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2 years ago

1.5kj of energy are used every second by a microwave oven, what is the power rating of the oven?

2 answers:

7 0

Power is the rate at which energy is consumed. When 1 J of energy is consumed in one second, the power consumed is 1 watt.

The microwave consumes 1.5 kJ of energy.

In joules, the energy consumed is 1500 J

$P=\frac{E}{t}$

Here, E is the energy consumed in time interval <em>t.</em>

The power rating of the microwave is therefore,

$P=\frac{1500J}{1 s} \\ =1500 W$

Thus, the power rating of the oven is 1500 W or 1.5 kW

Artemon [7]2 years ago

7 0

Power = work/ time
Power = 1.5 kj/ 1s
Power = 1.5kW

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A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

where, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda,$

and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

This is the de Broglie wavelength of the electron after scattering.

6 0

3 years ago

A train crosses a bridge that is 1300 m long in 1 minute and 5 seconds.

Nat2105 [25]

Answer:

I'm pretty sure it's 20m/s because 1300m divided by 65 seconds is 20 so I think it's 20m/s

Explanation:

3 0

2 years ago

What form of music has two parts and has two contrasting melodies? *

aleksandr82 [10.1K]

Answer: Ternary

Explanation: Ternary Form also has two parts, but is different from Binary Form. Relevance. Ternary form, sometimes called song form, is a three-part musical form where the first section (A) is repeated after the second section (B) ends. different words and same music.

Please mark as brainliest!

5 0

2 years ago

(HELP!! 10pts) a train travels 75km in 1hr, and then 68km in 2hrs. what is its average speed?

blondinia [14]

Average speed = total distance travelled ÷ total time taken

AS = (75km + 68km) ÷ (1hr + 2hr)

As = 143km ÷ 3hr

AS = 47.66667 km/hr

AS = 47.7 km/hr (3sf)

8 0

2 years ago

An object starts at rest. Its acceleration over 30 seconds is shown in the graph below:

ddd [48]

Answer:

The instantaneous speed of the object after the first five seconds is 12.5 m/s.

(C) is correct option.

Explanation:

Given that,

An object starts at rest. Its acceleration over 30 seconds.

We need to calculate the instantaneous speed of the object after the first five seconds

We know that,