What is an ip<br> Number

A fan that consumes 20 W of electric power when operating is claimed to discharge air from a ventilated room at a rate of 1.0 kg

solong [7]

Answer:

No, the claim is not reasonable for 20 W electric power consumption.

It is reasonable for 40 W electric power consumption.

Explanation:

Power = (1/2)*mass flow rate*(square of velocity)

mass flow rate = 1 kg/s

velocity = 8 m/s

square of velocity = 64 m^2 / s^2

Power = (1/2)*(1)*(64)

Power = 32 W

For a fan that consumes 20 W power it is not possible to deliver more power than 20 W but this one is delivering 32 W hence it is a false claim.

For a fan that consumes 40 W it is indeed possible to deliver 32 W considering the efficiency. Hence this claim is reasonable.

5 0

2 years ago

A utility generates electricity with a 36% efficient coal-fired power plant emitting the legal limit of 0.6 lb of SO2 per millio

Mama L [17]

Answer:

a) 570 kWh of electricity will be saved

b) the amount of SO₂ not be emitted or heat of electricity saved is 0.00162 ton/CLF

c) $1.296 can be earned by selling the SO₂ saved by a single CFL

Explanation:

Given the data in the question;

a) How many kilowatt-hours of electricity would be saved?

first, we determine the total power consumption by the incandescent lamp

$P_{incandescent}$ = 75 w × 10,000-hr = 750000 wh = 750 kWh

next, we also find the total power consumption by the fluorescent lamp

$P_{fluorescent}$ = 18 × 10000 = 180000 = 180 kWh

So the value of power saved will be;

$P_{saved}$ = $P_{incandescent}$ - $P_{fluorescent}$

$P_{saved}$ = 750 - 180

$P_{saved}$ = 570 kWh

Therefore, 570 kWh of electricity will be saved.

now lets find the heat of electricity saved in Bituminous

heat saved = energy saved per CLF / efficiency of plant

given that; the utility has 36% efficiency

we substitute

heat saved = 570 kWh/CLF / 36%

we know that; 1 kilowatt (kWh) = 3,412 btu per hour (btu/h)

so

heat saved = 570 kWh/CLF / 0.36 × (3412 Btu / kW-hr (

heat saved = 5.4 × 10⁶ Btu/CLF

i.e eat of electricity saved per CLF is 5.4 × 10⁶

b) How many 2,000-lb tons of SO₂ would not be emitted

2000 lb/tons = 5.4 × 10⁶ Btu/CLF

0.6 lb SO₂ / million Btu = x

so

x = [( 5.4 × 10⁶ Btu/CLF ) × ( 0.6 lb SO₂ / million Btu )] / 2000 lb/tons

x = [( 5.4 × 10⁶ Btu/CLF ) × ( 0.6 lb SO₂ )] / [ ( 10⁶) × ( 2000 lb/ton) ]

x = 3.24 × 10⁶ / 2 × 10⁹

x = 0.00162 ton/CLF

Therefore, the amount of SO₂ not be emitted or heat of electricity saved is 0.00162 ton/CLF

c) If the utility can sell its rights to emit SO2 at $800 per ton, how much money could the utility earn by selling the SO2 saved by a single CFL?

Amount = ( SO₂ saved per CLF ) × ( rate per CFL )

we substitute

Amount = 0.00162 ton/CLF × $800

= $1.296

Therefore; $1.296 can be earned by selling the SO₂ saved by a single CFL.

3 0

2 years ago

(SI units) Molten metal is poured into the pouring cup of a sand mold at a steady rate of 400 cm3/s. The molten metal overflows

maxonik [38]

Answer:

diameter of the sprue at the bottom is 1.603 cm

Explanation:

Given data;

Flow rate, Q = 400 cm³/s

cross section of sprue: Round

Diameter of sprue at the top $d_{top}$ = 3.4 cm

Height of sprue, h = 20 cm = 0.2 m

acceleration due to gravity g = 9.81 m/s²

Calculate the velocity at the sprue base

$V_{base}$ = √2gh

we substitute

$V_{base}$ = √(2 × 9.81 m/s² × 0.2 m )

$V_{base}$ = 1.98091 m/s

$V_{base}$ = 198.091 cm/s

diameter of the sprue at the bottom will be;

Q = AV = (π $d_{bottom}^2$ /4) × $V_{base}$

$d_{bottom}$ = √(4Q/π $V_{base}$ )

we substitute our values into the equation;

$d_{bottom}$ = √(4(400 cm³/s) / (π×198.091 cm/s))

$d_{bottom}$ = 1.603 cm

Therefore, diameter of the sprue at the bottom is 1.603 cm

6 0

2 years ago

Tech A says that bleeding an electronic brake control system is just like bleeding a non-electronic brake control system. Tech B

PtichkaEL [24]

Answer:

Tech A is correct.

Explanation:

An electric brake controller is a device that sends a signal to the trailer via vehicle's brakes. This reduces the wear and tear on the vehicle brakes. As a result, the vehicle stops.

Tech A says that bleeding an electronic brake control system is just like bleeding a non-electronic brake control system.

So,

Tech A is correct.

3 0

3 years ago

At what voltage would a () capacitor have the minimum energy to raise by ignoring all losses in the system? If needed, you may a

Charra [1.4K]

Answer:

V = 280.15 V

Explanation:

" The complete question is attached with figure"

Given:

- The capacitance of the capacitor C = 10 nF

- The amount of mass attached to motor m = 4 grams

- The amount of distance it is to be lifted h = 1 cm

- Ignore all other losses in the system

Find:

- The voltage required to lift the mass m through distance h?

Solution:

- The conservation of energy for the entire system is written as:

Work_gravity = U_c

Where,

Work_gravity: Work done by gravity on mass m

U_c: The amount of energy stored in a capacitor

m*g*h = 0.5*C*V^2

V^2 = 2*m*g*h / C

V = sqrt ( 2*m*g*h / C )

Plug in the values:

V = sqrt ( 2*0.004*9.81*0.01 / 10*10^-9 )

V = sqrt ( 78,480)

V = 28.15 V

8 0

3 years ago

What is an ip Number