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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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<span>When the Sun’s energy moves through space, it reaches Earth’s atmosphere and finally the surface.  This radiant solar energy warms the atmosphere and becomes heat energy. This heat energy is transferred throughout the planet’s systems in three ways: by radiation, conduction, and convection</span>
5 0
3 years ago
The potential difference between the surface of a 3.0-cm-diameter power line and a point 1.0 m distant is 3.9 kV. Find the line
Korolek [52]

Answer:

The linear charge density is 5.19 X 10⁻⁶ C/m

Explanation:

The potential difference between two cylinders, is given as

V = (λ/2πε)ln(b/a)

where;

λ is the line charge density on the power line.

b is the distance between the power line = 1 m

a is the radius of the wire = 1.5 cm = 0.015 m

ε is the permittivity of free space = 8.9 X 10⁻¹² C

V*2πε = λ* ln(b/a)

3900 *(2π*8.9 x10⁻¹²)= λ *ln(1/0.015)

2.1812 X 10⁻⁷ = 4.1997* λ

λ = 5.19 X 10⁻⁶ C/m

Therefore, the linear charge density is 5.19 X 10⁻⁶ C/m

6 0
3 years ago
A 56 kg sprinter, starting from rest, runs 49 m in 7.0 s at constant acceleration.what is the sprinter's power output at 2.0 s,
alexgriva [62]
The sprinter is in uniform accelerated motion, and its initial velocity is zero, so the relationship betwen space (S) and time (t) is
S= \frac{1}{2} a t^2
where a is the acceleration. Using the data of the problem, we can find a:
a= \frac{2S}{t^2} = \frac{2 \cdot 49 m}{(7.0 s)^2} =2.0 m/s^2
So now we can solve the 3 parts of the problem.

a) power output at t=2.0 s
The velocity at t=2.0 s is
v(t)=at=(2.0 m/s^2)(2.0 s)=4.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(4.0 m/s)^2=448 J

and so the power output is
P= \frac{E}{t} = \frac{448 J}{2.0 s} =224 W

b) power output at t=4.0s 
The velocity at t=4.0 s is
v(t)=at=(2.0 m/s^2)(4.0 s)=8.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(8.0 m/s)^2=1792 J

and so the power output is
P= \frac{E}{t} = \frac{1792 J}{4.0 s} =448 W

c) Power output at t=6.0 s
The velocity at t=2.0 s is
v(t)=at=(2.0 m/s^2)(6.0 s)=12.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(6.0 m/s)^2=4032 J

and so the power output is
P= \frac{E}{t} = \frac{4032 J}{6.0 s} =672 W
8 0
3 years ago
Psychology is a<br><br>A) science <br><br>B) philosophy <br><br>C) theory <br><br>D) opinion
yan [13]
Answer is A science
8 0
3 years ago
The sides of a rectangle are 6.01 meters and 12 meters. Taking the significant figures into account, what is the area of the rec
Contact [7]
Taking significant figures into account, 6.01 has 3 significant figures, and 12 meters has 2. In the rule of multiplication, the answer has to follow the least number of significant figures. This means that the answer we obtain must have only 2 significant figures. To find the area of the rectangle, the two sides are merely multiplied. This is shown below:

6.01 * 12 = 72.12 = 72

Therefore, the area is 72 sq. meters
3 0
3 years ago
Read 2 more answers
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