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IgorLugansk [536]
3 years ago
7

A low-resistance path in a circuit, commonly called a _____ can cause a circuit breaker to trip

Engineering
1 answer:
inessss [21]3 years ago
8 0

Answer:

b

Explanation:

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You are using a Geiger counter to measure the activity of a radioactive substance over the course of several minutes. If the rea
Reika [66]

Answer: 33.35 minutes

Explanation:

A(t) = A(o) *(.5)^[t/(t1/2)]....equ1

Where

A(t) = geiger count after time t = 100

A(o) = initial geiger count = 400

(t1/2) = the half life of decay

t = time between geiger count = 66.7 minutes

Sub into equ 1

100=400(.5)^[66.7/(t1/2)

Equ becomes

.25= (.5)^[66.7/(t1/2)]

Take log of both sides

Log 0.25 = [66.7/(t1/2)] * log 0.5

66.7/(t1/2) = 2

(t1/2) = (66.7/2 ) = 33.35 minutes

4 0
3 years ago
Liquid water is fed to a boiler at 24°C and 10 bar is converted at a constant pressure to saturated steam.
zepelin [54]

We can find the change in the enthalpy through the tables A5 for Saturated water, pressure table.

For 1bar=1000kPa:

T_{sat}=179.88\°c

H_{fg} = 2014.6kJ/kg

c_p=4.18 kJkg^{-1}{K^{-1}

\nu_g = 0.19436m^3/kg

Replacing,

\Delta h = h_{fg}+c_p(T_{sat}-T_{inlet})

\Delta h = 2014.6+4.18(179.88-24)

\Delta h=2666.17kJ/kg

With the specific volume we know can calculate the mass flow, that is

\dot{m}=\frac{\frac{15000}{3600}}{0.19436}

\dot{m} = 21.4378kg/s

Then the heat required in input is,

Q=\dot{m}\Delta h

Q=21.4378*2666.17

Q=57157.036kW

With the same value required of 15000m^3/h, we can calculate the velocity of the water, that is given by,

V= \frac{\dotV}{A}

V = \frac{\frac{15000}{3600}}{\pi /4 *(0.15)^2}

V=235.79m/s

Finally we can apply the steady flow energy equation, that is

\dot{m}(h_1+\frac{V^2}{2000})+Q = \dot{m}h_2

Re-arrange for Q,

Q=\dot{m}(h_2-h_1-\frac{V^2}{2000})

Q=\dot{m}(\Delta h-\frac{V^2}{2000})

Q= (21.4378)(2666.17-\frac{235.79^2}{2000})

Q= 56560.88kW

We can note that consider the Kinetic Energy will decrease the heat input.

4 0
3 years ago
At a certain location, wind is blowing steadily at 5 mph. Suppose that the mass density of air is 0.0796 lbm/ft3 and determine t
nlexa [21]

Answer:

The radius of a wind turbine is 691.1 ft

The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m

Explanation:

Given;

power generation potential (PGP) = 1000 kW

Wind speed = 5 mph = 2.2352 m/s

Density of air = 0.0796 lbm/ft³ = 1.275 kg/m³

Radius of the wind turbine r = ?

Wind energy per unit mass of air, e = E/m = 0.5 v² = (0.5)(2.2352)²

Wind energy per unit mass of air = 2.517 J/kg

PGP = mass flow rate * energy per unit mass

PGP = ρ*A*V*e

PGP = \rho *\frac{\pi r^2}{2} *V*e  \\\\r^2 = \frac{2*PGP}{\rho*\pi *V*e} , r=\sqrt{ \frac{2*PGP}{\rho*\pi *V*e}} = \sqrt{ \frac{2*10^6}{1.275*\pi *2.235*2.517}}

r = 210.64 m = 691.1 ft

Thus, the radius of a wind turbine is 691.1 ft

PGP = CVᵃ

For best design of wind turbine Betz limit (c) is taken between (0.35 - 0.45)

Let C = 0.4

PGP = Cvᵃ

take log of both sides

ln(PGP) = a*ln(CV)

a = ln(PGP)/ln(CV)

a = ln(1000)/ln(0.4 *2.2352) = 7.73

The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m

5 0
3 years ago
Guys do u know who is
avanturin [10]
I think so, I’ve seen her on here a couple of times. Though I’ve never spoken with said person

Hope this helps you (also why?)
6 0
4 years ago
Read 2 more answers
In the event of a crash, Personal Injury Protection (PIP) coverage pays toward the medical costs for _____.
jok3333 [9.3K]

Answer:

The answer is D. Your injuries, regardless of who's at fault.

8 0
3 years ago
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