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gulaghasi [49]
3 years ago
10

Marie lifts a 5 kg mass from the floor and puts in on a table one metre high. What is the work done?

Physics
1 answer:
sergey [27]3 years ago
8 0

Answer:

The work done was 49.05 N*m.

Explanation:

The work done by a a force is given by:

W = F*d

We first need to calculate the force necessary to lift the box from the ground, the force needed for is the weigh of the box, which is shown below:

F = m*g\\F = 5*9.81\\F = 49.05 \text{ N}

Therefore the work done by lifting the box and placing it on the table is:

W = 49.05*1 \\W = 49.05 \text{ N*m}

The work done was 49.05 N*m.

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On one end of the electromagnetic spectrum are radio waves, which have wavelengths billions of times longer than those of visible light. On the other end of the spectrum are gamma rays, with wavelengths billions of times smaller than those of visible light.

Explanation:

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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

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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.

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Which state of matter would be described as a highly energized charge particles with moving extremely fast
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A spring with spring constant of 34 N/m is stretched 0.12 m from its equilibrium position. How much work must be done to stretch
Nesterboy [21]

Answer:0.253Joules

Explanation:

First, we will calculate the force required to stretch the string. According to Hooke's law, the force applied to an elastic material or string is directly proportional to its extension.

F = ke where;

F is the force

k is spring constant = 34N/m

e is the extension = 0.12m

F = 34× 0.12 = 4.08N

To get work done,

Work is said to be done if the force applied to an object cause the body to move a distance from its initial position.

Work done = Force × Distance

Since F = 4.08m, distance = 0.062m

Work done = 4.08 × 0.062

Work done = 0.253Joules

Therefore, work done to stretch the string to an additional 0.062 m distance is 0.253Joules

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