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Zinaida [17]
3 years ago
15

You are cooking breakfast for yourself and a friend using a 1,140-W waffle iron and a 510-W coffeepot. Usually, you operate thes

e appliances from a 110-V outlet for 0.500 h each day. (a) At 12 cents per kWh, how much do you spend to cook breakfast during a 30.0 day period
Physics
1 answer:
RideAnS [48]3 years ago
7 0

Answer:

The cost is 297 cents.

Explanation:

Power of iron, P = 1140 W

Power of coffee pot, P' = 510 W

Voltage, V = 110 V

Time, t = 0.5 h each day

Cost = 12 cents per kWh

(a) Total energy

E = P x t + P' x t

E = 1140 x 0.5 x 60 x 60 + 510 x 0.5 x 60 x 60

E = 2052000 + 918000 = 2970000 J

1 kWh = 3.6 x 10^6 J

E = 0.825 kWh

For 30 days

E' = 0.825 x 30 = 24.75 kWh

So, the cost is

= 12 x 24.75 = 297 cents

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A metallic bond is a bond between a positively charged metal ion and
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Answer:

conduction electrons

Explanation:

Metallic bonding is a type of chemical bonding that rises from the electrostatic attractive force between conduction electrons and positively charged metal ions. It may be described as the sharing of free electrons among a structure of positively charged ions.

4 0
3 years ago
A motorist drives south at 20m/s for 3 min then turns west and travels 25 m
Over [174]
If the motorist travelled at 20 meters a second for three minutes, 3 minutes is the same as 180 seconds, you multiply 20 times 180, which equals 3600, add 25, and the answer is 3625.
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3 years ago
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a ball is dropped from rest at a height of 45.0 m above the ground. ignore the effects of air resistance. What is the speed of c
NemiM [27]

So, the final velocity of the ball when it is 10.0 m above the ground approximately <u>26.2 m/s</u>.

<h3>Introduction</h3>

Hi ! In this question, I will help you. This question uses the principle of final velocity in free fall. Free fall occurs only when an object is dropped (without initial velocity), so the falling object is only affected by the presence of gravity. In general, the final velocity in free fall can be expressed by this equation :

\boxed{\sf{\bold{v = \sqrt{2 \times g \times h}}}}

With the following condition :

  • v = final velocity (m/s)
  • h = height or any other displacement at vertical line (m)
  • g = acceleration of the gravity (m/s²)

<h3>Problem Solving</h3>

We know that :

  • \sf{h_1} = initial height = 45.0 m
  • \sf{h_2} = final height = 10.0 m
  • g = acceleration of the gravity = 9.8 m/s²

Note :

At this point 10 m above the ground, the object can still complete its movement up to exactly 0 m above the ground.

What was asked :

  • v = final velocity = ... m/s

Step by Step

\sf{v = \sqrt{2 \times g \times \Delta h}}

\sf{v = \sqrt{2 \times g \times (h_1 - h_2)}}

\sf{v = \sqrt{2 \times 9.8 \times (45 - 10)}}

\sf{v = \sqrt{19.6 \times 35}}

\sf{v = \sqrt{686}}

\boxed{\sf{v \approx 26.2 \: m/s}}

<h3>Conclusion</h3>

So, the final velocity of the ball when it is 10.0 m above the ground approximately 26.2 m/s.

<h3>See More :</h3>
  • The relationship between acceleration and the change in velocity and time in free fall brainly.com/question/26486625
5 0
2 years ago
A mass of air occupies a volume of 5.7 L at a pressure of 0.52 atm. What is the new pressure if the same mass of air at the same
Nikolay [14]

Apply Boyle's law:

PV = const.

P = pressure, V = volume, the product of P and V must stay constant

Our initial P and V values are:

P = 0.52atm, V = 5.7L

Our final P and V values are:

P = ?, V = 2.0L

Set the products of each set of PV values equal to each other and solve for the final P:

P(2.0) = 0.52(5.7)

P = 1.48atm

5 0
3 years ago
If a nearsighted person has a far point df that is 3.50m from the eye, what is the focal length f1 of the contact lenses that th
olga55 [171]

Answer:

f1= -350cm or -3.5m

f2= 22.1cm or 0.221m

Explanation:

A person is nearsighted when the person's far point is less than infinity. A diverging lens is normally used to correct this eye defect. A diverging lens has a negative focal length as seen in the solution attached.

Farsightedness is when a person's near point is farther than 25cm. This eye defect is corrected using a converging lens. The focal length of a converging lens is positive. This is evident in the solution attached. The near point is also referred to as the least distance of distinct vision.

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3 years ago
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