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3 years ago

If a device uses 280 watts of power a day, about how many kilowatt-hours will it use in 30 days?

Physics

1 answer:

Novay_Z [31]3 years ago

6 0

Answer:C

Explanation:

Power=280watts=280/1000 kilowatts

Power=0.28 kilowatts

Device use 0.28 kilowatts in 1 day

1day=24hours

Device use 0.28 kilowatts in 24hours

30days=30 x 24=720 hours

For 720 hours=(0.28x720) ➗ 24

For 720 hours=201.6 ➗ 24

For 720 hours=8.4

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Answer:

It reveals that light is a wave

Explanation:

Diffraction is the property of a wave in which there is a bending of the wave about the corners of an obstacle or aperture into the geometrical shadow of the obstacle or aperture.

This simply implies that a wave bends or spreads out when it passes through openings. Since the light diffracts through small slits and diffraction has been shown to occur in water waves and sound waves, this property of diffraction can only be characteristic of a wave and thus, this evidence reveals that light is a wave.

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3 years ago

Two blocks are placed at the ends of a horizontal massless board, as in the drawing. The board is kept from rotating and rests o

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Answer:

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3 years ago

The initial kinetic energy imparted to a 0.25 kg bullet is 1066 J. The acceleration of gravity is 9.81 m/s 2 . Neglecting air re

lubasha [3.4K]

Answer:

The range of the bullet is 0.435 kilometers.

Explanation:

According to the problem, maximum height is equal to the range of the bullet. That is:

$\Delta x = \Delta y$

Where:

$\Delta x$ - Range of the bullet, measured in meters.

$\Delta y$ - Maximum height of the bullet, measured in meters.

By the Principle of Energy Conservation, gravitational potential energy reaches its maximum at the expense of the initial kinetic energy. That is to say:

$K_{1} = U_{2}$

Where:

$K_{1}$ - Kinetic energy at point 1, measured in joules.

$U_{1}$ - Gravitational potential energy at point 2, measured in joules, and:

$U_{2} = m\cdot g \cdot \Delta y$

Where:

$m$ - Mass of the bullet, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

The maximum height is now cleared:

$K_{1} = m\cdot g \cdot \Delta y$

$\Delta y = \frac{K_{1}}{m\cdot g}$

If $K_{1} = 1066\,J$ , $m = 0.25\,kg$ and $g = 9.81\,\frac{m}{s^{2}}$ , the maximum height is now computed:

$\Delta y = \frac{1066\,J}{(0.25\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$\Delta y = 434.791\,m$

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3 years ago

How much time is needed to produce 720 Joules of work if 90 watts of power is used?

Tems11 [23]

Answer:

8 seconds

Explanation:

power (P) is defined as the rate at which work is done.

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by the definition of power,

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3 years ago

When energy is added to a wave, how can the wave change?.

nadya68 [22]

Answer:

When waves overlap in-phase (crest meets crest or trough meets trough) the waves energy is additive and the amplitude increases.

Explanation:

When waves overlap out-of-phase (crest meets trough) the waves cancel and the amplitude (energy) decreases. When two interfering waves cancel each other out.

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2 years ago