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3 years ago

How much money would be saved by turning off one 100.0-W lightbulb 3.0 h/day for 365 days if the

Physics

1 answer:

Pavlova-9 [17]3 years ago

3 0

Answer:

the money that would be saved is $13.14.

Explanation:

Given;

power consumed by the light bulb, P = 100 W = 0.1 kW

time of running the bulb, t = 3 hours for 365 days = 1,095 hours

cost rate of power consumption, C = $0.12 per kWh

Energy consumed by the light bulb for the given days;

E = Pt

E = 0.1 kW x 1,095 hr

E = 109.5 kWh

Cost of energy consumed = 109.5 kWh x $0.12 / kWh

= $13.14

Therefore, the money that would be saved is $13.14.

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A tank is 10 m long, 6 m wide, 3 m high, and contains kerosene with density 820 kg/m3 to a depth of 2.5 m. (Use 9.8 m/s2 for the

Valentin [98]

Answer:

Explanation:

a )

Pressure on the bottom

= hdg

= 2.5 x 820 x 9.8

= 20090 N / m²

b )

force = area x pressure

= 20090 x 10 x 6 N

= 1205400 N

c )

Force on 10m x 2.5 m face

center of gravity of the tank will lie at height 2 .5 / 2 = 1.25 m

Pressure = hdg

= 1.25 x 820 x 9.8

= 10045 N/m²

force = pressure x area

= 10045 x 10 x 2.5

= 251125 N .

7 0

3 years ago

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Why are multiple images seen when two plane mirrors are placed at an angle

Whitepunk [10]

Answer: Can I get a picture???

8 0

3 years ago

Why cooking pans don't have have metal handles?

Harrizon [31]

metals are conductors, meaning that heat travels through it better than insulators such as plastic (commonly used for pan handles)

Therefore, to avoid the heat burning hands, most pans don't have metal handles.

4 0

3 years ago

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A power plant produces 1000 MW to supply a city 40 km away. Current flows from the power plant on a single wire of resistance 0.

nikitadnepr [17]

Answer:

Current = 8696 A

Fraction of power lost = $\dfrac{80}{529}$ = 0.151

Explanation:

Electric power is given by

$P=IV$

where I is the current and V is the voltage.

$I=\dfrac{P}{V}$

Using values from the question,

$I=\dfrac{1000\times10^6 \text{ W}}{115\times10^3\text{ V}} = 8696 \text{ A}$

The power loss is given by

$P_\text{loss} = I^2R$

where R is the resistance of the wire. From the question, the wire has a resistance of $0.050\Omega$ per km. Since resistance is proportional to length, the resistance of the wire is

$R = 0.050\times40 = 2\Omega$

Hence,

$P_\text{loss} = \left(\dfrac{200000}{23}\right)^2\times2$

The fraction lost = $\dfrac{P_\text{loss}}{P}=\left(\dfrac{200000}{23}\right)^2\times2\div (1000\times10^6)=\dfrac{80}{529}=0.151$

3 0

3 years ago

The momentum of a falling rock is found to be 200 kg m/s. What is the mass of the rock if it falls with a velocity of 5.0 m/s

Snezhnost [94]

Answer:

$\boxed {\boxed {\sf 40 \ kilograms}}$

Explanation:

Momentum is the product of velocity and mass. The formula is:

$p=m*v$

We know the rock is falling. Its momentum is 200 kilograms meters per second and its velocity is 5 meters per second. Substitute the values into the formula.

$200 \ kg \ m/s = m * 5.0 \ m/s$

We are solving for m, the mass. We must isolate the variable. It is being multiplied by 5 meters per second. The inverse of multiplication is division, so we divided both sides by 5.0 m/s.

$\frac{200 \ kg \ m/s}{5.0 \ m/s}=\frac{ m* 5.0 \ m/s }{5.0 \ m/s}$

$\frac{200 \ kg \ m/s}{5.0 \ m/s}=m$

The units of meters per second (m/s) cancel.

$\frac{200 \ kg}{5.0 } =m$

$40 \ kg = m$

The falling rock has a mass of <u>40 kilograms.</u>

4 0

3 years ago