Calculate the amount of current flowing through a 75-watt light bulb that is connected to a 120-volt circuit in your home.

THE COMPUND INTEREST ON RS 30,000AT 7% PER ANNUM IS RS 4347 THE PERIOD IN YEARS

timofeeve [1]

Answer:

what wym

Explanation:

4 0

2 years ago

Read 2 more answers

A person is planning a bungee jump from a 40 meter high bridge. Under the bridge is a river with crocodiles, so the person does

Nonamiya [84]

Answer:

a. l = 19.7m, b. 18.55m, c. Impact Force = 3889.84 N

Explanation:

The total energy of the system when the person is at top of the bridge is

Potential energy = mgh, Kinetic energy = 0

The total energy of the system when the person reaches just above the surface

Potential energy = 0, Kinetic energy = 0, Spring energy = ½ K X2, where k is the spring constant and X is the deflection

Applying conservation of energy

mgh = 0 + 0 + ½ K X²

80 x 9.81 x 40 = ½ (3600/l) X²

31392 = ½ (3600/l) X²

We can also conclude that

l+ X + 1.75 = 40

l + X = 38.25

a. Substitute the value of x from above into the energy conversion expression

31392 = ½ (3600/l)(38.25 - l)²

31392 x 2/3600 = (38.25 + l² – 2l(38.25))/l

17.44l = l2 – 76.5l + 38.25²

l² – 76.5l – 17.44l +1463.0625 = 0

Solving for l we get

L = 19.7

Hence, length of the rope is 19.7m

b. The deflection is calculated by using the relation between l and X

L + X = 38.25

X = 38.25 – 19.7 = 18.55m

c. The impact force is calculated using the impact force formula which relates the impact force with the deflection

F = KX

F = (3600/l) . X

F = (3600/19.7) . (18.55) = 3889.84 N

Thus, the impact force is 3889.84 N

3 0

3 years ago

6. Find the heat flow in 24 hours through a refrigerator door 30.0" x 58.0" insulated with cellulose fiber 2.0" thick. The tempe

Ilia_Sergeevich [38]

Answer:

The heat flow in 24 hours through the refrigerator door is approximately 1,608.57 BTU

Explanation:

The given parameters are;

The duration of the heat transfer, t = 24 hours = 86,400 seconds

The area of the refrigerator door, A = 30.0" × 58.0" = 1,740 in.² = 1.122578 m²

The material of the insulator in the door = Cellulose fiber

The thickness of the insulator in the door, d = 2.0" = 0.0508 m

The temperature inside the fridge = 38° F = 276.4833 K

The temperature of the room = 78°F = 298.7056 K

The thermal conductivity of cellulose fiber = 0.040 W/(m·K)

By Fourier's law, the heat flow through a by conduction material is given by the following formula;

$\dfrac{Q}{t} = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d}$

$Q = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d} \times t$

Therefore, we have;

$Q = \dfrac{0.04 \times 1.122578 \times (298.7056 - 276.4833 ) }{0.0508} \times 86,400 =1,697,131.73522$

The heat flow in 24 hours through the refrigerator door, Q = 1,697,131.73522 J = 1,608.5705140685 BTU

7 0

2 years ago

A closed, rigid tank is filled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. If the gas

ch4aika [34]

Answer:

gauge pressure is 133 kPa

Explanation:

given data

initial temperature T1 = 27°C = 300 K

gauge pressure = 300 kPa = 300 × 10³ Pa

atmospheric pressure = 1 atm

final temperature T2 = 77°C = 350 K

to find out

final pressure

solution

we know that gauge pressure is = absolute pressure - atmospheric pressure so

P (gauge ) = 300 × 10³ Pa - 1 × $10^{5}$ Pa

P (gauge ) = 2 × $10^{5}$ Pa

so from idea gas equation

$\frac{P1*V1}{T1} = \frac{P2*V2}{T2}$ ................1

so ${P2} = \frac{P1*T2}{T1}$

${P2} = \frac{2*10^5*350}{300}$

P2 = 2.33 × $10^{5}$ Pa

so gauge pressure = absolute pressure - atmospheric pressure

gauge pressure = 2.33 × $10^{5}$ - 1.0 × $10^{5}$

gauge pressure = 1.33 × $10^{5}$ Pa

so gauge pressure is 133 kPa

4 0

3 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

3 years ago