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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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3 years ago
In which situation is work being done?
Keith_Richards [23]

Answer: D. The force and displacement are in the same direction.

Explanation:

The Work W done by a Force F refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path with distance d.  

Work is a scalar magnitude, and its unit in the International System of Units is the Joule (like energy).

Now, when the applied force is constant and the direction of the force and the direction of the displacement are <u>parallel</u>, the equation to calculate it is:  

W=(F)(d) (1)

When they are not parallel, both directions form an angle, let's call it \alpha. In that case the expression to calculate the Work is:  

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4 0
3 years ago
A 0.12 g honeybee acquires a charge of +24pC while flying. The earth's electric field near the surface is typically 100 N/C, dow
shusha [124]

Answer:

150000000

\dfrac{F_e}{F_g}=0.00000203873598369

49050000 N/C

Explanation:

q = Charge = 24 pC

m = Mass of honeybee = 0.12 g

E = Electric field = 100 N/C

g = Acceleration due to gravity = 9.81 m/s²

1\ C=6.25\times 10^{18}\ electrons

Number electrons is

n=24\times 10^{-12}\times 6.25\times 10^{18}\\\Rightarrow n=150000000

The number of electrons added or removed was 150000000

Force is given by

F_e=Eq\\\Rightarrow F_e=100\times 24\times 10^{-12}\\\Rightarrow F_e=2.4\times 10^{-9}\ N

The ratio is

\dfrac{F_e}{F_g}=\dfrac{2.4\times 10^{-9}}{0.12\times 10^{-3}\times 9.81}\\\Rightarrow \dfrac{F_e}{F_g}=0.00000203873598369

The ratio is \dfrac{F_e}{F_g}=0.00000203873598369

Balancing the forces we get

Eq=mg\\\Rightarrow E=\dfrac{mg}{q}\\\Rightarrow E=\dfrac{0.12\times 10^{-3}\times 9.81}{24\times 10^{-12}}\\\Rightarrow E=49050000\ N/C

The electric field required is 49050000 N/C

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