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boyakko [2]
2 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]2 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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A spacecraft travels at 1.5 X 108 m/s relative to Earth. A process onboard the
kvasek [131]

Answer:

73.6 minutes

Explanation:

relative time = time interval / √(1 - observer velocity² / speed of light²)

we have relative time. we want time interval.

rearrange

time interval = relative time x √(1 - observer velocity² / speed of light²)

convert 85 mins into seconds

85 x 60 = 5100

1.5 x 10⁸ as a number is 150000000

for c = 299 792 458

time interval = 5100 x √(1 - 150 000 000² / 299 792 458²)

for c = 3 x 10⁸

time interval = 5100 x √(1 - 150 000 000² / 300 000 000²)

time interval = 5100 x 0.866

time interval = 4415.71

divide by 60 for back into minutes

time = 73.6 minutes

4 0
2 years ago
Calculate the height from from which a body is released from rest if its velocity just before hitting the ground is30m\s
Kamila [148]

Answer:

height = 45 m

Explanation:

Initial velocity = 0 m/s

Final velocity = 30 m/s

Acceleration due to gravity = +10m/s^2 ( because ball is going down )

Using the eqn of motion

{v}^{2}  -  {u}^{2}  = 2gh

where v= final velocity ; u = initial velocity ; g = acceleration due to gravity ; h = height

So,

{30}^{2}  -  {0}^{2}  = 2 \times 10 \times h

=  > 20h = 900

=  > h =  \frac{900}{20}  = 45

5 0
3 years ago
Which of the following is a likely life cycle of a star?
melomori [17]

Answer:

D.

Explanation:

But this just happen for big stars, like more than 20x the Sun mass.

Shortly: A nebula is a cloud of gas and dust, the material starts to be acummuleted and became a protostar (is like a big planet, almost  a star). With enought mass this is a star, burn hydrogen and transform it in Helium.

This occurs in Main Sequence, is about almost all the life time of a star. Then starts the lack of hydrogen. Gravity compress everything, pressure goes up and heat all. Too much energy, Helium get burned and the star grews fast, became a Red Giant. Time pass and the fuel is over, no more making fusion, gravity compress the star, too much strenght, colapses, neutron star.

If it have pretty mass, ok. If have more than like 2x Sun mass, became a blackhole.

7 0
3 years ago
An inverse-time circuit breaker (CB) is used for branch-circuit short-circuit and ground-fault protection for a 30-horsepower, 2
Maru [420]

Answer:

Explanation:

Motor rating is given in horsepower (hp), it will be converted in watt (W).

Standard to install circuit breaker for an electric circuit is usually 20% ~ 25% more than the Rated Current of the circuit

while

Standard to install overload relay for an electric circuit is usually 20% ~ 25% more than the Running Current of the circuit.

So, to find the maximum capacity of the circuit breaker, rated current of the motor will be multiplied by 1.2 ~ 1.25

Step by Step Explanation:

30hp = 22371W (as 1hp = 745.7)

Assuming unity power factor (cosФ=1) and 208V phase to phase voltage:

Rated Power (watt) = √3 . V.I. cosФ

<em>{if 208V is phase to neutral voltage, then use following formula:</em>

<em> Rated Power (watt) = 3 . V.I. cosФ}</em>

\frac{22371}{\sqrt{3} * 208 * 1} = I

Rated Current = <u>62.169A</u>

So, required maximum rating for circuit breaker is:

20% to 25% of the rated current = 62.17*1.2 ~ 62.17*1.25

=74.6A ~ 77.7A

Hence, any breaker between the above mentioned rating will be appropriate.

3 0
3 years ago
A large electrical device is plugged into a 240-volt outlet and has a resistance of 160 ohms. How much power does the device use
lubasha [3.4K]
Yur answer is c.360 watts
5 0
3 years ago
Read 2 more answers
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