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nikdorinn [45]
3 years ago
7

What is the relationship between kinetic and potential energy of a falling object

Physics
1 answer:
borishaifa [10]3 years ago
5 0

Answer:

See the explanation below.

Explanation:

According to the law of energy conservation. This can be transformed from potential energy to kinetic or vice versa.

For example, when you have a body at rest at a height with respect to the ground level, this body possesses potential energy with respect to the ground, but the moment that the body falls the potential energy decreases as it loses height with respect to the ground, but as the potential energy decreases, the kinetic energy increases, that is to say, the potential energy is transformed into kinetic energy as the body is falling.

As the energy relationship is maintained we can say that the potential and kinetic energies are equal.

E_{pot}=m*g*h\\E_{kin}=\frac{1}{2} *m*v^{2} \\E_{pot}=E_{kin}

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Answer:

option (B) decreases

Explanation:

According to the Wein's displacement law, the minimum wavelength of the radiated emission is inversely proportional to the absolute temperature of the body which emits radiation.

\lambda_{m}\alpha \frac{1}{T}

Where, T is the absolute temperature of the body and λm is the minimum wavelength of heat radiated.

Here, as the temperature increases, the wavelength decreases.

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The diagram shows the trajectory of a ball that
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69 MPH

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What is the mass of an object which has a force of 600 N acting on it and is travelling
Paha777 [63]

Answer:

<em>The mass of the object is 40 Kg</em>

Explanation:

<u>Net Force</u>

According to the second Newton's law, the net force exerted by an external agent on an object is:

F = m.a

Where:

a = acceleration of the object.

m = mass of the object.

The mass can be calculated by solving for m:

\displaystyle m=\frac{F}{a}

The object has a net force of F=600 N acting on it and travels at a=15\ m/s^2, thus the mas is:

\displaystyle m=\frac{600}{15}

m = 40 Kg

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6 0
2 years ago
A mass of 10 g of oxygen fill a weighted piston–cylinder device at 20 kPa and 110°C. The device is now cooled until the tempe
mezya [45]

Answer:

The change of the volume of the device during this cooling is 14.3\times10^{-3}\ m^3

Explanation:

Given that,

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Final temperature = 0°C

We need to calculate the change of the volume of the device during this cooling

Using formula of change volume

\Delta V=V_{2}-V_{1}

\Delta V=\dfrac{mR}{P}(T_{2}-T_{1})

Put the value into the formula

\Delta V=\dfrac{0.3125\times0.0821}{2.0265\times10^{9}}(383-273)

\Delta V=14.297\ L

\Delta V=14.3\times10^{-3}\ m^3

Hence, The change of the volume of the device during this cooling is 14.3\times10^{-3}\ m^3

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3 years ago
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