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4 years ago

When a trebuchet counterweight is hoisted by soldiers, what form of energy is transferred into the machine as work?

Physics

2 answers:

kvasek [131]4 years ago

5 0

Answer: Potential energy to kinetic energy

Explanation:

Potential energy is the form of energy possessed by a body due to its position.

Kinetic energy is the form of energy possessed by a body due to its motion.

When soldiers use their energy to hoist the trebuchet counterweight, the potential energy stored in the counter weight converts to kinetic energy of the counterweight as it is released and falls.

kupik [55]4 years ago

3 0

And trade experts recreate a medieval throwing machine called a trebuchet. ... The potential energyfrom the raised counterweight is transformed into kinetic ... that was stored in the raised counterweight is transferred, in the form of kinetic energy, ... Design a working model of a trebuchet and demonstrate the power of a ...throwing machine called a trebuchet. To ... how well their designs will work. ... Where does the potential energy in the lever arm come from? ... To put a trebuchet into its “cocked” position, a team of people use their energy to hoist ... was stored in the raised counterweight is transferred, in the formof kinetic energy , to the.

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Sunny_sXe [5.5K]

Radius of Earth is given as

$R = 6.4 * 10^6 m$

now here the height from the surface of earth is same as that of radius

$h = 6.4 * 10^6 m$

now here the acceleration due to gravity at this height is given as

$g' = \frac{GM}{(R+h)^2}$

$g' = \frac{GM}{4R^2}$

$g' = \frac{g}{4}$

now the force of gravity on the given object will be

$F = mg'$

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$F = 1.225 N$

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3 years ago

If a sound wave is produced with a wavelength of 1.04m what is the waves frequency

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3 years ago

Q1- the speed will increase because the particles will have more energy which causes them to move faster

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<h3># Question </h3>

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3 years ago

What minimum speed must an electron have to excite the 492-nm-wavelength blue emission line in the hg spectrum?

mezya [45]

The energy required by the excitation of the line is:
ΔE = hν = hc / λ
where:
ΔE = energy difference
h = Planck constant
ν = line frequency
c = speed of light
λ = line wavelength

The energy difference must be supplied by the electron, supposing it transfers all its kinetic energy to excite the line:
$\Delta E = \frac{1}{2} m v^{2}$

Therefore,
$\frac{1}{2} m v^{2} = \frac{hc}{\lambda}$

And solving for v we get:
$v = \sqrt{ \frac{2hc}{m\lambda} }$

Plugging in numbers (after trasforing into the correct SI units of measurement):
$v = \sqrt{ \frac{(2)(6.6 \cdot 10^{-34})(3 \cdot 10^{8}) }{(9.11 \cdot 10^{-31})(4.92 \cdot 10^{-7}) } }$
=9.4 · 10⁵ m/s

Hence, the electron must have a speed of 9.4 · 10⁵ m/s in order to excite the 492nm line.

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4 years ago