\times 9.8}{37^2}\right)}{2}

\theta_1=15^o

And

\theta_2=90^o-15^o=75^o

5 0
3 years ago
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