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3 years ago

The volume of microbial culture is observed to increase according to the formula

Engineering

1 answer:

saw5 [17]3 years ago

3 0

The expression of V(m³)=e^(t(s)) to make V in in³ and t in minutes is;

V(in³) = (¹/₆₁₀₂₄)a $e^{\frac{1}{60}bt(h)$

We are given that;

Volume of microbial culture is observed to increase according to the formula;

V = e^(t)

where;

t is in seconds

V is in m³

We want to now express V in in³ and t in minutes.

Now, from conversions;

1 m³ = 61024 in³

Also; 1 second = 1/60 minutes

according to formula for exponential decay, we know that;

V = ae^(bt)

Thus, we have;

61024V = ae^(¹/₆₀b(t(h))

V(in³) = (¹/₆₁₀₂₄)a $e^{\frac{1}{60}bt(h)$

Read more about subject of formula at; brainly.com/question/790938

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2 years ago

Read 2 more answers

An industrial load with an operating voltage of 480/0° V is connected to the power system. The load absorbs 120 kW with a laggin

Leni [432]

Answer:

$Q=41.33 KVAR\ \\at\\\ 480 Vrms$

Explanation:

From the question we are told that:

Voltage $V=480/0 \textdegree V$

Power $P=120kW$

Initial Power factor $p.f_1=0.77 lagging$

Final Power factor $p.f_2=0.9 lagging$

Generally the equation for Reactive Power is mathematically given by

Q=P(tan \theta_2-tan \theta_1)

Since

$p.f_1=0.77$

$cos \theta_1 =0.77$

$\theta_1=cos^{-1}0.77$

$\theta_1=39.65 \textdegree$

And

$p.f_2=0.9$

$cos \theta_2 =0.9$

$\theta_2=cos^{-1}0.9$

$\theta_2=25.84 \textdegree$

Therefore

$Q=P(tan 25.84 \textdegree-tan 39.65 \textdegree)$

$Q=120*10^3(tan 25.84 \textdegree-tan 39.65 \textdegree)$

$Q=-41.33VAR$

Therefore

The size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging is

$Q=41.33 KVAR\ \\at\\\ 480 Vrms$

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3 years ago

Question 2) A material that is malleable can be defined as:

Reil [10]

Answer:

Explanation:

Able to return to it's original shape after distortion.

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3 years ago

An industrial system is rated at 600 v the system includes a partially exposed terminal block thats mounted on a bulkhead a new

Nastasia [14]

Answer:

10 ft or 3.048 m (for unqualified person)

1.09 ft or 0.33 m (for qualified person)

Explanation:

OSHA has prescribed "Minimum Approach Distance - MAD" to ensure safety of personnel working with electrical power lines.

For a unqualified person:

The minimum approach distance is 10 ft or 3.048 m

For a qualified person:

For a voltage in the range of 300V to 750V (Phase to Ground or Phase to Phase) the minimum approach distance is 1.09 ft or 0.33 m

Since 600V lies in the above range hence it is applicable on 600V.

OSHA has developed a full list of minimum approach distance according to the level of voltages (50V to 72000V)

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3 years ago

If a construction company is considering a new type of material to use in their construction, which factors would they

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I would honestly select every one of the given options. Gor a company evaluating this new material it would be very valuable to hit each of these factors.

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3 years ago