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4 years ago

Which of the following is the best example of kinetic energy being transformed into potential energy

2 answers:

3 0

A pendulum is probably the most common showing of this example. As the pendulum swings down, it converts its potential energy (height) into kinetic energy (velocity). At the lowest point the kinetic energy is the highest and the potential is the lowest. At the highest point in its swing the velocity is zero so the kinetic energy is zero and the potential energy is at a maximum (greatest height).

photoshop1234 [79]4 years ago

3 0

Answer:

kenetic enery

Explanation:

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A force of 660 n stretches a certain spring a distance of 0.300 m. what is the potential energy of the spring when a 70.0 kg mas

kkurt [141]

A force of 660 n stretches a certain spring a distance of 0.300 m. what is the potential energy of the spring when a 70.0 kg mass hangs vertically from it?

5 0

3 years ago

A meter stick is supported by a pivot at its center of mass. Assume that the meter stick is uniform and that the center of mass

iragen [17]

Answer:

64.5 cm

Explanation:

30 * 80 + x * 110 = 50* 190 => x = 64.5

5 0

3 years ago

An astronaut finds herself in a predicament in which she has become untethered from her shuttle. She figures that she could get

Blizzard [7]

In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.

The equation that defines the linear moment is given by

$mV_i = (m-m_O)V_f - m_OV_O$

where,

m=Total mass

$m_O =$ Mass of Object

$V_i =$ Velocity before throwing

$V_f =$ Final Velocity

$V_O =$ Velocity of Object

Our values are:

$m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg$

Solving to find the final speed, after throwing the object we have

$V_f=\frac{mV_0+m_TV_O}{m-m_O}$

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) $5.3Kg\rightarrow 15m/s$

$V_{f1}=\frac{mV_0+m_TV_O}{m-m_O}$

$V_{f1}=\frac{(98.7)*0+5.3*15}{98.7-5.3}$

$V_{f1}=0.8511m/s$

B) $7.9Kg\rightarrow 11.2m/s$

$V_{f2}=\frac{mV_{f1}+m_TV_O}{m-m_O}$

$V_{f2}=\frac{(98.7)(0.8511)+(7.9)(11.2)}{98.7-5.3-7.9}$

$V_{f2} = 2.0173m/s$

C) $10.5Kg\rightarrow 7m/s$

$V_{f3}=\frac{mV_{f2}+m_TV_O}{m-m_O}$

$V_{f3}=\frac{(98.7)(2.0173)+(10.5)(7)}{98.7-5.3-7.9-10.5}$

$V_{f3} = 3.63478m/s$

Therefore the final velocity of astronaut is 3.63m/s

7 0

3 years ago

A plastic box has an initial volume of 2.00 m 3 . It is then submerged below the surface of a liquid and its volume decreases to

nikitadnepr [17]

Answer:

Volume strain is 0.02

Explanation:

Volume strain is defined as the change in volume to the original volume.

It is given that,

Initial volume of the plastic box is 2 m³

It is then submerged below the surface of a liquid and its volume decreases to 1.96 m³

We need to find the volume strain on the box. It is defined as the change in volume divided by the original volume. So,

$\delta V=\dfrac{V_f-V_i}{V_i}\\\\\delta V=\dfrac{1.96-2}{2}\\\\\delta V=0.02$

So, the volume strain on the box is 0.02.

6 0

3 years ago

A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit

Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

$P = \frac{1}{2} \mu \omega^2 A^2 v$

Here,

$\mu$ = Linear mass density of the string

$\omega =$ Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

$v = \sqrt{\frac{T}{\mu}} \rightarrow$ Here T is the Period

For the linear mass density we have that

$\mu = \frac{m}{l}$

And the angular frequency can be written as

$\omega = 2\pi f$

Replacing this terms and the first equation we have that

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})$

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})$

$P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})$

PART A ) Replacing our values here we have that

$P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.2320W$

PART B) The new amplitude A' that is half ot the wavelength of the wave is

$A' = \frac{1.8*10^{-3}}{2}$

$A' = 0.9*10^{-3}$

Replacing at the equation of power we have that

$P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.058W$

8 0

3 years ago