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madam [21]
3 years ago
5

How are eras different from decades?

Engineering
1 answer:
ivolga24 [154]3 years ago
5 0

Answer:

decades are 10 years, while eras do not have set periods of time

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• Suppose that a particular algorithm has time complexity T(n) = 10 ∗ 2n, and that execution of the algorithm on a particular ma
elena-s [515]

Answer:

The number of inputs processed by the new machine is 64

Solution:

As per the question:

The time complexity is given by:

T(n) = 10\times 2n

where

n = number of inputs

T = Time taken by the machine for 'n' inputs

Also

The new machine is 65 times faster than the one currently in use.

Let us assume that the new machine takes the same time to solve k operations.

Then

T(k) = 64 T(n)

\frac{T(k)}{T(n)} = 64

\frac{20k}{20n} = 64

k = 64n

Thus the new machine will process 64 inputs in the time duration T

8 0
3 years ago
What Member function places a new node at the end of the linked list?
Nadusha1986 [10]
<h2>˜”*°•.˜”*°• Question •°*”˜.•°*”˜ </h2>

<em>What Member function places a new node at the end of the linked list? </em>

<h2>˜”*°•.˜”*°• Answer •°*”˜.•°*”˜</h2>

appendNode

<h2>˜”*°•.˜”*°• Explanation •°*”˜.•°*”˜</h2>

The appendNode() member function places a new node at the end of the linked list. The appendNode() requires an integer representing the current data of the node.

<h2>˜”*°•.˜”*°• Details •°*”˜.•°*”˜</h2>

Subject: Coding (?)

Grade: College

Keywords: Function, linked list, appendNode, integer

Hope this helped. <3

3 0
3 years ago
A steam turbine in a power plant receives 5 kg/s steam at 3000 kPa, 500°C. Twenty percent of the flow is extracted at 1000 kPa t
MrMuchimi

Answer:

The temperature of the first exit (feed to water heater) is at 330.15ºC. The second exit (exit of the turbine) is at 141ºC. The turbine Power output (if efficiency is %100) is 3165.46 KW

Explanation:

If we are talking of a steam turbine, the work done by the steam is done in an adiabatic process. To determine the temperature of the 2 exits, we have to find at which temperature of the steam with 1000KPa and 200KPa we have the same entropy of the steam entrance.

In this case for steam at 3000 kPa, 500°C, s= 7.2345Kj/kg K. i=3456.18 KJ/Kg

For steam at 1000 kPa and s= 7.2345Kj/kg K → T= 330.15ºC i=3116.48KJ/Kg

For steam at 200 kPa and s= 7.2345Kj/kg K → T= 141ºC i=2749.74KJ/Kg

For the power output, we have to multiply the steam flow with the enthalpic jump.

The addition of the 2 jumps is the total power output.

4 0
3 years ago
Turning operations that require heavy material removal typically use what setting on the
UkoKoshka [18]
Engine latte a mussi coffee chro man ok enign
3 0
2 years ago
The velocity of the 7.7-kg cylinder is 0.49 m/s at a certain instant. What is its speed v after dropping an additional 1.28 m? T
alex41 [277]

Answer:

the velocity of the cylinder after dropping an additional 1.28 m is 1.15 m/s

Explanation:

Using the work energy system

T_1 + U_{1-2} + T_2

The initial kinetic energy T_1 is ;

T_1 = \dfrac{1}{2}m_cv_1^2 + \dfrac{1}{2}I_o \omega^2

T_1 = \dfrac{1}{2}m_cv_1^2 + \dfrac{1}{2}(m_d \overline k^2)(\dfrac{v_1}{r_i})^2

where;

m_c = mass of the cylinder = 7.7 kg

v_1 = initial velocity of the cylinder = 0.49 m/s

I_o= moment of inertia of the drum about O

m_d =mass of the drum = 10.5 kg

r_i = radius of gyration = 0.3 m

\omega = angular velocity of the drum

T_1 = \dfrac{1}{2}(7.7)(0.49)^2 + \dfrac{1}{2}(10.5*(0.3)^2(\dfrac{0.49}{0.275})^2

T_1 = 2.426 \ J\\

The final kinetic energy is also calculated as:

T_2= \dfrac{1}{2}m_cv_2^2 + \dfrac{1}{2}(m_d \overline k^2)(\dfrac{v_2}{r_i})^2

T_2= \dfrac{1}{2}(7.7)(v_2^2)^2 + \dfrac{1}{2}(10.5*(0.3)^2(\dfrac{v_2}{0.275})^2

T_1 =10.10 v_2^2

Similarly, The workdone by all the forces on the cylinder can be expressed as:

U_{1-2} = m_cg(h) - (\dfrac{M}{r_i})h

where;

g = acceleration due to gravity

h = drop in height of the cylinder

M = frictional moment at O

U_{1-2} = 7.7*9.81*1.28 - 18.4(\dfrac{1.28}{0.275})

U_{1-2} =11.04 \ J

Finally, using the work energy application;

T_1 + U_{1-2} + T_2

2.426 + 11.04 = 10.10 v_2^2

13.466 = 10.10 v_2^2

v_2^2 = \dfrac{13.466}{10.10}

v_2^2 = 1.333

v_2 = \sqrt{1.333}

v_2 = 1.15 \ m/s

Thus, the velocity of the cylinder after dropping an additional 1.28 m is 1.15 m/s

8 0
3 years ago
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