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3 years ago

13

At the System Demo, each team demonstrates their own work but the system is not integrated. They say they underestimated the int

egration effort, but they cannot integrate every Iteration and meet the planned scope. What should the Scrum Master suggest?

1 answer:

Verdich [7]3 years ago

8 0

Answer:

The scrum master can suggest that:

Postpone the work on system integration for a while until the train is in sustain and enhance the section of road map of implementation

Explanation:

System Demo:

It is such a software that is used to measure the progress of Agile Release Train.

As the teams didn't expect that the integration of system will be a difficult task and they ultimately failed to accomplish the goal of integration so that scrum master asked them to stop working on the system integration until the Agile Release Train's sustainability and to improve and optimize the implementation of road map.

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To make the next delivery, Sam-I-Am goes 45 mi/h for 2 hours. How far did he travel in that time?

andrezito [222]

Hello =D

This problem is about cinematic

So

V = 45 mi/h

t = 2 h

Then

V= X/t

X = V*t

Then

X = (45)*(2)

X = 90 mi

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Why do bones weaken as a person gets older

Sonbull [250]

Answer:

The bones aren't as strong as a younger person, because an older person, ages due to time and it also depend on what they do in their past.

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2 years ago

The revolution of the earth around the sun demonstrate what motion?

olga2289 [7]

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1 year ago

What is the time constant of a series circuit where the capacitor is 0.330μF and the resistor is 10Ω ?

PtichkaEL [24]

Answer:

$\tau=3.3*10^{-6}s$

Explanation:

Take at look to the picture I attached you, using Kirchhoff's current law we get:

$C*\frac{dV}{dt}+\frac{V}{R}=0$

This is a separable first order differential equation, let's solve it step by step:

Express the equation this way:

$\frac{dV}{V}=-\frac{1}{RC}dt$

integrate both sides, the left side will be integrated from an initial voltage v to a final voltage V, and the right side from an initial time 0 to a final time t:

$\int\limits^V_v {\frac{dV}{V} } =-\int\limits^t_0 {\frac{1}{RC} } \, dt$

Evaluating the integrals:

$ln(\frac{V}{v})=e^{\frac{-t}{RC} }$

natural logarithm to both sides in order to isolate V:

$V(t)=ve^{-\frac{t}{RC} }$

Where the term RC is called time constant and is given by:

$\tau=R*C=10*(0.330*10^{-6})=3.3*10^{-6}s$

3 0

2 years ago

Suppose someone pours 0.250 kg of 20.0ºC water (about a cup) into a 0.500-kg aluminum pan with a temperature of 150ºC. Assume th

Troyanec [42]

Answer : The temperature when the water and pan reach thermal equilibrium short time later is, $59.10^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of aluminium = $0.90J/g^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_1$ = mass of aluminum = 0.500 kg = 500 g

$m_2$ = mass of water = 0.250 kg = 250 g

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of aluminum = $150^oC$

$T_2$ = initial temperature of water = $20^oC$

Now put all the given values in the above formula, we get:

$500g\times 0.90J/g^oC\times (T_f-150)^oC=-250g\times 4.184J/g^oC\times (T_f-20)^oC$

$T_f=59.10^oC$

Therefore, the temperature when the water and pan reach thermal equilibrium short time later is, $59.10^oC$

8 0

3 years ago