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3 years ago

A printer has a power of 100 W. It takes 30 seconds to print out a document. How much energy will it

Physics

2 answers:

Nikitich [7]3 years ago

8 0

Answer:

3000 Joules.

Explanation:

Black_prince [1.1K]3 years ago

3 0

100W translates to 100 watts consumed in 1 hour. Assuming the printer consumes electrical energy at a steady rate, then: 100 ÷ 60 = 1.6667 watts per minute. 1.6667 ÷ 2 = 0.8333 watts consumed in 30 seconds.

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A person walks 5.0kilometers north, then 5.0 kilometers east. His displacement is closest to ? A. 10 kilometers northwest B. 7.1

Rainbow [258]

Use vector analysis and calculate resultant vector using Pythagoras theorem:
5^2 + 5^2 = 50
Square root of 50 = approx 7.1 km NE
Therefore the answer is D

4 0

3 years ago

Which describes a scientist being creative?

AleksandrR [38]

Answer: Sara tries turning a test tube upside down to collect a gas.

A scientist is considered to be creative when he approaches a problem with new different ways. The conventional way is to design an experiment and take detailed notes, reading referenced studies previously done. Sara tries turning a test tube upside down to collect a gas is a creative way as Sara tries something different.

4 0

3 years ago

Read 2 more answers

An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr

Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

$F= mg$

Where, f = restoring force

$kx=mg$

$k=\dfrac{mg}{x}$

Put the value into the formula

$k=\dfrac{2.15\times9.8}{0.0895}$

$k=235.41\ N/m$

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Here, $v = A\omega$

$K.E=\dfrac{1}{2}m\times(A\omega)^2$

Here, $\omega=\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}kA^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2$

$K.E=0.06500\ J$

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0

3 years ago

Assignment S of write the Symbol Told, mercury and Cooper, Iron, Lead

m_a_m_a [10]

Answer:

Check explanation

Explanation:

Gold - Au (Aurum)

Mercury - Hg (Hydrargyrum)

Copper - Cu (Cuprum)

Iron - Fe (Ferrum)

Lead - Pb (Plumbum)

These elements in the periodic table are some of the elements represented by letters not in line with their names.

This is because, these elements were known in ancient times and therefore, they are represented by letters from their ancient names.

3 0

3 years ago

Consider the hydrogen atom. How does the energy difference between adjacent orbit radii change as the principal quantum number i

Kisachek [45]

Answer:

the energy difference between adjacent levels decreases as the quantum number increases

Explanation:

The energy levels of the hydrogen atom are given by the following formula:

$E=-E_0 \frac{1}{n^2}$

where

$E_0 = 13.6 eV$ is a constant

n is the level number

We can write therefore the energy difference between adjacent levels as

$\Delta E=-13.6 eV (\frac{1}{n^2}-\frac{1}{(n+1)^2})$

We see that this difference decreases as the level number (n) increases. For example, the difference between the levels n=1 and n=2 is

$\Delta E=-13.6 eV(\frac{1}{1^2}-\frac{1}{2^2})=-13.6 eV(1-\frac{1}{4})=-13.6 eV(\frac{3}{4})=-10.2 eV$

While the difference between the levels n=2 and n=3 is

$\Delta E=-13.6 eV(\frac{1}{2^2}-\frac{1}{3^2})=-13.6 eV(\frac{1}{4}-\frac{1}{9})=-13.6 eV(\frac{5}{36})=-1.9 eV$

And so on.

So, the energy difference between adjacent levels decreases as the quantum number increases.

5 0

3 years ago