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Alekssandra [29.7K]
3 years ago
10

A 100 watt incandescent light bulb is operated for 7 hours, and a 32 watt fluorescent light bulb is operated for the same period

of time. At 8 cents per kWh, what is the cost savings in using the fluorescent bulb?
Physics
1 answer:
Tema [17]3 years ago
4 0
<h2>Cost savings in using the fluorescent bulb is 3.81 cents</h2>

Explanation:

A 100 watt incandescent light bulb is operated for 7 hours.

Energy used by 100 watt incandescent light bulb = Power x Time

Energy used by 100 watt incandescent light bulb = 100 x 7 = 700 Wh

Energy used by 100 watt incandescent light bulb = 0.7 kWh

A 32 watt fluorescent light bulb is operated for 7 hours.

Energy used by 32 watt fluorescent light bulb = Power x Time

Energy used by 32 watt fluorescent light bulb = 32 x 7 = 224 Wh

Energy used by 32 watt fluorescent light bulb = 0.224 kWh

Cost of electricity per kWh = 8 cents

Cost of electricity by 100 watt incandescent light bulb = 0.7 x 8 = 5.6 cents

Cost of electricity by 32 watt fluorescent light bulb = 0.224 x 8 = 1.792 cents

Difference in costs = 5.6 - 1.792 = 3.808 cents

Cost savings in using the fluorescent bulb is 3.81 cents

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A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency of 0.680 Hz. The pendulum ha
sineoko [7]

Answer:

Therefore, the moment of inertia is:

I=0.37 \: kgm^{2}

Explanation:

The period of an oscillation equation of a solid pendulum is given by:

T=2\pi \sqrt{\frac{I}{Mgd}} (1)

Where:

  • I is the moment of inertia
  • M is the mass of the pendulum
  • d is the distance from the center of mass to the pivot
  • g is the gravity

Let's solve the equation (1) for I

T=2\pi \sqrt{\frac{I}{Mgd}}

I=Mgd(\frac{T}{2\pi})^{2}

Before find I, we need to remember that

T = \frac{1}{f}=\frac{1}{0.680}=1.47\: s

Now, the moment of inertia will be:

I=2*9.81*0.340(\frac{1.47}{2\pi})^{2}  

Therefore, the moment of inertia is:

I=0.37 \: kgm^{2}

I hope it helps you!

7 0
3 years ago
car was moving in a straight road of length 320 km it covered 240 km with an average velocity 75 km/hr then it ran out of fuel a
Stella [2.4K]

The average velocity of the car for the whole journey is 69.57 km/h.

The given parameters:

  • <em>Length of the road, L = 320 km</em>
  • <em>Distance covered = 240 km at 75 km/h</em>
  • <em>time spent refueling, t₂ = 0.6 hr</em>
  • <em>Final velocity, = 100 km/hr</em>

The time spent by the before refueling is calculated as follows;

t = \frac{d}{v} \\\\t_1 = \frac{240}{75} \\\\t_1 = 3.2 \ hours

The time spent by the car for the remaining journey;

t_3 = \frac{320 - 240}{100} \\\\t_3 = 0.8 \ hr

The total time of the journey is calculated as follows;

t = t_1 + t_2 + t_3\\\\t = 3.2 \ hr \ + \ 0.6 \ hr \ + \ 0.8 \ hr\\\\t = 4.6 \ hours

The average velocity of the car for the whole journey is calculated as follows;

v = \frac{total \ distance }{total \ time} \\\\v = \frac{320}{4.6} \\\\v = 69.57 \ km/h

Learn more about average velocity here: brainly.com/question/6504879

6 0
3 years ago
A block of mass m=9.0 kg and speed V and is behind a block of mass M= 27 kg and speed of .50 m/s. The surface is frictionless, a
sammy [17]

Answer:

2.06 m/s

Explanation:

From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.

Momentum before collision

Momentum is given by p=mv where m and v represent mass. The initial sum of momentum will be 9v+(27*0.5)=9v+13.5

Momentum after collision

The momentum after collision will be given by (9+27)*0.9=32.4

Relating the two then 9v+13.5=32.4

9v=18.5

V=2.055555555555555555555555555555555555555 m/s

Rounded off, v is approximately 2.06 m/s

5 0
3 years ago
A miniature quadcopter is located at x = -2.25 m and y, - 5.70 matt - 0 and moves with an average velocity having components Vv,
kupik [55]

Recall that average velocity is equal to change in position over a given time interval,

\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t}

so that the <em>x</em>-component of \vec v_{\rm ave} is

\dfrac{x_2 - (-2.25\,\mathrm m)}{1.60\,\mathrm s} = 2.70\dfrac{\rm m}{\rm s}

and its <em>y</em>-component is

\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}

Solve for x_2 and y_2, which are the <em>x</em>- and <em>y</em>-components of the copter's position vector after <em>t</em> = 1.60 s.

x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}

y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}

Note that I'm reading the given details as

x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

8 0
3 years ago
Which bar graph could represent the reaction rates of a reversible reaction<br> that has just begun?
Galina-37 [17]

The answer is A. Just did it.

8 0
3 years ago
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