Pls I need help for the problem solving part.

If the absolute pressure of a gas is 550.280 kPa, its gage pressure is

Gwar [14]

If the absolute pressure of a gas is 550.280 kPa, its gage pressure is <span> a. 101.325 kPa.
b. 448.955 kPa.
c. 651.605 kPa.
d. 277.280 kPa.</span>

The answer is B.

6 0

3 years ago

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If i punch my self and it will hurt am i weak or strong

Natasha2012 [34]

Answer:

You're strong.

Explanation:

I've been thinking of this for quite a while, and I realized that your body has a certain limit to how much pain it can take. So, punching yourself extremely hard will cause pain, because that's your body's reaction to immense pressure being put on it. But, the fact that you punched yourself so hard that it hurts, shows that you are capable of applying so much pressure; therefore, you are strong.

6 0

3 years ago

Because the pressure falls, water boils at a lower temperature with increasing altitude. Consequently, cake mixes and boiled egg

AlexFokin [52]

Answer:

1) The boiling point of water reduces by 3.28°C at 1,000 m above sea-level

2) The boiling point of water reduces by 6.56°C at 2,000 m above sea-level

Explanation:

The variation of the boiling point of water with elevation is given as follows

The boiling point reduces by 0.5°C for every 152.4 meter increase in elevation

At sea-level, the boiling point temperature of water = 100°C

1) At 1,000 m elevation, the boiling point temperature, T = 100 - (1,000/152.4) × 0.5 ≈ 96.72 °C

Therefore, the boiling point of water reduces by 100° - 96.72° = 3.28°C at 1,000 m above sea-level

2) At 2,000 m elevation, the boiling point temperature, T = 100 - (2,000/152.4) × 0.5 ≈ 93.44°C

The boiling point of water reduces by 100° - 93.44° = 6.56°C at 2,000 m above sea-level

7 0

3 years ago

Oscilloscope amplitude and frequency problem. Study the above graph. The volts/div dial is set to 2 volts/div and the time/div d

denis-greek [22]

Answer:

Amplitude = 8 Volts

Frequency = 0.067 kHz

Explanation:

Note: The missing picture in question is attached for your review.

Given:

Volts/Div = 2 V/div

Time/Div = 5 msec/div

Finding Amplitude:

Now, as you can see in the attached picture, there are 4 division between two peaks of the waveform, so,

$Amplitude = 4 div/volts * 2 volts/ div )\\Amplitude = 8 Volts$

(Multiplying by 2 V/div because oscilloscope dial is set at 2 V/div)

Finding Frequency:

As can be seen in attached picture, 3 division are there for one complete cycle of waveform,so,

$Time Period = 3 div * 5msec /div\\Time Perod = 15 msec$

Since,

$Frequency = \frac{1}{Time Period}\\Frequency = \frac{1}{15m}\\Frequency = 0.067 kHz$

8 0

3 years ago

For work to be accomplished we must have

Nat2105 [25]

The answer they are looking for is the last one. However the last two are technically correct but the third one would result in negative work.

8 0

3 years ago

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