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boyakko [2]
3 years ago
5

Compare the energy consumption of two commonly used items in the household. Calculate the energy used by a 1.40 kW toaster oven,

Wtoaster , which is used for 6.00 minutes, and then calculate the amount of energy that an 11.0 W compact fluorescent light (CFL) bulb, Wlight , uses when left on for 9.00 hours .
Physics
1 answer:
Nikolay [14]3 years ago
3 0

Explanation:

It is given that,

Power consumed by toaster oven, P=1.4\ kW

Time taken, t = 6 minutes = 0.1 hour

Energy used by toaster, W_{toaster}=P\times t

W_{toaster}=1.4\ kW\times 0.1\ h=0.14\ kWh

Similarly,

Power consumed by toaster oven, P=11\ W

Time taken, t = 9 minutes = 0.15 hour

Energy used by toaster, W_{toaster}=P\times t

W_{toaster}=11\ W\times 0.15\ h=1.65\ Wh=0.0016\ kWh

Hence, this is the required solution.

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Answer:

Part a)

E = \frac{\lambda}{2\pi \epsilon_0 r}

Part b)

E = \frac{\lambda}{2\pi \epsilon_0 r}

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

Explanation:

Part a)

By Guass law we know that

\int E. dA = \frac{q}{\epsilon_0}

E. 2\pi rL = \frac{\lambda L}{\epsilon_0}

E = \frac{\lambda}{2\pi \epsilon_0 r}

Part b)

Outside the outer cylinder we will again use Guass law

\int E. dA = \frac{q}{\epsilon_0}

E. 2\pi rL = \frac{\lambda L}{\epsilon_0}

E = \frac{\lambda}{2\pi \epsilon_0 r}

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

4 0
3 years ago
Convertir:<br> A. 3Km a m<br> B. 250 ma Km<br> C. 1000Cm a m<br> D. 10000 mm a Cm
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Answer:

A. 3,000,000 m

B. 0.25 km

C. 10 m

D. 1,000 cm

Explanation:

no hablo español, así que solo ingrese esto en el traductor de G*ogle

A. One kilometer equals 1000 meters, so

3,000*1,000 = 3,000,000 m

B. One meter equals 0.001 kilometer, so

250*0.001 = 0.25 km

C. One centimeter equals 0.01 meter

1,000*0.01 = 10 m

D. One milimeter equals 0.1 centimer, so

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A team of dogs accelerates a 290kg dogsled from 0 to 6.0m/s in 3.0 s. Assume that the acceleration is constant.Part AWhat is the
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Answer:

(a) a=2m/sec^2

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Time t = 3 sec

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(b) From second equation of motion we know that

s=ut+\frac{1}{2}at^2=0\times 3+\frac{1}{2}\times 2\times 3^2=9m

We know that work done is given by

W = F s = 580×9 =5220 j

(c) Time is given as t = 3 sec

We know that power is given as

P=\frac{W}{t}=\frac{5220}{3}=1740Watt

(d) Time t = 1.5 sec

So P=\frac{W}{t}=\frac{5220}{1.5}=3466.66Watt

5 0
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