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2 years ago

A ball dropped onto a trampoline returns to the same height after the rebound.

Physics

1 answer:

Jet001 [13]2 years ago

7 0

a)The ball's potential energy changes as it moves from the release point to the top of the rebound.

b)The ball's potential energy is at its highest and rests when it is in the top position.

<h3>What is potential energy?</h3>

The potential energy is due to the virtue of the position and the height. The unit for the potential energy is the joule.

The potential energy is mainly dependent upon the height of the object.

Potential energy = mgh

The kinetic energy of the body is due to the virtue of motion.

According to the Law of Conservation of Energy, energy can neither be created nor destroyed but can be transferred from one form to another.

The total energy is the sum of all the energies present in the system. The potential energy in a system is due to its position in the system.

A ball dropped onto a trampoline returns to the same height after the rebound.

Hence, the ball's potential energy changes as it moves from the release point to the top of the rebound. When the ball is at the top position the potential energy is maximum and the ball is at rest.

To learn more about the potential energy, refer to the link;

brainly.com/question/24284560

#SPJ1

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A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv

cestrela7 [59]

<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":

$P=\sigma A T^{4}$ (1)

Where:

$P=300J/min=5J/s=5W$ is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing $1W=\frac{1Joule}{second}=1\frac{J}{s}$

$\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}}$ is the Stefan-Boltzmann's constant.

$A=5cm^{2}=0.0005m^{2}$ is the Surface area of the body

$T=727\°C=1000.15K$ is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close. So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

$P=\sigma A \epsilon T^{4}$ (2)

Where $\epsilon$ is the body's emissivity

(the value we want to find)

Isolating $\epsilon$ from (2):

$\epsilon=\frac{P}{\sigma A T^{4}}$ (3)

Solving:

$\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}}$ (4)

Finally:

$\epsilon=0.17$ (5) This is the body's emissivity

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3 years ago

A compound consisting of only phosphorous and oxygen atoms is 43.64 % phosphorus by mass.The molecular mass/formula weight of th

DanielleElmas [232]

First, we calculate the mass of Phosphorous present:
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moles of P = mass / Ar
= 123.88 / 31
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We know that one mole of substance has 6.02 x 10²³ particles
Atoms of P = 4 x 6.02 x 10²³
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3 years ago

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Kazeer [188]

I'm pretty sure its a scale.

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3 years ago

An unknown material, m1 = 0.49 kg, at a temperature of T1 = 92 degrees C is added to a Dewer (an insulated container) which cont

erastova [34]

Answer:

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Explanation:

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$m_uc_u(T_{uf}-T_{ui})=-m_wc_w(T_{wf}-T_{wi})$

Since thermal equilibrium is reached we know that $T_{cf}=T_{wf}=T_f=31^{\circ}C=304^{\circ}K$ , where we have added $273^{\circ}$ to convert the temperature from Celsius to Kelvin, as <em>we must do</em>. Since we want the specific heat of the unknown material, we do: