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Cloud [144]
3 years ago
15

If an object has 100 J of PE, how much work can it do (with that PE)?

Physics
1 answer:
Stella [2.4K]3 years ago
4 0

Answer:

W = 100 J

Explanation:

The potential energy of an object is given by :

P=mgh

It is said to possessed by an object due to its position under the action of gravity.

The work done by an object is given by :

W = F×d , F is force and d is displacement

We know that, F = mg

So, if an object has 100 J of PE, it means that it can do the work of 100 J.          

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A guitar string with a linear mass density of 2.0 g/m is stretched between two vertical rods that are 65 cm apart. The string ho
Rudiy27

Answer:

717 Hz

Explanation:

<u>solution:</u>

The wave with three antinodes has m = 3. Thus, f_3 = 3f_1 and so the fundamental frequency of the string is  

f_1 =f_3/3

    =430 Hz/3

    =143 Hz

Thus, the frequency of the fifth harmonic is

f_5 = 5*f_1

      = 5*143 Hz

     = 717 Hz

7 0
3 years ago
Question 1 of 4 Attempt 4 The acceleration due to gravity, ???? , is constant at sea level on the Earth's surface. However, the
Evgen [1.6K]

Answer:

g(h) = g ( 1 - 2(h/R) )

<em>*At first order on h/R*</em>

Explanation:

Hi!

We can derive this expression for distances h small compared to the earth's radius R.

In order to do this, we must expand the newton's law of universal gravitation around r=R

Remember that this law is:

F = G \frac{m_1m_2}{r^2}

In the present case m1 will be the mass of the earth.

Additionally, if we remember Newton's second law for the mass m2 (with m2 constant):

F = m_2a

Therefore, we can see that

a(r) = G \frac{m_1}{r^2}

With a the acceleration due to the earth's mass.

Now, the taylor series is going to be (at first order in h/R):

a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R}

a(R) is actually the constant acceleration at sea level

and

a(R) =G \frac{m_1}{R^2} \\ \frac{da(r)}{dr}_{r=R} = -2 G\frac{m_1}{R^3}

Therefore:

a(R+h) \approx G\frac{m_1}{R^2} -2G\frac{m_1}{R^2} \frac{h}{R} = g(1-2\frac{h}{R})

Consider that the error in this expresion is quadratic in (h/R), and to consider quadratic correctiosn you must expand the taylor series to the next power:

a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R} + \frac{h^2}{2!} \frac{d^2a(r)}{dr^2}_{r=R}

6 0
4 years ago
Read 2 more answers
Which of the following is a good example of accelerated motion?
irakobra [83]

Answer:

an apple because its gaining speed as it falls

Explanation:

5 0
3 years ago
Read 2 more answers
Two children of mass 20.0 kg and 30.0 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If t
s2008m [1.1K]

Answer:

The distance from the pivot point that the small child will sit in order to maintain the balance is 1.8 m

Explanation:

Given;

mass of the bigger child, M = 30 kg

mass of the smaller child, m = 20 kg

distance between the two children, d = 3 m

This information can be represented diagrammatically;

                                    3m

         |<------------------------------------------------>|

----------------------------------------------------------------------------

         ↓             x            Δ            3-x           ↓

         20kg                                                 30kg

x is the distance from the pivot point that the small child will sit in order to maintain the balance

Take moment about the pivot;

Clockwise moment = anticlockwise moment

30(3-x) = 20x

90 -30x = 20x

90 = 20x + 30x

90 = 50x

x = 90 / 50

x = 1.8 m

Therefore, the distance from the pivot point that the small child will sit in order to maintain the balance is 1.8 m

7 0
3 years ago
Please help with this! The selected answer is the incorrect one, I'm making corrections.
elixir [45]

Answer:

The answer is C-Spin a magnet around the wire on the piece of iron. Hope that helps :)

6 0
3 years ago
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