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AleksandrR [38]
3 years ago
15

What major financial flop led to the end of the Sega Dreamcast and ultimately caused Sega to stop making game consoles altogethe

r?
1: The founder and CEO of Sega was found to be secretly skimming money off of the top of profits, leading to widespread distrust by the public and a sharp decline in sales until they ultimately had to shut down due to making no profit.

2: A small group of employees found a way to drain all of the Sega Corporation funding accounts and flee the country with all of the money, never to be heard from again.

3: A game called Shenmue, that cost more than $70 million to make, meant that everyone who owned a Dreamcast needed to buy two copies of the game just for Sega to make back the money they had spent to develop it-which didn't happen.

4: A game called Katamari Damacy that carried a virus that would infect any console that it was played on forced Sega to spend millions of dollars in refunds and bankrupted the company.​
Engineering
1 answer:
icang [17]3 years ago
6 0

Answer:

The major financial flop that led to the end of the Sega Dreamcast and ultimately caused Sega to stop making game consoles altogether is:

3: A game called Shenmue, that cost more than $70 million to make, meant that everyone who owned a Dreamcast needed to buy two copies of the game just for Sega to make back the money they had spent to develop it-which didn't happen.

Explanation:

Shenmue, released on December 29, 1999, was created for Dreamcast by Suzuki.  It was widely described as the most expensive video game ever produced.  It had an estimated production and marketing cost of between US$47 and $70 million, according to the latest available data.

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A cylindrical brass rod has a length of 5.00cm extending from a holder and a diameter of 4.50mm. Its Young's modulus is 98.0GPa.
Galina-37 [17]

Answer:

elongation of the brass rod is 0.01956 mm

Explanation:

given data

length = 5 cm = 50 mm

diameter = 4.50 mm

Young's modulus = 98.0 GPa

load = 610 N

to find out

what will be the elongation of the brass rod in mm

solution

we know here change in length formula that is express as

δ = \frac{PL}{AE}    ................1

here δ is change in length and P is applied load  and A id cross section area and E is Young's modulus and L is length

so all value in equation 1

δ = \frac{PL}{AE}  

δ = \frac{610*50}{\frac{\pi}{4} * 4.50^2 * 98*10^3}  

δ = 0.01956 mm

so elongation of the brass rod is 0.01956 mm

7 0
3 years ago
An automobile weighing 2500 lbf increases its gravitational potential energy by a magnitude of 2.25 × 104 Btu in going from an e
Mila [183]

Answer:

The elevation at the high point of the road is 12186.5 in ft.

Explanation:

The automobile weight is 2500 lbf.

The automobile increases its gravitational potential energy in 2.25 * 10^4 BTU. It means the mobile has increased its elevation.

The initial elevation is of 5183 ft.  

The first step is to convert Btu of potential energy to adequate units to work with data previously presented.

British Thermal Unit - 1 BTU = 778.17  lbf*ft

2.25 * 10^4 BTU (\frac{778.17 lbf*ft}{1BTU} ) = 1.75 * 10^7 lbf * ft

Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:  

Ep = m*g*(h_2 - h_1)\\ W = m*g  

Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.  

h_2 is the final elevation and h_1 is the initial elevation.

Replacing W in the Ep equation

Ep = W*(h_2 -h_1)\\(h_2 -h_1) = \frac{Ep}{W} \\h_2 = h_1 + \frac{Ep}{W}\\\\

Finally, the next step is to replace the variables of the problem.  

h_2 = 5183 ft + \frac{1.75 * 10^7 lbf*ft}{2500 lbf}\\h_2 = 5183 ft + 70003.5 ft\\h_2 = 12186.5 ft

The elevation at the high point of the road is 12186.5 in ft.  

3 0
3 years ago
What is this spray pattern defect most likely caused by:
DiKsa [7]

Answer:

fluid nozzle that is too large

6 0
2 years ago
Can anyone tell me what rock this is?
Oksi-84 [34.3K]

Answer: answer is B

Explanation:

8 0
3 years ago
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A single-phase transformer circuit feeds a motor and lighting load of 50 kilowatts. At a power factor of .8, the KVA rating of t
AveGali [126]

The KVA rating of the step down transformer at the given power factor would be 62.5 kVA.

<h3>What is power factor of a transformer?</h3>

Power factor (PF) is the ratio of working power, measured in kilowatts (kW), to apparent power, measured in kilovolt amperes (kVA).

PF = working power / apparent power

PF =  kW/kVA

kVA = kW/PF

kVA = 50 kW/0.8

kVA = 62.5 kVA

Thus, the KVA rating of the step down transformer at the given power factor would be 62.5 kVA.

Learn more about power factor here: brainly.com/question/7956945

#SPJ1

3 0
2 years ago
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